[R] my function decided to stop working, problem with scoping?
Hello list.
Until last week, I've been using my function without a problem. I can only
imagine what has changed (no package updates), but all of a sudden, a custom
function that uses raster::xyValues stopped working. For some reason, one of
the values is not found, even though it's there. Here is a snippet of code
from browser() call from within my custom function where everything goes
awry.
Browse[1]> effect.distance
[1] 220.2045 # just to make sure effect.distance is there
Browse[1]> unlist(xyValues(object = object, xy = origin.point, buffer =
effect.distance))
Error in .local(object, xy, ...) : object 'effect.distance' not found
Browse[1]> traceback()
2: stop(gettextf("no function '%s' could be found", f), domain = NA)
1: getS3method("xyValues", "method")
Browse[1]> environment()
Browse[1]> Q
> environment()
> effect.distance
Error: object 'effect.distance' not found
If I call the unlist(...) line in R_GlobalEnv, it seems to work fine. I'm
wondering why this behavior?
Cheers,
Roman
--
In God we trust, all others bring data.
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Re: [R] frequency, count rows, data for heat map
Please, reply to the r-help and not only to me personally. That way
others can can also help, or perhaps benefit from the answers.
You can use strplit to remove the last part of the strings. strplit
returns a list of character vectors from which you (if I understand
you correctly) only want to select the first element. I use laply from
the plyr library for this, although there are probably also other ways
of doing this.
library(plyr)
dat$V3 <- laply(strsplit(as.character(dat$V1), '_'), function(l) l[1])
After that you can use daply as I showed in my previous post
[daply(dat, V3 ~ V2, nrow)] or use the methods suggested by Dennis
Murphy to build your table.
Regards,
Jan
On Thu, Aug 26, 2010 at 1:41 AM, Trip Sweeney wrote:
> Jan,
> Thanks for responding to my post to listeserve about arranging data matrix
> for heat map.
> I am still a beginner, so the below is the code I used for the matrix and
> did not yet learn how to
> input 'data.frame' (which I need to know to use your code). The below code
> works
> and mock.txt file is attached. There is one thing, though. The input in
> column 1 is tricky
> in the mock.txt file. I need it to sum per unique ID based on character
> prior to the "_"
> So, for example the current script call 1079_17891 and 1079_14794 uniques
> when I want
> them to be tallied together since they are both part of same 1079 samples.
> Occasionally
> a sample has three characters before the "_", like 111_463428 etc in
> mock.txt. The substring
> after the "_" is variable length. In the end, it should be one row for 1079,
> one for 111, and one for 5576.
> Can you help me with this modification of the code? Any advice much
> appreciated. Sincerely, Trip
>
> dat<-read.table('mock.txt',sep="\t")
> sumData=matrix(NA,nrow=length(unique(dat[,1])),ncol=length(unique(dat[,2])))
> rownames(sumData)<-unique(dat[,1])
> colnames(sumData)<-unique(dat[,2])
>
> for (i in 1:dim(sumData)[1]){
> for(j in 1:dim(sumData)[2]){
> sumData[i,j]<-sum (dat[,1]==unique(dat[,1])[i] &
> dat[,2]==unique(dat[,2])[j])
> }
> }
>
> write.table(sumData,"SummarizedData.txt",sep="\t",col.names=NA)
>
On Wed, Aug 25, 2010 at 4:53 PM, rtsweeney wrote:
>
> Hi all,
> I have read posts of heat map creation but I am one step prior --
> Here is what I am trying to do and wonder if you have any tips?
> We are trying to map sequence reads from tumors to viral genomes.
>
> Example input file :
> 111 abc
> 111 sdf
> 111 xyz
> 1079 abc
> 1079 xyz
> 1079 xyz
> 5576 abc
> 5576 sdf
> 5576 sdf
>
> How may xyz's are there for 1079 and 111? How many abc's, etc?
> How many times did reads from sample (1079) align to virus xyz.
> In some cases there are thousands per virus in a give sample, sometimes one.
> The original file (two columns by tens of thousands of rows; 20 MB) is
> text file (tab delimited).
>
> Output file:
> abc sdf xyz
> 111 1 1 1
> 1079 1 0 2
> 5576 1 2 0
>
> Or, other ways to generate this data so I can then use it for heat map
> creation?
>
> Thanks for any help you may have,
>
> rtsweeney
> palo alto, ca
> --
> View this message in context:
> http://r.789695.n4.nabble.com/frequency-count-rows-data-for-heat-map-tp2338363p2338363.html
> Sent from the R help mailing list archive at Nabble.com.
>
> __
> R-help@r-project.org mailing list
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
1079_346281416490|ref|NC_013643.1|
1079_346281416323|ref|NC_013646.1|
1079_3789629367|ref|NC_001803.1|
1079_58830984428|ref|NC_004812.1|
1079_1292 9629367|ref|NC_001803.1|
1079_3956 9629357|ref|NC_001802.1|
1079_4736 9629357|ref|NC_001802.1|
1079_7732 21427641|ref|NC_004015.1|
1079_7855 118197620|ref|NC_008584.1|
1079_8618 32453484|ref|NC_004928.1|
1079_11540 10140926|ref|NC_002531.1|
1079_14794 9629367|ref|NC_001803.1|
1079_15738 109255272|ref|NC_008168.1|
1079_17891 299778956|ref|NC_014260.1|
1079_18414 157781212|ref|NC_009823.1|
1079_18414 157781216|ref|NC_009824.1|
1079_20312 9629367|ref|NC_001803.1|
1079_20497 9629357|ref|NC_001802.1|
1079_26750 9629367|ref|NC_001803.1|
1079_27926 9628113|ref|NC_001659.1|
1079_27926 9628113|ref|NC_001659.1|
1079_28033 84662653|ref|NC_007710.1|
1079_30020 47835019|ref|NC_004333.2|
1079_30371 9629367|ref|NC_001803.1|
1079_35750 50313241|ref|NC_001491.2|
1079_35750 50313241|ref|NC_001491.2|
111_463428 56694721|ref|NC_006560.1|
111_464636 114680053|ref|NC_008349.1|
111_464636 9627742|ref|NC_001623.1|
111_465190 9627186|ref|NC_001539.1|
111_467613 51557483|ref|NC_006151.1|
111_467613 51557483|ref|NC_006151.1|
111_467975 9627742|ref|NC_001623.1|
111_467975
Re: [R] Removing inter-bar spaces in barchart
Oops, small typo, should be: barchart_test_heights=sin(c(1:100)) barchart_test_bins=c(c(1:50),c(1:50)) groups=c(rep(1,50),rep(2,50)) # Wish below didn't have spaces! barchart(barchart_test_bins~barchart_test_heights,groups=groups) On Wed, Aug 25, 2010 at 4:46 PM, Jonathan Greenberg wrote: > I apologize, let me add some data to play with: > > barchart_test_heights=sin(c(1:100)) > barchart_test_bins=c(1:100) > groups=c(rep(1,50),rep(2,50)) > > I have pre-calculated histogram data and the bins they belong to. I > would like to plot the two histograms superimposed on one-another, > having group 1 bars for a given bin be next to group 2 bars, something > along the lines of: > > barchart(barchart_test_bins~barchart_test_heights,groups=groups) > > except with no space between the bars (and a key for the groups). > > --j > > P.S. And yes I was talking about barchart() from lattice -- however, I > am happy to use another function if that makes this easier. > > On Wed, Aug 25, 2010 at 1:55 PM, David Winsemius > wrote: >> >> On Aug 24, 2010, at 10:20 PM, Jonathan Greenberg wrote: >> >>> Rhelpers: >>> >>> I'm trying to make a barchart of a 2-group dataset >>> (barchart(x~y,data=data,groups=z,horizontal=FALSE)). My problem is >>> that I can't, for the life of me, seem to get rid of the inter-bar >>> space -- box.ratio set to 1 doesn't do much. Any ideas? I'd >>> ideally want zero space between the bars. Thanks! >> >> You didn't provide any data (nor did you illustrate with one of the >> available datasets that are used in examples.) >> >> Compare these two outputs: >> >> barchart(yield ~ year, data = barley, >> groups = variety, ylab = "Barley Yield (bushels/acre)", >> ) >> >> barchart(yield ~ variety, data = barley, >> groups = year, ylab = "Barley Yield (bushels/acre)", >> ) >> >> ... and notice that the variables in the "group" have no separation of their >> bars whereas the rhs variables do. This is the opposite of what I expected, >> so perhaps you think as I did and reversing the roles of"y" and "z" might >> help? >> >> -- >> >> David Winsemius, MD >> West Hartford, CT >> >> > > > > -- > Jonathan A. Greenberg, PhD > Assistant Project Scientist > Center for Spatial Technologies and Remote Sensing (CSTARS) > Department of Land, Air and Water Resources > University of California, Davis > One Shields Avenue > Davis, CA 95616 > Phone: 415-763-5476 > AIM: jgrn307, MSN: jgrn...@hotmail.com, Gchat: jgrn307 > -- Jonathan A. Greenberg, PhD Assistant Project Scientist Center for Spatial Technologies and Remote Sensing (CSTARS) Department of Land, Air and Water Resources University of California, Davis One Shields Avenue Davis, CA 95616 Phone: 415-763-5476 AIM: jgrn307, MSN: jgrn...@hotmail.com, Gchat: jgrn307 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Removing inter-bar spaces in barchart
I apologize, let me add some data to play with: barchart_test_heights=sin(c(1:100)) barchart_test_bins=c(1:100) groups=c(rep(1,50),rep(2,50)) I have pre-calculated histogram data and the bins they belong to. I would like to plot the two histograms superimposed on one-another, having group 1 bars for a given bin be next to group 2 bars, something along the lines of: barchart(barchart_test_bins~barchart_test_heights,groups=groups) except with no space between the bars (and a key for the groups). --j P.S. And yes I was talking about barchart() from lattice -- however, I am happy to use another function if that makes this easier. On Wed, Aug 25, 2010 at 1:55 PM, David Winsemius wrote: > > On Aug 24, 2010, at 10:20 PM, Jonathan Greenberg wrote: > >> Rhelpers: >> >> I'm trying to make a barchart of a 2-group dataset >> (barchart(x~y,data=data,groups=z,horizontal=FALSE)). My problem is >> that I can't, for the life of me, seem to get rid of the inter-bar >> space -- box.ratio set to 1 doesn't do much. Any ideas? I'd >> ideally want zero space between the bars. Thanks! > > You didn't provide any data (nor did you illustrate with one of the > available datasets that are used in examples.) > > Compare these two outputs: > > barchart(yield ~ year, data = barley, > groups = variety, ylab = "Barley Yield (bushels/acre)", > ) > > barchart(yield ~ variety, data = barley, > groups = year, ylab = "Barley Yield (bushels/acre)", > ) > > ... and notice that the variables in the "group" have no separation of their > bars whereas the rhs variables do. This is the opposite of what I expected, > so perhaps you think as I did and reversing the roles of"y" and "z" might > help? > > -- > > David Winsemius, MD > West Hartford, CT > > -- Jonathan A. Greenberg, PhD Assistant Project Scientist Center for Spatial Technologies and Remote Sensing (CSTARS) Department of Land, Air and Water Resources University of California, Davis One Shields Avenue Davis, CA 95616 Phone: 415-763-5476 AIM: jgrn307, MSN: jgrn...@hotmail.com, Gchat: jgrn307 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to plot y-axis on the right of x-axis
On Thu, 2010-08-26 at 06:19 +0800, elaine kuo wrote:
> Yes, I appreciated your answers which well hit my questions. (esp the
> perfect model parts).
>
>
> About the plot part, one more question.
> Is it possible to make the two plots (northern and southern richness)
> sharing the same Y axis (latitude = 0, the equator) ?
> In other words, southern richness would be on the left side of the Y
> axis,
> while northern one on the right side.
Yes, of course:
op <- par(mar = c(5.1,0,4.1,0), oma = c(0,4.1,0,4.1))
layout(matrix(1:2, nrow = 1))
plot(1:100)
plot(1:100, xlim = c(100,1), axes = FALSE)
axis(1)
axis(2, labels = FALSE, tcl = -0.5)
axis(2, labels = FALSE, tcl = 0.5)
axis(4)
box()
layout(1)
par(op)
Is one way doing things by hand. Notice the problem (or potential
problem) of having an axis in the middle of your data; it might obscure
data and the problem is exacerbated if you try to label it. Plus it is a
faff to fiddle around with the plotting margins, etc.
Can't you just code southern latitudes as -degrees N (minus degrees N)
and then plot all the data on the one plot, and fit your models to this
data rather that the original? After all, does it matter if you are at
0.1 degrees N or 0.1 degrees S and as such are in different hemispheres?
You are likely in the same biogeographical region. Here is an exmaple:
## dummy data
set.seed(123)
dat <- data.frame(richness = c(rpois(100, lambda = 5),
rpois(100, lambda = 2)),
latitudeN = c(sort(sample(seq(0.01, 90,
length = 1000),
100)),
sort(sample(seq(-0.01, -90,
length = 1000),
100
## fit model to N sites:
## ignoring that richness is a count:
## hint use glm( family = poisson) as a start
modN <- lm(richness ~ latitudeN, data = dat, subset = latitudeN > 0)
## fit model to S sites
modS <- lm(richness ~ latitudeN, data = dat, subset = latitudeN < 0)
# prediction data
pdatN <- data.frame(latitudeN = seq(0.01, 90, length = 100))
pdatS <- data.frame(latitudeN = seq(-0.01, -90, length = 100))
## predict
pdatN <- within(pdatN, richness <- predict(modN, newdata = pdatN))
pdatS <- within(pdatS, richness <- predict(modS, newdata = pdatS))
## plotting
plot(richness ~ latitudeN, data = dat,
xlab = expression(Latitude~(degree*N)),
ylab = "Species Richness")
lines(richness ~ latitudeN, data = pdatN, col = "red")
lines(richness ~ latitudeN, data = pdatS, col = "blue")
legend("topleft", ncol = 2, legend = c("N","S"),
col = c("red","blue"), lty = "solid", bty = "n")
HTH
G
> Elaine
>
>
>
>
> Two plot panels on the same device like:
>
> layout(1:2)
> plot(1:100)
> plot(1:100)
> layout(1)
>
>
> => Yes that's what I want
>
> The second panel for southern species richness, do you mean you want
> the
> plot to go like this:
>
> plot(1:100, xlim = c(100,1))
>
>
> => Yes, too.
>
>
>
>
>
>
--
%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%
Dr. Gavin Simpson [t] +44 (0)20 7679 0522
ECRC, UCL Geography, [f] +44 (0)20 7679 0565
Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk
Gower Street, London [w] http://www.ucl.ac.uk/~ucfagls/
UK. WC1E 6BT. [w] http://www.freshwaters.org.uk
%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%
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[R] Problem with terms of the form (a >1) and subsetting of a terms object
I have the following problem mydata <- data.frame(a=1:3) form <- ~ (a>1) - 1 model.matrix(form,mydata) a > 1FALSE a > 1TRUE 1 1 0 2 0 1 3 0 1 ... However model.matrix(update(terms(form)[1],~.-1),mydata) (Intercept) a > 1 - 1TRUE 1 1 1 2 1 1 3 1 1 ... Taking terms(form)[1] extracts the formula with only the first term, but the parentheses are thrown away, and after the update, I get a different formula. Is that supposed to be so? Thanks, Niels -- Niels Richard Hansen Web: www.math.ku.dk/~richard Associate Professor Email: niels.r.han...@math.ku.dk Department of Mathematical Sciences nielsrichardhan...@gmail.com University of Copenhagen Skype: nielsrichardhansen.dk Universitetsparken 5 Phone: +45 353 20783 (office) 2100 Copenhagen Ø +45 2859 0765 (mobile) Denmark __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] percentage sign in expression
>On Wed, 2010-08-25 at 09:32 +0100, e-letter wrote:
>> On 24/08/2010, David Winsemius wrote:
>> >
>> > On Aug 24, 2010, at 9:37 AM, e-letter wrote:
>> >
>> >> Readers,
>> >>
>> >> According to the documentation for the function 'plotmath' there is no
>> >> apparent possibility to add the percent sign (%) to a plot function,
>> >
>> > Where did you see an assertion made???
>> >
>> Within R I entered the command:
>>
>> ?plotmath
>
>Surely you don't expect the R help pages to document everything that
>*isn't* possible with a given function? They'd be infinitely long! ;-)
>
No, following previous advice it would be preferable to add the
following (or similar) to the documentation:
'syntax' 'meaning'
...
'%' %
...
>What you meant to say was, "?plotmath doesn't show me how to add a %
>symbol to a plot. How do I do it?"
>
>> Also accessed using:
>>
>> help.start(browser="opera")
>>
>> Navigating the web browser page:
>>
>> packages
>> packages in /usr/lib/R/library
>> grdevices
>> plotmath
>>
>> In the list headed 'syntax' and 'meaning' within the section 'details'.
>>
>
>> > > TestChars <- function(sign=1, font=1, ...)
>> > + {
>> > +if(font == 5) { sign <- 1; r <- c(32:126, 160:254)
>> > +} else if (l10n_info()$MBCS) r <- 32:126 else r <- 32:255
>> > +if (sign == -1) r <- c(32:126, 160:255)
>> > +par(pty="s")
>> > +plot(c(-1,16), c(-1,16), type="n", xlab="", ylab="",
>> > + xaxs="i", yaxs="i")
>> > +grid(17, 17, lty=1)
>> > +for(i in r) try(points(i%%16, i%/%16, pch=sign*i, font=font,...))
>> > + }
>> > > TestChars(font=5)
>> >
>> > Notice that the "%" sign is three characters to the right (i.e.
>> > higher) of the "forall" symbol that is produced by the example code
>>
>> I can't see 'forall' in the code above.
>
>You need to run it in R. Then you will see a plot of the glyphs
>available in the symbol font. 'forall' is a mathematical symbol:
>
>http://en.wikipedia.org/wiki/Table_of_mathematical_symbols
>
>> > they offer. (The numbering proceeds from bottom up which confused me
>> > at first.)
>> >
>> What numbering?
>
>The numbering of the glyphs so you can use the number to draw the symbol
>you want. They are on the plot. You did run the code provided by David?
>
I did not read any instruction that it was code to apply! Anyway,
below is the result:
> TestChars <- function(sign=1, font=1, ...)
+ + {
+ +if(font == 5) { sign <- 1; r <- c(32:126, 160:254)
+ +} else if (l10n_info()$MBCS) r <- 32:126 else r <- 32:255
Error: syntax error, unexpected '}' in:
"
"
> +if (sign == -1) r <- c(32:126, 160:255)
Error in sign == -1 : comparison (1) is possible only for atomic and list types
> +par(pty="s")
Error in +par(pty = "s") : invalid argument to unary operator
> +plot(c(-1,16), c(-1,16), type="n", xlab="", ylab="",
+ + xaxs="i", yaxs="i")
Error: syntax error, unexpected EQ_ASSIGN, expecting ',' in "+
plot(c(-1,16), c(-1,16), type="n", xlab="", ylab="","
> +grid(17, 17, lty=1)
Error in int_abline(a = a, b = b, h = h, v = v, untf = untf, ...) :
plot.new has not been called yet
> +for(i in r) try(points(i%%16, i%/%16, pch=sign*i, font=font,...))
Error: object "r" not found
> + }
Error: syntax error, unexpected '}' in "+ }"
> TestChars(font=5)
Error: could not find function "TestChars"
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] list of closures
Wow, I was aware of R's lazy evaluation but didn't realize that was the cause
here. If the function, adder(), was not simple as I'd defined, I suppose I
could also force the evaluation of "x" in a wrapping function before passing it
to adder() as an alternative:
> adder <- function(x) function(y) x + y> plus <- Map(function(x) {force(x);
> adder(x)},c(one=1,two=2))> plus$one(1)[1] 2> plus$two(1)[1] 3
Thanks Gabor!
> From: ggrothendi...@gmail.com
> Date: Thu, 26 Aug 2010 00:28:34 -0400
> Subject: Re: [R] list of closures
> To: obsessiv...@hotmail.com
> CC: r-help@r-project.org
>
> On Thu, Aug 26, 2010 at 12:04 AM, Stephen T. wrote:
> >
> > Hi, I wanted to create a list of closures. When I use Map(), mapply(),
> > lapply(), etc., to create this list, it appears that the wrong arguments
> > are being passed to the main function. For example:
> > Main function:
> >> adder <- function(x) function(y) x + y
> > Creating list of closures with Map():
> >> plus <- Map(adder,c(one=1,two=2))> plus$one(1)[1] 3> plus$two(1)[1] 3
> > Examining what value was bound to "x":
> >> Map(function(fn) get("x",environment(fn)),plus)$one[1] 2$two[1] 2
> >
> > This is what I had expected:
> >> plus <- list(one=adder(1),two=adder(2))> plus$one(1)[1] 2> plus$two(1)[1] 3
> >> Map(function(fn) get("x",environment(fn)),plus)$one[1] 1$two[1] 2
> >
> > Anyone know what's going on? Thanks much!
