Why aren't you using as.character instead of drop.levels?
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hello again,
i used the command
rbind(Mymatrix, t(as.data.frame(Z)))
and it works!
thanks very much for your replys!
marion
2011/10/11 David Winsemius
>
> On Oct 11, 2011, at 2:47 AM, Marion Wenty wrote:
>
> dear r-users,
>>
>> i have got a problem which i am trying to solve:
>>
>> i h
predict.gam is returning a matrix with a named column for each term.
Select the appropriate column. Example below
library(mgcv)
n<-200;sig <- 2
dat <- gamSim(1,n=n,scale=sig)
b <- gam(y~s(x0)+s(I(x1^2))+s(x2)+offset(x3),data=dat)
newd <- data.frame(x0=(0:30)/30,x1=(0:30)/30,x2=(0:30)/30,x3=(0:30
You haven't provided a reproducible example. You haven't even provided a subset
of your data, or the commands you used to read it in.
I might guess that
VAL <- DailyDiary[[1]]
might be what you wanted, and the "str" function might help you understand why.
Also, the "c" function does not "coerc
I'm using the package 'gdata' and 'drop.levels' in the following code as a
simplified example of what I want to do, pinpointing the issue of isolating
variable in a data frame:
/dfNamesAndHeight <- data.frame("Name" = c('Bill','Bob','Bo'), "Height" =
c(73,68,83))
name_Bob <- drop.levels(dfNamesAnd
pre<-predict(ozonea,groupA,type="terms",terms=NULL,newdata.guaranteed=FALSE,na.action=na.pass)
yeah!
but I don't know how to only show the value of s(ratio,bs="cr")
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I don't know the specifics of this package, but generally the C code
is called internally by R: it, however, requires compilation before R
can "talk" to it. You wont need to learn C though.
Look at the link David suggested for a precompiled version, but it may
be older than the development-version
I have a related question...I have a data frame similar with 74 rows that I
created with "header=TRUE", but when I try to coerce one of the data frame
columns into a vector, it shows up as having length 1, even though when I
print it, it shows 74 elements:
> VAL <- c(DailyDiary[1])
> VAL
> [1] 3 3
Dear Help-Rs,
I'm working with a file that contains large numbers and I need to export them
"as is". for example take:
x <-
c(27104010002005,27104020001805,27104090001810,90050013000140,90050013000120)
y <- c(1:5)
df <- data.frame(cbind(x,y))
When I then try a simple:
write.csv(df,file="d
Thanks, that lets me move to the next step in figuring out what's
wrong with my document :) Funny enough that's not the first problem
I've had with Sweave and non-ascii characters.
Peter
On Thu, Oct 13, 2011 at 7:30 PM, Yihui Xie wrote:
> OK, that is really helpful for diagnosis. In line 436 of
Hey,
Doesn't this give you a ridiculous Type 1 error? Maybe randomly select one
result and trust it.
Try bestglm or stepwise regression maybe.
Hope that's helpful,
Ken Hutchison
On Oct 14, 2554 BE, at 12:39 AM, "C.H." wrote:
> This is one solution
>
> ?sapply
>
> sa
This is one solution
?sapply
sapply(data.frame(iris$Sepal.Length, iris$Sepal.Width,
iris$Petal.Length, iris$Petal.Width), function(x)
(summary(aov(x~iris$Species
On Tue, Oct 11, 2011 at 5:10 PM, Joshua Wong wrote:
> Hi Guys,
>
> I have about 20 continous predictors and I want to do one-wa
Legends are built in columns. You need to find a graphics symbol to
put in the "points" column or you need to find something that the
lines paramater will turn into a dot (and I'm not sure what that might
be.) My suggestion would be to change the line type to dashed and use
" - - -" for the
It is about the legend.
As you see in the example the line is not above the points symbol.
I want the line in the symbol in the same column.
Thank you for you interest in helping me.
Have nice day!
El jue, 13-10-2011 a las 22:40 -0400, R. Michael Weylandt escribió:
> Looking at your provided e
Looking at your provided example (thank you!), I'm not entirely sure
what you want to put in the same column. Could you perhaps clarify --
is it something about the plot itself or the legend?
