Re: [R] How to use lm() output for systemfit() 'Seemingly unrelated regression'

2010-09-06 Thread zbynek.jano...@gmail.com

Thanks, it works fine now.
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[R] representing NULL values in a vector

2010-09-06 Thread raje...@cse.iitm.ac.in

Hi,

I have a vector who contents should look like this,

"c" "d" NULL "e" "f" etc
or 
4 5 6 NULL 7 8 9

how can I represent the null value?
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[R] max limit of list size and vector size?

2010-09-06 Thread raje...@cse.iitm.ac.in

Hi,

Is it possible for me to store a list of vectors of 1 million entries?
like,
cc<-list(c(1,2,1million),c(1,2,1million))

also

what is the length of the longest string in R? I keep getting info from a 
socket and keep pasting on a string...when will this start becoming a problem?
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[R] Correct coefficients from treatment contrasts?

2010-09-06 Thread B W
Hello,Iam trying 
to take the information from the summary of my best fit logisticregression 
model for the occurrence of a high elevation plant spp. and createthe 
appropriate equation that will calculate probability of occurrence, given 
thedata. My predictors include both continuous variables (slope and a second 
orderpolynomial of elevation) and a discrete variable for aspect (warm and 
cool). Ihave left unchanged the default contrasts option, so I believe that 
thefollowing coefficients were created using treatment contrasts.  Myquestion 
how can I take this summary output and create the logistic equationthat will 
allow me to calculate probability of occurrence. My interests are touse this to 
spatially display this info in a GIS environment. I have made adraft equation 
(shown below) that uses the coefficients from this summaryoutput, but this 
appears to be incorrect – values always return zeroprobabilities. Presumably I 
need to adjust the values in some way – but I amunclear as to how to 
proceed. Anyguidance would be appreciated!  >summary 
(model.Slope.Elevation.Aspect) Call:glm(formula= Po ~ Slope + poly(Elevation, 
2) + Aspect_2, family = quasibinomial) DevianceResiduals:     Min  1Q   
Median   3Q Max  -1.0532  -0.4167 -0.2760  -0.1823   
3.3376  Coefficients:  Estimate Std. Error t 
valuePr(>|t|)    (Intercept)  -4.577707   0.222406 -20.583  < 2e-16 
***Slope 0.039959   0.003593 11.121  < 2e-16 
***poly(Elevation,2)1   8.050898   5.601956  1.437   0.1508    
poly(Elevation,2)2 -37.694521   6.297806  -5.985 2.39e-09 
***Aspect_2w 0.429229   0.174760  2.456   0.0141 *  ---   (1/ (1 +  
exp(-1 * (-4.577707 + 0.039959*Slope + 8.050898 * poly(Elevation, 2)1 + 
-37.694521 * poly(Elevation, 2)2 + 0.429229* Aspect_2w) 

Brendan Wilson
2530 Alexis Road
Shoreacres BC 
Canada  V1N 4P6
Ph: 1.250.359.5905



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Re: [R] colorRamp of image to span larger range than dataset

2010-09-06 Thread Barry Rowlingson
On Mon, Sep 6, 2010 at 2:21 AM,   wrote:
> colorRampPalette returns a function, not a list of colours.  You might want 
> to try something like:
>
> ascols <-
>    colorRampPalette(c("gray", "yellow", "darkgoldenrod1", "orange", "red"), 
> interpolate="spline")
> x <- 10*(1:nrow(volcano))
> y <- 10*(1:ncol(volcano))
> image(x, y, volcano, col = ascols(300), axes = FALSE)
>
> ### spot the difference...

 That doesn't work - all it does is use a palette of 300 colours to
span the range of the data. At least the plot will actually show
something now rather than failing, but its not what the poster wanted.

 I wrote a package to handle this kind of thing, called
"colourschemes" - not yet on CRAN, get from R-forge:

install.packages("colourschemes", repos="http://R-Forge.R-project.org";)

the package lets you define colour scheme functions that map values to
colours. By using the same function to set colours for different plots
you can ensure a consistent mapping of value to colour, even for data
with wildly varying data ranges.

The examples in the vignette show how to use them with image() which
is a bit tricky since image() can only plot against a palette and not
individual colours.

 Here's how you'd create the palette using colourschemes:

 > rs = 
 > rampInterpolate(c(0,300),c("gray","yellow","darkgoldenrod1","orange","red"))

 'rs' is now a colour scheme function. Try rs(0), rs(195) or rs(300).
You get back the colours that correspond to those values. Now to see
the whole palette try:

 > plot(0:300,col=rs(0:300))

 (assuming you want the colour scheme to start at 0). See the vignette
for more examples.

 Of course if you are dealing with lots of raster data then you
probably want the raster package anyway, which is something else
entirely...

Barry

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Re: [R] representing NULL values in a vector

2010-09-06 Thread Patrick Burns

Perhaps you mean NA rather than NULL.

If NA is not what you want, then I
think you'll need to explain your
application.

On 06/09/2010 06:00, raje...@cse.iitm.ac.in wrote:


Hi,

I have a vector who contents should look like this,

"c" "d" NULL "e" "f" etc
or
4 5 6 NULL 7 8 9

how can I represent the null value?
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--
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http://www.burns-stat.com
(home of 'Some hints for the R beginner'
and 'The R Inferno')

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Re: [R] how do I transform this to a for loop

2010-09-06 Thread Karl Brand

Hi Bill,

I didn't make the original post, but its pretty similar to some thing i 
would have queried the list about. But, as an R dilatante i find more 
curious your question-


"...but why would you want to do so?"

Is this because you'd typically use the given nine lines of explicit 
code to carve up a single dataset into nine symmetrical variants ? Or 
that some contextual information may affect how you would write the 
for() loop?


As i lack the experience to know any better, i perceive your for() loop 
as de rigour in efficient use of R, and the preferance of all 
experienced R user's. But not having any formal education in R or role 
models as such, its only an assumption (compeletely ignoring for the 
moment processing efficiency/speed, rounding error and such).


But which i now question! Explicit, simple crude looking code; or, 
something which demands a little more proficiency with the language?


cheers,

Karl



On 9/6/2010 6:16 AM, bill.venab...@csiro.au wrote:


sseq<- c(1, seq(5, 40, by = 5))
for(i in 1:length(sseq))
assign(paste("arima", i, sep=""), arima(data.ts[sseq[i]:(sseq[i]+200)], 
order=c(1,1,1)))

...but why would you want to do so?


-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On 
Behalf Of lord12
Sent: Monday, 6 September 2010 10:57 AM
To: r-help@r-project.org
Subject: [R] how do I transform this to a for loop


arima1<- arima(data.ts[1:201], order = c(1,1,1))
arima2<- arima(data.ts[5:205], order = c(1,1,1))
arima3<- arima(data.ts[10:210], order = c(1,1,1))
arima4<- arima(data.ts[15:215], order = c(1,1,1))
arima5<- arima(data.ts[20:220], order = c(1,1,1))
arima6<- arima(data.ts[25:225], order = c(1,1,1))
arima7<- arima(data.ts[30:230], order = c(1,1,1))
arima8<- arima(data.ts[35:235], order = c(1,1,1))
arima9<- arima(data.ts[40:240], order = c(1,1,1))



--
Karl Brand 
Department of Genetics
Erasmus MC
Dr Molewaterplein 50
3015 GE Rotterdam
P +31 (0)10 704 3409 | F +31 (0)10 704 4743 | M +31 (0)642 777 268

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Re: [R] representing NULL values in a vector

2010-09-06 Thread Jim Lemon

On 09/06/2010 03:00 PM, raje...@cse.iitm.ac.in wrote:


Hi,

I have a vector who contents should look like this,

"c" "d" NULL "e" "f" etc
or
4 5 6 NULL 7 8 9

how can I represent the null value?


Hi rajesh,
For character vectors, "" will probably suffice, but for numbers, you 
are probably stuck with NA.


Jim

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[R] replacing functions

2010-09-06 Thread Karen Sargsyan

 Dear All,

Is it possible to replace function with my own? I want to apply pca 
clustering, but
to use some strange correlation function. I'm asking about replacing, 
say, mean() with new content of mean() and use standard other functions, 
which might use mean() as part.


karsar

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Re: [R] how do I transform this to a for loop

2010-09-06 Thread Paul Hiemstra

Hi Karl,

The "why do it like this" is probably direct towards creating 9 new 
objects for the arima results (Is this right Bill?). A better option 
would be to create a list with nine entries. This is much easier for any 
subsequent analyses. An example that uses lapply (an efficient syntax 
for loops):


sseq <- c(1, seq(5, 40, by = 5))
result_list = lapply(sseq, function(num) {
arima(data.ts[num:(num+200)], order=c(1,1,1))
})

cheers,
Paul

On 09/06/2010 10:46 AM, Karl Brand wrote:

Hi Bill,

I didn't make the original post, but its pretty similar to some thing 
i would have queried the list about. But, as an R dilatante i find 
more curious your question-


"...but why would you want to do so?"

Is this because you'd typically use the given nine lines of explicit 
code to carve up a single dataset into nine symmetrical variants ? Or 
that some contextual information may affect how you would write the 
for() loop?


As i lack the experience to know any better, i perceive your for() 
loop as de rigour in efficient use of R, and the preferance of all 
experienced R user's. But not having any formal education in R or role 
models as such, its only an assumption (compeletely ignoring for the 
moment processing efficiency/speed, rounding error and such).


But which i now question! Explicit, simple crude looking code; or, 
something which demands a little more proficiency with the language?


cheers,

Karl



On 9/6/2010 6:16 AM, bill.venab...@csiro.au wrote:


sseq<- c(1, seq(5, 40, by = 5))
for(i in 1:length(sseq))
assign(paste("arima", i, sep=""), 
arima(data.ts[sseq[i]:(sseq[i]+200)], order=c(1,1,1)))


...but why would you want to do so?


-Original Message-
From: r-help-boun...@r-project.org 
[mailto:r-help-boun...@r-project.org] On Behalf Of lord12

Sent: Monday, 6 September 2010 10:57 AM
To: r-help@r-project.org
Subject: [R] how do I transform this to a for loop


arima1<- arima(data.ts[1:201], order = c(1,1,1))
arima2<- arima(data.ts[5:205], order = c(1,1,1))
arima3<- arima(data.ts[10:210], order = c(1,1,1))
arima4<- arima(data.ts[15:215], order = c(1,1,1))
arima5<- arima(data.ts[20:220], order = c(1,1,1))
arima6<- arima(data.ts[25:225], order = c(1,1,1))
arima7<- arima(data.ts[30:230], order = c(1,1,1))
arima8<- arima(data.ts[35:235], order = c(1,1,1))
arima9<- arima(data.ts[40:240], order = c(1,1,1))






--
Drs. Paul Hiemstra
Department of Physical Geography
Faculty of Geosciences
University of Utrecht
Heidelberglaan 2
P.O. Box 80.115
3508 TC Utrecht
Phone:  +3130 253 5773
http://intamap.geo.uu.nl/~paul
http://nl.linkedin.com/pub/paul-hiemstra/20/30b/770

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Re: [R] replacing functions

2010-09-06 Thread Paul Hiemstra

Hi Karsar,

To replace mean you can make a new function with the same name:

l = runif(10)
mean(l)
mean = function(x) return(1)
mean(l)

But there must be a better way...

cheers,
Paul

On 09/06/2010 11:52 AM, Karen Sargsyan wrote:

 Dear All,

Is it possible to replace function with my own? I want to apply pca 
clustering, but
to use some strange correlation function. I'm asking about replacing, 
say, mean() with new content of mean() and use standard other 
functions, which might use mean() as part.


karsar

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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.