>
> R uses lazy evaluation of function arguments. Try forcing x:
>
> > adder <- function(x) { force(x); function(y) x + y }
> > plus <- Map(adder,c(one=1,two=2))
> > plus$one(1)
> [1] 2
> > plus$two(1)
> [1] 3
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Re: [R] percentage sign in expression
>It's possible that my help page is different than yours. Right after
>the syntax/meaning description on mine (which is a Mac OSX system) is
>a paragraph:
>
>"The symbol font uses Adobe Symbol encoding so, for example, a lower
>case mu can be obtained either by the special symbol mu or by
>symbol("m"). This provides access to symbols that have no special
>symbol name, for example, the universal, or forall, symbol is
>symbol("\042"). To see what symbols are available in this way
>useTestChars(font=5) as given in the examples for points: some are
>only available on some devices."
>
>(In this case I would be surprised if the help pages were different
>because this makes a cross-reference to the examples in points. I am
>not surprised about cross-platform differences in descriptions of
>graphical devices and would have included a caveat if I were
>corresponding on rhelp about such. I suppose the font issues could be
>platform specific so if you want to correct me on this point, I will
>try to file it away. I did, however, give you the code needed to to
>display Symbols and it sounds further on that it succeeded)
>
I agree and seems that in this example that the documentation for
graphical devices is platform dependent. This should be stated in the
manual if not already. Please see previous post for the errors trying
to apply the commands provided.
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Re: [R] my function decided to stop working, problem with scoping?
To clarify, when I run the unlist(...) in R_GlobalEnv, I create all the necessary variables (including effect.distance). -- View this message in context: http://r.789695.n4.nabble.com/my-function-decided-to-stop-working-problem-with-scoping-tp2339228p2339263.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] list of closures
Hi Philippe, thanks for the suggestion - for my smaller problems I find that
closures are quicker to define and deploy. At larger scales, I've implemented
S4 objects with methods and attributes - though while elegant, I find that OO
(apart from what's built-in) adds significant biolerplate in defining classes,
so try to avoid it when I can... but I don't know much about the proto package
except that it loads with gsubfn() - might look into it. Thanks!Stephen
> Date: Thu, 26 Aug 2010 06:51:50 +0200
> From: phgrosj...@sciviews.org
> To: obsessiv...@hotmail.com
> CC: ggrothendi...@gmail.com; r-help@r-project.org
> Subject: Re: [R] list of closures
>
> Unless for learning R, you should really consider R.oo or proto packages
> that may be more convenient for you (but you don't provide enough
> context to tell).
> Best,
>
> Philippe
>
> On 26/08/10 06:28, Gabor Grothendieck wrote:
> > On Thu, Aug 26, 2010 at 12:04 AM, Stephen T.
> > wrote:
> >>
> >> Hi, I wanted to create a list of closures. When I use Map(), mapply(),
> >> lapply(), etc., to create this list, it appears that the wrong arguments
> >> are being passed to the main function. For example:
> >> Main function:
> >>> adder<- function(x) function(y) x + y
> >> Creating list of closures with Map():
> >>> plus<- Map(adder,c(one=1,two=2))> plus$one(1)[1] 3> plus$two(1)[1] 3
> >> Examining what value was bound to "x":
> >>> Map(function(fn) get("x",environment(fn)),plus)$one[1] 2$two[1] 2
> >>
> >> This is what I had expected:
> >>> plus<- list(one=adder(1),two=adder(2))> plus$one(1)[1] 2>
> >>> plus$two(1)[1] 3
> >>> Map(function(fn) get("x",environment(fn)),plus)$one[1] 1$two[1] 2
> >>
> >> Anyone know what's going on? Thanks much!
> >
> > R uses lazy evaluation of function arguments. Try forcing x:
> >
> >> adder<- function(x) { force(x); function(y) x + y }
> >> plus<- Map(adder,c(one=1,two=2))
> >> plus$one(1)
> > [1] 2
> >> plus$two(1)
> > [1] 3
> >
> > __
> > R-help@r-project.org mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
> >
> >
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Secant Method Convergence (Method to replicate Excel XIRR/IRR)
I couldn't resist the temptation.
Here is the example using the R package nleqslv.
library(nleqslv)
npv <- function (irr, cashFlow, times) sum(cashFlow / (1 + irr)^times)
CF <- c(-1000,500,500,500,500,500)
dates <-
c("1/1/2001","2/1/2002","3/1/2003","4/1/2004","5/1/2005","6/1/2006")
cfDate <- as.Date(dates,format="%m/%d/%Y")
times <- as.numeric(difftime(cfDate, cfDate[1], units="days"))/365.24
irr.start <- 0.05
# default Broyden ==> secant like method
nleqslv(irr.start, npv, control=list(trace=1), cashFlow=CF, times=times)
/Berend
--
View this message in context:
http://r.789695.n4.nabble.com/Secant-Method-Convergence-Method-to-replicate-Excel-XIRR-IRR-tp2338672p2339187.html
Sent from the R help mailing list archive at Nabble.com.
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Re: [R] Problem with terms of the form (a >1) and subsetting of a terms object
I think you need an I(), i.e., form <- ~ I(a > 1) - 1 I hope it helps. Best, Dimitris On 8/26/2010 9:17 AM, Niels Richard Hansen wrote: I have the following problem mydata <- data.frame(a=1:3) form <- ~ (a>1) - 1 model.matrix(form,mydata) a > 1FALSE a > 1TRUE 1 1 0 2 0 1 3 0 1 ... However model.matrix(update(terms(form)[1],~.-1),mydata) (Intercept) a > 1 - 1TRUE 1 1 1 2 1 1 3 1 1 ... Taking terms(form)[1] extracts the formula with only the first term, but the parentheses are thrown away, and after the update, I get a different formula. Is that supposed to be so? Thanks, Niels -- Dimitris Rizopoulos Assistant Professor Department of Biostatistics Erasmus University Medical Center Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands Tel: +31/(0)10/7043478 Fax: +31/(0)10/7043014 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to plot y-axis on the right of x-axis
On Thu, 2010-08-26 at 06:11 +0800, elaine kuo wrote:
> Thank you.
> The answers provides the right direction and a code was written accordingly.
>
> One more request, if the label of axis X wants to be drawn from 5 to 1 (left
> to right)
> rather than 1 to 5, is it fine to change axis (4, at = NULL) ?
> If so, which value should be input ?
If I understand you correctly, No. The data have been plotted assuming
xlim of (more or less) c(1, 5). All axis can do is change the labels and
where the marks are placed. You need to manually set the xlim argument
to plot to get the data plotted the other way round.
plot(1:5, axes = FALSE, xlim = c(5,1))
axis(1)
axis(4)
box()
Notice that all we've done here is change the limits for the x axis in
the plot() call, we haven't had to change any of the other code.
Note also that axis(side = 4) is on the one on the right so doesn't
pertain to the x-axis. If you want to know how to use axis more
flexibly, read ?axis as it explains there what 'at' is a requires as
input from you. For example, we might want to label at different points,
not 1,2,3,4,5 etc:
plot(1:5, axes = FALSE, xlim = c(5,1))
axis(1)
axis(4, at = seq(1.5, 4.5, by = 1))
box()
HTH
G
>
> Thanks again.
>
> Elaine
>
> code
> plot(1:5,axes=FALSE)
> axis(1)
> axis(4)
> box()
>
>
>
>
> On Wed, Aug 25, 2010 at 5:39 PM, Jim Lemon wrote:
>
> > On 08/25/2010 09:12 AM, elaine kuo wrote:
> >
> >> Dear List,
> >>
> >> I have a richness data distributing across 20 N to 20 S latitude. (120
> >> E-140
> >> E longitude).
> >>
> >>
> >> I would like to draw the richness in the north hemisphere and a regression
> >> line in the plot
> >> (x-axis: latitude, y-axis: richness in the north hemisphere).
> >> The above demand is done using plot.
> >>
> >> Then, south hemisphere richness and regression are required to be
> >> generated
> >> using
> >> the same y-axis above but an x-axis on the left side of the y-axis.
> >> (The higher latitude in the south hemisphere, the left it would move)
> >>
> >> Please kindly share how to design the south plot and regression line for
> >> richness.
> >> Also, please advise if any more info is in need.
> >>
> >> Hi Elaine,
> > Changing the position of the axes is easily done by changing the "side" and
> > "pos" arguments of the "axis" function. If you want to move the y-axis to
> > the right (or as you state it, the x axis to the left):
> >
> > # y axis on the left
> > plot(1:5,axes=FALSE)
> > axis(1)
> > axis(2)
> > # add a y axis one third of the way to the right
> > xylim<-par("usr")
> > axis(2,pos=xylim[1]+diff(xylim[1:2])/3)
> > # add another y axis two thirds of the way
> > axis(4,pos=xylim[2]-diff(xylim[1:2])/3)
> > # add one more on the right
> > axis(4)
> >
> > You can move the x axis up and down using the same tricks.
> >
> > Jim
> >
>
> [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
--
%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%
Dr. Gavin Simpson [t] +44 (0)20 7679 0522
ECRC, UCL Geography, [f] +44 (0)20 7679 0565
Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk
Gower Street, London [w] http://www.ucl.ac.uk/~ucfagls/
UK. WC1E 6BT. [w] http://www.freshwaters.org.uk
%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] About choosing file
Hi folks, Following command only works on Windows > Test01=read.table(file.choose(), header=TRUE) It popup open a dialog box for choosing file. But it doesn't work on Linux (Ubuntu); > Test01=read.table(file.choose(), header=TRUE) Enter file name: Entering file name doesn't work. Please advise. TIA B.R. Stephen L __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] lattice help required
On Wed, 2010-08-25 at 11:30 -0400, RICHARD M. HEIBERGER wrote:
> Kay,
>
> doe this do what you want?
>
>
> dotplot(y1+y2 ~ facs$Treatment|facs$Sites,
> outer=TRUE,
> scales = list(x = list(rot = 90, tck=c(1,0))),
> ylab=c("y1", "y2"),
> xlab=c("Site 1", "Site 2"),
> strip=FALSE)
This would be better, to avoid abusing the poor R formula handling
sugar:
dotplot(y1 + y2 ~ Treatment | Sites, ## no $
## Supply a data argument
data = facs,
## as before
outer=TRUE,
scales = list(x = list(rot = 90, tck=c(1,0))),
ylab=c("y1", "y2"),
xlab=c("Site 1", "Site 2"),
strip=FALSE)
G
>
> On Wed, Aug 25, 2010 at 11:04 AM, Kay Cichini wrote:
>
> >
> > hello,
> >
> > i want to stack two lattice plots beneath each other using one x-axis and
> > sharing the same text-panels,
> > like:
> > #
> > library(lattice)
> >
> > y1 <- rnorm(100,100,10)
> > y2 <- rnorm(100,10,1)
> > facs<-expand.grid(Sites=rep(c("Site I","Site II"),25),Treatment=c("A","B"))
> > pl1<-dotplot(y1 ~ facs$Treatment|facs$Sites,
> > scales = list(x = list(rot = 90, tck=c(1,0
> > pl2<-dotplot(y2 ~ facs$Treatment|facs$Sites,
> > scales = list(x = list(rot = 90, tck=c(1,0
> >
> > print(pl1, split=c(1,2,1,2), more=TRUE)
> > print(pl2, split=c(1,1,1,2))
> > #
> >
> > but as said, ideally the plots should be stacked with only the lower plot
> > giving the x-axis annotation
> > and only the upper plot with text-panels.
> >
> > thanks a lot,
> > kay
> >
> > -
> >
> > Kay Cichini
> > Postgraduate student
> > Institute of Botany
> > Univ. of Innsbruck
> >
> >
> > --
> > View this message in context:
> > http://r.789695.n4.nabble.com/lattice-help-required-tp2338382p2338382.html
> > Sent from the R help mailing list archive at Nabble.com.
> >
> > __
> > R-help@r-project.org mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide
> > http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
> >
>
> [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
--
%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%
Dr. Gavin Simpson [t] +44 (0)20 7679 0522
ECRC, UCL Geography, [f] +44 (0)20 7679 0565
Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk
Gower Street, London [w] http://www.ucl.ac.uk/~ucfagls/
UK. WC1E 6BT. [w] http://www.freshwaters.org.uk
%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] reliability of R-Forge?
On Thu, 2010-08-26 at 02:30 -0400, David Kane wrote: > How reliable is R-Forge? http://r-forge.r-project.org/ > > It is down now (for me). Reporting "R-Forge Could Not Connect to Database: " > > I have just started to use it for a project. It has been down for > several hours (at least) on different occasions over the last couple > of days. Is that common? Will it be more stable soon? > > Apologies if this is not an appropriate question for R-help. > > Dave Kane In my experience, very. Although in the past few days there have been two periods of downtime (error as you report). The resource has been going through lots of changes over the past several months so some of the features of R-Forge have been offline (like the full build process) from time to time, but this is the first full outage I've been aware of for sometime. R-Forge hosts hundreds of projects and packages, which also attests to general reliability. G -- %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% Dr. Gavin Simpson [t] +44 (0)20 7679 0522 ECRC, UCL Geography, [f] +44 (0)20 7679 0565 Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk Gower Street, London [w] http://www.ucl.ac.uk/~ucfagls/ UK. WC1E 6BT. [w] http://www.freshwaters.org.uk %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Change value of a slot of an S4 object within a method.
OK, I admit. It will never win a beauty price, but I won't either so we go pretty well together. Seriously, I am aware this is about the worst way to solve such a problem. But as I said, rewriting the class definitions wasn't an option any more. I'll definitely take your advice for the next project. Cheers Joris On Wed, Aug 25, 2010 at 9:36 PM, Steve Lianoglou wrote: > Howdy, > My eyes start to gloss over on their first encounter of `substitute` > ... add enough `eval`s and (even) an `expression` (!) to that, and > you'll probably see me running for the hills ... but hey, if it makes > sense to you, more power to you ;-) > > Glad you found a fix that works, > > -steve > > -- > Steve Lianoglou > Graduate Student: Computational Systems Biology > | Memorial Sloan-Kettering Cancer Center > | Weill Medical College of Cornell University > Contact Info: http://cbio.mskcc.org/~lianos/contact > -- Joris Meys Statistical consultant Ghent University Faculty of Bioscience Engineering Department of Applied mathematics, biometrics and process control tel : +32 9 264 59 87 joris.m...@ugent.be --- Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] percentage sign in expression
On Thu, 2010-08-26 at 08:17 +0100, e-letter wrote:
> >On Wed, 2010-08-25 at 09:32 +0100, e-letter wrote:
> >> On 24/08/2010, David Winsemius wrote:
> >> >
> >> > On Aug 24, 2010, at 9:37 AM, e-letter wrote:
> >> >
> >> >> Readers,
> >> >>
> >> >> According to the documentation for the function 'plotmath' there is no
> >> >> apparent possibility to add the percent sign (%) to a plot function,
> >> >
> >> > Where did you see an assertion made???
> >> >
> >> Within R I entered the command:
> >>
> >> ?plotmath
> >
> >Surely you don't expect the R help pages to document everything that
> >*isn't* possible with a given function? They'd be infinitely long! ;-)
> >
> No, following previous advice it would be preferable to add the
> following (or similar) to the documentation:
>
> 'syntax' 'meaning'
> ...
> '%' %
> ...
Should we also get the R devs to add
'a' a
'b' b
?
This is just literal text. And as such is not what plotmath is all
about, which is providing notation for common mathematical symbology.
'%' is not, it is plain text. And that is handled by expression() or
bquote() etc. Nothing to do with plotmath.
Note that it does go into details of how to view the glyphs available
with the symbol font on your machine (using the TestChars() function
available in the example section of ?points ) and then plot these glyphs
in both the main body of the help page and in the example. So it is
documented there, but again, do you want the devs to do:
'syntax' 'Meaning'
symbol("\042")FOO
symbol("\043")BAR
How do they handle different availability of glyphs on the various OSes
that there might be etc? Hence the TestChars() and generic instructions
on how to draw all glyphs in your systems symbol font. It just requires
a little bit more effort on your part to find the correct \xxx code.
G
> >What you meant to say was, "?plotmath doesn't show me how to add a %
> >symbol to a plot. How do I do it?"
> >
> >> Also accessed using:
> >>
> >> help.start(browser="opera")
> >>
> >> Navigating the web browser page:
> >>
> >> packages
> >> packages in /usr/lib/R/library
> >> grdevices
> >> plotmath
> >>
> >> In the list headed 'syntax' and 'meaning' within the section 'details'.
> >>
> >
> >> > > TestChars <- function(sign=1, font=1, ...)
> >> > + {
> >> > +if(font == 5) { sign <- 1; r <- c(32:126, 160:254)
> >> > +} else if (l10n_info()$MBCS) r <- 32:126 else r <- 32:255
> >> > +if (sign == -1) r <- c(32:126, 160:255)
> >> > +par(pty="s")
> >> > +plot(c(-1,16), c(-1,16), type="n", xlab="", ylab="",
> >> > + xaxs="i", yaxs="i")
> >> > +grid(17, 17, lty=1)
> >> > +for(i in r) try(points(i%%16, i%/%16, pch=sign*i, font=font,...))
> >> > + }
> >> > > TestChars(font=5)
> >> >
> >> > Notice that the "%" sign is three characters to the right (i.e.
> >> > higher) of the "forall" symbol that is produced by the example code
> >>
> >> I can't see 'forall' in the code above.
> >
> >You need to run it in R. Then you will see a plot of the glyphs
> >available in the symbol font. 'forall' is a mathematical symbol:
> >
> >http://en.wikipedia.org/wiki/Table_of_mathematical_symbols
> >
> >> > they offer. (The numbering proceeds from bottom up which confused me
> >> > at first.)
> >> >
> >> What numbering?
> >
> >The numbering of the glyphs so you can use the number to draw the symbol
> >you want. They are on the plot. You did run the code provided by David?
> >
> I did not read any instruction that it was code to apply! Anyway,
> below is the result:
>
> > TestChars <- function(sign=1, font=1, ...)
> + + {
> + +if(font == 5) { sign <- 1; r <- c(32:126, 160:254)
> + +} else if (l10n_info()$MBCS) r <- 32:126 else r <- 32:255
> Error: syntax error, unexpected '}' in:
> "
> "
> > +if (sign == -1) r <- c(32:126, 160:255)
> Error in sign == -1 : comparison (1) is possible only for atomic and list
> types
> > +par(pty="s")
> Error in +par(pty = "s") : invalid argument to unary operator
> > +plot(c(-1,16), c(-1,16), type="n", xlab="", ylab="",
> + + xaxs="i", yaxs="i")
> Error: syntax error, unexpected EQ_ASSIGN, expecting ',' in "+
> plot(c(-1,16), c(-1,16), type="n", xlab="", ylab="","
> > +grid(17, 17, lty=1)
> Error in int_abline(a = a, b = b, h = h, v = v, untf = untf, ...) :
> plot.new has not been called yet
> > +for(i in r) try(points(i%%16, i%/%16, pch=sign*i, font=font,...))
> Error: object "r" not found
> > + }
> Error: syntax error, unexpected '}' in "+ }"
> > TestChars(font=5)
> Error: could not find function "TestChars"
>
> __
> R-help@r-project.org mailing list
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
--
%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%
Dr. Gavin Simpson [t] +44 (0)20 7679 0522
ECR
Re: [R] Removing inter-bar spaces in barchart
On Thu, Aug 26, 2010 at 5:22 AM, Jonathan Greenberg wrote: > Oops, small typo, should be: > > > barchart_test_heights=sin(c(1:100)) > barchart_test_bins=c(c(1:50),c(1:50)) > groups=c(rep(1,50),rep(2,50)) > > # Wish below didn't have spaces! > barchart(barchart_test_bins~barchart_test_heights,groups=groups) barchart(barchart_test_bins~barchart_test_heights,groups=groups, box.width = 1) (see ?panel.barchart.) -Deepayan __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Looking for an image (R 64-bit on Linux 64-bit) on Amazon EC2
It would presumably help you get a good answer if you explained what Amazon EC2 was, and what limitations are involved in that platform. You got a very good answer for standard 64-bit Linux, because that was the recognizable part of your query. To go beyond that we need more information. *Can* you install software from source? What version of linux is it running? Are repos for that version available? Etc. In other words, how do you install any software? Sarah On Wed, Aug 25, 2010 at 9:12 PM, noclue_ wrote: > > >>> You have a 64 bit Linux? If so... > >>>Dowload the sources > > Do you mean download Linux kernel source code and then compile it on Amazon > EC2? > > > -- > View this message in context: > http://r.789695.n4.nabble.com/Looking-for-an-image-R-64-bit-on-Linux-64-bit-on-Amazon-EC2-tp2338938p2339072.html > Sent from the R help mailing list archive at Nabble.com. > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. > -- Sarah Goslee http://www.stringpage.com http://www.astronomicum.com http://www.functionaldiversity.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Surprising behaviour survey-package with missing values
[I already sent this message to the list almost a day ago using an
other email address, but that message does not seem to get through. In
case it finally does get through, I appologize for the the double
posting]
Dear list,
I got some surprising results when using the svytotal routine from the
survey package with data containing missing values.
Some example code demonstrating the behaviour is included below.
I have a stratified sampling design where I want to estimate the total
income. In some strata some of the incomes are missing. I want to
ignore these missing incomes. I would have expected that
svytotal(~income, design=mydesign, na.rm=TRUE) would do the trick.
However, when calculating the estimates 'by hand' the estimates were
different from those obtained from svytotal. The estimated mean
incomes do agree with each other. It seems that using the na.rm option
with svytotal is the same as replacing the missing values with zero's,
which is not what I would have expected, especially since this
behaviour seems to differ from that of svymean. Is there a reason for
this behaviour?