Michael
>
> On Thu, Oct 13, 2011 at 8:00 AM, Kenneth Roy Cabrera Torres
> wrote:
>> Dear R users:
>>
>> I
OK, that is really helpful for diagnosis. In line 436 of your tex
file, there is an NA, but it should really be this:
}% this brace ends the effect of ‘include external’
So the reason for LaTeX failure was this missing bracket }. And the
reason for R to output NA here is most likely to be that in
On Oct 13, 2011, at 7:47 PM, malhomidi wrote:
Hi again,
I've looked at the links above
No links visible to the rest of us non-Nabble users. When will nabble-
users ever learn to include context? (or to post with new Subject:
lines when the topic changes?)
and I see the develop
I'm not sure what the "column to identify other the other columns"
maps to graphically, but perhaps something like this will get you
started
V <- read.table(textConnection("
45678site
235641 4563 C
218942 10 6
source(FILE, print.eval = TRUE)
Hope this helps & good work on getting the next round of R enthusiasts
up and going!
Michael
On Thu, Oct 13, 2011 at 8:59 PM, Stephen Davies wrote:
> (Apologies for the n00b question.)
>
> Hello, I'm teaching R in an introductory programming course and am walking
Hi Rolf and others,
Once again, I think a misunderstanding. My inexperience has led me to
use the
term "broken stick" where "non-linear segments" might be more
appropriate.
This is all experimental data (t and seg_an), and yes I am fitting a
model to this data (see the "crv" function in
my exam
Thank you both. For my purposes I need the for statement. I looked into merge
but what I wound up doing was installing the package 'compare' and using the
if statement /if((compare(dfCity[k,1], dfState[j,1],
dropLevels=TRUE)==TRUE){do something}/. This seemed to behave the way I
wanted.
--
View t
(Apologies for the n00b question.)
Hello, I'm teaching R in an introductory programming course and am walking
the students through the baby steps. One thing I'd like to be able to do is
have them copy the commands they type at the R console into a text file,
and then execute the text file to see t
Hi Rolf and Others,
Thanks Rolf for your suggestion.
I've had a look at segmented, but can't see how it can help.
My model is the solution to an ODE of a physical system. The fitted
parameters from
this model are the answers I require.
I'm not sure how segmented can help -- it is
a linear re
Hi again,
I've looked at the links above and I see the development version of
the igraph library. I see the src folder implemented in C. Are these source
codes available in R or I just would have to use the C code? The reason is
that I just started learning R and I really want to stay awa
On Oct 13, 2011, at 9:31 PM, R. Michael Weylandt wrote:
A more detailed description is available in Circle 7 of the R Inferno
(among other places), but the short answer is you type methods(plot)
to see all the different plot functions and then just type the name of
the specific method to get th
Sure, here's the generated .tex file:
http://pastebin.com/b9fTsr0h
Peter
On Thu, Oct 13, 2011 at 6:14 PM, Yihui Xie wrote:
> Seems to be weird. I can run the example smoothly under Ubuntu with R
> 2.13.2. Could you also post your .tex file?
>
> Regards,
> Yihui
> --
> Yihui Xie
> Phone: 515-294
A more detailed description is available in Circle 7 of the R Inferno
(among other places), but the short answer is you type methods(plot)
to see all the different plot functions and then just type the name of
the specific method to get the code. If you know which method you are
going for, you can
Dear R People:
How do you find the code underneath a generic function, please?
Sorry for the dumb question.
Thanks,
Sincerely,
Erin
--
Erin Hodgess
Associate Professor
Department of Computer and Mathematical Sciences
University of Houston - Downtown
mailto: erinm.hodg...@gmail.com
__
Seems to be weird. I can run the example smoothly under Ubuntu with R
2.13.2. Could you also post your .tex file?
Regards,
Yihui
--
Yihui Xie
Phone: 515-294-2465 Web: http://yihui.name
Department of Statistics, Iowa State University
2215 Snedecor Hall, Ames, IA
On Thu, Oct 13, 2011 at 7:01 PM,
I'm trying to get pgfSweave up and running. Hopefully I can get it
working from within LyX, but first I'm just trying to get the simplest
possible thing (compiling one of the example files in the pgfSweave
package) to work. I'm using the example that comes in the pgfSweave
package unmodified, but f
On 14/10/11 11:39, Redding, Matthew wrote:
Hi Rolf and Others,
Thanks Rolf for your suggestion.