--
Drs. Paul Hiemstra
Department of Physical Geography
Faculty of Geosciences
University of Utrecht
Heidelberglaan 2
P.O. Box 80.115
3508 TC Utrecht
Phone:  +3130 253 5773
http://intamap.geo.uu.nl/~paul
http://nl.linkedin.com/pub/paul-hiemstra/20/30b/770

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Re: [R] how do I transform this to a for loop

2010-09-06 Thread Ivan Calandra
  Hi Karl,

I think the question here is why would you want to create different 
objects in the loop using assign().
Usually, using lists is better (more efficient?), although I sometimes 
use assign() too in this context. I do it when I want to export the 
object as separate files (xls, Rbin, svg, etc). If I don't need separate 
files, I use lists.
I'm no expert so I'm not even sure I use the "correct" approach, but it 
might help you get a better understanding.

It would be something like:
sseq<- c(1, seq(5, 40, by = 5))
arima <- vector(mode="list", length=length(sseq)) ##not sure it is 
necessary, but it might be
for(i in 1:length(sseq)) {
arima[[i]] <- arima(data.ts[sseq[i]:(sseq[i]+200)], order=c(1,1,1))
}

HTH,
Ivan


Le 9/6/2010 10:46, Karl Brand a écrit :
> Hi Bill,
>
> I didn't make the original post, but its pretty similar to some thing 
> i would have queried the list about. But, as an R dilatante i find 
> more curious your question-
>
> "...but why would you want to do so?"
>
> Is this because you'd typically use the given nine lines of explicit 
> code to carve up a single dataset into nine symmetrical variants ? Or 
> that some contextual information may affect how you would write the 
> for() loop?
>
> As i lack the experience to know any better, i perceive your for() 
> loop as de rigour in efficient use of R, and the preferance of all 
> experienced R user's. But not having any formal education in R or role 
> models as such, its only an assumption (compeletely ignoring for the 
> moment processing efficiency/speed, rounding error and such).
>
> But which i now question! Explicit, simple crude looking code; or, 
> something which demands a little more proficiency with the language?
>
> cheers,
>
> Karl
>
>
>
> On 9/6/2010 6:16 AM, bill.venab...@csiro.au wrote:
>>
>> sseq<- c(1, seq(5, 40, by = 5))
>> for(i in 1:length(sseq))
>> assign(paste("arima", i, sep=""), 
>> arima(data.ts[sseq[i]:(sseq[i]+200)], order=c(1,1,1)))
>>
>> ...but why would you want to do so?
>>
>>
>> -Original Message-
>> From: r-help-boun...@r-project.org 
>> [mailto:r-help-boun...@r-project.org] On Behalf Of lord12
>> Sent: Monday, 6 September 2010 10:57 AM
>> To: r-help@r-project.org
>> Subject: [R] how do I transform this to a for loop
>>
>>
>> arima1<- arima(data.ts[1:201], order = c(1,1,1))
>> arima2<- arima(data.ts[5:205], order = c(1,1,1))
>> arima3<- arima(data.ts[10:210], order = c(1,1,1))
>> arima4<- arima(data.ts[15:215], order = c(1,1,1))
>> arima5<- arima(data.ts[20:220], order = c(1,1,1))
>> arima6<- arima(data.ts[25:225], order = c(1,1,1))
>> arima7<- arima(data.ts[30:230], order = c(1,1,1))
>> arima8<- arima(data.ts[35:235], order = c(1,1,1))
>> arima9<- arima(data.ts[40:240], order = c(1,1,1))
>>
>

-- 
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

**
http://www.for771.uni-bonn.de
http://webapp5.rrz.uni-hamburg.de/mammals/eng/mitarbeiter.php


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[R] Finding the two most recent dates

2010-09-06 Thread Newbie19_02

Dear R help,

I have the following data frame:

structure(list(prochi = c("ind_1", "ind_1", "ind_1", 
"ind_1", "ind_1", "ind_1", "ind_1", "ind_1", 
"ind_1", "ind_1"), date_1st_event = structure(c(14784, 
14784, 14784, 14784, 14784, 14784, 14784, 14784, 14784, 14784
), class = "Date"), bp_date = structure(c(12660, 14571, 13392, 
13080, 12012, 13080, 13894, 14622, 12654, 13894), class = "Date"), 
SBP = c(135L, 160L, 135L, 153L, 150L, 153L, 151L, 126L, 150L, 
151L), DBP = c(85L, 80L, NA, 79L, 82L, 79L, 76L, 60L, 82L, 
91L)), .Names = c("prochi", "date_1st_event", "bp_date", "SBP", 
"DBP"), row.names = 108:117, class = "data.frame")

It consists of repeated measures for the same individual.  What I want to do
is find the two most recent blood pressure readings (SBP and DBP) using
date_1st_event and bp_date.  What I would do to find the most recent date is
to subtract date_1st_event-bp_date and then aggregate by min.  I'm not sure
how to find the two most recent dates.  

Are there some functions that can help me or will I have to write a function
from scratch.  Any help just to point me in the right direction.

Thanks,
Natalie
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Re: [R] R time series analysis

2010-09-06 Thread matteodefelice


lord12 wrote:
> 
> I have a data file with a given time series of price data and I would like
> to split the time series into a test set and training set. I would then
> like to build an ARIMA model on the training set and apply this model on
> test set. 
> 
I had recently the same problem and, after checking documentation and
mailing list archives, I discovered that it is not possible to apply the
same model on a different data set. Of course you can create the model on a
part of the dataset and then check the prediction with the remaining part,
as a testing set. But, if you have new data you and you want to apply the
same model on them...nothing! I checked the source code of ARIMA functions
but it was too complex and I hadn't enough time to learn all that stuff.
However I found a little workaround:

1. I calibrate the model on the "training part" 
2. I create a new model with the same parameters, using "fixed" (check arima
documentation) on the new data
3. go to step 2. every time you have new data

It worked for me. 
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Re: [R] representing NULL values in a vector

2010-09-06 Thread Duncan Murdoch

On 06/09/2010 1:00 AM, raje...@cse.iitm.ac.in wrote:

Hi,

I have a vector who contents should look like this,

"c" "d" NULL "e" "f" etc
or 
4 5 6 NULL 7 8 9


how can I represent the null value?


As others have said, you probably want NA rather than NULL.  If you 
really want NULL, then use a list (a generic vector).  So


x <- list("c", "d", NULL, "e", "f")
y <- list(4,5,6,NULL, 7,8,9)

You need to be careful when setting values, because

y[[1]] <- NULL

will *remove* element 1, not set it to NULL.  To set it to NULL, use

y[1] <- list(NULL)

Duncan Murdoch

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[R] size limit of string/parse a string and convert to vector

2010-09-06 Thread raje...@cse.iitm.ac.in

Hi,
I have a loop as follows,

dataStr <- character(0)
 repeat{
  fstr<-read.socket(sockfd)
  if(fstr=="")
  break
  dataStr<-paste(dataStr,fstr)
 }

at what point does dataStr stop accepting(gets full)? I'm sending millions of 
records over the socket and need to know if all of it can go into dataStr.

Also, Incase all of it cannot go into dataStr, I need to parse each 
read.socket. In such a case, 
I have a string as follows,
"|1,ab,2.34|2,cd,3.44|" how can I parse this to become a list of 2 string 
vectors, namely,
list(c("1","ab","2.34"),c("2","cd","3.44"))

Any help is appreciated
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[R] Strange behavior of interval values in optimize()

2010-09-06 Thread Michael Bernsteiner

Hi all,

I'm using optimize() to find the minimum of the following function f, and 
minimize it (without 

f<-function(delta,P,U){
minimiz<-P+delta*U
x<-minimiz[1]
y<-minimiz[2]
z<-100*(y-x^2)^2+(1-x)^2
return(z)
}

result<-optimize(f, interval=c(-1, 1), P=c(0.99,1.01), U=c(1,0))
result


Nothing unexpected in the output so far:

> result<-optimize(f, interval=c(-1, 1), P=c(0.99,1.01), U=c(1,0))
> result
$minimum
[1] 0.01498144

$objective
[1] 2.482991e-05


But, when I choose a larger Interval in the optimization method:

result<-optimize(f, interval=c(-100, 10), P=c(0.99,1.01), U=c(1,0))
result

> result<-optimize(f, interval=c(-100, 10), P=c(0.99,1.01), U=c(1,0))
> result
$minimum
[1] -1.989997

$objective
[1] 4.01


The result gets worse (even though the old interval is included in the new one).

In fact I don't want any restrictons for the interval values (something like 
interval=(-Inf, Inf) or at least the interval should be as large as possible.

Does anyone know this strange behavior? Where does it come from?

Thank you,
Fabian
  
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Re: [R] representing NULL values in a vector

2010-09-06 Thread raje...@cse.iitm.ac.in
NA is good.thanks

- Original Message -
From: Patrick Burns 
To: r-help@r-project.org, raje...@cse.iitm.ac.in
Sent: Mon, 06 Sep 2010 13:55:34 +0530 (IST)
Subject: Re: [R] representing NULL values in a vector

Perhaps you mean NA rather than NULL.

If NA is not what you want, then I
think you'll need to explain your
application.

On 06/09/2010 06:00, raje...@cse.iitm.ac.in wrote:
>
> Hi,
>
> I have a vector who contents should look like this,
>
> "c" "d" NULL "e" "f" etc
> or
> 4 5 6 NULL 7 8 9
>
> how can I represent the null value?
>   [[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.
>

-- 
Patrick Burns
pbu...@pburns.seanet.com
http://www.burns-stat.com
(home of 'Some hints for the R beginner'
and 'The R Inferno')

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Re: [R] Installing rJava fails on Gentoo (amd64) with Sun JDK - checking JNI data types... error

2010-09-06 Thread Helgi Tomasson
Try the following:

as super-user

java-config -L 

java-config -S (put a number pointing to sun-jdk-1.6)

R CMD javareconf

export JAVA_HOME=/usr/lib/jvm/sun-jdk-1.6/jre

export R_JAVA_LD_LIBRARY_PATH=${JAVA_HOME}/lib/amd64/server:
${JAVA_HOME}/lib/amd64:
${JAVA_HOME}/../lib/amd64::/usr/java/packages/lib/amd64:/usr/lib64:/lib64:/lib:/usr/lib

start R

install.packages("rJava")
install.packages("JGR")

then as a normal user 

C_NUMERIC=C
export LC_NUMERIC

/usr/lib/jvm/sun-jdk-1.6/jre/bin/java -Xmx512m -Djgr.load.pkgs=yes
-Djgr.loader.ver="${JGR_LOADER_VER}" -Djava.library.path=".:
${LD_LIBRARY_PATH}" -classpath "/usr/lib64/R/library/rJava/java/boot"
-Drjava.path="/usr/lib64/R/library/rJava"
-Drjava.class.path="/usr/lib64/R/library/rJava/jri/JRI.jar:/usr/lib64/R/library/iplots/java/iplots.jar:/usr/lib64/R/library/JGR/java/JGR.jar"
 -Dmain.class=org.rosuda.JGR.JGR RJavaClassLoader $*


library(JGR)
JGR()

This got JGR started on my computer. I still have problems with
dyn.load.