I can of course remove the missing values myself before creating the
survey object. However, with many different variables with different
missing values, this is not very practical. Is there an easy way to
get the behaviour I want?
Thanks for your help.
With regards,
Jan van der Laan
=== EXAMPLE ===
library(survey)
library(plyr)
# generate some data
data <- data.frame(
id = 1:20,
stratum = rep(c("a", "b"), each=10),
income = rnorm(20, 100),
n = rep(c(100, 200), each=10)
)
data$income[5] <- NA
# calculate mean and total income for every stratum using survey package
des <- svydesign(ids=~id, strata=~stratum, data=data, fpc=~n)
svyby(~income, by=~stratum, FUN=svytotal, design=des, na.rm=TRUE)
mn <- svyby(~income, by=~stratum, FUN=svymean, design=des, na.rm=TRUE)
mn
n <- svyby(~n, by=~stratum, FUN=svymean, design=des)
# total does not equal mean times number of persons in stratum
mn[2] * n[2]
# calculate mean and total income 'by hand'. This does not give the same total
# as svytotal, but it does give the same mean
ddply(data, .(stratum), function(d) {
data.frame(
mean = mean(d$income, na.rm=TRUE),
n = mean(d$n),
total = mean(d$income, na.rm=TRUE) * mean(d$n)
)
})
# when we set income to 0 for missing cases and repeat the previous estimation
# we get the same answer as svytotal (but not svymean)
data2 <- data
data2$income[is.na(data$income )] <- 0
ddply(data2, .(stratum), function(d) {
data.frame(
mean = mean(d$income, na.rm=TRUE),
n = mean(d$n),
total = mean(d$income, na.rm=TRUE) * mean(d$n)
)
})
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Re: [R] reliability of R-Forge?
David Kane kanecap.com> writes: > > How reliable is R-Forge? http://r-forge.r-project.org/ > > It is down now (for me). Reporting "R-Forge Could Not Connect to Database: " > > I have just started to use it for a project. It has been down for > several hours (at least) on different occasions over the last couple > of days. Is that common? Will it be more stable soon? > > Apologies if this is not an appropriate question for R-help. > Dave, This is rather a subject for R-devel. Ignoring this inappropriateness: yes, indeed, R-Forge has been flaky lately. The database was disconnected for the whole weekend, came back on Monday, and is gone again. It seems that mailing lists and email alerts of commits were not working even when the basic R-Forge was up. I have sent two messages to r-fo...@r-project.org on these problems. I haven't got no response, but soon after the first message the Forge woke up, and soon after the second message it went down. Since I'm not Bayeasian, I don't know what to say about the effect of my messages. Cheers, Jari Oksanen __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Puzzle
I have data similar to this: Location Surveyor Result A1 83 A2 76 A3 45 B1 71 B4 67 C2 23 C5 12 D3 34 E4 75 F4 46 G5 90 etc (5 million records in total) I need to divide the data to many subsets then randomly select 5 different locations and 5 different surveyors (one at each of the 5 randomly selected locations) for each subset. The function I have written basically picks five locations and then 1 surveyor in each location, checks that there are five different surveyors and if there isn't tries again. The problem is that for some subsets this doesn't work. Some subsets don't have enough locations/surveyors or both, but this can be checked for easily. The problem subsets do have enoughs locations and surveyors but still cannot produce 5 locations each with a different surveyor. The matrix below demonstrates such a subset: locations A B C D E 1 1 0 0 0 0 Surveyors 2 1 0 0 0 0 3 1 0 0 0 0 4 1 0 0 0 0 5 1 1 1 1 1 I cannot think of a way to check for such a situation and therefore I have simply programmed the function to give up after 100 attempts if it can't find a solution. This is not very satisfactory however as the analysis takes a very long time to run and it would also be very useful useful for me to know how many suitable solution there are. I reckon some of you clever folk out there must be able to think of a better solution. Any advice appreciated, Ben __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problem with terms of the form (a >1) and subsetting of a terms object
On 26/08/10 09.30, Dimitris Rizopoulos wrote: I think you need an I(), i.e., form <- ~ I(a > 1) - 1 Yes, it solves the concrete problem, but does not really answer the question. Let me rephrase. Should the use of terms like (a>1) be discouraged in R and replaced by I(a>1) systematically, or is the behavior below unexpected? Best, Niels I hope it helps. Best, Dimitris On 8/26/2010 9:17 AM, Niels Richard Hansen wrote: I have the following problem mydata <- data.frame(a=1:3) form <- ~ (a>1) - 1 model.matrix(form,mydata) a > 1FALSE a > 1TRUE 1 1 0 2 0 1 3 0 1 ... However model.matrix(update(terms(form)[1],~.-1),mydata) (Intercept) a > 1 - 1TRUE 1 1 1 2 1 1 3 1 1 ... Taking terms(form)[1] extracts the formula with only the first term, but the parentheses are thrown away, and after the update, I get a different formula. Is that supposed to be so? Thanks, Niels -- Niels Richard Hansen Web: www.math.ku.dk/~richard Associate Professor Email: niels.r.han...@math.ku.dk Department of Mathematical Sciences nielsrichardhan...@gmail.com University of Copenhagen Skype: nielsrichardhansen.dk Universitetsparken 5 Phone: +45 353 20783 (office) 2100 Copenhagen Ø +45 2859 0765 (mobile) Denmark __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] print method for str?
Hello useRs and guRus, I was trying to add the support of str() in the ascii package, and I realized that str() does not have a print method. It uses cat() from inside the function and returns nothing. Since it is not usual in R, I wonder why? Is there a particular reason? Thank you very much for your attention on... well, this unimportant question. david [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Problems when Apply a script to a list
Dear users,
***I have a function f to simulate data from a model (example below used
only to show my problems)
f<-function(n,mean1){
a<-matrix(rnorm(n, mean1 , sd = 1),ncol=5)
b<-matrix(runif(n),ncol=5)
data<-rbind(a,b)
out<-data
out}
*I want to simulate 1000 datasets (here only 5) so I use
S<-list()
for (i in 1:5){
S[[i]]<-f(n=10,mean1=0)}
**I have a very complicated function for estimation of a model which I
want to apply to Each one of the above simulated datasets
fun<-function(data){data<-as.matrix(data)
sink(' Example.txt',append=TRUE)
cat("\n***\nEstimation
\n\nDataset Sim : ",
i )
d<-data%*%t(data)
s<-solve(d)
print(s)
out<-s
out
}
results<-list()
for (i in 1:5){results[[i]]<-fun(data=S[[i]])}
My questions are:
1) for some datasets system is computational singular and this causes
execution of the for to stop.By this way I have only results until this
problem happens.How can I pass over the execution for this step and have
results for All other datasets for which function fun is applicable?
2) After some runs to my program, I receive this error message someError in
sink("output.txt") : sink stack is full . How can I solve this problem, as I
want to have results of my program for 1000 datasets.
3) Using for is the correct way to run my proram for a list
Thanks alot
Evgenia
--
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http://r.789695.n4.nabble.com/Problems-when-Apply-a-script-to-a-list-tp2339403p2339403.html
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Re: [R] Showing a line function on the ploat area
On 08/26/2010 01:06 AM, Neta wrote: I have a collection of results. I use R to get the linearization presented. how can I get R to show the equation and R^2 value on the plot area next to the graph? Hi Neta, You can use functions like textbox or boxed.labels (plotrix package) to display arbitrary text: par(mar=c(6,4,4,2)) plot(1:5) par(xpd=TRUE) library(plotrix) boxed.labels(4,0, "response~x1+x2\nr-squared = 0.7326",FALSE) par(xpd=FALSE Jim __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Removing inter-bar spaces in barchart
On 08/26/2010 06:36 AM, Dennis Murphy wrote: ... It looks like he's using barchart() in lattice, which does have a box.ratio argument. That's why I didn't suggest using barp with width=0.5 Jim __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] using an objects contents in a text string
Thankyou!!! I knew this was so simple and I had tried paste but with the wrong syntax so I moved on. This opens a whole new world to me!! -- View this message in context: http://r.789695.n4.nabble.com/using-an-objects-contents-in-a-text-string-tp2338061p2339437.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] frequency table for a list matching some conditions
Dear all,
I have a list that contains 3 sublists( x1, x2, x3)
mylist<-list(x1=c("A","A","A","B","C","Z","Y"),x2=c("D","D","E","E","F","Z","X"),x3=c("A","A","A","B","Y","Z"))
mylist
$x1
[1] "A" "A" "A" "B" "C" "Z" "Y"
$x2
[1] "D" "D" "E" "E" "F" "Z" "X"
$x3
[1] "A" "A" "A" "B" "Y" "Z"
I also have another list, that contains some specific elements.
newlist<-c("A","B","C","D","E","F")
newlist
[1] "A" "B" "C" "D" "E" "F"
I want to match mylist[1], mylist[2], etc with newlist and I am looking for the
following output:
A B C D E F
x1 3 1 1 NA NA NA
x2 NA NA NA 2 2 1
x3 3 1 NA NA NA NA
Any thoughts?
Many thanks
Olga
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Re: [R] vector allocation error
The answer is indeed a LINUX cluster with more memory. Thanks, Josquin -- View this message in context: http://r.789695.n4.nabble.com/vector-allocation-error-tp2333299p2339451.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] frequency table for a list matching some conditions
one way is the following:
mylist <- list(x1 = c("A","A","A","B","C","Z","Y"),
x2 = c("D","D","E","E","F","Z","X"),
x3 = c("A","A","A","B","Y","Z"))
newlist <- c("A","B","C","D","E","F")
tab <- t(sapply(mylist, function (x)
table(factor(x, levels = newlist
tab[tab == 0] <- NA
tab
I hope it helps.
Best,
Dimitris
On 8/26/2010 11:47 AM, Olga Lyashevska wrote:
Dear all,
I have a list that contains 3 sublists( x1, x2, x3)
mylist<-list(x1=c("A","A","A","B","C","Z","Y"),x2=c("D","D","E","E","F","Z","X"),x3=c("A","A","A","B","Y","Z"))
mylist
$x1
[1] "A" "A" "A" "B" "C" "Z" "Y"
$x2
[1] "D" "D" "E" "E" "F" "Z" "X"
$x3
[1] "A" "A" "A" "B" "Y" "Z"
I also have another list, that contains some specific elements.
newlist<-c("A","B","C","D","E","F")
newlist
[1] "A" "B" "C" "D" "E" "F"
I want to match mylist[1], mylist[2], etc with newlist and I am looking for the
following output:
A B C D E F
x1 3 1 1 NA NA NA
x2 NA NA NA 2 2 1
x3 3 1 NA NA NA NA
Any thoughts?
Many thanks
Olga
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--
Dimitris Rizopoulos
Assistant Professor
Department of Biostatistics
Erasmus University Medical Center
Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands
Tel: +31/(0)10/7043478
Fax: +31/(0)10/7043014
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[R] ggplot inside cycle
Dear all
I want to save several ggplots in one pdf document. I tried this
for (i in names(iris)[2:4]) {
p<-ggplot(iris, aes(x=Sepal.Length, y=iris[,i], colour=Species))
p+geom_point(aes(size=3))
}
with different variations of y input but was not successful. In past I
used qplot in similar fashion which worked
for(i in names(mleti)[7:15]) print(qplot(sito, mleti1[,i],
facets=~typ,ylab=i, geom=c("point", "line"), colour=ordered(minuty),
data=mleti1))
So I wonder if anybody used ggplot in cycle and how to solve input of
variables throughout cycle
Thank you
Petr
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Re: [R] ggplot inside cycle
You haven't wrapped p in the print command, which is one of the ways to
make sure the plot gets printed when we need it.
print(p+geom_point(aes(size=3))) does the trick
On 08/26/2010 06:08 AM, Petr PIKAL wrote:
Dear all
I want to save several ggplots in one pdf document. I tried this
for (i in names(iris)[2:4]) {
p<-ggplot(iris, aes(x=Sepal.Length, y=iris[,i], colour=Species))
p+geom_point(aes(size=3))
}
with different variations of y input but was not successful. In past I
used qplot in similar fashion which worked
for(i in names(mleti)[7:15]) print(qplot(sito, mleti1[,i],
facets=~typ,ylab=i, geom=c("point", "line"), colour=ordered(minuty),
data=mleti1))
So I wonder if anybody used ggplot in cycle and how to solve input of
variables throughout cycle
Thank you
Petr
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Re: [R] Problems when Apply a script to a list
Answers below.
On Thu, Aug 26, 2010 at 11:20 AM, Evgenia wrote:
>
> Dear users,
>
> ***I have a function f to simulate data from a model (example below used
> only to show my problems)
>
> f<-function(n,mean1){
> a<-matrix(rnorm(n, mean1 , sd = 1),ncol=5)
> b<-matrix(runif(n),ncol=5)
> data<-rbind(a,b)
> out<-data
> out}
>
> *I want to simulate 1000 datasets (here only 5) so I use
> S<-list()
>
> for (i in 1:5){
> S[[i]]<-f(n=10,mean1=0)}
>
> **I have a very complicated function for estimation of a model which I
> want to apply to Each one of the above simulated datasets
>
> fun<-function(data){data<-as.matrix(data)
> sink(' Example.txt',append=TRUE)
> cat("\n***\nEstimation
> \n\nDataset Sim : ",
> i )
> d<-data%*%t(data)
> s<-solve(d)
> print(s)
> out<-s
> out
> }
> results<-list()
> for (i in 1:5){results[[i]]<-fun(data=S[[i]])}
>
>
> My questions are:
> 1) for some datasets system is computational singular and this causes
> execution of the for to stop.By this way I have only results until this
> problem happens.How can I pass over the execution for this step and have
> results for All other datasets for which function fun is applicable?
see ?try, or ?tryCatch.
I'd do something in the line of
for(i in 1:5){
tmp <- try(fun(data=S[[i]]))
results[[i]] <- ifelse(is(tmp,"try-error"),NA,tmp)
}
Alternatively, you could also use lapply :
results <- lapply(S,function(x{
tmp <- try(fun(data=x))
ifelse(is(tmp,"try-error"),NA,tmp)
})
>
> 2) After some runs to my program, I receive this error message someError in
> sink("output.txt") : sink stack is full . How can I solve this problem, as I
> want to have results of my program for 1000 datasets.
That is because you never empty the sink. add sink() after the last
line you want in the file. This will empty the sink buffer to the
file. Otherwise R keeps everything in the memory, and that gets too
full after a while.
>
> 3) Using for is the correct way to run my proram for a list
See the lapply solution.
>
> Thanks alot
>
> Evgenia
>
> --
> View this message in context:
> http://r.789695.n4.nabble.com/Problems-when-Apply-a-script-to-a-list-tp2339403p2339403.html
> Sent from the R help mailing list archive at Nabble.com.
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
Joris Meys
Statistical consultant
Ghent University
Faculty of Bioscience Engineering
Department of Applied mathematics, biometrics and process control
tel : +32 9 264 59 87
joris.m...@ugent.be
---
Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php
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Re: [R] frequency table for a list matching some conditions
Thanks Dimitris,
It works nicely!
Regards,
Olga
On Thu, 2010-08-26 at 11:55 +0200, Dimitris Rizopoulos wrote:
> one way is the following:
>
> mylist <- list(x1 = c("A","A","A","B","C","Z","Y"),
> x2 = c("D","D","E","E","F","Z","X"),
> x3 = c("A","A","A","B","Y","Z"))
> newlist <- c("A","B","C","D","E","F")
>
>
> tab <- t(sapply(mylist, function (x)
> table(factor(x, levels = newlist
> tab[tab == 0] <- NA
> tab
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Re: [R] ggplot inside cycle
Thanks. I knew I forgot something obvious.
Petr
r-help-boun...@r-project.org napsal dne 26.08.2010 12:29:52:
> You haven't wrapped p in the print command, which is one of the ways to
> make sure the plot gets printed when we need it.
> print(p+geom_point(aes(size=3))) does the trick
> On 08/26/2010 06:08 AM, Petr PIKAL wrote:
> > Dear all
> >
> > I want to save several ggplots in one pdf document. I tried this
> >
> > for (i in names(iris)[2:4]) {
> > p<-ggplot(iris, aes(x=Sepal.Length, y=iris[,i], colour=Species))
> > p+geom_point(aes(size=3))
> > }
> >
> > with different variations of y input but was not successful. In past I
> > used qplot in similar fashion which worked
> >
> > for(i in names(mleti)[7:15]) print(qplot(sito, mleti1[,i],
> > facets=~typ,ylab=i, geom=c("point", "line"), colour=ordered(minuty),
> > data=mleti1))
> >
> > So I wonder if anybody used ggplot in cycle and how to solve input of
> > variables throughout cycle
> >
> > Thank you
> >
> > Petr
> >
> > __
> > R-help@r-project.org mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
> >
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
__
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[R] equalize function with zero and convert it
Dear all, I want to equalize a symbolic derivative (of a function with two variables) with zero and convert it to a variable, e.g. x. I'm computing the derivative by: (found it in the archive) library(Ryacas) f <- function(x,y) (100-x-y)*x-10*x yacas(f) # register f with yacas Df <- f body(Df) <- yacas(expression(deriv(f(x,y),x)))[[1]] Df R-Output: function (x, y) 100 - x - y - x - 10 Questions : 1. How can I equalize the derivative(140-x-y-x-20) with zero? 2. How can I convert it to x? I want something like: 45-0.5*y=x With only one variable or two non-symbolic variables it's working with uniroot(f, c(-10, 10))$root. Thank you! Immanuel Seeger [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] relimp
I am trying to use Rcmdr 1.5_4 with R-2.11.1 (order to run to run the new version of misdist-0.5.3 which is built under R-2.11.1). however relimp is required for Rcmdr and the version of relimp_1.0.1 downloaded from CRAN will not work with the latest version of Rcmdr (I get error message telling me to reload it). Is there any way round this problem or will there be a new version relimp that is compatible? Any advice would be gratefully received. many thanks John Prof. Barrett Univ. of Aberystwyth [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Quick GREP challenge
> grep("f[0-9]+=", "f1=5,f22=3,", value = T)
[1] "f1=5,f22=3,"
How do I make the line output c("f1", "f22") instead? (Actually, c(1,22)
would be even better).
Thank you.
--
View this message in context:
http://r.789695.n4.nabble.com/Quick-GREP-challenge-tp2339486p2339486.html
Sent from the R help mailing list archive at Nabble.com.
__
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Re: [R] how to plot y-axis on the right of x-axis
On 08/26/2010 08:11 AM, elaine kuo wrote: ... One more request, if the label of axis X wants to be drawn from 5 to 1 (left to right) rather than 1 to 5, is it fine to change axis (4, at = NULL) ? If so, which value should be input ? Hi Elaine, There are a couple of ways to do this. One is to use the revaxis function (plotrix) that reverses the conventional direction of the axes by silently changing the sign of the values. If you use this, you must remember to change the sign of the position of your y axes. Another way is the plot in the normal direction but print the tick labels backwards: plot(1:5,xaxt="n") axis(1,1:5,5:1) Using this method, you have to remember that your labels don't correspond to the values, so if you wanted a y axis positioned at x=4, you would have to use: axis(2,1:5,pos=2) Jim __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] lattice help required
Hi
Coming a bit late to the thread another way to reduce the space between the
panels may be along the lines of Deepayan's reply with the layout.widths in
http://finzi.psych.upenn.edu/R/Rhelp02/archive/43626.html
I regularly use this and the latticeExtra addition will be welcome - I
have not downloaded it yet.
Also the 'useOuterStrips' from the latticeExtra package.may be of
assistance in the setting the strips.
Regards
Duncan
Duncan Mackay
Department of Agronomy and Soil Science
University of New England
ARMIDALE NSW 2351
Email home: mac...@northnet.com.au
At 17:34 26/08/2010, you wrote:
On Wed, 2010-08-25 at 11:30 -0400, RICHARD M. HEIBERGER wrote:
> Kay,
>
> doe this do what you want?
>
>
> dotplot(y1+y2 ~ facs$Treatment|facs$Sites,
> outer=TRUE,
> scales = list(x = list(rot = 90, tck=c(1,0))),
> ylab=c("y1", "y2"),
> xlab=c("Site 1", "Site 2"),
> strip=FALSE)
This would be better, to avoid abusing the poor R formula handling
sugar:
dotplot(y1 + y2 ~ Treatment | Sites, ## no $
## Supply a data argument
data = facs,
## as before
outer=TRUE,
scales = list(x = list(rot = 90, tck=c(1,0))),
ylab=c("y1", "y2"),
xlab=c("Site 1", "Site 2"),
strip=FALSE)
G
>
> On Wed, Aug 25, 2010 at 11:04 AM, Kay Cichini
wrote:
>
> >
> > hello,
> >
> > i want to stack two lattice plots beneath each other using one x-axis and
> > sharing the same text-panels,
> > like:
> > #
> > library(lattice)
> >
> > y1 <- rnorm(100,100,10)
> > y2 <- rnorm(100,10,1)
> > facs<-expand.grid(Sites=rep(c("Site I","Site
II"),25),Treatment=c("A","B"))
> > pl1<-dotplot(y1 ~ facs$Treatment|facs$Sites,
> > scales = list(x = list(rot = 90, tck=c(1,0
> > pl2<-dotplot(y2 ~ facs$Treatment|facs$Sites,
> > scales = list(x = list(rot = 90, tck=c(1,0
> >
> > print(pl1, split=c(1,2,1,2), more=TRUE)
> > print(pl2, split=c(1,1,1,2))
> > #
> >
> > but as said, ideally the plots should be stacked with only the lower plot
> > giving the x-axis annotation
> > and only the upper plot with text-panels.