I've had a look at segmented, but can't see how it can help.
My model is the solution to an ODE of a physical system. The fitted
parameters from
this model are the answers I require.
I'm not sure ho
hello
I want to make a boxplot with diferents rows and also include a column to sort
into two groups to each of the other columns
my date set looks like this:
4 5 6 7 8 site
23 56 41 45 63 C
21 89 42 10 63 E
Hi,
i had same problem with xlsReadWrite. this function loads a required
package. try this:
>xls.getshlib()
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Hi all,
I'd like to plot the Real and Imaginary parts of some f(z) as two
different surfaces in wireframe (the row/column axes are the real and
imag axes). I know I can do it by, roughly speaking, something like
plotz <- expand.grid(x={range of Re(z)}, y={range of Im(z), groups=1:2)
plotz$fun
There are lots of options since you did not tell us what you want on the
axis (or what you have tried).
For example if you want more than 6 tick marks/labels, replace xlim=c(0,
3000) with xaxp=c(0, 3000, 12) to get labels every 250 meters instead of
500. Depending on the size of the graph window y
Very nice! I am quite impressed at how flexible data.table is.
On Thu, Oct 13, 2011 at 1:05 AM, Matthew Dowle wrote:
> Using Josh's nice example, with data.table's built-in 'by' (optimised
> grouping) yields a 6 times speedup (100 seconds down to 15 on
> my netbook).
>
>> system.time(all.2b <- l
One package that you can use is Rcartogram from Omegahat, although it
took me a long long time to figure out how to use it for real maps. I
noticed there was another unpublished package named cart in R-Forge,
and I have never tried it.
I also want to know if there are other R packages that have as
Hi All,
I am a relative newbie to R and have the following problem I was trying to
solve. I had taken a look at the 'sample selection' package but was having
trouble applying it to my use case and was wondering if anyone out there had
done something similar and could share code or documentation
lincoln hotmail.com> writes:
>
> Hi all,
>
> I have run a (glm) analysis where the dependent variable is the gender
> (family=binomial) and the predictors are percentages.
> I get a warning saying "fitted probabilities numerically 0 or 1 occurred"
> that is indicating that quasi-separation or s
D_Tomas hotmail.com> writes:
>
> Hi userRs!
>
> I am trying to fit some GLM-poisson and neg.binomial. The neg. Binomial
> model is to account for over-dispersion.
>
> When I fit the poisson model i get:
> (Dispersion parameter for poisson family taken to be 1)
>
> However, if I estimate the d
Hi,
just put it in the formula:
aggregate(Number ~ Letter+Test,data=dtf,max)
cheers
Am 13.10.2011 19:30, schrieb syrvn:
> Hello again,
>
>
> dtf<-read.table(textConnection("Letter Test Number
> a b 1
>
Woohoo! Thank you Sarah and Michael. You are rock stars!
Daniel
-Original Message-
From: Sarah Goslee [mailto:sarah.gos...@gmail.com]
Sent: Thursday, October 13, 2011 11:54 AM
To: Bailey, Daniel
Cc: r-help@r-project.org
Subject: Re: [R] getting data associated with coordinates in a spa
On Thu, Oct 13, 2011 at 2:44 PM, Bailey, Daniel wrote:
> Michael, that's half of the problem solved (whew!!). Now how do I change the
> data at that location?
You assign it a new value, just as for any assignment. Using the
example from my previous email:
> data(meuse.grid)
> m = SpatialPixels
Michael, that's half of the problem solved (whew!!). Now how do I change the
data at that location?
This is not an intuitive way to manipulate data.
-Original Message-
From: R. Michael Weylandt [mailto:michael.weyla...@gmail.com]
Sent: Thursday, October 13, 2011 11:35 AM
To: Bailey, Dan
OT question: can R produce Cartograms?