Best regards


Helgi Tomasson





On Fri, 2010-08-13 at 23:31 +0100, Anthony Staines wrote:
> Installing rJava fails consistently, whether installed via
> the command line as below, or through "install.packages(
> 'rJava' )", and whether 0.84 or 0.85 is used, with the
> message :-
> "checking JNI data types... configure: error: One or more
> JNI types differ from the corresponding native type. You may
> need to use non-standard compiler flags or a different
> compiler in order to fix this."
> 
> I've tried to fix it using suggestions from the r-help list,
> and the Gentoo java documentation, but had no joy so far.
> Other Java stuff works fine on this box.
> 
> # R CMD INSTALL rJava_0.8-5.tar.gz
> * installing to library ‘/usr/lib64/R/library’
> * installing *source* package ‘rJava’ ...
> checking for gcc... x86_64-pc-linux-gnu-gcc -std=gnu99
> checking for C compiler default output file name... a.out
> checking whether the C compiler works... yes
> checking whether we are cross compiling... no
> checking for suffix of executables...
> checking for suffix of object files... o
> checking whether we are using the GNU C compiler... yes
> checking whether x86_64-pc-linux-gnu-gcc -std=gnu99 accepts
> -g... yes
> checking for x86_64-pc-linux-gnu-gcc -std=gnu99 option to
> accept ISO C89... none needed
> checking how to run the C preprocessor...
> x86_64-pc-linux-gnu-gcc -std=gnu99 -E
> checking for grep that handles long lines and -e... /bin/grep
> checking for egrep... /bin/grep -E
> checking for ANSI C header files... yes
> checking for sys/wait.h that is POSIX.1 compatible... yes
> checking for sys/types.h... yes
> checking for sys/stat.h... yes
> checking for stdlib.h... yes
> checking for string.h... yes
> checking for memory.h... yes
> checking for strings.h... yes
> checking for inttypes.h... yes
> checking for stdint.h... yes
> checking for unistd.h... yes
> checking for string.h... (cached) yes
> checking sys/time.h usability... yes
> checking sys/time.h presence... yes
> checking for sys/time.h... yes
> checking for unistd.h... (cached) yes
> checking for an ANSI C-conforming const... yes
> checking whether time.h and sys/time.h may both be
> included... yes
> configure: checking whether x86_64-pc-linux-gnu-gcc
> -std=gnu99 supports static inline...
> yes
> checking whether setjmp.h is POSIX.1 compatible... yes
> checking whether sigsetjmp is declared... yes
> checking whether siglongjmp is declared... yes
> checking Java support in R... present:
> interpreter : '/usr/bin/java'
> archiver: '/usr/bin/jar'
> compiler:
> '/home/astaines/.gentoo/java-config-2/current-user-vm/bin/javac'
> header prep.: '/usr/bin/javah'
> cpp flags   : '-I/opt/sun-jdk-1.6.0.20/jre/../include
> -I/opt/sun-jdk-1.6.0.20/jre/../include/linux'
> java libs   : '-L/opt/sun-jdk-1.6.0.20/jre/lib/amd64/server
> -L/opt/sun-jdk-1.6.0.20/jre/lib/amd64
> -L/opt/sun-jdk-1.6.0.20/jre/../lib/amd64 -L
> -L/usr/java/packages/lib/amd64 -L/usr/lib64 -L/lib64 -L/lib
> -L/usr/lib -ljvm'
> checking whether JNI programs can be compiled... yes
> checking JNI data types... configure: error: One or more JNI
> types differ from the corresponding native type. You may
> need to use non-standard compiler flags or a different
> compiler in order to fix this.
> ERROR: configuration failed for package ‘rJava’
> * removing ‘/usr/lib64/R/library/rJava’
> 
> 
> # R --version
> R version 2.11.1 (2010-05-31)
> 
> # java -showversion
> java version "1.6.0_20"
> Java(TM) SE Runtime Environment (build 1.6.0_20-b02)
> Java HotSpot(TM) 64-Bit Server VM (build 16.3-b01, mixed  mode)
> 
> The compiler mentioned in the R output :-
>   /home/astaines/.gentoo/java-config-2/current-user-vm
> is a link pointing to
>   /usr/lib/jvm/sun-jdk-1.6
> which in turn is a link pointing to
>   /opt/sun-jdk-1.6.0.20
> however, any of the three will work.
> 
> # R CMD javareconf
> *** JAVA_HOME is not a valid path, ignoring
> Java interpreter : /usr/bin/java
> Java version : 1.6.0_20
> Jav

Re: [R] how to cluster vectors of factors

2010-09-06 Thread Rafael Björk
If I understand you correctly and each factor consists of binary data, you
may want to check out monothethic analysis, available in the package
'cluster'.

For a simple example and short description of the method to get you started,
just type in:

require(cluster)
?mona

As far as i know there's nothing strictly theoretically invalid in using
hierarchical clustering on binary data, and you may want to see how results
may differ by trying the example data (and your own data) with other
clustering methods. For example, compare:

require(cluster)
data(animals)

plot(mona(animals))
plot(agnes(animals))


2010/9/2 tueken 

>
> Hello all
>
> I wonder what can i use to cluster vectors which composed of several
> factors.
> lets say around 30 different factors compose a vector, and if the factor is
> present then it encoded as 1, if not presented then it will be encoded as
> 0.
> I was thinking of using hierarchical clustering, as i know the distance
> between two vector were calculated through euclidean distance function, but
> i dont think this distance would be correct to separate the data, cause two
> vector with different composition, could end up having similar distance to
> another vector.
> hope someone could give me some clue!
> --
> View this message in context:
> http://r.789695.n4.nabble.com/how-to-cluster-vectors-of-factors-tp2514654p2514654.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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and provide commented, minimal, self-contained, reproducible code.


[R] Creating named.list from two matrix columns

2010-09-06 Thread Viki S

Hi Friends,
I am new to R.

On R utility class pages, creating "named.list" is described with this command :
new("named.list",a=1,b=2)


For large matrix having two columns, such as :

"row1"   2334
"row2"   347
"row3"   379
...

I want to create a named.list like :
$row1
[1] 2334

$row2
[1] 347

...

Can anyone explain how "named.list" variable can be created by using two 
specified columns of a dataframe or matrix object, where one of the two columns 
is assigned as a name (string) and
other as its corresponding value ?

Thanks
  
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and provide commented, minimal, self-contained, reproducible code.


Re: [R] Finding the two most recent dates

2010-09-06 Thread Paul Hiemstra

Hi Natalie,

By far the easiest thing to do is to convert the date to a special date 
class. See as.POSIXct for example. I'm not sure that 14784 means, nor 
what the data says in the bp_date column. Probably the two combine into 
a specific date?


Once you've converted the columns into a POSIXct object, you can use the 
min() function to find the minimum.


cheers,
Paul

On 09/06/2010 12:45 PM, Newbie19_02 wrote:

Dear R help,

I have the following data frame:

structure(list(prochi = c("ind_1", "ind_1", "ind_1",
"ind_1", "ind_1", "ind_1", "ind_1", "ind_1",
"ind_1", "ind_1"), date_1st_event = structure(c(14784,
14784, 14784, 14784, 14784, 14784, 14784, 14784, 14784, 14784
), class = "Date"), bp_date = structure(c(12660, 14571, 13392,
13080, 12012, 13080, 13894, 14622, 12654, 13894), class = "Date"),
 SBP = c(135L, 160L, 135L, 153L, 150L, 153L, 151L, 126L, 150L,
 151L), DBP = c(85L, 80L, NA, 79L, 82L, 79L, 76L, 60L, 82L,
 91L)), .Names = c("prochi", "date_1st_event", "bp_date", "SBP",
"DBP"), row.names = 108:117, class = "data.frame")

It consists of repeated measures for the same individual.  What I want to do
is find the two most recent blood pressure readings (SBP and DBP) using
date_1st_event and bp_date.  What I would do to find the most recent date is
to subtract date_1st_event-bp_date and then aggregate by min.  I'm not sure
how to find the two most recent dates.

Are there some functions that can help me or will I have to write a function
from scratch.  Any help just to point me in the right direction.

Thanks,
Natalie
   



--
Drs. Paul Hiemstra
Department of Physical Geography
Faculty of Geosciences
University of Utrecht
Heidelberglaan 2
P.O. Box 80.115
3508 TC Utrecht
Phone:  +3130 253 5773
http://intamap.geo.uu.nl/~paul
http://nl.linkedin.com/pub/paul-hiemstra/20/30b/770

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[R] Two images functions

2010-09-06 Thread Alaios
Hello everyone.
I would like to ask you what happens when two functions with the same name 
exist. I discovered this today when I wrote
?images (I was trying to understand how it works)

?images gave me the following output: 

Help on topic 'image' was found in the following packages:
Image
 (in package raster in library 
/home/apa/R/x86_64-unknown-linux-gnu-library/2.11)
Display a Color Image
 (in package graphics in library /usr/lib64/R/library)


How can I be sure which function is called when I write Image(x,y,f) or Image(f)

I would like to thank you in advance for your help.

Best Regards
Alex



  
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Re: [R] Two images functions

2010-09-06 Thread Duncan Murdoch

On 06/09/2010 8:14 AM, Alaios wrote:

Hello everyone.
I would like to ask you what happens when two functions with the same name 
exist. I discovered this today when I wrote

?images (I was trying to understand how it works)

?images gave me the following output: 


Help on topic 'image' was found in the following packages:
Image
 (in package raster in library 
/home/apa/R/x86_64-unknown-linux-gnu-library/2.11)

Display a Color Image
 (in package graphics in library /usr/lib64/R/library)


How can I be sure which function is called when I write Image(x,y,f) or Image(f)


R is case sensitive, so you need to write image, not Image.  The one 
found will be the one that is earliest on the search list (which you can 
see by running search()).


If you want to be sure to get a particular one regardless of the search 
list, prefix the name with the package name, e.g. raster::image will 
find the one in the raster package.


Duncan Murdoch



I would like to thank you in advance for your help.

Best Regards
Alex



  
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Re: [R] Finding the two most recent dates

2010-09-06 Thread Dieter Menne

Nathalie,

your method of sending sample data is fine. 

dt = structure(list(prochi = c("ind_1", "ind_1", "ind_1",
"ind_1", "ind_1", "ind_1", "ind_1", "ind_1",
"ind_1", "ind_1"), date_1st_event = structure(c(14784,
14784, 14784, 14784, 14784, 14784, 14784, 14784, 14784, 14784
), class = "Date"), bp_date = structure(c(12660, 14571, 13392,
13080, 12012, 13080, 13894, 14622, 12654, 13894), class = "Date"),
SBP = c(135L, 160L, 135L, 153L, 150L, 153L, 151L, 126L, 150L,
151L), DBP = c(85L, 80L, NA, 79L, 82L, 79L, 76L, 60L, 82L,
91L)), .Names = c("prochi", "date_1st_event", "bp_date", "SBP",
"DBP"), row.names = 108:117, class = "data.frame")

# The most recent date
iRecent =  which.max(dt$bp_date) # index of record with most recent date
dt[iRecent,]
# ind_1 2010-06-24 2010-01-13 126  60

# To Paul Hiemstra: we have the integer representation of date here
as.integer(dt$bpdate[iRecent])

I believe that the description of your problem is not complete (homework?),
because there is only on subject and one 1st event. So please try to restate
the rest of the problem.