> >
> > thanks a lot,
> > kay
> >
> > -
> >
> > Kay Cichini
> > Postgraduate student
> > Institute of Botany
> > Univ. of Innsbruck
> >
> >
> > --
> > View this message in context:
> >
http://r.789695.n4.nabble.com/lattice-help-required-tp2338382p2338382.html
> > Sent from the R help mailing list archive at Nabble.com.
> >
> > __
> > R-help@r-project.org mailing list
> > https://stat.ethz.ch/mailman/listinfo/r-help
> > PLEASE do read the posting guide
> >
http://www.R-project.org/posting-guide.html
> > and provide commented, minimal, self-contained, reproducible code.
> >
>
> [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
--
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Dr. Gavin Simpson [t] +44 (0)20 7679 0522
ECRC, UCL Geography, [f] +44 (0)20 7679 0565
Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk
Gower Street, London [w] http://www.ucl.ac.uk/~ucfagls/
UK. WC1E 6BT. [w] http://www.freshwaters.org.uk
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and provide commented, minimal, self-contained, reproducible code.
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Re: [R] SEM : Warning : Could not compute QR decomposition of Hessian
Dear John,
Thank a lot for your answer. Indeed, i misunderstood the place of
indicator variables in the path diagram, i thank they were considered
as exogeneous variables. Thanks for your answer, i changed variances
of latente variables to NA, and remove indicator variables from
fixed.x argument. Unfortunately, it doesn't change the warning message.
I place the path diagram of the model here : http://yfrog.com/n7pathdiagj
Because of the diagram complexity, i used some colors for the
different paths :
- green for environnental variable
- brown for landscape variables
- orange for municipalities variables
- blue for inter-municipalitie variables
- red for departemental variable
- grey for landuse variables
- and in dark the relations between indicator and latente variables,
errors, variances and covariances.
There must be other mistakes in the model, i can't find them.
Thanks again,
Regards,
Anne
John Fox a écrit :
Dear Anne,
I started to diagram your model but stopped when I noticed some problems:
(1) Some variables, such as pays_alti, are clearly endogenous, since they
have arrows pointing to them, yet are declared as fixed exogenous variables;
that clearly doesn't make sense. (2) You've placed conflicting constraints
on factor loadings and the variances of latent variables, for example
setting both the path from the factor type_paysage to the indicator
pays_alti and the variance of type_paysage to 1; again, that doesn't make
sense.
Do you have a path diagram for the model that you're trying to fit? It would
in particular be helpful to have a diagram of the structural part of the
model.
Regards,
John
John Fox
Senator William McMaster
Professor of Social Statistics
Department of Sociology
McMaster University
Hamilton, Ontario, Canada
web: socserv.mcmaster.ca/jfox
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
On
Behalf Of Anne Mimet
Sent: August-25-10 10:27 AM
To: r-help@r-project.org
Subject: [R] SEM : Warning : Could not compute QR decomposition of Hessian
Hi useRs,
I'm trying for the first time to use a sem. The model finally runs,
but gives a warning saying :
"In sem.default(ram = ram, S = S, N = N, param.names = pars, var.names
= vars, : Could not compute QR decomposition of Hessian.
Optimization probably did not converge. "
I found in R-help some posts on this warning, but my attemps to modify
the code didn't change the warning message (i tried to give an error
of 1 to the latente variables). I can't figure what the problem is.
Here is the code :
tab<-read.table("F:/Mes documents/stats/sem/donnees_corr.txt",
header=T, sep="",na.strings = "NA")
tab[,46]<-as.factor(tab[,46])
tab[,24]<-as.factor(tab[,24])
tab[,40]<-as.factor(tab[,40])
fct_cor<-hetcor(tab, ML=T)
cor_tab<- fct_cor$correlations
moment_tab<-read.moments(diag=F, names=c('c1','c2', 'c3','c4','c5',
'c6','c7', 'c8', 'c9', 'ind_plando', 'long_sup15', 'long_inf15',
'pente', 'est', 'sud','ouest', 'nord' ,'reg_hydriq', 'prof_sol',
'pierro', 'efferv', 'struct','drainage','texture', 'route1_pond',
'route2_pond',
'pourcactif', 'tx_chomage', 'pourcagric', 'pourc_jeunes', 'pop99',
'rev_imp_foyer','eq_CONC', 'eq_sante', 'eq_edu', 'sold_nat',
'sold_mig', 'tps_dom_emp','TXEMPLOI','ORIECO','dist_paris','axe1',
'axe2', 'axe3', 'dist_protect','urbanisation','pays_incli','pays_alti'))
# after comes the moment matrix (triangular)
ram_tab<-specify.model()
type_paysage->pays_alti,NA,1
type_paysage->pays_incli, pays2, NA
pedo->reg_hydriq, NA, 1
pedo->prof_sol, ped8, NA
pedo->pierro, ped9, NA
pedo->efferv, ped10, NA
pedo->struct, ped11, NA
pedo->drainage, ped12, NA
pedo->texture, ped13, NA
adj_99->c1, NA,1
adj_99->c2, adj2,NA
adj_99->c3, adj3,NA
adj_99->c4, adj4,NA
adj_99->c5, adj5,NA
adj_99->c6, adj6,NA
adj_99->c7, adj7,NA
adj_99->c8, adj8,NA
adj_99->c9, adj9,NA
etat_hexa->axe1, NA, 1
etat_hexa->axe2, et2, NA
etat_hexa->axe3, et3, NA
socioBV->sold_mig, BV1, NA
socioBV->sold_nat, BV2, NA
socioBV->TXEMPLOI, BV3, NA
socioBV->ORIECO, BV4, NA
socioBV->tps_dom_emp, NA, 1
eqBV->eq_CONC, NA, 1
eqBV->eq_sante, eq2, NA
eqBV->eq_edu, eq3, NA
socio_com->pourcactif , NA, 1
socio_com->tx_chomage, com2, NA
socio_com->pourcagric, com3, NA
socio_com->pourc_jeunes, com4, NA
socio_com->pop99, com5, NA
socio_com->rev_imp_foyer, com7, NA
access_hexa->route1_pond, NA, 1
access_hexa->route2_pond, acc2, NA
hydro->ind_plando, NA, 1
hydro->long_sup15, eau2, NA
hydro->long_inf15, eau3, NA
topog->pente, NA, 1
topog->est, top2, NA
topog->sud, top3, NA
topog->nord, top4, NA
topog->ouest, top5, NA
dist_protect-> urbanisation, cor1,NA
dist_protect-> adj_99, cor2, NA
dist_protect-> etat_hexa, cor3, NA
topog-> urbanisation, cor4, NA
topog-> adj_99, cor5, NA
topog-> etat_hexa, cor6, NA
topog-> access_hexa, cor7, NA
topog<->hydro, cor8, NA
topog<->pedo, cor9, NA
pedo-> urbanisation, cor10, NA
pedo-> adj_99, cor11, NA
pedo-> etat_hexa, cor12, NA
pedo<->hydro, cor1,
Re: [R] Quick GREP challenge
It isn't entirely clear what you are trying to do - your grep() statement
simply returns the entire string you started with.
But to turn that string into a vector, you will need some combination
of gsub(), strsplit(), and as.numeric() with the exact values depending
on the exact form of the output returned by grep. By c(1, 22) do you
mean a vector (my first assumption), or do you actually mean the
string "c(1, 22)"?
If the former, maybe something like this:
> x <- "f1=5,f22=3"
> x <- gsub("=[0-9]+", "", x)
> x
[1] "f1,f22"
> x <- gsub("f", "", x)
> x
[1] "1,22"
> x <- strsplit(x, ",")
> x
[[1]]
[1] "1" "22"
> x <- unlist(x)
> x
[1] "1" "22"
> x <- as.numeric(x)
> x
[1] 1 22
Sarah
On Thu, Aug 26, 2010 at 6:16 AM, Dimitri Shvorob
wrote:
>
>> grep("f[0-9]+=", "f1=5,f22=3,", value = T)
> [1] "f1=5,f22=3,"
>
> How do I make the line output c("f1", "f22") instead? (Actually, c(1,22)
> would be even better).
>
> Thank you.
>
>
--
Sarah Goslee
http://www.functionaldiversity.org
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Quick GREP challenge
I too am unsure of what is required but another way is
gsub(",$","",gsub("[f53=]+","",a))
[1] "1,22"
Regards
Duncan
Duncan Mackay
Department of Agronomy and Soil Science
University of New England
ARMIDALE NSW 2351
Email home: mac...@northnet.com.au
At 21:47 26/08/2010, you wrote:
It isn't entirely clear what you are trying to do - your grep() statement
simply returns the entire string you started with.
But to turn that string into a vector, you will need some combination
of gsub(), strsplit(), and as.numeric() with the exact values depending
on the exact form of the output returned by grep. By c(1, 22) do you
mean a vector (my first assumption), or do you actually mean the
string "c(1, 22)"?
If the former, maybe something like this:
> x <- "f1=5,f22=3"
> x <- gsub("=[0-9]+", "", x)
> x
[1] "f1,f22"
> x <- gsub("f", "", x)
> x
[1] "1,22"
> x <- strsplit(x, ",")
> x
[[1]]
[1] "1" "22"
> x <- unlist(x)
> x
[1] "1" "22"
> x <- as.numeric(x)
> x
[1] 1 22
Sarah
On Thu, Aug 26, 2010 at 6:16 AM, Dimitri Shvorob
wrote:
>
>> grep("f[0-9]+=", "f1=5,f22=3,", value = T)
> [1] "f1=5,f22=3,"
>
> How do I make the line output c("f1", "f22") instead? (Actually, c(1,22)
> would be even better).
>
> Thank you.
>
>
--
Sarah Goslee
http://www.functionaldiversity.org
__
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and provide commented, minimal, self-contained, reproducible code.
__
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[R] Random slopes in lmer
Hi I want to extract the random slopes from a lmer (I am doing a random regression), but are the answers obtained from ranef or coef? My model is: mod1<-lmer(B~ A +(A|bird), family=quasibinomial) And I want to obtain a slope for each individual bird but am not sure which output I need and can't find the answer anywhere. Thanks Sam Dr Samantha Patrick EU INTERREG Post Doc Davy 618 Marine Biology & Ecology Research Centre University of Plymouth Plymouth PL4 8AA T: 01752 586165 M: 07740472719 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Quick GREP challenge
On Thu, Aug 26, 2010 at 6:16 AM, Dimitri Shvorob
wrote:
>
>> grep("f[0-9]+=", "f1=5,f22=3,", value = T)
> [1] "f1=5,f22=3,"
>
> How do I make the line output c("f1", "f22") instead? (Actually, c(1,22)
> would be even better).
>
strapply in gsubfn extracts matches based on content rather than
delimiters. See home page for more info http://gsubfn.googlecode.com
> library(gsubfn)
> x <- "f1=5,f22=3,"
> strapply(x, "f\\d+")[[1]]
[1] "f1" "f22"
--
GKX Group
GKX Associates Inc.
1-877-GKX-GROUP
ggrothendieck at gmail.com
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] print method for str?
On Thu, Aug 26, 2010 at 4:42 AM, David Hajage wrote: > Hello useRs and guRus, > > I was trying to add the support of str() in the ascii package, and I > realized that str() does not have a print method. It uses cat() from inside > the function and returns nothing. > > Since it is not usual in R, I wonder why? Is there a particular reason? > > Thank you very much for your attention on... well, this unimportant > question. > It still dispatches to S3 methods though. See str.zoo in the zoo package as an example. -- GKX Group GKX Associates Inc. 1-877-GKX-GROUP ggrothendieck at gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Passing data to aov
Hello Again Gurus and Lurkers:
I¹m trying to build a very user-friendly function which does aov without
having the user type in a formula (which would be tedious in this case).
The idea is to take the response from a PCA score matrix, and the factors
from a list. A simple example is the function given below, along with test
data and a sample call to the function.
I'm certainly having trouble understanding the proper ways to work with
formulas and related items, but I think what I do in the function should
work (it's built on ideas dug out of the archives). However, when the data
is passed to aov (directly or via manova), something in the bowels of aov
complains with the following error:
Error in model.frame.default(formula = form, drop.unused.levels = TRUE) :
object is not a matrix
To me, the formula looks legitimate, and the variables in the formula are
all in the environment (I think: The way I am doing this is basically that
described in ?summary.manova where only a formula is passed, no data
argument). Based upon reading the archives, the problem might arise with
one of the deparse statements in aov, but I can't resolve it. It might also
be one of scoping/environment, but again, this is only an idea.
TIA for any assistance. Bryan
*
Bryan Hanson
Professor of Chemistry & Biochemistry
DePauw University, Greencastle IN USA
hypTestScores <-
function(mylist, score.matrix, pcs = 1:3, fac = NULL, ...) {
scores <- score.matrix[,pcs]
#str(scores)# looks correct to me
form <- as.formula(paste("scores", "~", paste(fac, collapse = "*")),
env = parent.frame())
#str(form) # looks correct to me
attach(mylist)
if (length(pcs) > 1) out <- manova(formula = form, ...)
if (length(pcs) == 1) out <- aov(formula = form, ...)
print(summary(out))
detach(mylist)
invisible(out)
}
# test data
td1 <- matrix(rnorm(50), ncol = 5) # like PCA scores
td2 <- list(
f1 = as.factor(sample(c("A", "B"), 10, replace = TRUE)),
f2 = as.factor(sample(c("C", "D"), 10, replace = TRUE)))
# test call
hypTestScores(mylist = td2, score.matrix = td1,
fac = c("f1", "f2"))
detach("mylist") # needed if there is an error
> sessionInfo()
R version 2.11.0 (2010-04-22)
x86_64-apple-darwin9.8.0
locale:
[1] en_US.UTF-8/en_US.UTF-8/C/C/en_US.UTF-8/en_US.UTF-8
attached base packages:
[1] datasets tools grid graphics grDevices utils stats
[8] methods base
other attached packages:
[1] faraway_1.0.4 GGally_0.2 xtable_1.5-6
[4] mvbutils_2.5.1 ggplot2_0.8.8 proto_0.3-8
[7] reshape_0.8.3 ChemoSpec_1.45 R.utils_1.4.0
[10] R.oo_1.7.2 R.methodsS3_1.2.0 rgl_0.91
[13] lattice_0.18-5 mvoutlier_1.4 plyr_1.0.3
[16] RColorBrewer_1.0-2 chemometrics_0.8 som_0.3-5
[19] robustbase_0.5-0-1 rpart_3.1-46 pls_2.1-0
[22] pcaPP_1.8-1mvtnorm_0.9-9 nnet_7.3-1
[25] mclust_3.4.4 MASS_7.3-5 lars_0.9-7
[28] e1071_1.5-23 class_7.3-2
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Re: [R] relimp
Hi John,
I'm having some trouble sorting out your request, for the following reasons:
1) I don't know what misdist is (can't find it on CRAN or google) or why you
need Rcmdr to run it. Presumably you meant mixdist?
2) relimp is not required by Rcmdr, merely suggested. Rcmdr will work fine
without it.
3) I hope that by "downloaded from CRAN" you do not mean that you manually
downloaded it. R has package management, use it! What happens exactly when
you try
install.packages("Rcmdr" dep=TRUE)
?
4) I had no trouble installing Rcmdr (and all dependencies etc., including
relimp) using the command above. My session information is below. Please
include yours (as well as the actual errors you get, if any) in your reply.
> sessionInfo()
R version 2.11.1 (2010-05-31)
i486-pc-linux-gnu
locale:
[1] LC_CTYPE=en_US.utf8 LC_NUMERIC=C
[3] LC_TIME=en_US.utf8LC_COLLATE=en_US.utf8
[5] LC_MONETARY=C LC_MESSAGES=en_US.utf8
[7] LC_PAPER=en_US.utf8 LC_NAME=C
[9] LC_ADDRESS=C LC_TELEPHONE=C
[11] LC_MEASUREMENT=en_US.utf8 LC_IDENTIFICATION=C
attached base packages:
[1] splines tcltk stats graphics grDevices utils datasets
[8] methods base
other attached packages:
[1] Rcmdr_1.6-0 car_2.0-2 survival_2.35-8 nnet_7.3-1
[5] MASS_7.3-7
loaded via a namespace (and not attached):
[1] tools_2.11.1
Best,
Ista
On Thu, Aug 26, 2010 at 6:39 AM, John Barrett [jzb] wrote:
> I am trying to use Rcmdr 1.5_4 with R-2.11.1 (order to run to run the new
> version of misdist-0.5.3 which is built under R-2.11.1). however relimp is
> required for Rcmdr and the version of relimp_1.0.1 downloaded from CRAN will
> not work with the latest version of Rcmdr (I get error message telling me to
> reload it). Is there any way round this problem or will there be a new
> version relimp that is compatible? Any advice would be gratefully received.
> many thanks
> John
>
> Prof. Barrett
> Univ. of Aberystwyth
>
>
>
>[[alternative HTML version deleted]]
>
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> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
Ista Zahn
Graduate student
University of Rochester
Department of Clinical and Social Psychology
http://yourpsyche.org
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Re: [R] equalize function with zero and convert it
On Thu, Aug 26, 2010 at 6:57 AM, Immanuel Seeger wrote:
> Dear all,
> I want to equalize a symbolic derivative (of a function with two
> variables) with zero and convert it to a variable, e.g. x.
>
> I'm computing the derivative by: (found it in the archive)
> library(Ryacas)
> f <- function(x,y) (100-x-y)*x-10*x
> yacas(f) # register f with yacas
> Df <- f
> body(Df) <- yacas(expression(deriv(f(x,y),x)))[[1]]
> Df
>
> R-Output:
> function (x, y)
> 100 - x - y - x - 10
>
> Questions :
> 1. How can I equalize the derivative(140-x-y-x-20) with zero?
> 2. How can I convert it to x?
>
> I want something like:
> 45-0.5*y=x
>
> With only one variable or two non-symbolic variables it's working with
> uniroot(f, c(-10, 10))$root.
>
This solves for x:
> library(Ryacas)
Loading required package: XML
> x <- Sym("x")
> y <- Sym("y")
> Solve((100-x-y)*x-10*x == 0, "x")
[1] "Starting Yacas!"
expression(list(x == -((10 - (100 - y) + root((100 - y - 10)^2,
2))/2), x == -((10 - (100 - y) - root((100 - y - 10)^2, 2))/2)))
--
GKX Group
GKX Associates Inc.
1-877-GKX-GROUP
ggrothendieck at gmail.com
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Re: [R] Problem with terms of the form (a >1) and subsetting of a terms object
On Aug 26, 2010, at 10:35 AM, Niels Richard Hansen wrote: > > > On 26/08/10 09.30, Dimitris Rizopoulos wrote: >> I think you need an I(), i.e., >> >> form <- ~ I(a > 1) - 1 > > Yes, it solves the concrete problem, but does not really answer > the question. Let me rephrase. Should the use of terms like (a>1) > be discouraged in R and replaced by I(a>1) systematically, or is the > behavior below unexpected? I'd say "yes", and "maybe". I think the only constructs that are guaranteed to protect arithmetic expressions are function calls, including I(). So it is more or less coincidental that ~ (a>1) - 1 works in the first place; when updating it, you probably should "expect the unexpected". There could still be a bug -- it looks somewhat wrong that a>1 isn't retained as a call to "<" with arguments a and 1, parentheses or not, but the model formula expansion code involves deep and dark magic, and I'd just avoid going too near it. -- Peter Dalgaard Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] approxfun-problems (yleft and yright ignored)
Hi Greg,
thanks for the suggestion:
I have attached some small dataset that can be used to reproduce the
odd behavior of the approxfun-function.
If it gets stripped off my email, it can also be downloaded at:
http://bioinf.gen.tcd.ie/approx.data.Rdata
Strangely, the problem seems specific to the data structure in my
expression set, when I use simulated data, everything worked fine.
Here is some code that I run and resulted in the strange output that I
have described in my initial post:
> ### load the data: a list called approx.data
> load(file="approx.data.Rdata")
> ### contains the slots "x", "y", "input"
> names(approx.data)
[1] "x" "y" "input"
> ### with y ranging between 0 and 1
> range(approx.data$y)
[1] 0 1
> ### compare ranges of x and input-x values (the latter is a small subset of
> 500 data points):
> range(approx.data$x)
[1] 3.098444 7.268812
> range(approx.data$input)
[1] 3.329408 13.026700
>
>
> ### generate the interpolation function (warning message benign)
> interp <- approxfun(approx.data$x, approx.data$y, yleft=1, yright=0, rule=2)
Warning message:
In approxfun(approx.data$x, approx.data$y, yleft = 1, yright = 0, :
collapsing to unique 'x' values
>
> ### apply to input-values
> y.out <- sapply(approx.data$input, interp)
>
> ### still I find output values >1, even though yleft=1:
> range(y.out)
[1] 0.00 7.207233
> hist(y.out)
>
> ### and the input-data points for which strange interpolation does occur have
> no unusual distribution (however, they lie close to max(x)):
> hist(approx.data$input[which(y.out>1)])
The session info can be found below, thanks a million for any help.