Here's an example of World Population:
http://www.worldmapper.org/display.php?selected=2
This might make Texas smaller and Rhode Island larger
Robert Farley
LACMTA
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@
Ah yes, my eternal nemesis the S4 class...
You were basically there with
e[e$coordinates==(0,17),]
but for some access stuff that comes from the SpatialDataPointsFrame class.
You'll probably want to do this in two steps:
coords = coordinates(e)
## Use the access function coordinates to get a 2
Hi,
On Thu, Oct 13, 2011 at 2:05 PM, Bailey, Daniel wrote:
> Thank you Sarah. I tried your suggestion, and if I coerce it into a normal
> data.frame, that method works. But if you've already made the data into a
> SpatialPixelsDataFrame and run coordinates (both from the package "sp") so
> tha
Unless your audience is mainly interested in Texas and California and is
completely content to ignore Rhode Island, then I would suggest that you look
at the state.vbm map in the TeachingDemos package that works with the maptools
package. The example there shows coloring based on a variable.
-
Because SpatialPointsDataFrame is S4 object, you may try index by @
e@coords
or coordinates(e)
Weidong Gu
On Thu, Oct 13, 2011 at 2:18 PM, Bailey, Daniel wrote:
> Michael,
> Thank you for the tips. The suggestion didn't work though. Here is the output
> of str(e):
> Formal class 'SpatialPointsD
Michael,
Thank you for the tips. The suggestion didn't work though. Here is the output
of str(e):
Formal class 'SpatialPointsDataFrame' [package "sp"] with 5 slots
..@ data :'data.frame': 168 obs. of 2 variables:
.. ..$ catchcandata: num [1:168] 47.2 50.4 53.7 58 69.8 ...
.. ..$ secti
In your plot call, you can use xaxt = "n" to turn off the default x
axis tick marks, then add
axis(1, at = VARIABLEWHEREYOUWANTTICKMARKS) # If you want ticks at
the x you put in, its just axis(1, at = x)
to get ticks where you want them. There's also a label= argument if
you want them to be labl
It's going to depend how the coordinates are stored within the data
frame. Do you perhaps know if they are factors or character strings?
(I'm not familiar with the package). If you don't know, type
str(NAMEOFYOUROBJECT) and we can help interpret the output.
Untested, I think this would actually wo
Hi userRs!
I am trying to fit some GLM-poisson and neg.binomial. The neg. Binomial
model is to account for over-dispersion.
When I fit the poisson model i get:
(Dispersion parameter for poisson family taken to be 1)
However, if I estimate the dispersion coefficient by means of:
sum(residuals(fi
The easiest work-around I've found for this problem is to create a vector in
your data frame just using numbers to order them how you want, create a
separate "labeling" data frame with those numbers and corresponding text
labels, and then enter the vector with the grouping names from the labeling
f
Dear R users,
I am quite desperate for help. I haven't used R in a couple of years and I'm
currently finishing a masters project and running out of time to figure out my
problem. I have read and tried the examples on many websites, in your forums
and R Help yet still can't manage to change the
Thank you Sarah. I tried your suggestion, and if I coerce it into a normal
data.frame, that method works. But if you've already made the data into a
SpatialPixelsDataFrame and run coordinates (both from the package "sp") so that
the columns "x" and "y" become a single column "coordinates" with t
Hello again,
dtf<-read.table(textConnection("Letter Test Number
a b 1
a b 1
b b 1
Thanks for your answers! Will check them now :)
Yes, sorry, I was wrong.
Letter Number
d 0
d 0
should be:
Letter Number
d 0
after applying the algorithm!
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Sent fro
syrvn wrote on 10/13/2011 11:42:44 AM:
>
> Hi,
>
>
> imagine the following matrix/data.frame
>
> Letter Number
> a 1
> a 1
> b 1
> b 0
> c 0
> c 1
> d 0
> d 0
>
> If the numbers for two identical letters are also identical then I want
to
> remove either the first or the
> second row of that
Why would you end up with d1 in your output if you don't have a d1 in your
original data frame? Are you saying that, when both letters have a zero after
them, you want to replace one of them with a 1?