Dieter





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Re: [R] Finding the two most recent dates

2010-09-06 Thread jim holtman
Here is one way of doing it:

> x
prochi date_1st_eventbp_date SBP DBP
108  ind_1 2010-06-24 2004-08-30 135  85
109  ind_1 2010-06-24 2009-11-23 160  80
110  ind_1 2010-06-24 2006-09-01 135  NA
111  ind_1 2010-06-24 2005-10-24 153  79
112  ind_1 2010-06-24 2002-11-21 150  82
113  ind_1 2010-06-24 2005-10-24 153  79
114  ind_1 2010-06-24 2008-01-16 151  76
115  ind_1 2010-06-24 2010-01-13 126  60
116  ind_1 2010-06-24 2004-08-24 150  82
117  ind_1 2010-06-24 2008-01-16 151  91
> # find the two most recent date for an individual
> mostRecent <- lapply(split(x, x$prochi), function(.ind){
+ # get two most recent (or 1 if only one exists)
+ index <- order(.ind$bp_date, decreasing=TRUE)[1:2]
+ .ind[index,]  # return
+ })
> # put back into a dataframe
> do.call(rbind, mostRecent)
  prochi date_1st_eventbp_date SBP DBP
ind_1.115  ind_1 2010-06-24 2010-01-13 126  60
ind_1.109  ind_1 2010-06-24 2009-11-23 160  80
>


On Mon, Sep 6, 2010 at 6:45 AM, Newbie19_02  wrote:
>
> Dear R help,
>
> I have the following data frame:
>
> structure(list(prochi = c("ind_1", "ind_1", "ind_1",
> "ind_1", "ind_1", "ind_1", "ind_1", "ind_1",
> "ind_1", "ind_1"), date_1st_event = structure(c(14784,
> 14784, 14784, 14784, 14784, 14784, 14784, 14784, 14784, 14784
> ), class = "Date"), bp_date = structure(c(12660, 14571, 13392,
> 13080, 12012, 13080, 13894, 14622, 12654, 13894), class = "Date"),
>    SBP = c(135L, 160L, 135L, 153L, 150L, 153L, 151L, 126L, 150L,
>    151L), DBP = c(85L, 80L, NA, 79L, 82L, 79L, 76L, 60L, 82L,
>    91L)), .Names = c("prochi", "date_1st_event", "bp_date", "SBP",
> "DBP"), row.names = 108:117, class = "data.frame")
>
> It consists of repeated measures for the same individual.  What I want to do
> is find the two most recent blood pressure readings (SBP and DBP) using
> date_1st_event and bp_date.  What I would do to find the most recent date is
> to subtract date_1st_event-bp_date and then aggregate by min.  I'm not sure
> how to find the two most recent dates.
>
> Are there some functions that can help me or will I have to write a function
> from scratch.  Any help just to point me in the right direction.
>
> Thanks,
> Natalie
> --
> View this message in context: 
> http://r.789695.n4.nabble.com/Finding-the-two-most-recent-dates-tp2528185p2528185.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.
>



-- 
Jim Holtman
Cincinnati, OH
+1 513 646 9390

What is the problem that you are trying to solve?

__
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Re: [R] anova of glm output

2010-09-06 Thread Dieter Menne


francogrex wrote:
> 
> out <- glm(response~Var1+Var2+Var3..,family=binomial,data=mydata)
> summary(out)
> stepAIC(out)
> anova(out, test='Chisq')
> I understand that stepAIC is used to select the model with the lowest AIC
> (the best model) but can someone explain what is the purpose of doing the
> anova: anova(out, test='Chisq')? What extra information does it bring?
> Thanks
> 

"out" is of class glm (and lm, if that matters). So we have anova working on
a glm class, which is documented in anova.glm.

You should be aware that this anova depends on the order of you Varx, so be
cautious.

Dieter


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Re: [R] Creating named.list from two matrix columns

2010-09-06 Thread jim holtman
Is this what you want:

> x
V1   V2
1 row1 2334
2 row2  347
3 row3  379
> x.list <- as.list(x$V2)
> names(x.list) <- x$V1
> x.list
$row1
[1] 2334

$row2
[1] 347

$row3
[1] 379



On Mon, Sep 6, 2010 at 7:55 AM, Viki S  wrote:
>
> Hi Friends,
> I am new to R.
>
> On R utility class pages, creating "named.list" is described with this 
> command :
> new("named.list",a=1,b=2)
>
>
> For large matrix having two columns, such as :
>
> "row1"   2334
> "row2"   347
> "row3"   379
> ...
>
> I want to create a named.list like :
> $row1
> [1] 2334
>
> $row2
> [1] 347
>
> ...
>
> Can anyone explain how "named.list" variable can be created by using two 
> specified columns of a dataframe or matrix object, where one of the two 
> columns is assigned as a name (string) and
> other as its corresponding value ?
>
> Thanks
>
>        [[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.
>



-- 
Jim Holtman
Cincinnati, OH
+1 513 646 9390

What is the problem that you are trying to solve?

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[R] dataframe row names from list

2010-09-06 Thread raje...@cse.iitm.ac.in

Hi,
I have a list which looks like this...
> str(y)
List of 10
 $ : chr [1:4] "ABCD" "5" "0" "1"
 $ : chr [1:4] "DEF" "15" "1" "16"
 $ : chr [1:4] "AAA" "2" "17" "8"
 $ : chr [1:4] "SSS" "15" "25" "1"
 $ : chr [1:4] "III" "15" "26" "4"
 $ : chr [1:4] "OPQ" "7" "30" "4"
 $ : chr [1:4] "TYR" "14" "34" "8"
 $ : chr [1:4] "IRTS" "15" "42" "1"
 $ : chr [1:4] "LLL" "15" "43" "2"
 $ : chr [1:4] "AQW" "3" "45" "4"
 
I need to create a dataframe whose row names are chr[1] of each vector..ie 
ABCD,DEF,AAA ETC. how can I do this?
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Re: [R] replacing functions

2010-09-06 Thread Dieter Menne


Karen Sargsyan wrote:
> 
> Is it possible to replace function with my own? I want to apply pca 
> clustering, but to use some strange correlation function. I'm asking about
> replacing,  say, mean() with new content of mean() and use standard other
> functions,  which might use mean() as part.
> 

The usual way would be to get the source code for the function you are
trying to change (or, fast way, with getAnywhere(pca)), copy it to an
editor, and make the changes. However, better always rename it before using,
e.g. pcaKaren <- function

If the code you are trying to change is not directly called, but indirectly,
things are more complicated. In most cases, it's easiest to get the whole
package, put it into a different namespace, and make the changes. Easiest
does not mean not easy, though.

Dieter


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[R] inserting a vector as a row in a data.frame

2010-09-06 Thread raje...@cse.iitm.ac.in

Hi,

is it possible to insert a vector as a row in a data.frame?
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Re: [R] Strange behavior of interval values in optimize()

2010-09-06 Thread Dieter Menne


Michael Bernsteiner wrote:
> 
> 
> I'm using optimize() to find the minimum of the following function f, and
> minimize it (without 
> .
> But, when I choose a larger Interval in the optimization method:
> 
> The result gets worse (even though the old interval is included in the new
> one).
> 
> In fact I don't want any restrictons for the interval values (something
> like interval=(-Inf, Inf) or at least the interval should be as large as
> possible.
> 
> 

The most likely cause is a second minimum.

Dieter

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[R] anova of glm output

2010-09-06 Thread francogrex

Hi, this is more related to understanding some statistics while using R; I've
see such output in a paper:
out <- glm(response~Var1+Var2+Var3..,family=binomial,data=mydata)
summary(out)
stepAIC(out)
anova(out, test='Chisq')
I understand that stepAIC is used to select the model with the lowest AIC
(the best model) but can someone explain what is the purpose of doing the
anova: anova(out, test='Chisq')? What extra information does it bring?
Thanks
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[R] How to get "mypkg-manual.pdf"

2010-09-06 Thread Juliet Ndukum
I am building a package say mypkg. Five months ago, when I built the package I 
got the mypkg manual in pdf format. 

Today, after making updates, I build the same package, same name, and steps; 
unfortunately I do not get the manual in pdf format. 
Rather I get the following message: 
cd: can't cd to /cygdrive/c/Documents saving output to 'mypkg-manual.pdf' 
...Done

Could any one help me on how to get the "mypkg-manual.pdf".
Thank you in advance for your help.
JN



  
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Re: [R] size limit of string/parse a string and convert to vector

2010-09-06 Thread jim holtman
try this:

> x <- "|1,ab,2.34|2,cd,3.44|"
> # split by the "|" and remove vectors of zero characters
> x.sp <- strsplit(x, '|', fixed = TRUE)[[1]]
> x.sp <- x.sp[nchar(x.sp) > 0]
> # now split by comma
> x.comma <- strsplit(x.sp, ',')
> # you can now access you data
> x.comma
[[1]]
[1] "1""ab"   "2.34"

[[2]]
[1] "2""cd"   "3.44"



On Mon, Sep 6, 2010 at 6:06 AM, raje...@cse.iitm.ac.in
 wrote:
>
> Hi,
> I have a loop as follows,
>
> dataStr <- character(0)
>  repeat{
>  fstr<-read.socket(sockfd)
>  if(fstr=="")
>  break
>  dataStr<-paste(dataStr,fstr)
>  }
>
> at what point does dataStr stop accepting(gets full)? I'm sending millions of 
> records over the socket and need to know if all of it can go into dataStr.
>
> Also, Incase all of it cannot go into dataStr, I need to parse each 
> read.socket. In such a case,
> I have a string as follows,
> "|1,ab,2.34|2,cd,3.44|" how can I parse this to become a list of 2 string 
> vectors, namely,
> list(c("1","ab","2.34"),c("2","cd","3.44"))
>
> Any help is appreciated
>        [[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.
>



-- 
Jim Holtman
Cincinnati, OH
+1 513 646 9390

What is the problem that you are trying to solve?

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Re: [R] dataframe row names from list

2010-09-06 Thread Ivan Calandra
  Hi!

I'm sure there's an easier way, but that works for me:

test_list <- list(c("ABC","5","0"), c("DEF","10","1")) ##just a part of 
your example, think about using dput() to create a copy/pastable example
test_df <- t(as.data.frame(test_list)[-1,])
rownames(test_df) <- t(as.data.frame(test_list)[1,])

HTH,
Ivan

Le 9/6/2010 13:41, raje...@cse.iitm.ac.in a écrit :
> Hi,
> I have a list which looks like this...
>> str(y)
> List of 10
>   $ : chr [1:4] "ABCD" "5" "0" "1"
>   $ : chr [1:4] "DEF" "15" "1" "16"
>   $ : chr [1:4] "AAA" "2" "17" "8"
>   $ : chr [1:4] "SSS" "15" "25" "1"
>   $ : chr [1:4] "III" "15" "26" "4"
>   $ : chr [1:4] "OPQ" "7" "30" "4"
>   $ : chr [1:4] "TYR" "14" "34" "8"
>   $ : chr [1:4] "IRTS" "15" "42" "1"
>   $ : chr [1:4] "LLL" "15" "43" "2"
>   $ : chr [1:4] "AQW" "3" "45" "4"
>
> I need to create a dataframe whose row names are chr[1] of each vector..ie 
> ABCD,DEF,AAA ETC. how can I do this?
>   [[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.
>

-- 
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

**
http://www.for771.uni-bonn.de
http://webapp5.rrz.uni-hamburg.de/mammals/eng/mitarbeiter.php


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Re: [R] max limit of list size and vector size?