Sam
On 25 August 2010 19:31, Greg Snow wrote:
> The plots did not come through, see the posting guide for which attachments
> are allowed. It will be easier for us to help if you can send reproducible
> code (we can copy and paste to run, then examine, edit, etc.). Try finding a
> subset of your data for which the problem still occurs, then send the data if
> possible, or similar simulated data if you cannot send original data.
>
> --
> Gregory (Greg) L. Snow Ph.D.
> Statistical Data Center
> Intermountain Healthcare
> greg.s...@imail.org
> 801.408.8111
>
>
>> -Original Message-
>> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
>> project.org] On Behalf Of Samuel Wuest
>> Sent: Wednesday, August 25, 2010 8:20 AM
>> To: r-help@r-project.org
>> Subject: [R] approxfun-problems (yleft and yright ignored)
>>
>> Dear all,
>>
>> I have run into a problem when running some code implemented in the
>> Bioconductor panp-package (applied to my own expression data), whereby
>> gene
>> expression values of known true negative probesets (x) are interpolated
>> onto
>> present/absent p-values (y) between 0 and 1 using the *approxfun -
>> function*{stats}; when I have used R version 2.8, everything had
>> worked fine,
>> however, after updating to R 2.11.1., I got unexpected output
>> (explained
>> below).
>>
>> Please correct me here, but as far as I understand, the yleft and
>> yright
>> arguments set the extreme values of the interpolated y-values in case
>> the
>> input x-values (on whose approxfun is applied) fall outside range(x).
>> So if
>> I run approxfun with yleft=1 and yright=0 with y-values between 0 and
>> 1,
>> then I should never get any values higher than 1. However, this is not
>> the
>> case, as this code-example illustrates:
>>
>> > ### define the x-values used to construct the approxfun, basically
>> these
>> are 2000 expression values ranging from ~ 3 to 7:
>> > xNeg <- NegExprs[, 1]
>> > xNeg <- sort(xNeg, decreasing = TRUE)
>> >
>> > ### generate 2000 y-values between 0 and 1:
>> > yNeg <- seq(0, 1, 1/(length(xNeg) - 1))
>> > ### define yleft and yright as well as the rule to clarify what
>> should
>> happen if input x-values lie outside range(x):
>> > interp <- approxfun(xNeg, yNeg, yleft = 1, yright = 0, rule=2)
>> Warning message:
>> In approxfun(xNeg, yNeg, yleft = 1, yright = 0, rule = 2) :
>> collapsing to unique 'x' values
>> > ### apply the approxfun to expression data that range from ~2.9 to
>> 13.9
>> and can therefore lie outside range(xNeg):
>> > PV <- sapply(AllExprs[, 1], interp)
>> > range(PV)
>> [1] 0.000 6208.932
>> > summary(PV)
>> Min. 1st Qu. Median Mean 3rd Qu. Max.
>> 0.000e+00 0.000e+00 2.774e-03 1.299e+00 3.164e-01 6.209e+03
>>
>> So the resulting output PV object contains data ranging from 0 to 6208,
>> the
>> latter of which lies outside yleft and is not anywhere close to extreme
>> y-values that were used to set up the interp-function. This seems wrong
>> to
>> me, and from what I understand, yleft and yright are simply ignored?
>>
>> I have attached a few histograms that visualize the data distributions
>> of
>> the objects I xNeg, yNeg, AllExprs[,1] (== input x-values) and PV (the
>> output), so that it is easier to make sense of the data structures...
>>
>> Does anyone have an explanation for this or
[R] Passing arguments between S4 methods fails within a function:bug? example with raster package.
Dear all,
This problem came up initially while debugging a function, but it
seems to be a more general problem of R. I hope I'm wrong, but I can't
find another explanation. Let me illustrate with the raster package.
For an object "RasterLayer" (which inherits from Raster), there is a
method xyValues defined with the signature
(object="RasterLayer",xy="matrix"). There is also a method with
signature (object="Raster",xy="vector"). The only thing this method
does, is change xy into a matrix and then pass on to the next method
using callGeneric again. Arguments are passed.
Now this all works smoothly, as long as you stay in the global environment :
require(raster)
a <- raster()
a[] <- 1:ncell(a)
origin <- c(-80,50)
eff.dist <- 10
unlist(xyValues(a,xy=origin,buffer=eff.dist))
[1] 14140 14141 14500 14501
Now let's make a very basic test function :
test <- function(x,orig.point){
eff.distance <- 10
p <- unlist(xyValues(x,xy=orig.point,buffer=eff.distance))
return(p)
}
This gives the following result :
> test(a,origin)
Error in .local(object, xy, ...) : object 'eff.distance' not found
huh? Apparently, eff.distance got lost somewhere in the parsetree (am
I saying this correctly?)
The funny thing is when we change origin to a matrix :
> origin <- matrix(origin,ncol=2)
> unlist(xyValues(a,xy=origin,buffer=eff.dist))
[1] 14140 14141 14500 14501
> test(a,origin)
[1] 14140 14141 14500 14501
It all works again! So something goes wrong with passing the arguments
from one method to another using callGeneric. Is this a bug in R or am
I missing something obvious?
The relevant code from the raster package :
setMethod("xyValues", signature(object='Raster', xy='vector'),
function(object, xy, ...) {
if (length(xy) == 2) {
callGeneric(object, matrix(xy, ncol=2), ...)
} else {
stop('xy coordinates should be a two-column matrix or
data.frame,
or a vector of two numbers.')
}
} )
setMethod("xyValues", signature(object='RasterLayer', xy='matrix'),
function(object, xy, method='simple', buffer=NULL, fun=NULL,
na.rm=TRUE) {
if (dim(xy)[2] != 2) {
stop('xy has wrong dimensions; it should have 2
columns' )
}
if (! is.null(buffer)) {
return( .xyvBuf(object, xy, buffer, fun, na.rm=na.rm) )
}
if (method=='bilinear') {
return(.bilinearValue(object, xy))
} else if (method=='simple') {
cells <- cellFromXY(object, xy)
return(.readCells(object, cells))
} else {
stop('invalid method argument. Should be simple or
bilinear.')
}
}
)
--
Joris Meys
Statistical consultant
Ghent University
Faculty of Bioscience Engineering
Department of Applied mathematics, biometrics and process control
tel : +32 9 264 59 87
joris.m...@ugent.be
---
Disclaimer : http://helpdesk.ugent.be/e-maildisclaimer.php
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Re: [R] Quick GREP challenge
On Aug 26, 2010, at 8:15 AM, Gabor Grothendieck wrote:
On Thu, Aug 26, 2010 at 6:16 AM, Dimitri Shvorob
wrote:
grep("f[0-9]+=", "f1=5,f22=3,", value = T)
[1] "f1=5,f22=3,"
How do I make the line output c("f1", "f22") instead? (Actually,
c(1,22)
would be even better).
strapply in gsubfn extracts matches based on content rather than
delimiters. See home page for more info http://gsubfn.googlecode.com
library(gsubfn)
x <- "f1=5,f22=3,"
strapply(x, "f\\d+")[[1]]
[1] "f1" "f22"
I read the request as for c(1,22) offered without quotes so t emeaning
was obscure but perhaps this is what OP wanted:
as.numeric( strapply(x, "f(\\d+)=")[[1]] )
[1] 1 22
# or
> as.numeric(unlist(strapply(x, "f(\\d+)=")))
[1] 1 22
--
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1-877-GKX-GROUP
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Re: [R] Passing data to aov
Hi,
First, some toy data would have helped a lot, and that can explain why
you got so few answers...
Second, I have maybe some thoughts about it, not sure if this can help.
The error tells you that the data passed to aov is not a matrix. In your
function, you don't specify explicitly the data object to be used, that
might be a problem. And that object has to be a matrix, it might not be.
Maybe something like:
form<- as.formula(paste("pcs", "~", paste(fac, collapse = "*")),
...
if (length(pcs) == 1) out<- aov(formula = form, data=score.matrix)
That would work if I understood what you're doing (hum...) and if both "pcs"
and "fac" are parts of "score.matrix". But I've always done aov from a
dataframe, which makes sense if you have both numerical variables and factors.
How can you do it from a matrix? Maybe my understanding is limited.
As I said, just some thoughts...
HTH,
Ivan
Le 8/26/2010 14:47, Bryan Hanson a écrit :
> Hello Again Gurus and Lurkers:
>
> I¹m trying to build a very user-friendly function which does aov without
> having the user type in a formula (which would be tedious in this case).
> The idea is to take the response from a PCA score matrix, and the factors
> from a list. A simple example is the function given below, along with test
> data and a sample call to the function.
>
> I'm certainly having trouble understanding the proper ways to work with
> formulas and related items, but I think what I do in the function should
> work (it's built on ideas dug out of the archives). However, when the data
> is passed to aov (directly or via manova), something in the bowels of aov
> complains with the following error:
>
> Error in model.frame.default(formula = form, drop.unused.levels = TRUE) :
>object is not a matrix
>
> To me, the formula looks legitimate, and the variables in the formula are
> all in the environment (I think: The way I am doing this is basically that
> described in ?summary.manova where only a formula is passed, no data
> argument). Based upon reading the archives, the problem might arise with
> one of the deparse statements in aov, but I can't resolve it. It might also
> be one of scoping/environment, but again, this is only an idea.
>
> TIA for any assistance. Bryan
> *
> Bryan Hanson
> Professor of Chemistry& Biochemistry
> DePauw University, Greencastle IN USA
>
>
> hypTestScores<-
> function(mylist, score.matrix, pcs = 1:3, fac = NULL, ...) {
>
> scores<- score.matrix[,pcs]
> #str(scores)# looks correct to me
> form<- as.formula(paste("scores", "~", paste(fac, collapse = "*")),
> env = parent.frame())
> #str(form) # looks correct to me
> attach(mylist)
> if (length(pcs)> 1) out<- manova(formula = form, ...)
> if (length(pcs) == 1) out<- aov(formula = form, ...)
> print(summary(out))
> detach(mylist)
> invisible(out)
> }
>
> # test data
> td1<- matrix(rnorm(50), ncol = 5) # like PCA scores
> td2<- list(
> f1 = as.factor(sample(c("A", "B"), 10, replace = TRUE)),
> f2 = as.factor(sample(c("C", "D"), 10, replace = TRUE)))
>
> # test call
> hypTestScores(mylist = td2, score.matrix = td1,
> fac = c("f1", "f2"))
> detach("mylist") # needed if there is an error
>
>> sessionInfo()
> R version 2.11.0 (2010-04-22)
> x86_64-apple-darwin9.8.0
>
> locale:
> [1] en_US.UTF-8/en_US.UTF-8/C/C/en_US.UTF-8/en_US.UTF-8
>
> attached base packages:
> [1] datasets tools grid graphics grDevices utils stats
> [8] methods base
>
> other attached packages:
> [1] faraway_1.0.4 GGally_0.2 xtable_1.5-6
> [4] mvbutils_2.5.1 ggplot2_0.8.8 proto_0.3-8
> [7] reshape_0.8.3 ChemoSpec_1.45 R.utils_1.4.0
> [10] R.oo_1.7.2 R.methodsS3_1.2.0 rgl_0.91
> [13] lattice_0.18-5 mvoutlier_1.4 plyr_1.0.3
> [16] RColorBrewer_1.0-2 chemometrics_0.8 som_0.3-5
> [19] robustbase_0.5-0-1 rpart_3.1-46 pls_2.1-0
> [22] pcaPP_1.8-1mvtnorm_0.9-9 nnet_7.3-1
> [25] mclust_3.4.4 MASS_7.3-5 lars_0.9-7
> [28] e1071_1.5-23 class_7.3-2
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
Ivan CALANDRA
PhD Student
University of Hamburg
Biozentrum Grindel und Zoologisches Museum
Abt. Säugetiere
Martin-Luther-King-Platz 3
D-20146 Hamburg, GERMANY
+49(0)40 42838 6231
ivan.calan...@uni-hamburg.de
**
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http://webapp5.rrz.uni-hamburg.de/mammals/eng/mitarbeiter.php
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Re: [R] equalize function with zero and convert it
On Thu, Aug 26, 2010 at 9:45 AM, Immanuel Seeger wrote:
> Thanks, but my problem is that I have my formulas in functions.
> How can I do that for functions or e.g. for the body(Df) below?
Try this:
> library(Ryacas)
> Solve(deparse(body(f)), "x")
[1] "Starting Yacas!"
expression(list(x == -((10 - (100 - y) + root((100 - y - 10)^2,
2))/2), x == -((10 - (100 - y) - root((100 - y - 10)^2, 2))/2)))
>
> out <- yacas(Solve(deparse(body(f)), "x"))
> out$text[[1]][[2]]
x == -((10 - (100 - y) + root((100 - y - 10)^2, 2))/2)
> And: How can I refer to the first solution of expression?
> Thanks a lot!
>
> -Ursprüngliche Nachricht-
> Von: Gabor Grothendieck [mailto:ggrothendi...@gmail.com]
> Gesendet: Donnerstag, 26. August 2010 14:56
> An: Immanuel Seeger
> Cc: r-help@r-project.org
> Betreff: Re: [R] equalize function with zero and convert it
>
> On Thu, Aug 26, 2010 at 6:57 AM, Immanuel Seeger
> wrote:
>> Dear all,
>> I want to equalize a symbolic derivative (of a function with two
>> variables) with zero and convert it to a variable, e.g. x.
>>
>> I'm computing the derivative by: (found it in the archive)
>> library(Ryacas)
>> f <- function(x,y) (100-x-y)*x-10*x
>> yacas(f) # register f with yacas
>> Df <- f
>> body(Df) <- yacas(expression(deriv(f(x,y),x)))[[1]]
>> Df
>>
>> R-Output:
>> function (x, y)
>> 100 - x - y - x - 10
>>
>> Questions :
>> 1. How can I equalize the derivative(140-x-y-x-20) with zero?
>> 2. How can I convert it to x?
>>
>> I want something like:
>> 45-0.5*y=x
>>
>> With only one variable or two non-symbolic variables it's working with
>> uniroot(f, c(-10, 10))$root.
>>
>
> This solves for x:
>
>> library(Ryacas)
> Loading required package: XML
>> x <- Sym("x")
>> y <- Sym("y")
>> Solve((100-x-y)*x-10*x == 0, "x")
> [1] "Starting Yacas!"
> expression(list(x == -((10 - (100 - y) + root((100 - y - 10)^2,
> 2))/2), x == -((10 - (100 - y) - root((100 - y - 10)^2, 2))/2)))
>
> --
> GKX Group
> GKX Associates Inc.
> 1-877-GKX-GROUP
> ggrothendieck at gmail.com
>
>
--
GKX Group, GKX Associates Inc.
Statistics & Software Consulting
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Passing data to aov
On Aug 26, 2010, at 8:47 AM, Bryan Hanson wrote:
Hello Again Gurus and Lurkers:
I’m trying to build a very user-friendly function which does aov
without
having the user type in a formula (which would be tedious in this
case).
The idea is to take the response from a PCA score matrix, and the
factors
from a list. A simple example is the function given below, along
with test
data and a sample call to the function.
I'm certainly having trouble understanding the proper ways to work
with
formulas and related items, but I think what I do in the function
should
work (it's built on ideas dug out of the archives). However, when
the data
is passed to aov (directly or via manova), something in the bowels
of aov
complains with the following error:
Error in model.frame.default(formula = form, drop.unused.levels =
TRUE) :
object is not a matrix
Actually I got a different error (also on a Mac. but with a version
that is two month later than yours and with quite a smaller number of
packages loaded):
> hypTestScores(mylist = td2, score.matrix = td1,
+fac = c("f1", "f2"))
Error in eval(expr, envir, enclos) : object 'scores' not found
Which went away if I removed the "env=" argument. My guess is that you
were telling manova to look further up the lexical tree than it should
have (or that attach() operation messed up the environment somehow):
> hypTestScores(mylist = td2, score.matrix = td1,
+fac = c("f1", "f2"))
Df Pillai approx F num Df den Df Pr(>F)
f1 1 0.31663 0.61777 3 4 0.6392
f2 1 0.36652 0.77145 3 4 0.5673
f1:f2 1 0.15687 0.24808 3 4 0.8593
Residuals 6
> detach("mylist") # needed if there is an error
Error in detach("mylist") : invalid 'name' argument
You might want to do inside the function something like
res =try( your function )
if (res="try-error") {stop() } else{
}
To me, the formula looks legitimate, and the variables in the
formula are
all in the environment (I think: The way I am doing this is
basically that
described in ?summary.manova where only a formula is passed, no data
argument). Based upon reading the archives, the problem might arise
with
one of the deparse statements in aov, but I can't resolve it. It
might also
be one of scoping/environment, but again, this is only an idea.
TIA for any assistance. Bryan
*
Bryan Hanson
Professor of Chemistry & Biochemistry
DePauw University, Greencastle IN USA
hypTestScores <-
function(mylist, score.matrix, pcs = 1:3, fac = NULL, ...) {
scores <- score.matrix[,pcs]
#str(scores)# looks correct to me
form <- as.formula(paste("scores", "~", paste(fac, collapse =
"*")),
env = parent.frame())
#str(form) # looks correct to me
attach(mylist)
if (length(pcs) > 1) out <- manova(formula = form, ...)
if (length(pcs) == 1) out <- aov(formula = form, ...)
print(summary(out))
detach(mylist)
invisible(out)
}
# test data
td1 <- matrix(rnorm(50), ncol = 5) # like PCA scores
td2 <- list(
f1 = as.factor(sample(c("A", "B"), 10, replace = TRUE)),
f2 = as.factor(sample(c("C", "D"), 10, replace = TRUE)))
# test call
hypTestScores(mylist = td2, score.matrix = td1,
fac = c("f1", "f2"))
detach("mylist") # needed if there is an error
sessionInfo()
R version 2.11.0 (2010-04-22)
x86_64-apple-darwin9.8.0
locale:
[1] en_US.UTF-8/en_US.UTF-8/C/C/en_US.UTF-8/en_US.UTF-8
attached base packages:
[1] datasets tools grid graphics grDevices utils stats
[8] methods base
other attached packages:
[1] faraway_1.0.4 GGally_0.2 xtable_1.5-6
[4] mvbutils_2.5.1 ggplot2_0.8.8 proto_0.3-8
[7] reshape_0.8.3 ChemoSpec_1.45 R.utils_1.4.0
[10] R.oo_1.7.2 R.methodsS3_1.2.0 rgl_0.91
[13] lattice_0.18-5 mvoutlier_1.4 plyr_1.0.3
[16] RColorBrewer_1.0-2 chemometrics_0.8 som_0.3-5
[19] robustbase_0.5-0-1 rpart_3.1-46 pls_2.1-0
[22] pcaPP_1.8-1mvtnorm_0.9-9 nnet_7.3-1
[25] mclust_3.4.4 MASS_7.3-5 lars_0.9-7
[28] e1071_1.5-23 class_7.3-2
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
David Winsemius, MD
West Hartford, CT
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] daisy(): space allocation issue
Hi, I'm trying to apply the function daisy() to a data.frame 1x10 but I have not enough space (error message: cannot allocate vector of length 1476173280). I didn't imagine I was not able to work with a matrix of just 1 observations... I have setted in Rgui --max-mem-size=2G (I'm not able to set more space..) How can I solve this issue? Separating observations depending on some rules? thanks -- View this message in context: http://r.789695.n4.nabble.com/daisy-space-allocation-issue-tp2339844p2339844.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] About plot graphs
Hi folks, Following command prints 2 graphs side-by-side:- layout(matrix(1:2, nrow=1)) plot(Date,Input_No.) plot(Test01$Date, Test01$Input_No.) However each is a square graph I need a rectangular layout. Pls advise how to make it. TIA B.R. satimis __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] About plot graphs
On Aug 26, 2010, at 10:44 AM, Stephen Liu wrote: Hi folks, Following command prints 2 graphs side-by-side:- layout(matrix(1:2, nrow=1)) ?layout #Read the details for the widths and heights arguments more carefully and (as it suggests) see the examples. plot(Date,Input_No.) plot(Test01$Date, Test01$Input_No.) However each is a square graph I need a rectangular layout. Pls advise how to make it. TIA -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] SEM : Warning : Could not compute QR decomposition of Hessian
Dear Anne,
> -Original Message-
> From: Anne Mimet [mailto:ami...@mnhn.fr]
> Sent: August-26-10 7:45 AM
> To: John Fox
> Cc: r-help@r-project.org
> Subject: RE: [R] SEM : Warning : Could not compute QR decomposition of
> Hessian
>
> Dear John,
>
> Thank a lot for your answer. Indeed, i misunderstood the place of
> indicator variables in the path diagram, i thank they were considered
> as exogeneous variables. Thanks for your answer, i changed variances
> of latente variables to NA, and remove indicator variables from
> fixed.x argument. Unfortunately, it doesn't change the warning message.
> I place the path diagram of the model here : http://yfrog.com/n7pathdiagj
>
> Because of the diagram complexity, i used some colors for the
> different paths :
> - green for environnental variable
> - brown for landscape variables
> - orange for municipalities variables
> - blue for inter-municipalitie variables
> - red for departemental variable
> - grey for landuse variables
> - and in dark the relations between indicator and latente variables,
> errors, variances and covariances.
>
> There must be other mistakes in the model, i can't find them.