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-projec
try this:
> x <- read.table(textConnection("Letter Number
+ a 1
+ a 1
+ b 1
+ b 0
+ c 0
+ c 1
+ d 0
+ d 0"), as.is = TRUE, header = TRUE)
> closeAllConnections()
> # following assumes that there are pairs of numbers
> result <- do.call(rbind, lapply(split(x, x$Letter), function(.pair){
+ if (a
Hi Syrvn,
how about this
dtf<-read.table(textConnection("Letter Number
a 1
a 1
b 1
b 0
c 0
c 1
d 0
d 0"),header=T)
aggregate(Number~Letter,data=dtf,max)
cheers.
Am 13.10.2011 18:42, schrieb syrvn:
> Hi,
>
>
> imagine the following matrix/data.frame
>
> Letter Number
> a 1
> a 1
> b 1
> b 0
> c
Andrey wrote on 10/13/2011 08:40:21 AM:
>
> Dear All,
>
> For a vector, I use this
>
> xu<-1:20
> t<-rep((1:4),each=5)
> tapply(xu,t,mean)
> 1 2 3 4
> 3 8 13 18
>
> and for a matrix the only way I may guess is:
> > xu
> [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
> [1,]
On Oct 13, 2011, at 12:42 PM, syrvn wrote:
Hi,
imagine the following matrix/data.frame
Letter Number
a 1
a 1
b 1
b 0
c 0
c 1
d 0
d 0
If the numbers for two identical letters are also identical then I
want to
remove either the first or the
second row of that letter. If for a letter the nu
Hi,
imagine the following matrix/data.frame
Letter Number
a 1
a 1
b 1
b 0
c 0
c 1
d 0
d 0
If the numbers for two identical letters are also identical then I want to
remove either the first or the
second row of that letter. If for a letter the numbers are 1 and 0 I want to
remove the row with t
On Oct 13, 2011, at 11:20 AM, pigpigmeow wrote:
I'm confused...
I type ..
predict.gam(ozonea,type=s(ratio,bs="cr"))
pressure maxtemp s(avetemp) s(ratio)
1 -0.0459102290 -0.185178463 0.263358446 -0.164558673
2 -0.0286464652 -0.194731320 0.199315027 0.727823293
30.04780734
Functions with prototypes of the form
SEXP myfunc(SEXP, SEXP, ..., SEXP)
must be called via .Call(), not .C().
Also, you declared myfunction as returning
SEXP but returned nothing. Try ending the
function with
return R_NilValue;
You should change the default compiler flags
to report all warni
Would it be worthwhile to update the read.spss implementation using the
more recent discoveries from the PSPP group? I don't mean to copy their
code; but to use the ideas in their code. Is anyone working on this? I
wouldn't want the effort to be duplicated.
On Thu, 2011-10-13 at 16:22 +0200, Uwe L
On Oct 13, 2011, at 12:12 PM, Alejandro Coca Castro wrote:
Hi, somebody knows a package for running Growing Neural Gas in R.
> require(sos)
> findFn("Growing Neural Gas")
found 0 matches
x has zero rows; nothing to display.
Warning message:
In findFn("Growing Neural Gas") :
HIT not found i
Besides being a much better solution, it displays ties (which I see as a
benefit). For example, if I ask for 5 I get 8 for top values since 12 occurs
3 times.
Here is the same thing David posted with slight mods to generalize it a bit
for cnt:
x <- swiss$Education[1:25]
dat = matrix(x,5,5)
colnam
Hi, somebody knows a package for running Growing Neural Gas in R.
Thanks in advance,
--
Alejandro Coca
UN
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PLEASE do read the posting guide http://www.R-project.org/posti
On Oct 13, 2011, at 10:42 AM, Ben qant wrote:
Here is a more R'sh solution (speed unknown).
Really? The intermediate, potentially large, objects seem to be
proliferating.
Courtesy of Mark Leeds (I
modified it a bit to generalize it for a cnt input and get min and
max).
Again, getting
On Thu, 2011-10-13 at 08:20 -0700, pigpigmeow wrote:
> I'm confused...
> I type ..