2010-09-06 Thread jim holtman
It is easy to store a list of that size:

> x <- list(1:1e6, 1:1e6, 1:1e6)
> object.size(x)
12000112 bytes
> str(x)
List of 3
 $ : int [1:100] 1 2 3 4 5 6 7 8 9 10 ...
 $ : int [1:100] 1 2 3 4 5 6 7 8 9 10 ...
 $ : int [1:100] 1 2 3 4 5 6 7 8 9 10 ...

Now it really depends on how you are processing the data.  If you are
appending to a vector each time:

myData <- c(myData, newData)

This is not efficient since it may be making copies of the vector each
time to extend it.  It is best to preallocate the vector and then
store the result:

myData <- numeric(100)  # hold 1 million values
for (i in 1:1e6)  myData[i] <- newData

My rule of thumb is that a single object should take up no more than
25% of the available RAM so that copies can be made during the
processing.

HTH

On Mon, Sep 6, 2010 at 2:54 AM, raje...@cse.iitm.ac.in
 wrote:
>
> Hi,
>
> Is it possible for me to store a list of vectors of 1 million entries?
> like,
> cc<-list(c(1,2,1million),c(1,2,1million))
>
> also
>
> what is the length of the longest string in R? I keep getting info from a 
> socket and keep pasting on a string...when will this start becoming a problem?
>        [[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.
>



-- 
Jim Holtman
Cincinnati, OH
+1 513 646 9390

What is the problem that you are trying to solve?

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Re: [R] inserting a vector as a row in a data.frame

2010-09-06 Thread Ivan Calandra
  Hi again,

see ?rbind

Ivan

Le 9/6/2010 14:11, raje...@cse.iitm.ac.in a écrit :
> Hi,
>
> is it possible to insert a vector as a row in a data.frame?
>   [[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.
>

-- 
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

**
http://www.for771.uni-bonn.de
http://webapp5.rrz.uni-hamburg.de/mammals/eng/mitarbeiter.php


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Re: [R] Correct coefficients from treatment contrasts?

2010-09-06 Thread David Winsemius


On Sep 6, 2010, at 4:03 AM, B W wrote:



->Hello,I am trying to take the information from the summary of my  
best fit logisticregression model for the occurrence of a high  
elevation plant spp. and create the appropriate equation that will  
calculate probability of occurrence, given the data. My predictors  
include both continuous variables (slope and a second  
orderpolynomial of elevation) and a discrete variable for aspect  
(warm and cool). I have left unchanged the default contrasts option,  
so I believe that thefollowing coefficients were created using  
treatment contrasts.  My question how can I take this summary output  
and create the logistic equation that will allow me to calculate  
probability of occurrence. My interests are touse this to spatially  
display this info in a GIS environment.


I think you should:

-- Read the Posting Guide where you should learn that this is a plain  
text mailing list and that you need to change the configuration of  
your mail client.


-- Read the help page and read other documentation regarding the use  
of the predict function.


I have made adraft equation (shown below) that uses the coefficients  
from this summaryoutput, but this appears to be incorrect – values  
always return zeroprobabilities. Presumably I need to adjust the  
values in some way – but I am unclear as to how to proceed.  
Anyguidance would be appreciated!



 >summary (


Call:glm(formula= Po ~ Slope + poly(Elevation, 2) + Aspect_2, family  
= quasibinomial) DevianceResiduals: Min  1Q   Median
3Q Max  -1.0532  -0.4167 -0.2760  -0.1823   3.3376


Coefficients:  Estimate Std. Error t valuePr(>| 
t|)(Intercept)  -4.577707   0.222406 -20.583  < 2e-16 ***



Slope 0.039959   0.003593 11.121  < 2e-16 ***



poly(Elevation,2)1   8.050898   5.601956  1.437   0.1508



poly(Elevation,2)2 -37.694521   6.297806  -5.985 2.39e-09 ***



Aspect_2w 0.429229   0.174760  2.456   0.0141 *  ---


You may get predictions at the original data points with:

pred < predict(model.Slope.Elevation.Aspect)

 (1/ (1 +  exp(-1 * (-4.577707 + 0.039959*Slope + 8.050898 *  
poly(Elevation, 2)1 + -37.694521 * poly(Elevation, 2)2 + 0.429229*  
Aspect_2w)


Brendan Wilson
2530 Alexis Road
Shoreacres BC
Canada  V1N 4P6
Ph: 1.250.359.5905



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David Winsemius, MD
West Hartford, CT

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Re: [R] extracting x,y coordinates from a contour plot

2010-09-06 Thread Charles Annis, P.E.
Thank you, David:

I obviously didn't look hard enough.  This is exactly what I need.

Charles Annis, P.E.

charles.an...@statisticalengineering.com
561-352-9699
http://www.StatisticalEngineering.com

-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of David Winsemius
Sent: Monday, September 06, 2010 12:43 AM
To: charles.an...@statisticalengineering.com
Cc: r-help@r-project.org
Subject: Re: [R] extracting x,y coordinates from a contour plot


On Sep 5, 2010, at 11:48 PM, Charles Annis, P.E. wrote:

> Requisite info: R version 2.11.1 (2010-05-31) running on a 64 bit HP  
> Windows
> 7 machine.
>
Doubt that makes much of a difference here.

> I have used contour() for several years.  Now I would like to  
> extract from a
> contour plot the x, y coordinates of a contour z=constant.  This  
> seems as
> though it would be straight-forward but I've been unsuccessful in my
> searches of CRAN.

Suggest you read the help page for contour and the pages to which it  
links as well as working the examples. The answer is illustrated in  
the examples on that page.

> Can anyone provide a hint?

 > x <- 10*1:nrow(volcano)
 > y <- 10*1:ncol(volcano)
 > xy160 <- contourLines(x, y, volcano, nlevels=1, levels=160)
 > str(xy160)
List of 2
  $ :List of 3
   ..$ level: num 160
   ..$ x: num [1:165] 110 108 105 102 103 ...
   ..$ y: num [1:165] 295 300 310 320 330 ...
  $ :List of 3
   ..$ level: num 160
   ..$ x: num [1:31] 270 263 262 260 260 ...
   ..$ y: num [1:31] 310 320 330 340 350 ...

-- 
David.

David Winsemius, MD
West Hartford, CT

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[R] nlme Output

2010-09-06 Thread Edward Patzelt
Everyone -

What do the NaN's mean here?  Is this analysis a problem?


Linear mixed-effects model fit by maximum likelihood
 Data: tmp.dat
   AIC  BIClogLik
  1611.251 1638.363 -797.6253

Random effects:
 Formula: ~1 | group_id
 (Intercept) Residual
StdDev: 0.0003077668 9.236715

Fixed effects: AvgTrials ~ time + factor(group_id) + time *
factor(group_id)
   Value Std.Error  DF   t-value p-value
(Intercept)18.159722  3.576664 213  5.077279  0.
time4.192708  1.655674 213  2.532327  0.0121
factor(group_id)2  -6.929563  5.235700   0 -1.323522 NaN
factor(group_id)3  -1.654554  4.189575   0 -0.394922 NaN
time:factor(group_id)2  1.729911  2.423658 213  0.713760  0.4762
time:factor(group_id)3 -2.555111  1.939396 213 -1.317478  0.1891
 Correlation:
   (Intr) time   fc(_)2 fc(_)3 t:(_)2
time   -0.926
factor(group_id)2  -0.683  0.632
factor(group_id)3  -0.854  0.790  0.583
time:factor(group_id)2  0.632 -0.683 -0.926 -0.540
time:factor(group_id)3  0.790 -0.854 -0.540 -0.926  0.583

Standardized Within-Group Residuals:
   Min Q1Med Q3Max
-1.8842754 -0.6979785 -0.3370998  0.5666704  3.0943948

Number of Observations: 219
Number of Groups: 3
Warning message:
In pt(q, df, lower.tail, log.p) : NaNs produced

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[R] Help on write.xlsx library(xlsx)

2010-09-06 Thread Ravi S. Shankar
Hi Adrian,

 

dat=data.frame(matrix(0,3,3))

 

write.xlsx(dat,"z:/dat.xlsx",sheetName="sheet1",append=F)

write.xlsx(dat,"z:/dat.xlsx",sheetName="sheet2",append=F)

 

The above code works and creates new worksheets. But if I want to append
to an existing worksheet I seem to get an error.

 

write.xlsx(dat,"z:/dat.xlsx",sheetName="sheet2",append=T) - This gives
an error saying "The workbook already contains a sheet of this name"

 

Could you please let me know how to append data to  an existing sheet
and not create a new sheet each time?

 

Thank you

Ravi

This e-mail may contain confidential and/or privileged i...{{dropped:13}}

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Re: [R] how do I transform this to a for loop

2010-09-06 Thread Karl Brand

Hi Paul, Ivan,

Hartstikke bedankt and thanks alot for sharing these thoughts. I can see 
'listing up' multiple symmetrical data sets makes a lot of sense. As 
does using lapply() on them which i understand to be more 
efficient/faster than for().


Goodo- with your concensus (and helpful examples) i can tell myself 
investing the extra time to use lapply on lists /will/ pay off in the 
long run vs. copying and pasting (nearly) the same line of code 10 times 
for every data manipulation...


thanks again,

Karl



On 9/6/2010 12:09 PM, Paul Hiemstra wrote:

Hi Karl,

The "why do it like this" is probably direct towards creating 9 new
objects for the arima results (Is this right Bill?). A better option
would be to create a list with nine entries. This is much easier for any
subsequent analyses. An example that uses lapply (an efficient syntax
for loops):

sseq <- c(1, seq(5, 40, by = 5))
result_list = lapply(sseq, function(num) {
arima(data.ts[num:(num+200)], order=c(1,1,1))
})

cheers,
Paul

On 09/06/2010 10:46 AM, Karl Brand wrote:

Hi Bill,

I didn't make the original post, but its pretty similar to some thing
i would have queried the list about. But, as an R dilatante i find
more curious your question-

"...but why would you want to do so?"

Is this because you'd typically use the given nine lines of explicit
code to carve up a single dataset into nine symmetrical variants ? Or
that some contextual information may affect how you would write the
for() loop?

As i lack the experience to know any better, i perceive your for()
loop as de rigour in efficient use of R, and the preferance of all
experienced R user's. But not having any formal education in R or role
models as such, its only an assumption (compeletely ignoring for the
moment processing efficiency/speed, rounding error and such).

But which i now question! Explicit, simple crude looking code; or,
something which demands a little more proficiency with the language?

cheers,

Karl



On 9/6/2010 6:16 AM, bill.venab...@csiro.au wrote:


sseq<- c(1, seq(5, 40, by = 5))
for(i in 1:length(sseq))
assign(paste("arima", i, sep=""),
arima(data.ts[sseq[i]:(sseq[i]+200)], order=c(1,1,1)))

...but why would you want to do so?