You haven't sent me your updated R code, so I'm just going by the path
diagram, but, yes, I think that there are more problems. I assume that the
circles represent error variables, which in the RAM formulation used by
sem() would appear as self-directed double-headed arrows. If that's the
case, why do observed exogenous variables like dist.protect have an error
component? As well, if the absence of double-headed arrows between pairs of
exogenous variables implies that they are uncorrelated, then your model has
many such constraints. Did you really intend that? I'm also not convinced
that the model is identified -- there are, I believe, just 6 exogenous
variables -- but the (possibly inadvertent) constraints placed on error
correlations may indentify it. Just to be clear, there is no guarantee that
sem() will be able to estimate your model even if it is specified as you
intend and even if it is identified.
At this point, I'd make two suggestions: First, if you want to continue to
try to use the sem package, and if you haven't already done so, I'd suggest
that you read my 2006 article "Structural equation modeling with the sem
package in R", Structural Equation Modeling, 13:465486, which is referenced
in ?sem. Second, there are apparently sufficient problems here that you
would do well to find someone to talk to about the formulation of your
model.
Regards,
John
>
> Thanks again,
>
> Regards,
>
> Anne
>
>
>
> John Fox a écrit :
>
> > Dear Anne,
> >
> > I started to diagram your model but stopped when I noticed some
problems:
> > (1) Some variables, such as pays_alti, are clearly endogenous, since
they
> > have arrows pointing to them, yet are declared as fixed exogenous
> variables;
> > that clearly doesn't make sense. (2) You've placed conflicting
constraints
> > on factor loadings and the variances of latent variables, for example
> > setting both the path from the factor type_paysage to the indicator
> > pays_alti and the variance of type_paysage to 1; again, that doesn't
make
> > sense.
> >
> > Do you have a path diagram for the model that you're trying to fit? It
> would
> > in particular be helpful to have a diagram of the structural part of the
> > model.
> >
> > Regards,
> > John
> >
> >
> > John Fox
> > Senator William McMaster
> > Professor of Social Statistics
> > Department of Sociology
> > McMaster University
> > Hamilton, Ontario, Canada
> > web: socserv.mcmaster.ca/jfox
> >
> >
> >> -Original Message-
> >> From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org]
> > On
> >> Behalf Of Anne Mimet
> >> Sent: August-25-10 10:27 AM
> >> To: r-help@r-project.org
> >> Subject: [R] SEM : Warning : Could not compute QR decomposition of
Hessian
> >>
> >> Hi useRs,
> >>
> >> I'm trying for the first time to use a sem. The model finally runs,
> >> but gives a warning saying :
> >> "In sem.default(ram = ram, S = S, N = N, param.names = pars, var.names
> >> = vars, : Could not compute QR decomposition of Hessian.
> >> Optimization probably did not converge. "
> >>
> >> I found in R-help some posts on this warning, but my attemps to modify
> >> the code didn't change the warning message (i tried to give an error
> >> of 1 to the latente variables). I can't figure what the problem is.
> >> Here is the code :
> >>
> >>
> >> tab<-read.table("F:/Mes documents/stats/sem/donnees_corr.txt",
> >> header=T, sep="",na.strings = "NA")
> >>
> >> tab[,46]<-as.factor(tab[,46])
> >> tab[,24]<-as.factor(tab[,24])
> >> tab[,40]<-as.factor(tab[,40])
> >>
> >> fct_cor<-hetcor(tab, ML=T)
> >> cor_tab<- fct_cor$correlations
> >>
> >> moment_tab<-read.moments(diag=F, names=c('c1','c2', 'c3','c4','c5',
> >> 'c6','c7', 'c8', 'c9', 'ind_plando', 'long_sup15', 'long_inf15',
> >> 'pente', 'est', 'sud','ouest',
Re: [R] Passing data to aov
Hi!
I didn't see the toy data, my bad.
I guess David's solution fixed the problem.
Ivan
Le 8/26/2010 16:10, Bryan Hanson a écrit :
> Hi Ivan, there is toy data given in the original post.
>
> The object passed to aov is a matrix, it is called scores, and it is passed
> via the formula that is created in the sense that the formula specifies
> objects in the calling environment. At least that is what is supposed to
> happen.
>
> I've not been able to find an example of manova using a data frame - aov,
> yes. I think that is because the response vector in manova must be a
> matrix, and you can't put a matrix in a data frame as a separate object.
> You could put the columns of the matrix into the data frame, but then it is
> not seen as a single entity, it's seen as separate columns, and specifying
> this in the formula looks impossible as I understand it.
>
> But, obviously there is something I don't understand! Thanks, Bryan
>
>
> On 8/26/10 9:50 AM, "Ivan Calandra" wrote:
>
>>Hi,
>>
>> First, some toy data would have helped a lot, and that can explain why
>> you got so few answers...
>>
>> Second, I have maybe some thoughts about it, not sure if this can help.
>>
>> The error tells you that the data passed to aov is not a matrix. In your
>> function, you don't specify explicitly the data object to be used, that
>> might be a problem. And that object has to be a matrix, it might not be.
>> Maybe something like:
>>
>> form<- as.formula(paste("pcs", "~", paste(fac, collapse = "*")),
>> ...
>> if (length(pcs) == 1) out<- aov(formula = form, data=score.matrix)
>>
>> That would work if I uderstood what you're doing (hum...) and if both "pcs"
>> and "fac" are parts of "score.matrix". But I've always done aov from a
>> dataframe, which makes sense if you have both numerical variables and
>> factors.
>> How can you do it from a matrix? Maybe my understanding is limited.
>>
>> As I said, just some thoughts...
>> HTH,
>> Ivan
>>
>>
>> Le 8/26/2010 14:47, Bryan Hanson a écrit :
>>> Hello Again Gurus and Lurkers:
>>>
>>> I¹m trying to build a very user-friendly function which does aov without
>>> having the user type in a formula (which would be tedious in this case).
>>> The idea is to take the response from a PCA score matrix, and the factors
>>> from a list. A simple example is the function given below, along with test
>>> data and a sample call to the function.
>>>
>>> I'm certainly having trouble understanding the proper ways to work with
>>> formulas and related items, but I think what I do in the function should
>>> work (it's built on ideas dug out of the archives). However, when the data
>>> is passed to aov (directly or via manova), something in the bowels of aov
>>> complains with the following error:
>>>
>>> Error in model.frame.default(formula = form, drop.unused.levels = TRUE) :
>>> object is not a matrix
>>>
>>> To me, the formula looks legitimate, and the variables in the formula are
>>> all in the environment (I think: The way I am doing this is basically that
>>> described in ?summary.manova where only a formula is passed, no data
>>> argument). Based upon reading the archives, the problem might arise with
>>> one of the deparse statements in aov, but I can't resolve it. It might also
>>> be one of scoping/environment, but again, this is only an idea.
>>>
>>> TIA for any assistance. Bryan
>>> *
>>> Bryan Hanson
>>> Professor of Chemistry& Biochemistry
>>> DePauw University, Greencastle IN USA
>>>
>>>
>>> hypTestScores<-
>>> function(mylist, score.matrix, pcs = 1:3, fac = NULL, ...) {
>>>
>>> scores<- score.matrix[,pcs]
>>> #str(scores)# looks correct to me
>>> form<- as.formula(paste("scores", "~", paste(fac, collapse = "*")),
>>> env = parent.frame())
>>> #str(form) # looks correct to me
>>> attach(mylist)
>>> if (length(pcs)> 1) out<- manova(formula = form, ...)
>>> if (length(pcs) == 1) out<- aov(formula = form, ...)
>>> print(summary(out))
>>> detach(mylist)
>>> invisible(out)
>>> }
>>>
>>> # test data
>>> td1<- matrix(rnorm(50), ncol = 5) # like PCA scores
>>> td2<- list(
>>> f1 = as.factor(sample(c("A", "B"), 10, replace = TRUE)),
>>> f2 = as.factor(sample(c("C", "D"), 10, replace = TRUE)))
>>>
>>> # test call
>>> hypTestScores(mylist = td2, score.matrix = td1,
>>> fac = c("f1", "f2"))
>>> detach("mylist") # needed if there is an error
>>>
sessionInfo()
>>> R version 2.11.0 (2010-04-22)
>>> x86_64-apple-darwin9.8.0
>>>
>>> locale:
>>> [1] en_US.UTF-8/en_US.UTF-8/C/C/en_US.UTF-8/en_US.UTF-8
>>>
>>> attached base packages:
>>> [1] datasets tools grid graphics grDevices utils stats
>>> [8] methods base
>>>
>>> other attached packages:
>>>[1] faraway_1.0.4 GGally_0.2 xtable_1.5-6
>>>[4] mvbutils_2.5.1 ggplot2_0.8.8 proto_0.3-8
>>>[7] reshape_0.8.3 ChemoSpec_1.45 R.utils_1.4.0
>>> [10] R.oo_1.7
Re: [R] Random slopes in lmer
coef(mod1)$bird will give you a matrix with two columns. The first column is the intercept for each bird and the second column is the slope for each bird. ranef(mod1) will also give you a matrix of two columns. These represent the random effects. That is, how much the intercept (or slope) is shifted from overall mean. HTH, Darin On Thu, Aug 26, 2010 at 01:15:10PM +0100, Samantha Patrick wrote: > Hi > > I want to extract the random slopes from a lmer (I am doing a random > regression), but are the answers obtained from ranef or coef? > > My model is: mod1<-lmer(B~ A +(A|bird), family=quasibinomial) > > And I want to obtain a slope for each individual bird but am not sure which > output I need and can't find the answer anywhere. > > Thanks > > Sam > > > Dr Samantha Patrick > EU INTERREG Post Doc > Davy 618 > Marine Biology & Ecology Research Centre > University of Plymouth > Plymouth > PL4 8AA > > T: 01752 586165 > M: 07740472719 > > > [[alternative HTML version deleted]] > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Passing data to aov
A big thanks David! Eliminating the env = parent.frame() seems to fix
everything. I hate it when I'm that close... I included that because from
?as.formula it seemed to be the default, but on re-read, I guess it doesn't
really say that. In addition, I had included it even though it was the
default as I felt at some point I would have to change it so I wanted it
there as a reminder. Instead, it led me astray...
Thanks so much! Bryan (coming to you from the R-Inferno today)
On 8/26/10 10:14 AM, "David Winsemius" wrote:
>
> On Aug 26, 2010, at 8:47 AM, Bryan Hanson wrote:
>
>> Hello Again Gurus and Lurkers:
>>
>> I¹m trying to build a very user-friendly function which does aov
>> without
>> having the user type in a formula (which would be tedious in this
>> case).
>> The idea is to take the response from a PCA score matrix, and the
>> factors
>> from a list. A simple example is the function given below, along
>> with test
>> data and a sample call to the function.
>>
>> I'm certainly having trouble understanding the proper ways to work
>> with
>> formulas and related items, but I think what I do in the function
>> should
>> work (it's built on ideas dug out of the archives). However, when
>> the data
>> is passed to aov (directly or via manova), something in the bowels
>> of aov
>> complains with the following error:
>>
>> Error in model.frame.default(formula = form, drop.unused.levels =
>> TRUE) :
>> object is not a matrix
>
> Actually I got a different error (also on a Mac. but with a version
> that is two month later than yours and with quite a smaller number of
> packages loaded):
>> hypTestScores(mylist = td2, score.matrix = td1,
> +fac = c("f1", "f2"))
> Error in eval(expr, envir, enclos) : object 'scores' not found
>
> Which went away if I removed the "env=" argument. My guess is that you
> were telling manova to look further up the lexical tree than it should
> have (or that attach() operation messed up the environment somehow):
>
>> hypTestScores(mylist = td2, score.matrix = td1,
> +fac = c("f1", "f2"))
>Df Pillai approx F num Df den Df Pr(>F)
> f1 1 0.31663 0.61777 3 4 0.6392
> f2 1 0.36652 0.77145 3 4 0.5673
> f1:f2 1 0.15687 0.24808 3 4 0.8593
> Residuals 6
>
>> detach("mylist") # needed if there is an error
> Error in detach("mylist") : invalid 'name' argument
>
> You might want to do inside the function something like
>
> res =try( your function )
> if (res="try-error") {stop() } else{
> }
>
>>
>> To me, the formula looks legitimate, and the variables in the
>> formula are
>> all in the environment (I think: The way I am doing this is
>> basically that
>> described in ?summary.manova where only a formula is passed, no data
>> argument). Based upon reading the archives, the problem might arise
>> with
>> one of the deparse statements in aov, but I can't resolve it. It
>> might also
>> be one of scoping/environment, but again, this is only an idea.
>>
>> TIA for any assistance. Bryan
>> *
>> Bryan Hanson
>> Professor of Chemistry & Biochemistry
>> DePauw University, Greencastle IN USA
>>
>>
>> hypTestScores <-
>>function(mylist, score.matrix, pcs = 1:3, fac = NULL, ...) {
>>
>>scores <- score.matrix[,pcs]
>> #str(scores)# looks correct to me
>>form <- as.formula(paste("scores", "~", paste(fac, collapse =
>> "*")),
>>env = parent.frame())
>> #str(form) # looks correct to me
>>attach(mylist)
>>if (length(pcs) > 1) out <- manova(formula = form, ...)
>>if (length(pcs) == 1) out <- aov(formula = form, ...)
>>print(summary(out))
>>detach(mylist)
>>invisible(out)
>>}
>>
>> # test data
>> td1 <- matrix(rnorm(50), ncol = 5) # like PCA scores
>> td2 <- list(
>>f1 = as.factor(sample(c("A", "B"), 10, replace = TRUE)),
>>f2 = as.factor(sample(c("C", "D"), 10, replace = TRUE)))
>>
>> # test call
>> hypTestScores(mylist = td2, score.matrix = td1,
>>fac = c("f1", "f2"))
>> detach("mylist") # needed if there is an error
>>
>>> sessionInfo()
>> R version 2.11.0 (2010-04-22)
>> x86_64-apple-darwin9.8.0
>>
>> locale:
>> [1] en_US.UTF-8/en_US.UTF-8/C/C/en_US.UTF-8/en_US.UTF-8
>>
>> attached base packages:
>> [1] datasets tools grid graphics grDevices utils stats
>> [8] methods base
>>
>> other attached packages:
>> [1] faraway_1.0.4 GGally_0.2 xtable_1.5-6
>> [4] mvbutils_2.5.1 ggplot2_0.8.8 proto_0.3-8
>> [7] reshape_0.8.3 ChemoSpec_1.45 R.utils_1.4.0
>> [10] R.oo_1.7.2 R.methodsS3_1.2.0 rgl_0.91
>> [13] lattice_0.18-5 mvoutlier_1.4 plyr_1.0.3
>> [16] RColorBrewer_1.0-2 chemometrics_0.8 som_0.3-5
>> [19] robustbase_0.5-0-1 rpart_3.1-46 pls_2.1-0
>> [22] pcaPP_1.8-1mvtnorm_0.9-9 nnet_7.3-1
>> [25] mclust_3.4.4 MASS_7.3-5 lars_0.9-7
>> [28] e1071_1.5-23 class_7.3-2
>>
>> ___
[R] reliability of R-Forge?
On Thu, 26 Aug 2010, Gavin Simpson wrote: On Thu, 2010-08-26 at 02:30 -0400, David Kane wrote: How reliable is R-Forge? http://r-forge.r-project.org/ It is down now (for me). Reporting "R-Forge Could Not Connect to Database: " late to chime in, so had tossed the first piece. As this relates to 'reliability of R-Forge' in the sense of possible process issues, rather than availability of the archive, I wanted to 'tag into' this thread I 'mirror' r-forge, so I have not seen this ... One thing I note, mirroring r-forge, and processing 'diffs' netween successive days, is that the md5sums of some packages regularly change without version number bumps. From this morning's report in my email: Thu Aug 26 04:30:01 EDT 2010 --- /tmp/rforge-pre.txt 2010-08-26 04:30:33.0 -0400 +++ /tmp/rforge-post.txt2010-08-26 04:38:03.0 -0400 @@ -8,18 +8,18 @@ AquaEnv_1.0-1.tar.gz 615059a5369d1aba149e6142fedffdde ArvoRe_0.1.6.tar.gzc955ae7c64c4270740172ad2219060ff BB_2010.7-1.tar.gz 4f85093ab24fac5c0b91539ec6efb8b7 -BCE_2.0.tar.gz 5a3fe3ecabbe2b2e278f6a48fc19d18d -BIOMOD_1.1-5.tar.gzd2f74f21bc8858844f8d71627fd8e687 +BCE_2.0.tar.gz 65a968c586e729a1c1ca34a37f5c293a +BIOMOD_1.1-5.tar.gz6929e5ad6a14709de7065286ec684942 ... -BTSPAS_2010.08.tar.gz 16b8f265846a512c329f0b52ba1924ab +BTSPAS_2010.08.tar.gz 809a96b11f1094e95b217af113abd0ac ... -BayesR_0.1-1.tar.gz72bd41c90845032eb9d15c4c6d086dec +BayesFactorPCL_0.5.tar.gz 173ab741c399309314eff240a4c3cd6f +BayesR_0.1-1.tar.gz9560b511f1b955a60529599672d58fea ... -BiplotGUI_0.0-6.tar.gz 594b3a275cde018eaa74e1ef974dd522 +BiplotGUI_0.0-6.tar.gz 857a484fdba6cb97be4e42e38bb6d0fd ... -IsoGene_1.0-18.tar.gz 679a5aecb7182474ed6a870fa52ca2e3 +IsoGene_1.0-18.tar.gz f37572957b2a9846a8d738ec88ac8690 and so forth. I've not taken the trime to understand why seemingly new versions are appearing without version bumps yet. Is anyone aware of explanations, other than a release process that does not require unique versioning of differing content? [it seems pretty basic to me that a 'receiver' of new content could do the checks I do, and decline to push conflicting md5sums over an identically named prior candidate in archive] -- Russ herrold __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Find classes for each column of a data.frame
Hello,
Is there a simple way to get the class type for each column of a
data.frame? I am in the situation where I would like to get all the
columns of a data.frame that are factors.
I have tried:
apply(df,2,class)
but all the columns come back as class "character".
Thanks
Dan
--
**
Daniel Brewer, Ph.D.
Institute of Cancer Research
Molecular Carcinogenesis
Email: daniel.bre...@icr.ac.uk
**
The Institute of Cancer Research: Royal Cancer Hospital, a charitable Company
Limited by Guarantee, Registered in England under Company No. 534147 with its
Registered Office at 123 Old Brompton Road, London SW7 3RP.
This e-mail message is confidential and for use by the a...{{dropped:2}}
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Re: [R] Find classes for each column of a data.frame
On Aug 26, 2010, at 11:31 AM, Daniel Brewer wrote: Hello, Is there a simple way to get the class type for each column of a data.frame? I am in the situation where I would like to get all the columns of a data.frame that are factors. I have tried: apply(df,2,class) lapply(df, class) but all the columns come back as class "character". Agree that is not what I would have expected. -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Find classes for each column of a data.frame
That's because apply works on arrays/matrices, not data.frames. It therefore coerces your data.frame to a matrix of type character, since you have factors, thus the result. You want sapply or lapply, since a data.frame is actually a list. sapply(df, class) and then to get what you want: df[sapply(df, is.factor)] Daniel Brewer wrote: Hello, Is there a simple way to get the class type for each column of a data.frame? I am in the situation where I would like to get all the columns of a data.frame that are factors. I have tried: apply(df,2,class) but all the columns come back as class "character". Thanks Dan __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Find classes for each column of a data.frame
On Aug 26, 2010, at 11:31 AM, Daniel Brewer wrote: Hello, Is there a simple way to get the class type for each column of a data.frame? I am in the situation where I would like to get all the columns of a data.frame that are factors. I have tried: apply(df,2,class) but all the columns come back as class "character". Just realized it happens because of matrix coercion. -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Find classes for each column of a data.frame
Since a data.frame is a list, you should use lapply() or sapply(): > df <- data.frame(a=1:5, b=LETTERS[1:5]) > lapply(df, class) $a [1] "integer" $b [1] "factor" > sapply(df, class) a b "integer" "factor" HTH, Ivan Le 8/26/2010 17:31, Daniel Brewer a écrit : > Hello, > > Is there a simple way to get the class type for each column of a > data.frame? I am in the situation where I would like to get all the > columns of a data.frame that are factors. > > I have tried: > apply(df,2,class) > but all the columns come back as class "character". > > Thanks > > Dan > -- Ivan CALANDRA PhD Student University of Hamburg Biozentrum Grindel und Zoologisches Museum Abt. Säugetiere Martin-Luther-King-Platz 3 D-20146 Hamburg, GERMANY +49(0)40 42838 6231 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] print method for str?