> predict.gam(ozonea,type=s(ratio,bs="cr"))
That is not a valid 'type'; normally you'd use `type = "terms"` or `type
= "iterms"`, depending on whether you want (any) standard errors to
include the uncertainty about
Be careful with the idiom
x[, -which(columnIsBad)]
If no columns are bad this leads to
x[, -integer(0)]
which is a data.rame with no columns,
exactly the opposite of what you want.
x[, !columnIsBad]
doesn't have that problem. However, if
you can't tell if a column is bad or not
(i.e., col
This is R, not S-Plus.
In the first two lines you have
expr <- as.expression(substitute(fun))
nvals <- length(eval(expr, envir = as.list(meanval)))
Simplified example:
y <- 0
fn1 <- function(){
y <- 1
fn1sub <- function() print(y)
fn1sub()
}
fn2sub <- function() print(y)
fn2 <- fun
I think you made it very clearly. thx
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I'm confused...
I type ..
predict.gam(ozonea,type=s(ratio,bs="cr"))
pressure maxtemp s(avetemp) s(ratio)
1 -0.0459102290 -0.185178463 0.263358446 -0.164558673
2 -0.0286464652 -0.194731320 0.199315027 0.727823293
30.0478073459 -0.013227033 0.002228896 0.342373202
4 -0.00
Hi all,
I have run a (glm) analysis where the dependent variable is the gender
(family=binomial) and the predictors are percentages.
I get a warning saying "fitted probabilities numerically 0 or 1 occurred"
that is indicating that quasi-separation or separation is occurring.
This makes sense given
You can use the DEoptim function in DEoptim package and to include a line of
code within your objective function that assigns a very high value when the
constraints are not satisfied. I have tried that and it works.
-
Juan David Ospina Arango
School of Statistics
Universidad Nacional de Colo
hi: if you make the design matrix correctly, you can fit that using the
systemfit function
in Arne Henningsen's systemfit package. You need to construct the response
by making one long column response of R1 on top of R2 on top of R3. Then you
need to make a diagonal X matrix with X1, X2 and X3 bein
On 12.10.2011 20:13, forget_f1 wrote:
Hi,
I hope someone can help me with the following issue.
I need find the minimum beta that satisfies the following:
inf{beta>0 | f(x+beta*f(x))*f(x)<=0}
where f() is a function and x is a sample statistic.
Functions such as "nlminb" and "constrOptim" m
Here is a more R'sh solution (speed unknown). Courtesy of Mark Leeds (I
modified it a bit to generalize it for a cnt input and get min and max).
Again, getting cnt highest and lowest values in the entire matrix and
display the data point row and column names with each:
> x <- swiss$Education[1:25]
On 11.10.2011 10:25, taby gathoni wrote:
Dear r user,
please find my attached sample of the dataset i am using to create a
crosstable and eventually plot a histogram from the output.
I am using the cut2 function to create bins, about 7 of them using the code
after reading the data:
cluste
On 13.10.2011 15:50, guoshicheng2005 wrote:
Dear All,
Can I use lm() to fit more than one response in
single expression. e.g data is a matrix of these
variables
R1 R2 R3 X1 X2 X3
1 2 1 1 2 3
Now i wnat to fit
R1~X1
R2~X2
R3~X3
in turn, and I don't want to do it use loops
of couse it i
On 11.10.2011 12:07, Smart Guy wrote:
Hi,
I have one doubt about one of the parameter of 'read.spss()' from
'foreign' package.
Here is the syntax :-
read.spss ( file,
use.value.labels = TRUE,
to.data.frame = FALSE,
max.value.labels = Inf,
trim.factor.names = FALSE,
Dear all,
I'm trying to replicate the panel function behaviour that Deepayan and
others discuss here:
https://stat.ethz.ch/pipermail/r-help/2007-April/130779.html
and Oscar Lamigueiro refers to here and provides sample code for:
http://www.r-bloggers.com/confidence-bands-with-lattice-and-r/
whe
Dear All,
Can I use lm() to fit more than one response in
single expression. e.g data is a matrix of these
variables
R1 R2 R3 X1 X2 X3
1 2 1 1 2 3
Now i wnat to fit
R1~X1
R2~X2
R3~X3
in turn, and I don't want to do it use loops
of couse it it easy to make it using loops,but the proceed is
Dear All,
For a vector, I use this
xu<-1:20
t<-rep((1:4),each=5)
tapply(xu,t,mean)
1 2 3 4
3 8 13 18
and for a matrix the only way I may guess is:
> xu
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8]
[1,]14321432
[2,]21432
Answering my own question.