-Original Message-
From: r-help-boun...@r-project.org
[mailto:r-help-boun...@r-project.org] On Behalf Of lord12
Sent: Monday, 6 September 2010 10:57 AM
To: r-help@r-project.org
Subject: [R] how do I transform this to a for loop


arima1<- arima(data.ts[1:201], order = c(1,1,1))
arima2<- arima(data.ts[5:205], order = c(1,1,1))
arima3<- arima(data.ts[10:210], order = c(1,1,1))
arima4<- arima(data.ts[15:215], order = c(1,1,1))
arima5<- arima(data.ts[20:220], order = c(1,1,1))
arima6<- arima(data.ts[25:225], order = c(1,1,1))
arima7<- arima(data.ts[30:230], order = c(1,1,1))
arima8<- arima(data.ts[35:235], order = c(1,1,1))
arima9<- arima(data.ts[40:240], order = c(1,1,1))








--
Karl Brand 
Department of Genetics
Erasmus MC
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Re: [R] nlme Output

2010-09-06 Thread ONKELINX, Thierry
Dear Edward,

You have no degrees of freedom left to estimate those p-values. Your
design does not allows for the model your implemented. We need a brief
summary of your design in order to help you further.

HTH,

Thierry



ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek
team Biometrie & Kwaliteitszorg
Gaverstraat 4
9500 Geraardsbergen
Belgium

Research Institute for Nature and Forest
team Biometrics & Quality Assurance
Gaverstraat 4
9500 Geraardsbergen
Belgium

tel. + 32 54/436 185
thierry.onkel...@inbo.be
www.inbo.be

To call in the statistician after the experiment is done may be no more
than asking him to perform a post-mortem examination: he may be able to
say what the experiment died of.
~ Sir Ronald Aylmer Fisher

The plural of anecdote is not data.
~ Roger Brinner

The combination of some data and an aching desire for an answer does not
ensure that a reasonable answer can be extracted from a given body of
data.
~ John Tukey
  

> -Oorspronkelijk bericht-
> Van: r-help-boun...@r-project.org 
> [mailto:r-help-boun...@r-project.org] Namens Edward Patzelt
> Verzonden: maandag 6 september 2010 15:43
> Aan: r-help@r-project.org
> Onderwerp: [R] nlme Output
> 
> Everyone -
> 
> What do the NaN's mean here?  Is this analysis a problem?
> 
> 
> Linear mixed-effects model fit by maximum likelihood
>  Data: tmp.dat
>AIC  BIClogLik
>   1611.251 1638.363 -797.6253
> 
> Random effects:
>  Formula: ~1 | group_id
>  (Intercept) Residual
> StdDev: 0.0003077668 9.236715
> 
> Fixed effects: AvgTrials ~ time + factor(group_id) + time *
> factor(group_id)
>Value Std.Error  DF   t-value p-value
> (Intercept)18.159722  3.576664 213  5.077279  0.
> time4.192708  1.655674 213  2.532327  0.0121
> factor(group_id)2  -6.929563  5.235700   0 -1.323522 NaN
> factor(group_id)3  -1.654554  4.189575   0 -0.394922 NaN
> time:factor(group_id)2  1.729911  2.423658 213  0.713760  0.4762
> time:factor(group_id)3 -2.555111  1.939396 213 -1.317478  0.1891
>  Correlation:
>(Intr) time   fc(_)2 fc(_)3 t:(_)2
> time   -0.926
> factor(group_id)2  -0.683  0.632
> factor(group_id)3  -0.854  0.790  0.583
> time:factor(group_id)2  0.632 -0.683 -0.926 -0.540
> time:factor(group_id)3  0.790 -0.854 -0.540 -0.926  0.583
> 
> Standardized Within-Group Residuals:
>Min Q1Med Q3Max
> -1.8842754 -0.6979785 -0.3370998  0.5666704  3.0943948
> 
> Number of Observations: 219
> Number of Groups: 3
> Warning message:
> In pt(q, df, lower.tail, log.p) : NaNs produced
> 
>   [[alternative HTML version deleted]]
> 
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> 

Druk dit bericht a.u.b. niet onnodig af.
Please do not print this message unnecessarily.

Dit bericht en eventuele bijlagen geven enkel de visie van de schrijver weer 
en binden het INBO onder geen enkel beding, zolang dit bericht niet bevestigd is
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[R] Aggregating the matrices

2010-09-06 Thread Sergey Goriatchev
Hello everyone.

Say we have the following:

a <- matrix(c(-75, 3, 5, 9, 2, 3, 5), nrow=1, dim=list("06092010",
c("ES", "PT", "Z ", "CF", "GX", "ST", "EO")))
b <- matrix(c(-5, 2, 4, 12, 5), nrow=1, dim=list("06092010", c("PT",
"CF", "AT", "EM", "ST")))
d <- cbind(a, b)

I want to calculate sums of the columns that have similar column names
and then output this summary
What I want to have is an array that looks like:

ES  PT Z  CF...
-75  -2  5  11...

I tried the following, but it did not work:
aggregate(d, list(colnames(d)), sum)

How can I achieve my objective?

Thank you in advance.

Sergey

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Re: [R] Finding the two most recent dates

2010-09-06 Thread Newbie19_02

Dear all,

Thanks very much for the replies and for the help.  

This whole  data set consists of about 7000 individuals who have had
multiple blood pressure measures taken over time so I just used one
individual as an example. I'm sorry if it looked like homework...it isn't.

Jim your solution worked really well thanks.  I was thinking of things in
terms of diff time and not order, so thanks.

Natalie


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Re: [R] Aggregating the matrices

2010-09-06 Thread Gabor Grothendieck
On Mon, Sep 6, 2010 at 9:56 AM, Sergey Goriatchev  wrote:
> Hello everyone.
>
> Say we have the following:
>
> a <- matrix(c(-75, 3, 5, 9, 2, 3, 5), nrow=1, dim=list("06092010",
> c("ES", "PT", "Z ", "CF", "GX", "ST", "EO")))
> b <- matrix(c(-5, 2, 4, 12, 5), nrow=1, dim=list("06092010", c("PT",
> "CF", "AT", "EM", "ST")))
> d <- cbind(a, b)
>
> I want to calculate sums of the columns that have similar column names
> and then output this summary
> What I want to have is an array that looks like:
>
> ES  PT Z  CF...
> -75  -2  5  11...
>
> I tried the following, but it did not work:
> aggregate(d, list(colnames(d)), sum)
>
> How can I achieve my objective?
>
> Thank you in advance.
>


Try this:

> library(plyr)
> colSums(rbind.fill(as.data.frame(a), as.data.frame(b)), na.rm = TRUE)
 ES  PT  Z   CF  GX  ST  EO  AT  EM
-75  -2   5  11   2   8   5   4  12




-- 
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Re: [R] Aggregating the matrices

2010-09-06 Thread David Winsemius


On Sep 6, 2010, at 9:56 AM, Sergey Goriatchev wrote:


Hello everyone.

Say we have the following:

a <- matrix(c(-75, 3, 5, 9, 2, 3, 5), nrow=1, dim=list("06092010",
c("ES", "PT", "Z ", "CF", "GX", "ST", "EO")))
b <- matrix(c(-5, 2, 4, 12, 5), nrow=1, dim=list("06092010", c("PT",
"CF", "AT", "EM", "ST")))
d <- cbind(a, b)

I want to calculate sums of the columns that have similar column names
and then output this summary
What I want to have is an array that looks like:

ES  PT Z  CF...
-75  -2  5  11...

I tried the following, but it did not work:
aggregate(d, list(colnames(d)), sum)


ES is not in the duplicated column names so perhaps your English  
specification is not what you meant:

> d
  ES PT Z  CF GX ST EO PT CF AT EM ST
06092010 -75  3  5  9  2  3  5 -5  2  4 12  5

> dupled <- colnames(d)[duplicated(colnames(d))]
> sapply(dupled, function(x) sum( d[, x]))
PT CF ST
 3  9  3

If you wanted simple a sum over unique column names then it would have  
been somewhat simpler (no need to construct a duplicated set):


> sapply(unique(colnames(d)), function(x) sum( d[, x]))
 ES  PT  Z   CF  GX  ST  EO  AT  EM
-75   3   5   9   2   3   5   4  12



How can I achieve my objective?

Thank you in advance.

Sergey

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David Winsemius, MD
West Hartford, CT

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[R] Time Series

2010-09-06 Thread trb1

Hi 

How would I analyse time series with 
- different lengths (i.e. one has 9 entries and the other has 14 entries)
- different frequency (i.e. dates are random - no repeated length)
- multiple values for the same time entry (e.g. 2009-10-23 below)

i.e. my data takes the form:
1st time series

2009-10-07  0.009378
2009-10-19  0.014790
2009-10-23  -0.005946
2009-10-23  0.009096
2009-11-08  0.004189
2009-11-10  -0.004592
2009-11-17  0.009397
2009-11-24  0.003411
2009-12-02  0.003300
2010-01-15  0.010873
2010-01-20  0.010712
2010-01-20  0.022237

2nd time series

2009-09-23  0.076253
2009-10-07  0.039255
2010-02-17  0.039045
2010-03-09  0.024201
2010-03-25  -0.039810
2010-04-13  -0.012428

I am unable to get any functions to work.
A simple plot would be nice! 


Thanks.

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Re: [R] Aggregating the matrices

2010-09-06 Thread Sergey Goriatchev
Gabor, David, thank you.

David, your last suggestion is what I need.

Regards,
Sergey

On Mon, Sep 6, 2010 at 16:12, David Winsemius  wrote:
>
> On Sep 6, 2010, at 9:56 AM, Sergey Goriatchev wrote:
>
>> Hello everyone.
>>
>> Say we have the following:
>>
>> a <- matrix(c(-75, 3, 5, 9, 2, 3, 5), nrow=1, dim=list("06092010",
>> c("ES", "PT", "Z ", "CF", "GX", "ST", "EO")))
>> b <- matrix(c(-5, 2, 4, 12, 5), nrow=1, dim=list("06092010", c("PT",
>> "CF", "AT", "EM", "ST")))
>> d <- cbind(a, b)
>>
>> I want to calculate sums of the columns that have similar column names
>> and then output this summary
>> What I want to have is an array that looks like:
>>
>> ES  PT Z  CF...
>> -75  -2  5  11...
>>
>> I tried the following, but it did not work:
>> aggregate(d, list(colnames(d)), sum)
>
> ES is not in the duplicated column names so perhaps your English
> specification is not what you meant:
>> d
>          ES PT Z  CF GX ST EO PT CF AT EM ST
> 06092010 -75  3  5  9  2  3  5 -5  2  4 12  5
>
>> dupled <- colnames(d)[duplicated(colnames(d))]
>> sapply(dupled, function(x) sum( d[, x]))
> PT CF ST
>  3  9  3
>
> If you wanted simple a sum over unique column names then it would have been
> somewhat simpler (no need to construct a duplicated set):
>
>> sapply(unique(colnames(d)), function(x) sum( d[, x]))
>  ES  PT  Z   CF  GX  ST  EO  AT  EM
> -75   3   5   9   2   3   5   4  12
>
>>
>> How can I achieve my objective?
>>
>> Thank you in advance.
>>
>> Sergey
>>
>> __
>> 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
>
>



-- 
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Re: [R] mac: lib/gtk.pkg

2010-09-06 Thread David Winsemius


On Sep 5, 2010, at 10:32 AM, Daniele Sluijters wrote:


Hello,

I'm sorry to just pop-up on the mailing list like this and ask a  
relatively non-R related question but I had no idea whom else to  
contact on this matter.
I'm working on a completely different port of an application to OS X  
which requires GTK and through Google'ing stumbled on a rather  
recent GTK installer for Mac at: http://r.research.att.com/


I was wondering if anyone here knows how GTK and its dependencies  
were packaged into that pkg? It'd be a lifesaver if someone could  
point me in the right direction.


Again, sorry for the non-R related question but this place seemed  
like the only option.