On Thu, 2010-08-26 at 10:42 +0200, David Hajage wrote: > Hello useRs and guRus, > > I was trying to add the support of str() in the ascii package, and I > realized that str() does not have a print method. It uses cat() from inside > the function and returns nothing. > > Since it is not usual in R, I wonder why? Is there a particular reason? The Value section of ?str has a (the?) reason. Efficiency. str is an S3 generic though, so you can write your own methods for it. HTH G > Thank you very much for your attention on... well, this unimportant > question. > > david > > [[alternative HTML version deleted]] > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. -- %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% Dr. Gavin Simpson [t] +44 (0)20 7679 0522 ECRC, UCL Geography, [f] +44 (0)20 7679 0565 Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk Gower Street, London [w] http://www.ucl.ac.uk/~ucfagls/ UK. WC1E 6BT. [w] http://www.freshwaters.org.uk %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Find classes for each column of a data.frame
On Thu, Aug 26, 2010 at 4:31 PM, Daniel Brewer wrote: > Hello, > > Is there a simple way to get the class type for each column of a > data.frame? I am in the situation where I would like to get all the > columns of a data.frame that are factors. > > I have tried: > apply(df,2,class) > but all the columns come back as class "character". apply is treating it as a matrix, and so converts it to the lowest common form. try lapply and use is.factor, with some unlist for good measure: z=data.frame(x=1:10,l=factor(1:10)) unlist(lapply(z,is.factor)) x l FALSE TRUE Barry __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Passing arguments between S4 methods fails within a function:bug? example with raster package.
Joris,
I looked at the problem. Here is a minimal example reproducing
the error
setGeneric("myplus",function(x,y,...) standardGeneric("myplus"))
setMethod("myplus",c(x="numeric",y="numeric"),
function(x,y,z=0) x+y+z
)
setMethod("myplus",c(x="numeric",y="list"),
function(x,y,...) callGeneric(x,unlist(y),...)
)
myplus(1,1) ## 2; OK
myplus(1,1,1) ## 3; OK
myplus(1,list(1)) ## 2; OK
myplus(1,list(1),z=1) ## 3; OK
a.c1 <- 1
myplus(1,list(1),z=a.c1) ## 3; OK
test <- function(x,y) {
a.c <- 1## Problem occurs when using '.' in variable name
myplus(a,y,z=a.c)
}
test(1,list(1)) ## error
test(1,1) ## 3; OK
It looks like a bug. I went through all environments from inside
the call of myplus("numeric","numeric") and a.c is nicely located
in parent.frame(6) with the value 1. However, when it has a '.' (dot)
in the variable name, the variable is not found. If it has a name
without a '.' (dot) there is no problem:
test <- function(x,y) {
a <- 1
myplus(a,y,z=a)
}
test(1,list(1)) ## 3; OK
- Niels
On 26/08/10 15.32, Joris Meys wrote:
Dear all,
This problem came up initially while debugging a function, but it
seems to be a more general problem of R. I hope I'm wrong, but I can't
find another explanation. Let me illustrate with the raster package.
For an object "RasterLayer" (which inherits from Raster), there is a
method xyValues defined with the signature
(object="RasterLayer",xy="matrix"). There is also a method with
signature (object="Raster",xy="vector"). The only thing this method
does, is change xy into a matrix and then pass on to the next method
using callGeneric again. Arguments are passed.
Now this all works smoothly, as long as you stay in the global environment :
require(raster)
a<- raster()
a[]<- 1:ncell(a)
origin<- c(-80,50)
eff.dist<- 10
unlist(xyValues(a,xy=origin,buffer=eff.dist))
[1] 14140 14141 14500 14501
Now let's make a very basic test function :
test<- function(x,orig.point){
eff.distance<- 10
p<- unlist(xyValues(x,xy=orig.point,buffer=eff.distance))
return(p)
}
This gives the following result :
test(a,origin)
Error in .local(object, xy, ...) : object 'eff.distance' not found
huh? Apparently, eff.distance got lost somewhere in the parsetree (am
I saying this correctly?)
The funny thing is when we change origin to a matrix :
origin<- matrix(origin,ncol=2)
unlist(xyValues(a,xy=origin,buffer=eff.dist))
[1] 14140 14141 14500 14501
test(a,origin)
[1] 14140 14141 14500 14501
It all works again! So something goes wrong with passing the arguments
from one method to another using callGeneric. Is this a bug in R or am
I missing something obvious?
The relevant code from the raster package :
setMethod("xyValues", signature(object='Raster', xy='vector'),
function(object, xy, ...) {
if (length(xy) == 2) {
callGeneric(object, matrix(xy, ncol=2), ...)
} else {
stop('xy coordinates should be a two-column matrix or
data.frame,
or a vector of two numbers.')
}
} )
setMethod("xyValues", signature(object='RasterLayer', xy='matrix'),
function(object, xy, method='simple', buffer=NULL, fun=NULL,
na.rm=TRUE) {
if (dim(xy)[2] != 2) {
stop('xy has wrong dimensions; it should have 2
columns' )
}
if (! is.null(buffer)) {
return( .xyvBuf(object, xy, buffer, fun, na.rm=na.rm) )
}
if (method=='bilinear') {
return(.bilinearValue(object, xy))
} else if (method=='simple') {
cells<- cellFromXY(object, xy)
return(.readCells(object, cells))
} else {
stop('invalid method argument. Should be simple or
bilinear.')
}
}
)
--
Niels Richard Hansen Web: www.math.ku.dk/~richard
Associate Professor Email: niels.r.han...@math.ku.dk
Department of Mathematical Sciences nielsrichardhan...@gmail.com
University of Copenhagen Skype: nielsrichardhansen.dk
Universitetsparken 5 Phone: +45 353 20783 (office)
2100 Copenhagen Ø +45 2859 0765 (mobile)
Denmark
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Re: [R] Find classes for each column of a data.frame
On Aug 26, 2010, at 10:36 AM, David Winsemius wrote: > > On Aug 26, 2010, at 11:31 AM, Daniel Brewer wrote: > >> Hello, >> >> Is there a simple way to get the class type for each column of a >> data.frame? I am in the situation where I would like to get all the >> columns of a data.frame that are factors. >> >> I have tried: >> apply(df,2,class) > > lapply(df, class) > >> but all the columns come back as class "character". > > Agree that is not what I would have expected. > sapply(iris, class) Sepal.Length Sepal.Width Petal.Length Petal.Width Species "numeric""numeric""numeric""numeric" "factor" > apply(iris, 2, class) Sepal.Length Sepal.Width Petal.Length Petal.Width Species "character" "character" "character" "character" "character" The second occurs because apply() will coerce the data frame to a matrix: > str(as.matrix(iris)) chr [1:150, 1:5] "5.1" "4.9" "4.7" "4.6" "5.0" "5.4" ... - attr(*, "dimnames")=List of 2 ..$ : NULL ..$ : chr [1:5] "Sepal.Length" "Sepal.Width" "Petal.Length" "Petal.Width" ... Since a matrix can only contain a single class of objects (recall that a matrix is a vector with dim attributes), 'iris' becomes a character matrix. HTH, Marc Schwartz __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] daisy(): space allocation issue
On Thu, 2010-08-26 at 07:35 -0700, abanero wrote: > Hi, > > I'm trying to apply the function daisy() to a data.frame 1x10 but I have > not enough space (error message: cannot allocate vector of length > 1476173280). > > I didn't imagine I was not able to work with a matrix of just 1 > observations... I have setted in Rgui --max-mem-size=2G (I'm not able to set > more space..) You are trying make a 10,000 x 10,000 matrix of dissimilarities. > How can I solve this issue? Separating observations depending on some rules? Get/use a machine with more RAM? I doubt separating observations into chunks and doing the dissimilarity computations on those chunks then recombining will work as the end result will still be the 10k x 10k matrix. (If that is what you meant.) What do you want to do with the dissimilarities? If clustering, try the clara() function in the same package (cluster) as daisy. But then you'd need to work out whether clustering such a large number of observations is a useful activity... If something else, perhaps let us know what you want to do with the dissimilarities or what you are trying to achieve as there may be other things that you can do instead. > thanks HTH G -- %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% Dr. Gavin Simpson [t] +44 (0)20 7679 0522 ECRC, UCL Geography, [f] +44 (0)20 7679 0565 Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk Gower Street, London [w] http://www.ucl.ac.uk/~ucfagls/ UK. WC1E 6BT. [w] http://www.freshwaters.org.uk %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] R: daisy(): space allocation issue
Hi, Gavin Simpson wrote: > What do you want to do with the dissimilarities? Clustering. Gavin Simpson wrote: > If clustering, try the clara() function I can't because the variables are mixed (numeric and categorical) and (I suppose..) I should use "gower" distance in advance, shouldn't I? Thanks Giuseppe -Messaggio originale- Da: Gavin Simpson [mailto:gavin.simp...@ucl.ac.uk] Inviato: 26 August 2010 17:47 A: abanero Cc: r-help@r-project.org Oggetto: Re: [R] daisy(): space allocation issue On Thu, 2010-08-26 at 07:35 -0700, abanero wrote: > Hi, > > I'm trying to apply the function daisy() to a data.frame 1x10 but I have > not enough space (error message: cannot allocate vector of length > 1476173280). > > I didn't imagine I was not able to work with a matrix of just 1 > observations... I have setted in Rgui --max-mem-size=2G (I'm not able to set > more space..) You are trying make a 10,000 x 10,000 matrix of dissimilarities. > How can I solve this issue? Separating observations depending on some rules? Get/use a machine with more RAM? I doubt separating observations into chunks and doing the dissimilarity computations on those chunks then recombining will work as the end result will still be the 10k x 10k matrix. (If that is what you meant.) What do you want to do with the dissimilarities? If clustering, try the clara() function in the same package (cluster) as daisy. But then you'd need to work out whether clustering such a large number of observations is a useful activity... If something else, perhaps let us know what you want to do with the dissimilarities or what you are trying to achieve as there may be other things that you can do instead. > thanks HTH G -- %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% Dr. Gavin Simpson [t] +44 (0)20 7679 0522 ECRC, UCL Geography, [f] +44 (0)20 7679 0565 Pearson Building, [e] gavin.simpsonATNOSPAMucl.ac.uk Gower Street, London [w] http://www.ucl.ac.uk/~ucfagls/ UK. WC1E 6BT. [w] http://www.freshwaters.org.uk %~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~%~% __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Random slopes in lmer
Hi Thanks for your help - however if I have a model of: mod1<-lmer(B~ A+C+(A|bird), family=quasibinomial) coef then gives me an individual slope for factors A and C. However the random effect is only nested within factor - so I am only trying to allow the slope to vary in relation to effect A Many Thanks Sam Dr Samantha Patrick EU INTERREG Post Doc Davy 618 Marine Biology & Ecology Research Centre University of Plymouth Plymouth PL4 8AA T: 01752 586165 M: 07740472719 -Original Message- From: Darin A. England [mailto:engl...@cs.umn.edu] Sent: 26 August 2010 16:12 To: Samantha Patrick Cc: r-help@R-project.org Subject: Re: [R] Random slopes in lmer coef(mod1)$bird will give you a matrix with two columns. The first column is the intercept for each bird and the second column is the slope for each bird. ranef(mod1) will also give you a matrix of two columns. These represent the random effects. That is, how much the intercept (or slope) is shifted from overall mean. HTH, Darin On Thu, Aug 26, 2010 at 01:15:10PM +0100, Samantha Patrick wrote: > Hi > > I want to extract the random slopes from a lmer (I am doing a random > regression), but are the answers obtained from ranef or coef? > > My model is: mod1<-lmer(B~ A +(A|bird), family=quasibinomial) > > And I want to obtain a slope for each individual bird but am not sure which > output I need and can't find the answer anywhere. > > Thanks > > Sam > > > Dr Samantha Patrick > EU INTERREG Post Doc > Davy 618 > Marine Biology & Ecology Research Centre > University of Plymouth > Plymouth > PL4 8AA > > T: 01752 586165 > M: 07740472719 > > > [[alternative HTML version deleted]] > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] reliability of R-Forge?
On 26 August 2010 at 11:28, R P Herrold wrote: | On Thu, 26 Aug 2010, Gavin Simpson wrote: | | > On Thu, 2010-08-26 at 02:30 -0400, David Kane wrote: | >> How reliable is R-Forge? http://r-forge.r-project.org/ | >> | >> It is down now (for me). Reporting "R-Forge Could Not Connect to Database: " | | late to chime in, so had tossed the first piece. As this | relates to 'reliability of R-Forge' in the sense of possible | process issues, rather than availability of the archive, I | wanted to 'tag into' this thread | | I 'mirror' r-forge, so I have not seen this ... | | One thing I note, mirroring r-forge, and processing 'diffs' | netween successive days, is that the md5sums of some packages | regularly change without version number bumps. From this | morning's report in my email: | | Thu Aug 26 04:30:01 EDT 2010 | | --- /tmp/rforge-pre.txt 2010-08-26 04:30:33.0 -0400 | +++ /tmp/rforge-post.txt2010-08-26 04:38:03.0 | -0400 | @@ -8,18 +8,18 @@ | AquaEnv_1.0-1.tar.gz 615059a5369d1aba149e6142fedffdde | ArvoRe_0.1.6.tar.gzc955ae7c64c4270740172ad2219060ff | BB_2010.7-1.tar.gz 4f85093ab24fac5c0b91539ec6efb8b7 | -BCE_2.0.tar.gz 5a3fe3ecabbe2b2e278f6a48fc19d18d | -BIOMOD_1.1-5.tar.gzd2f74f21bc8858844f8d71627fd8e687 | +BCE_2.0.tar.gz 65a968c586e729a1c1ca34a37f5c293a | +BIOMOD_1.1-5.tar.gz6929e5ad6a14709de7065286ec684942 | ... | -BTSPAS_2010.08.tar.gz 16b8f265846a512c329f0b52ba1924ab | +BTSPAS_2010.08.tar.gz 809a96b11f1094e95b217af113abd0ac | ... | -BayesR_0.1-1.tar.gz72bd41c90845032eb9d15c4c6d086dec | +BayesFactorPCL_0.5.tar.gz 173ab741c399309314eff240a4c3cd6f | +BayesR_0.1-1.tar.gz9560b511f1b955a60529599672d58fea | ... | -BiplotGUI_0.0-6.tar.gz 594b3a275cde018eaa74e1ef974dd522 | +BiplotGUI_0.0-6.tar.gz 857a484fdba6cb97be4e42e38bb6d0fd | ... | -IsoGene_1.0-18.tar.gz 679a5aecb7182474ed6a870fa52ca2e3 | +IsoGene_1.0-18.tar.gz f37572957b2a9846a8d738ec88ac8690 | | and so forth. I've not taken the trime to understand why | seemingly new versions are appearing without version bumps | yet. | | Is anyone aware of explanations, other than a release process | that does not require unique versioning of differing content? | [it seems pretty basic to me that a 'receiver' of new content | could do the checks I do, and decline to push conflicting | md5sums over an identically named prior candidate in archive] Version numbers change only when DESCRIPTION/Version gets updated. Content (of the tarball) and thusly md5sum changes whenever _any_ file in the archive changes. Methinks you tricked yourself into assuming tarballs have to be constant because they are on CRAN _where changes happen only with new releases_. Dirk -- Dirk Eddelbuettel | e...@debian.org | http://dirk.eddelbuettel.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Find classes for each column of a data.frame
On Thu, Aug 26, 2010 at 11:31 AM, Daniel Brewer wrote: > Hello, > > Is there a simple way to get the class type for each column of a > data.frame? I am in the situation where I would like to get all the > columns of a data.frame that are factors. > > I have tried: > apply(df,2,class) > but all the columns come back as class "character". > Try this where iris is a builtin data frame with one several numeric columns and one factor column, Species: > onlyFactors <- Filter(is.factor, iris) > str(onlyFactors) 'data.frame': 150 obs. of 1 variable: $ Species: Factor w/ 3 levels "setosa","versicolor",..: 1 1 1 1 1 1 1 1 1 1 ... -- Statistics & Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Quick GREP challenge
Many thanks! -- View this message in context: http://r.789695.n4.nabble.com/Quick-GREP-challenge-tp2339486p2339818.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] equalize function with zero and convert it
Thanks, but my problem is that I have my formulas in functions.
How can I do that for functions or e.g. for the body(Df) below?
And: How can I refer to the first solution of expression?
Thanks a lot!
-Ursprüngliche Nachricht-
Von: Gabor Grothendieck [mailto:ggrothendi...@gmail.com]
Gesendet: Donnerstag, 26. August 2010 14:56
An: Immanuel Seeger
Cc: r-help@r-project.org
Betreff: Re: [R] equalize function with zero and convert it
On Thu, Aug 26, 2010 at 6:57 AM, Immanuel Seeger
wrote:
> Dear all,
> I want to equalize a symbolic derivative (of a function with two
> variables) with zero and convert it to a variable, e.g. x.
>
> I'm computing the derivative by: (found it in the archive)
> library(Ryacas)
> f <- function(x,y) (100-x-y)*x-10*x
> yacas(f) # register f with yacas
> Df <- f
> body(Df) <- yacas(expression(deriv(f(x,y),x)))[[1]]
> Df
>
> R-Output:
> function (x, y)
> 100 - x - y - x - 10
>
> Questions :
> 1. How can I equalize the derivative(140-x-y-x-20) with zero?
> 2. How can I convert it to x?
>
> I want something like:
> 45-0.5*y=x
>
> With only one variable or two non-symbolic variables it's working with
> uniroot(f, c(-10, 10))$root.
>
This solves for x:
> library(Ryacas)
Loading required package: XML
> x <- Sym("x")
> y <- Sym("y")
> Solve((100-x-y)*x-10*x == 0, "x")
[1] "Starting Yacas!"
expression(list(x == -((10 - (100 - y) + root((100 - y - 10)^2,
2))/2), x == -((10 - (100 - y) - root((100 - y - 10)^2, 2))/2)))
--
GKX Group
GKX Associates Inc.
1-877-GKX-GROUP
ggrothendieck at gmail.com
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[R] mailing list
Hello, I was wondering if I could be taken off the list to get emails? Thanks Sonia [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Puzzle
Ben Holt bio.ku.dk> writes: > I have data similar to this: > > Location Surveyor Result > A1 83 > A2 76 > A3 45 > B1 71 > B4 67 > C2 23 > C5 12 > D3 34 > E4 75 > F4 46 > G5 90 > etc (5 million records in total) > > I need to divide the data to many subsets then randomly select 5 different > locations and 5 different surveyors (one at each of the 5 randomly selected > locations) for each subset. > > The function I have written basically picks five locations and then 1 > surveyor in each location, checks that there are five different surveyors > and if there isn't tries again. The problem is that for some subsets this > doesn't work. > > Some subsets don't have enough locations/surveyors or both, but this can be > checked for easily. The problem subsets do have enoughs locations and > surveyors but still cannot produce 5 locations each with a different > surveyor. The matrix below demonstrates such a subset: > > locations > A B C D E > 1 1 0 0 0 0 > Surveyors 2 1 0 0 0 0 > 3 1 0 0 0 0 > 4 1 0 0 0 0 > 5 1 1 1 1 1 > > I cannot think of a way to check for such a situation and therefore I have > simply programmed the function to give up after 100 attempts if it can't > find a solution. This is not very satisfactory however as the analysis takes > a very long time to run and it would also be very useful useful for me to > know how many suitable solution there are. If I understand you correctly: The task of finding a maximal number of 'independent' pairs (of surveyors and locations) is called the "generalized (or: maximum) linear assignment" problem, see http://en.wikipedia.org/wiki/Generalized_assignment_problem and there appears not to exist an efficient algorithm. One can think of procedures to find at least 5 such pairs (or to report infeasibility), but I think with millions of records in one of those subsets any approach (using loops) will be intolerable slow. Perhaps you can rethink your original intention or try to provide additional information during the process of generating the data. Of course, I would be glad to hear more positive news as this kind of problem does show up in many discrete optimization tasks. Hans Werner > I reckon some of you clever folk out there must be able to think of a better > solution. > > Any advice appreciated, > > Ben > __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Random slopes in lmer
??? You were provided exactly what you requested. I think you either need to read up on what random effects models mean or more clearly communicate what YOU mean. (It's unclear to me, anyway). -- Bert Gunter Genentech Nonclinical Statistics On Thu, Aug 26, 2010 at 8:57 AM, Samantha Patrick wrote: > Hi > Thanks for your help - however if I have a model of: > > mod1<-lmer(B~ A+C+(A|bird), family=quasibinomial) > > coef then gives me an individual slope for factors A and C. However the > random effect is only nested within factor - so I am only trying to allow the > slope to vary in relation to effect A > > Many Thanks > > Sam > > Dr Samantha Patrick > EU INTERREG Post Doc > Davy 618 > Marine Biology & Ecology Research Centre > University of Plymouth > Plymouth > PL4 8AA > > T: 01752 586165 > M: 07740472719 > > > -Original Message- > From: Darin A. England [mailto:engl...@cs.umn.edu] > Sent: 26 August 2010 16:12 > To: Samantha Patrick > Cc: r-help@R-project.org > Subject: Re: [R] Random slopes in lmer > > coef(mod1)$bird will give you a matrix with two columns. The first > column is the intercept for each bird and the second column is the > slope for each bird. > > ranef(mod1) will also give you a matrix of two columns. These > represent the random effects. That is, how much the intercept (or > slope) is shifted from overall mean. > > HTH, > Darin > > On Thu, Aug 26, 2010 at 01:15:10PM +0100, Samantha Patrick wrote: >> Hi >> >> I want to extract the random slopes from a lmer (I am doing a random >> regression), but are the answers obtained from ranef or coef? >> >> My model is: mod1<-lmer(B~ A +(A|bird), family=quasibinomial) >> >> And I want to obtain a slope for each individual bird but am not sure which >> output I need and can't find the answer anywhere. >> >> Thanks >> >> Sam >> >> >> Dr Samantha Patrick >> EU INTERREG Post Doc >> Davy 618 >> Marine Biology & Ecology Research Centre >> University of Plymouth >> Plymouth >> PL4 8AA >> >> T: 01752 586165 >> M: 07740472719 >> >> >> [[alternative HTML version deleted]] >> >> __ >> R-help@r-project.org mailing list >> https://stat.ethz.ch/mailman/listinfo/r-help >> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html >> and provide commented, minimal, self-contained, reproducible code. > > __ > R-help@r-project.org mailing list > https://stat.ethz.ch/mailman/listinfo/r-help > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] mailing list
Certainly. The link at the bottom of each and every email to this list has the web address at which you may do so. For your convenience: https://stat.ethz.ch/mailman/listinfo/r-help Sarah On Thu, Aug 26, 2010 at 9:05 AM, Sonia Spirling wrote: > Hello, > > I was wondering if I could be taken off the list to get emails? > > Thanks > Sonia > -- Sarah Goslee http://www.functionaldiversity.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Find classes for each column of a data.frame
On Thu, Aug 26, 2010 at 4:44 PM, Marc Schwartz wrote:
>> sapply(iris, class)
> Sepal.Length Sepal.Width Petal.Length Petal.Width Species
> "numeric" "numeric" "numeric" "numeric" "factor"
Note that comparing the result of class(foo) is a bad way of telling
if something is of a particular class, because of inheritance -
class(foo) can be a vector if the object belongs to more than one
class.