?sample (!)
y <- by(x, x$TrSeasYr, function(x) mean(sample(x[,1], size=999, replace =
TRUE)))
>>> Tim Howard 10/13/2011 9:42 AM >>>
All -
I have an uneven set of replicates and would like to sample from this set X
number of times to generate a mean for each groupi
Dear Krishnan,
This behaviour isn't particular to scatterplot() in car. Try setting
options(scipen=10) and see ?options.
I hope this helps,
John
John Fox
Senator William McMaster
Professor of Social Statistics
Department of Sociology
McMaster University
Hamilt
All -
I have an uneven set of replicates and would like to sample from this set X
number of times to generate a mean for each grouping variable. I was thinking
the boot package would be the thing to use, but now I'm not so sure ... given
the discussion here:
http://finzi.psych.upenn.edu/Rhel
try this:
> x <- read.table(textConnection("A B C D
> E
+ 12 33 Error1 71 Error2
+ 12 33 Error1 71 Error2
+ 12 33 Error1 71 Error2
+ 12 33
Hello everyone,
I'd like to search for certain "expressions" (characters) in my data.frame
and delete the containing columns.
So, for example in the below table, I'd like to delete all columns which
contain the expression "Error". That is, R should delete column C and E from
my data. Any id
I'd be inclined to use predict(ozonea,type="terms") to extract the
estimates of s(ratio,bs = "cr")
that you need. But do you really want newozone - s(ratio,bs="cr") when
you've used a log link?
best,
Simon
On 10/13/2011 09:05 AM, pigpigmeow wrote:
hi! I hope all of you can help me this ques
Sören,
have a look at package snowfall which provides sfIsRunning.
HTH
Claudia
Am 13.10.2011 06:34, schrieb Søren Højsgaard:
Is there a 'proper' way of checking if cluster is active. For example, I create
a cluster called .PBcluster
str(.PBcluster)
List of 4
$ :List of 3
..$ con :Cl
Kathie wrote on 10/13/2011 06:33:59 AM:
>
> Dear R users,
>
> I'd like to count the number of integers in a vector y.
>
> Here is an example.
>
> y <- c(0,1,1,3,3,3,5,5,6)
>
> In fact, I know how to count the number of specific number in y.
>
> sum(y==0) -> 1
> sum(y==1) -> 2
> sum(y==2) -> 0
try this:
> y <- c(0,1,1,3,3,3,5,5,6)
> x <- tabulate(y+1)
> names(x) <- seq(from = 0, by = 1, length = length(x))
> x
0 1 2 3 4 5 6
1 2 0 3 0 2 1
>
On Thu, Oct 13, 2011 at 7:33 AM, Kathie wrote:
> Dear R users,
>
> I'd like to count the number of integers in a vector y.
>
> Here is an example.
I believe the lag() function can be used to rig this up.
Something like sma.365 <- SMA(lag(data, 12), n=120) # Untested, but seems right
Michael
On Thu, Oct 13, 2011 at 6:31 AM, Laura wrote:
> Hello,
>
> I used the TTR package in R to calculate moving averages. I have a monthly
> time series a
I am not an expert on this, but there is a way to check this. You can predict
from a gam using predict(ozonea, newdata=...). In the "newdata" argument you
can specify the X-values of interest to you. Thus, you can compare if your
predictions are the same when predicted directly from the gam or when
Slight addendum, tabulate() ignores zeros so you'll need to do tabulate(y+1).
Table will handle zeros but won't look for values that never appear
(in your example 2 & 4).
Michael
On Thu, Oct 13, 2011 at 8:51 AM, R. Michael Weylandt
wrote:
> Table() or more generally tabulate()
>
> Though, as a
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