The is an R-SIG-Mac mailing list. It's webpage is at the top of the  
SIG entries on:

http://www.r-project.org/mail.html

I suspect that Simon Urbanek, who maintains the AT&T webpages and very  
probably created that package, sometimes reads rhelp but I'm not sure  
on what schedule. You might see if he makes his email address  
available on those pages.


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[R] Aggregate certain rows in a matrix

2010-09-06 Thread Kennedy

Hi,

I have a matrix that looks like this 

  a <- c(1,1,1,1,2,2,3,3,3,3)
  b <- c(2,2,2,3,4,4,4,5,5,6)
  c <- c(1,2,3,4,5,6,7,8,9,10)
  M <- matrix(nr=10,nc=3)
  M[,1] <- a
  M[,2] <- b
  M[,3] <- c

> M
  [,1] [,2] [,3]
 [1,]121
 [2,]122
 [3,]123
 [4,]134
 [5,]245
 [6,]246
 [7,]347
 [8,]358
 [9,]359
[10,]36   10

I want to reduce the matrix according to the following: If the values of the
two first columns are the same in two or more rows the values in the third
column of the corresponding rows should be added and only one of the rows
should be keept. Hence the matrix M above should look like this

  1 2 6
  1 3 4
  2 4 11
  3 4 7
  3 5 17
  3 6 10


Thank you

Henrik



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[R] calculating area between plot lines

2010-09-06 Thread A. Marcia BARBOSA
Hi everyone. I have these data:

probClass<-seq(0,0.9,0.1)
prob1<-c(0.0070,0.0911,0.1973,0.2949,0.3936,0.5030,0.5985,0.6869,0.7820,0.8822)
prob2<-c(0.0066,0.0791,0.2358,0.3478,0.3714,0.3860,0.6667,0.6400,0.7000,1.)

# which I'm plotting as follows:

plot(probClass,prob1,xlim=c(0,1),ylim=c(0,1),xaxs='i',yaxs='i',type="n")
lines(probClass,prob1)
lines(probClass,prob2)
polygon(c(probClass,rev(probClass)),c(prob2,rev(prob1)),col="red",border=NA)


Given that the total area of the plot is 1, how can I calculate the
area between the plotted lines (red polygon)? I have only found the
areapl function in the splancs package, but it doesn't work for
self-intersecting polygons..

Any help will be gratefully received. Cheers,
Márcia

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Re: [R] Aggregate certain rows in a matrix

2010-09-06 Thread Dimitris Rizopoulos

one way is the following:

M <- cbind(c(1,1,1,1,2,2,3,3,3,3), c(2,2,2,3,4,4,4,5,5,6),
c(1,2,3,4,5,6,7,8,9,10))

ind <- do.call(paste, c(as.data.frame(M[, 1:2], sep = "\r")))
M[, 3] <- ave(M[, 3], ind, FUN = "sum")
unique(M)


I hope it helps.

Best,
Dimitris


On 9/6/2010 4:29 PM, Kennedy wrote:


Hi,

I have a matrix that looks like this

   a<- c(1,1,1,1,2,2,3,3,3,3)
   b<- c(2,2,2,3,4,4,4,5,5,6)
   c<- c(1,2,3,4,5,6,7,8,9,10)
   M<- matrix(nr=10,nc=3)
   M[,1]<- a
   M[,2]<- b
   M[,3]<- c


M

   [,1] [,2] [,3]
  [1,]121
  [2,]122
  [3,]123
  [4,]134
  [5,]245
  [6,]246
  [7,]347
  [8,]358
  [9,]359
[10,]36   10

I want to reduce the matrix according to the following: If the values of the
two first columns are the same in two or more rows the values in the third
column of the corresponding rows should be added and only one of the rows
should be keept. Hence the matrix M above should look like this

   1 2 6
   1 3 4
   2 4 11
   3 4 7
   3 5 17
   3 6 10


Thank you

Henrik





--
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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Re: [R] Aggregate certain rows in a matrix

2010-09-06 Thread Barry Rowlingson
On Mon, Sep 6, 2010 at 3:29 PM, Kennedy  wrote:

> I want to reduce the matrix according to the following: If the values of the
> two first columns are the same in two or more rows the values in the third
> column of the corresponding rows should be added and only one of the rows
> should be keept. Hence the matrix M above should look like this
>
>  1 2 6
>  1 3 4
>  2 4 11
>  3 4 7
>  3 5 17
>  3 6 10

Use library(plyr), convert to data frame, do ddply, convert back to
matrix if you want. I'm surprised mmply doesn't do it, but I dont
think it does:

 > Md=data.frame(M)
 > ddply(Md,c(1,2),function(r){sum(r[,3])})
  X1 X2 V1
1  1  2  6
2  1  3  4
3  2  4 11
4  3  4  7
5  3  5 17
6  3  6 10

plyr is on CRAN and that's the third time today I've told someone to use it.

Barry

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[R] How R converts data between objects

2010-09-06 Thread Alaios
Hello everyone. 
I would kindly request your help concerning how R converts data between 
different structrures.

In the following example please keep attention on the following two
1) 
I create 
f <- GaussRF(x=x, y=y, model=model, grid=TRUE,param=c(mean, variance, nugget, 
scale, alpha))
with 
image(x,y,f)
   and
image(f)
I get exactly the same image.

then
2)I try to store f inside a raster layer using:
r <- setValues(r,as.matrix(f))

then comes the tricky part. I am trying to use image again and get the same 
output as the ouput I got by image(f).
I tried
image(as.matrix(getValues(r)))
which give a completely different output. 
image function expects a matrix... but for some reason as.matrix(getValues(r)) 
returns a huge vector :(
also tried
as.matrix(getValues(r),ncol=ncol(r),nrow=nrow(r)) to force as.matrix to return 
the appropriate matrix but this also failed and I only got back a vector again

Could you please let me understand why this might happening?
Best Regards
Alex

P.S Below you will find part of my code.



...(lines omitted, declarations)
x <- seq(1, dimx, step)
y <- seq(1, dimy, step)
f <- GaussRF(x=x, y=y, model=model, grid=TRUE,param=c(mean, variance, nugget, 
scale, alpha))

# image(x, y, f) Displays the matrix
# f is a 2x2 matrix of dimension (x,y)

r <- raster(nrow=dimx, ncol=dimy)
r <- setValues(r,as.matrix(f))
# getValues(r)



  
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Re: [R] How to get "mypkg-manual.pdf"

2010-09-06 Thread Duncan Murdoch

On 06/09/2010 9:19 AM, Juliet Ndukum wrote:
I am building a package say mypkg. Five months ago, when I built the package I 
got the mypkg manual in pdf format. 

Today, after making updates, I build the same package, same name, and steps; 
unfortunately I do not get the manual in pdf format. 
Rather I get the following message: 
cd: can't cd to /cygdrive/c/Documents saving output to 'mypkg-manual.pdf' 
...Done


Could any one help me on how to get the "mypkg-manual.pdf".


It looks as though you are trying to put it into a file path with spaces 
in it, and the version of R you're using doesn't like the spaces.  Try 
saving it somewhere else.


I believe the current release doesn't care about spaces, but I'm not 
sure of that.  Building a pdf depends on external tools that are 
sometimes out of our control.


Duncan Murdoch


Thank you in advance for your help.
JN



  
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Re: [R] Time Series

2010-09-06 Thread Gabor Grothendieck
On Mon, Sep 6, 2010 at 10:24 AM, trb1  wrote:
>
> Hi
>
> How would I analyse time series with
> - different lengths (i.e. one has 9 entries and the other has 14 entries)
> - different frequency (i.e. dates are random - no repeated length)
> - multiple values for the same time entry (e.g. 2009-10-23 below)
>
> i.e. my data takes the form:
> 1st time series
>
> 2009-10-07      0.009378
> 2009-10-19      0.014790
> 2009-10-23      -0.005946
> 2009-10-23      0.009096
> 2009-11-08      0.004189
> 2009-11-10      -0.004592
> 2009-11-17      0.009397
> 2009-11-24      0.003411
> 2009-12-02      0.003300
> 2010-01-15      0.010873
> 2010-01-20      0.010712
> 2010-01-20      0.022237
>
> 2nd time series
>
> 2009-09-23      0.076253
> 2009-10-07      0.039255
> 2010-02-17      0.039045
> 2010-03-09      0.024201
> 2010-03-25      -0.039810
> 2010-04-13      -0.012428
>
> I am unable to get any functions to work.
> A simple plot would be nice!
>


Try this.  We read in each series taking the last point in the event
that there are multiple points with the same date.  Then we plot each.

If we wish to plot them on the same plot then we merge each series
(for producing points) together with linear interpolations of both
series (for producing lines) and plot.

Lines1 <- "2009-10-07  0.009378
2009-10-19  0.014790
2009-10-23  -0.005946
2009-10-23  0.009096
2009-11-08  0.004189
2009-11-10  -0.004592
2009-11-17  0.009397
2009-11-24  0.003411
2009-12-02  0.003300
2010-01-15  0.010873
2010-01-20  0.010712
2010-01-20  0.022237"

Lines2 <- "2009-09-23  0.076253
2009-10-07  0.039255
2010-02-17  0.039045
2010-03-09  0.024201
2010-03-25  -0.039810
2010-04-13  -0.012428"

library(zoo)

z1 <- read.zoo(textConnection(Lines1), aggregate = function(x) tail(x, 1))
z2 <- read.zoo(textConnection(Lines2), aggregate = function(x) tail(x, 2))

plot(z1, type = "o")
plot(z2, type = "o")

# or together on the same plot
zz <- merge(z1, z2)
zz <- merge(z1, z2, na.approx(zz, na.rm = FALSE))
plot(zz, type = c("p", "p", "l", "l"), screen = 1:2)


-- 
Statistics & Software Consulting
GKX Group, GKX Associates Inc.
tel: 1-877-GKX-GROUP
email: ggrothendieck at gmail.com

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[R] rbind() overwriting data.frame()

2010-09-06 Thread rajesh j
Hi,

first off, I wanna ask how do I declare a data.frame of 0 rows and n
columns?

Coming to my problem,

I have a data.frame of 22 columns by dynamic rows which I insert using
rbind. The total number of rows could go upto 2,00,000. The problem is that
after about 800 or 900 get inserted rbind starts overwriting the data.frame
and I end up with a total of 800-900 rows. What is up with that?
The 22 columns are all strings each having about 10 characters
-- 
Rajesh.J

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Re: [R] rbind() overwriting data.frame()

2010-09-06 Thread Ivan Calandra
  Hi again!

I'm trying to follow your general goal from your questions today but 
it's not easy.

First, declaring a data.frame of 0 rows is a bad idea. It is much faster 
to define the length and number of rows from the beginning and to fill 
it then.

Second, I don't know how to do it! What I know is that, to my knowledge 
(maybe I overlooked some posts in the archive), there is no easy way to 
do it, such as lists or vectors. The easiest might be to create a list 
with the correct length with list(), fill it with whatever data and then 
convert it to a data.frame with as.data.frame() when it's finished.

Third, for your problem, maybe do.call() can help you. I don't know what 
you did up to now, but it sounds that you tried to do it iteratively (in 
a loop) instead of vectorizing it (though I don't know if do.call() can 
be really called vectorized). There was a post yesterday/today on 
do.call(). You'll surely find it if you look with RSiteSearch().

Last, I don't know if it is relevant for you, but I've read on the list 
many times that matrices are faster to deal with. If all your columns 
and rows have the same type, then you can use matrices.