Always test using is.whatever(foo), which returns TRUE if foo is a whatever:
> f=factor(letters)
> class(f)
[1] "factor"
> class(f)=c("factor","alphabet")
> f
[1] a b c d e f g h i j k l m n o p q r s t u v w x y z
Levels: a b c d e f g h i j k l m n o p q r s t u v w x y z
> is.factor(f)
[1] TRUE
but now:
> class(f)=="factor"
[1] TRUE FALSE
Barry
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Re: [R] Find classes for each column of a data.frame
try
lapply(df, class)
Steve E
On Thu, Aug 26, 2010 at 4:31 PM, Daniel Brewer
wrote:
> Hello,
>
> Is there a simple way to get the class type for each column of a
> data.frame? I am in the situation where I would like to get all the
> columns of a data.frame that are factors.
>
> I have tried:
> apply(df,2,class)
> but all the columns come back as class "character".
apply is treating it as a matrix, and so converts it to the lowest
common form.
try lapply and use is.factor, with some unlist for good measure:
z=data.frame(x=1:10,l=factor(1:10))
unlist(lapply(z,is.factor))
x l
FALSE TRUE
Barry
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Re: [R] reliability of R-Forge?
On Thu, 26 Aug 2010, Dirk Eddelbuettel wrote: On 26 August 2010 at 11:28, R P Herrold wrote: | Is anyone aware of explanations, other than a release process | that does not require unique versioning of differing content? | [it seems pretty basic to me that a 'receiver' of new content | could do the checks I do, and decline to push conflicting | md5sums over an identically named prior candidate in archive] Version numbers change only when DESCRIPTION/Version gets updated. Content (of the tarball) and thusly md5sum changes whenever _any_ file in the archive changes. Methinks you tricked yourself into assuming tarballs have to be constant because they are on CRAN _where changes happen only with new releases_. It is good to know that my tool was looking at the wrong datum. Thank you, Dirk. I shall amend my tool. Probably I'll add a 'unroller' and diff tool so I may seem the changes pushing through Doesn't this cause your builder some heartburn in Debian package md5sum verification phase of autobuilding, or do you simply auto-bump the release field within a version series? -- Russ herrold __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Passing arguments between S4 methods fails within a function:bug? example with raster package.
On 8/26/2010 8:43 AM, Niels Richard Hansen wrote:
setGeneric("myplus",function(x,y,...) standardGeneric("myplus"))
setMethod("myplus",c(x="numeric",y="numeric"),
function(x,y,z=0) x+y+z
)
setMethod("myplus",c(x="numeric",y="list"),
function(x,y,...) callGeneric(x,unlist(y),...)
)
myplus(1,1) ## 2; OK
myplus(1,1,1) ## 3; OK
myplus(1,list(1)) ## 2; OK
myplus(1,list(1),z=1) ## 3; OK
a.c1 <- 1
myplus(1,list(1),z=a.c1) ## 3; OK
test <- function(x,y) {
a.c <- 1## Problem occurs when using '.' in variable name
myplus(a,y,z=a.c)
}
test(1,list(1)) ## error
test(1,1) ## 3; OK
I get an error in both instances, because in fact there is no variable
'a' defined (anywhere) in the session. I do get your results if I define
a global variable a <- 2
The reason for the original problem can be seen by looking at how the
methods are implemented (I added parentheses in the setMethod calls for
clarity)
> showMethods(myplus, includeDef=TRUE)
Function: myplus (package .GlobalEnv)
x="numeric", y="list"
function (x, y, ...)
{
callGeneric(x, unlist(y), ...)
}
x="numeric", y="numeric"
function (x, y, ...)
{
.local <- function (x, y, z = 0)
{
x + y + z
}
.local(x, y, ...)
}
and in particular the c("numeric", "numeric") method has an extra
argument z, and so has a nested .local function. This is how the methods
package deals with method signatures that differ from the generic.
I think, roughly, that when c("numeric", "numeric") tries finally to
resolve it's argument 'x', it looks for a variable 'a' in the
environment two levels up (method --> generic) from where it is being
resolved. This is fine if the method has no .local environment (e.g., if
c("numeric", "numeric") where defined as function(x, y, ...) { x +y }).
I think that there is no general solution to this; the user could
callGeneric or evaluate symbols from arbitrarily nested functions within
either the calling or called methods themselves, and could arrive at a
particular method through callGeneric or other routes (e.g.,
callNextMethod) that are not consistent in terms of the implied frame of
evaluation. I could be mistaken.
A simple solution is to provided named arguments to callGeneric, e.g.,
callGeneric(x=x, y=unlist(y), ...).
These issues are likely to cause problems for eval / substitute
expressions like the one posted by Joris yesterday, where there are
implicit assumptions about the environment in which the experession is
being evaluated.
Martin
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Re: [R] Random slopes in lmer
I'm sure this has appeared before on this list, but the biggest help to me has been: "Gelman and Hill", Data Analysis and Regression Using Multilevel/Hierarchical Models, which contains clear explanations and the R code to go along. Also check out http://lme4.r-forge.r-project.org/book/ Looking at Sam's model specification, C is a fixed effect. On Thu, Aug 26, 2010 at 09:15:14AM -0700, Bert Gunter wrote: > ??? > You were provided exactly what you requested. I think you either need > to read up on what random effects models mean or more clearly > communicate what YOU mean. (It's unclear to me, anyway). > > -- > Bert Gunter > Genentech Nonclinical Statistics > > On Thu, Aug 26, 2010 at 8:57 AM, Samantha Patrick > wrote: > > Hi > > Thanks for your help - however if I have a model of: > > > > mod1<-lmer(B~ A+C+(A|bird), family=quasibinomial) > > > > coef then gives me an individual slope for factors A and C. ?However the > > random effect is only nested within factor - so I am only trying to allow > > the slope to vary in relation to effect A > > > > Many Thanks > > > > Sam > > > > Dr Samantha Patrick > > EU INTERREG Post Doc > > Davy 618 > > Marine Biology & Ecology Research Centre > > University of Plymouth > > Plymouth > > PL4 8AA > > > > T: 01752 586165 > > M: 07740472719 > > > > > > -Original Message- > > From: Darin A. England [mailto:engl...@cs.umn.edu] > > Sent: 26 August 2010 16:12 > > To: Samantha Patrick > > Cc: r-help@R-project.org > > Subject: Re: [R] Random slopes in lmer > > > > coef(mod1)$bird will give you a matrix with two columns. The first > > column is the intercept for each bird and the second column is the > > slope for each bird. > > > > ranef(mod1) will also give you a matrix of two columns. These > > represent the random effects. That is, how much the intercept (or > > slope) is shifted from overall mean. > > > > HTH, > > Darin > > > > On Thu, Aug 26, 2010 at 01:15:10PM +0100, Samantha Patrick wrote: > >> Hi > >> > >> I want to extract the random slopes from a lmer (I am doing a random > >> regression), but are the answers obtained from ranef or coef? > >> > >> My model is: mod1<-lmer(B~ A +(A|bird), family=quasibinomial) > >> > >> And I want to obtain a slope for each individual bird but am not sure > >> which output I need and can't find the answer anywhere. > >> > >> Thanks > >> > >> Sam > >> > >> > >> Dr Samantha Patrick > >> EU INTERREG Post Doc > >> Davy 618 > >> Marine Biology & Ecology Research Centre > >> University of Plymouth > >> Plymouth > >> PL4 8AA > >> > >> T: 01752 586165 > >> M: 07740472719 > >> > >> > >> ? ? ? [[alternative HTML version deleted]] > >> > >> __ > >> R-help@r-project.org mailing list > >> https://stat.ethz.ch/mailman/listinfo/r-help > >> PLEASE do read the posting guide > >> http://www.R-project.org/posting-guide.html > >> and provide commented, minimal, self-contained, reproducible code. > > > > __ > > R-help@r-project.org mailing list > > https://stat.ethz.ch/mailman/listinfo/r-help > > PLEASE do read the posting guide http://www.R-project.org/posting-guide.html > > and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Quick GREP challenge
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Dimitri Shvorob
> Sent: Thursday, August 26, 2010 3:16 AM
> To: r-help@r-project.org
> Subject: [R] Quick GREP challenge
>
>
> > grep("f[0-9]+=", "f1=5,f22=3,", value = T)
> [1] "f1=5,f22=3,"
>
> How do I make the line output c("f1", "f22") instead?
> (Actually, c(1,22)
> would be even better).
If you had S+ you could use subpattern and keep arguments
to strsplit (which otherwise acts like R's strsplit):
> strings <- c("f1=5,f22=3,", "g4,f55,f66", "hello")
> strsplit(strings, "f([[:digit:]]+)", subpattern=1, keep=TRUE)
[[1]]:
[1] "1" "22"
[[2]]:
[1] "55" "66"
[[3]]:
character(0)
keep=TRUE means to split the string by what is not in
the pattern (keeping what is in the pattern) instead
of the usual splitting by what is in the pattern and
keeping what is not in the pattern.
subpattern=n means to match by the whole pattern but
to use only the n'th parenthesized subpattern when
deciding what to split by or keep. In this case the
whole pattern matches 'f' followed by 1 or more digits
but we only want the keep the digits. (subpattern=0,
the default, means to use the entire pattern.)
You still need to use as.integer() or as.numeric() on
the output elements.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
>
>
> Thank you.
>
>
> --
> View this message in context:
> http://r.789695.n4.nabble.com/Quick-GREP-challenge-tp2339486p2
339486.html
> Sent from the R help mailing list archive at Nabble.com.
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] reliability of R-Forge?
On Aug 26, 2010, at 11:01 AM, Dirk Eddelbuettel wrote: > > On 26 August 2010 at 11:28, R P Herrold wrote: > | On Thu, 26 Aug 2010, Gavin Simpson wrote: > | > | > On Thu, 2010-08-26 at 02:30 -0400, David Kane wrote: > | >> How reliable is R-Forge? http://r-forge.r-project.org/ > | >> > | >> It is down now (for me). Reporting "R-Forge Could Not Connect to > Database: " > | > | late to chime in, so had tossed the first piece. As this > | relates to 'reliability of R-Forge' in the sense of possible > | process issues, rather than availability of the archive, I > | wanted to 'tag into' this thread > | > | I 'mirror' r-forge, so I have not seen this ... > | > | One thing I note, mirroring r-forge, and processing 'diffs' > | netween successive days, is that the md5sums of some packages > | regularly change without version number bumps. From this > | morning's report in my email: > | > | Thu Aug 26 04:30:01 EDT 2010 > | > | --- /tmp/rforge-pre.txt 2010-08-26 04:30:33.0 -0400 > | +++ /tmp/rforge-post.txt2010-08-26 04:38:03.0 > | -0400 > | @@ -8,18 +8,18 @@ > | AquaEnv_1.0-1.tar.gz 615059a5369d1aba149e6142fedffdde > | ArvoRe_0.1.6.tar.gzc955ae7c64c4270740172ad2219060ff > | BB_2010.7-1.tar.gz 4f85093ab24fac5c0b91539ec6efb8b7 > | -BCE_2.0.tar.gz 5a3fe3ecabbe2b2e278f6a48fc19d18d > | -BIOMOD_1.1-5.tar.gzd2f74f21bc8858844f8d71627fd8e687 > | +BCE_2.0.tar.gz 65a968c586e729a1c1ca34a37f5c293a > | +BIOMOD_1.1-5.tar.gz6929e5ad6a14709de7065286ec684942 > | ... > | -BTSPAS_2010.08.tar.gz 16b8f265846a512c329f0b52ba1924ab > | +BTSPAS_2010.08.tar.gz 809a96b11f1094e95b217af113abd0ac > | ... > | -BayesR_0.1-1.tar.gz72bd41c90845032eb9d15c4c6d086dec > | +BayesFactorPCL_0.5.tar.gz 173ab741c399309314eff240a4c3cd6f > | +BayesR_0.1-1.tar.gz9560b511f1b955a60529599672d58fea > | ... > | -BiplotGUI_0.0-6.tar.gz 594b3a275cde018eaa74e1ef974dd522 > | +BiplotGUI_0.0-6.tar.gz 857a484fdba6cb97be4e42e38bb6d0fd > | ... > | -IsoGene_1.0-18.tar.gz 679a5aecb7182474ed6a870fa52ca2e3 > | +IsoGene_1.0-18.tar.gz f37572957b2a9846a8d738ec88ac8690 > | > | and so forth. I've not taken the trime to understand why > | seemingly new versions are appearing without version bumps > | yet. > | > | Is anyone aware of explanations, other than a release process > | that does not require unique versioning of differing content? > | [it seems pretty basic to me that a 'receiver' of new content > | could do the checks I do, and decline to push conflicting > | md5sums over an identically named prior candidate in archive] > > Version numbers change only when DESCRIPTION/Version gets updated. > > Content (of the tarball) and thusly md5sum changes whenever _any_ file > in the archive changes. > > Methinks you tricked yourself into assuming tarballs have to be constant > because they are on CRAN _where changes happen only with new releases_. > > Dirk I might also point out that the same process is in place for the daily tarballs of R itself, available via: ftp://ftp.stat.math.ethz.ch/Software/R/ The R version does not change with each new daily tarball, but the svn rev number will change with each new commit. Thus, each day, the checksum will be different, since the tarball is generated each day with the new svn commits from the prior 24 hours, unless of course, there have been no new commits in that time frame. In the case of the R tarballs, the generating script also includes a file called SVN-REVISION, which will contain something like: Revision: 52804 Last Changed Date: 2010-08-25 You don't get that with the R-Forge tarballs, since that file/info is not included in R source package tarballs. The same would occur BTW, if you did an svn checkout rather than using the tarball directly. Your checkout would reflect the current rev state of the svn repo at that time, which will be more granular than a daily source tarball and might be different 5 minutes later, if a new commit occurred in that time frame. This approach is part and parcel of using an svn repo. You can have main trunks (in the case of R, R-Devel), along with defined branches (eg. R 2.x.y), each of which may get hundreds or thousands of commits over some window of time, without a version bump during that time frame. R-Forge packages will not have the same frequency of commits, but the same approach can be applied. So you need to differentiate between the ongoing development/commit process and the versioned release process. HTH, Marc Schwartz __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problem with clusterCall, "Error in checkForRemoteErrors(lapply(cl, recvResult)) : "
Did anyone manage to have a look at this? I have now tried foreach with doMC backen and multicore, but still no luck. If someone can help that would be much appreciated! -- View this message in context: http://r.789695.n4.nabble.com/Problem-with-clusterCall-Error-in-checkForRemoteErrors-lapply-cl-recvResult-tp2338375p2340076.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Data Manipulations and SQL
Greetings Gabor, is it possible to open a channel to a data frame in the default environment? there are cases when using a sql update statement is the simplest alternative, so instead of dumping the df and then updating and then reimporting it I would like to update the df directly in R. Thank you. Stephen -- View this message in context: http://r.789695.n4.nabble.com/Data-Manipulations-and-SQL-tp860420p2340098.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] anova for plm objects
Dear All,
I'm looking to perform an ANOVA between two nested panel fixed effects models.
I tried with anova, as well as with waldtest. anova tells me there is no method
available for plm objects, while waldtest tells me my models are not nested. I
think they are instead. The difference between the two models is that in the
second I let the regression coefficients of a given variable variate according
to a factor.
MODEL 1
depFE.plm = plm(urate ~ l1_urate + month, model = "within", data =
unempl_data1, effect = "individual")
MODEL 2
depLDFE.plm = plm(urate ~ factor(ID):l1_urate + month, model = "within", data =
unempl_data1, effect = "individual")
anova
> anova(depFE.plm, depLDFE.plm)
Errore in UseMethod("anova") :
no applicable method for 'anova' applied to an object of class "c('plm',
'panelmodel')"
waldtest
> waldtest(depFE.plm, depLDFE.plm)
Errore in modelCompare(objects[[i - 1]], objects[[i]], vfun = vcov.) :
models are not nested
Did anyone code a plm method for anova?
OR
Does anyone know why according to waldtest my models are not nested?
Thanks
Roberto
Roberto Patuelli, Ph.D.
Istituto Ricerche Economiche (IRE) (Institute for Economic Research)
Università della Svizzera Italiana (University of Lugano)
via Maderno 24, CP 4361
CH-6904 Lugano
Switzerland
Phone: +41-(0)58-666-4166
Fax: +39-02-700419665
Email: roberto.patue...@usi.ch
Homepage: http://www.people.lu.unisi.ch/patuellr
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Re: [R] Data Manipulations and SQL
On Thu, Aug 26, 2010 at 1:18 PM, stephenb wrote: > is it possible to open a channel to a data frame in the default environment? > there are cases when using a sql update statement is the simplest > alternative, so instead of dumping the df and then updating and then > reimporting it I would like to update the df directly in R. Assuming that this is a question about sqldf see these links. The first has an example of update and the second discusses connections that persist across sqldf commands: http://code.google.com/p/sqldf/#8._Why_am_I_having_problems_with_update? http://code.google.com/p/sqldf/#Example_10._Persistent_Connections You may also wish to use the RSQLite package directly. -- Statistics & Software Consulting GKX Group, GKX Associates Inc. tel: 1-877-GKX-GROUP email: ggrothendieck at gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] non-linear plot parameters
I need the parameters estimated for a non-linear equation, an example of the data is below. rm(list=ls()) Time<-c( 0, 0, 0, 0, 0, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 5, 5, 5, 5, 5, 8, 8, 8, 8, 8) Level<-c( 100, 110, 90, 95, 87, 60, 65, 61, 55, 57, 40, 41, 50, 47, 44, 44, 42, 38, 40, 37, 37, 35, 40, 34, 32, 20, 22, 25, 27, 29) plot(Time,Level,pch=16) Keith -- M. Keith Cox, Ph.D. Alaska NOAA Fisheries, National Marine Fisheries Service Auke Bay Laboratories 17109 Pt. Lena Loop Rd. Juneau, AK 99801 keith@noaa.gov marlink...@gmail.com U.S. (907) 789-6603 [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Puzzle
Look for a vectorized solution such as ?table or ?aggregate to obtain a list of combination counts, followed by logical indexing or ?subset to get a set of valid combinations, and then use ?sample to get your random selections of locations/surveyors and then process only those combinations from your original data set. Keep in mind that constraints that you apply late in the selection process that can be applied earlier or on less data will usually become more efficient. If you provide actual sample statements that you have already tried, you may get a less abstract response. See the posting guide mentioned at the end of every posting for more guidance. "Ben Holt" wrote: >I have data similar to this: > >Location Surveyor Result >A1 83 >A2 76 >A3 45 >B1 71 >B4 67 >C2 23 >C5 12 >D3 34 >E4 75 >F4 46 >G5 90 >etc (5 million records in total) > >I need to divide the data to many subsets then randomly select 5 different >locations and 5 different surveyors (one at each of the 5 randomly selected >locations) for each subset. > >The function I have written basically picks five locations and then 1 surveyor >in each location, checks that there are five different surveyors and if there >isn't tries again. The problem is that for some subsets this doesn't work. > >Some subsets don't have enough locations/surveyors or both, but this can be >checked for easily. The problem subsets do have enoughs locations and >surveyors but still cannot produce 5 locations each with a different surveyor. > The matrix below demonstrates such a subset: > > locations > A B C D E >1 1 0 0 0 0 >Surveyors 2 1 0 0 0 0 >3 1 0 0 0 0 >4 1 0 0 0 0 >5 1 1 1 1 1 > >I cannot think of a way to check for such a situation and therefore I have >simply programmed the function to give up after 100 attempts if it can't find >a solution. This is not very satisfactory however as the analysis takes a >very long time to run and it would also be very useful useful for me to know >how many suitable solution there are. > >I reckon some of you clever folk out there must be able to think of a better >solution. > >Any advice appreciated, > >Ben > >__ >R-help@r-project.org mailing list >https://stat.ethz.ch/mailman/listinfo/r-help >PLEASE do read the posting guide http://www.R-project.org/posting-guide.html >and provide commented, minimal, self-contained, reproducible code. --- Jeff NewmillerThe . . Go Live... DCN:Basics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/BatteriesO.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