There are surely guys that know more about this stuff somewhere on the 
list, but I hope it can get you started.
Ivan

Le 9/6/2010 16:56, rajesh j a écrit :
> Hi,
>
> first off, I wanna ask how do I declare a data.frame of 0 rows and n
> columns?
>
> Coming to my problem,
>
> I have a data.frame of 22 columns by dynamic rows which I insert using
> rbind. The total number of rows could go upto 2,00,000. The problem is that
> after about 800 or 900 get inserted rbind starts overwriting the data.frame
> and I end up with a total of 800-900 rows. What is up with that?
> The 22 columns are all strings each having about 10 characters

-- 
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

**
http://www.for771.uni-bonn.de
http://webapp5.rrz.uni-hamburg.de/mammals/eng/mitarbeiter.php


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Re: [R] Aggregate certain rows in a matrix

2010-09-06 Thread David Winsemius


On Sep 6, 2010, at 10:47 AM, Dimitris Rizopoulos wrote:


one way is the following:

M <- cbind(c(1,1,1,1,2,2,3,3,3,3), c(2,2,2,3,4,4,4,5,5,6),
   c(1,2,3,4,5,6,7,8,9,10))

ind <- do.call(paste, c(as.data.frame(M[, 1:2], sep = "\r")))
M[, 3] <- ave(M[, 3], ind, FUN = "sum")
unique(M)


I had been working on a similar approach with ave( ,paste(), sum)  
inside a datafrmae, but I liked your approach of setting up the  
results of the paste operation as a vector outside of M. (Skips the  
dataframe operation I was using.) The above solution is "destructive",  
so I constructed this similar alternative that returns the results  
without altering M:


> cbind(M, ave(M[ , 3], list(M[,1], M[,2]), FUN=sum))[
   !duplicated(M[,1:2]),  
c(1,2,4)]

 [,1] [,2] [,3]
[1,]126
[2,]134
[3,]24   11
[4,]347
[5,]35   17
[6,]36   10





I hope it helps.

Best,
Dimitris


On 9/6/2010 4:29 PM, Kennedy wrote:


Hi,

I have a matrix that looks like this

  a<- c(1,1,1,1,2,2,3,3,3,3)
  b<- c(2,2,2,3,4,4,4,5,5,6)
  c<- c(1,2,3,4,5,6,7,8,9,10)
  M<- matrix(nr=10,nc=3)
  M[,1]<- a
  M[,2]<- b
  M[,3]<- c


M

  [,1] [,2] [,3]
 [1,]121
 [2,]122
 [3,]123
 [4,]134
 [5,]245
 [6,]246
 [7,]347
 [8,]358
 [9,]359
[10,]36   10

I want to reduce the matrix according to the following: If the  
values of the
two first columns are the same in two or more rows the values in  
the third
column of the corresponding rows should be added and only one of  
the rows

should be keept. Hence the matrix M above should look like this

  1 2 6
  1 3 4
  2 4 11
  3 4 7
  3 5 17
  3 6 10


Thank you

Henrik





--
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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David Winsemius, MD
West Hartford, CT

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Re: [R] rbind() overwriting data.frame()

2010-09-06 Thread ONKELINX, Thierry
This will give a matrix with 0 rows.

data.frame(matrix(nrow = 0,  ncol = 22, dimnames = list(NULL,
LETTERS[1:22])))

But you should avoid growing dataframes is the final dataframe is going
to be large. You are very likely to get memory problems. It is much to
better to create a large enough dataframe and then overwrite the rows.
And it is faster too...

> nrows <- 2000
> ncols <- 22
> system.time({
+ tmp <- data.frame(matrix(nrow = 0,  ncol = ncols))
+ for(i in seq_len(nrows)){
+ tmp <- rbind(tmp, rnorm(ncols))
+ }
+ })
   user  system elapsed 
   7.830.027.86 
> system.time({
+ tmp <- data.frame(matrix(nrow = nrows,  ncol = ncols))
+ for(i in seq_len(nrows)){
+ tmp[i, ] <- rnorm(ncols)
+ }
+ })
   user  system elapsed 
   3.750.003.76 

#In this case an apply construction was even faster

> system.time({
+ tmp <- t(sapply(seq_len(nrows), function(i){
+ rnorm(ncols)
+ }))
+ })
   user  system elapsed 
   0.020.000.02 






ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek
team Biometrie & Kwaliteitszorg
Gaverstraat 4
9500 Geraardsbergen
Belgium

Research Institute for Nature and Forest
team Biometrics & Quality Assurance
Gaverstraat 4
9500 Geraardsbergen
Belgium

tel. + 32 54/436 185
thierry.onkel...@inbo.be
www.inbo.be

To call in the statistician after the experiment is done may be no more
than asking him to perform a post-mortem examination: he may be able to
say what the experiment died of.
~ Sir Ronald Aylmer Fisher

The plural of anecdote is not data.
~ Roger Brinner

The combination of some data and an aching desire for an answer does not
ensure that a reasonable answer can be extracted from a given body of
data.
~ John Tukey
  

> -Oorspronkelijk bericht-
> Van: r-help-boun...@r-project.org 
> [mailto:r-help-boun...@r-project.org] Namens rajesh j
> Verzonden: maandag 6 september 2010 16:57
> Aan: r-help@r-project.org
> Onderwerp: [R] rbind() overwriting data.frame()
> 
> Hi,
> 
> first off, I wanna ask how do I declare a data.frame of 0 
> rows and n columns?
> 
> Coming to my problem,
> 
> I have a data.frame of 22 columns by dynamic rows which I 
> insert using rbind. The total number of rows could go upto 
> 2,00,000. The problem is that after about 800 or 900 get 
> inserted rbind starts overwriting the data.frame and I end up 
> with a total of 800-900 rows. What is up with that?
> The 22 columns are all strings each having about 10 characters
> --
> Rajesh.J
> 
>   [[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.
> 

Druk dit bericht a.u.b. niet onnodig af.
Please do not print this message unnecessarily.

Dit bericht en eventuele bijlagen geven enkel de visie van de schrijver weer 
en binden het INBO onder geen enkel beding, zolang dit bericht niet bevestigd is
door een geldig ondertekend document. The views expressed in  this message 
and any annex are purely those of the writer and may not be regarded as stating 
an official position of INBO, as long as the message is not confirmed by a duly 
signed document.

__
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Re: [R] Strange behavior of interval values in optimize()

2010-09-06 Thread Michael Bernsteiner

that was my first idea as well, but as the result shows, the minimized function 
value of the wider interval is greater.

In
 addidtion, the problem also exists, if the minimized parameter in the 
case of the larger interval also already lies within the smaller 
interval:


f<-function(delta,P,U){
minimiz<-P+delta*U
x<-minimiz[1]
y<-minimiz[2]
z<-100*(y-x^2)^2+(1-x)^2
return(z)
}


> result<-optimize(f, interval=c(-10, 1), P=c(0.99,1.01), U=c(1,0))
> result
$minimum
[1] 0.01496807

$objective
[1] 2.483522e-05

> 
> result<-optimize(f, interval=c(-100, 1), P=c(0.99,1.01), U=c(1,0))
> result
$minimum
[1] -1.989994

$objective
[1] 4.01

This does still make no sense for me

Fabian


> Date: Mon, 6 Sep 2010 06:18:16 -0700
> From: dieter.me...@menne-biomed.de
> To: r-help@r-project.org
> Subject: Re: [R] Strange behavior of interval values in optimize()
> 
> 
> 
> Michael Bernsteiner wrote:
> > 
> > 
> > I'm using optimize() to find the minimum of the following function f, and
> > minimize it (without 
> > .
> > But, when I choose a larger Interval in the optimization method:
> > 
> > The result gets worse (even though the old interval is included in the new
> > one).
> > 
> > In fact I don't want any restrictons for the interval values (something
> > like interval=(-Inf, Inf) or at least the interval should be as large as
> > possible.
> > 
> > 
> 
> The most likely cause is a second minimum.
> 
> Dieter
> 
> -- 
> View this message in context: 
> http://r.789695.n4.nabble.com/Strange-behavior-of-interval-values-in-optimize-tp2528208p2528366.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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[R] path analysis

2010-09-06 Thread Guy rotem
Hi.

which package i need to install to be able to run "Path analysis" using r?

many thanks, Guy

-- 
Guy Rotem
Department of Life Sciences
The Spatial Ecology Lab
Ben Gurion University of the Negev
P.O.B. 653   Beer-Sheva 84105
ISRAEL

+972-52-3354485 (mobile)
+972-8-6461350 (lab)

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[R] boxplot knowing Q1, Q3, median, upper and lower whisker value

2010-09-06 Thread David A.

Dear list,

I am using a external program that outputs Q1, Q3, median, upper and lower 
whisker values for various datasets simultaneously in a tab delimited format. 
After importing this text file into R, I would like to plot a boxplot using 
these given values and not the original series of data points, i.e. not using 
something like boxplot(mydata).

Is there an easy way for doing this? If I am not wrong, boxplot() does not 
accept these values as parameters.

Cheers,

Dave 
  
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Re: [R] rbind() overwriting data.frame()

2010-09-06 Thread rajesh j
But If I do that how will I resize later?

On Mon, Sep 6, 2010 at 8:54 PM, ONKELINX, Thierry
wrote:

> This will give a matrix with 0 rows.
>
> data.frame(matrix(nrow = 0,  ncol = 22, dimnames = list(NULL,
> LETTERS[1:22])))
>
> But you should avoid growing dataframes is the final dataframe is going
> to be large. You are very likely to get memory problems. It is much to
> better to create a large enough dataframe and then overwrite the rows.
> And it is faster too...
>
> > nrows <- 2000
> > ncols <- 22
> > system.time({
> + tmp <- data.frame(matrix(nrow = 0,  ncol = ncols))
> + for(i in seq_len(nrows)){
> + tmp <- rbind(tmp, rnorm(ncols))
> + }
> + })
>   user  system elapsed
>   7.830.027.86
> > system.time({
> + tmp <- data.frame(matrix(nrow = nrows,  ncol = ncols))
> + for(i in seq_len(nrows)){
> + tmp[i, ] <- rnorm(ncols)
> + }
> + })
>   user  system elapsed
>   3.750.003.76
>
> #In this case an apply construction was even faster
>
> > system.time({
> + tmp <- t(sapply(seq_len(nrows), function(i){
> + rnorm(ncols)
> + }))
> + })
>   user  system elapsed
>   0.020.000.02
>
>
>
>
> 
> 
> ir. Thierry Onkelinx
> Instituut voor natuur- en bosonderzoek
> team Biometrie & Kwaliteitszorg
> Gaverstraat 4
> 9500 Geraardsbergen
> Belgium
>
> Research Institute for Nature and Forest
> team Biometrics & Quality Assurance
> Gaverstraat 4
> 9500 Geraardsbergen
> Belgium
>
> tel. + 32 54/436 185
> thierry.onkel...@inbo.be
> www.inbo.be
>
> To call in the statistician after the experiment is done may be no more
> than asking him to perform a post-mortem examination: he may be able to
> say what the experiment died of.
> ~ Sir Ronald Aylmer Fisher
>
> The plural of anecdote is not data.
> ~ Roger Brinner
>
> The combination of some data and an aching desire for an answer does not
> ensure that a reasonable answer can be extracted from a given body of
> data.
> ~ John Tukey
>
>
> > -Oorspronkelijk bericht-
> > Van: r-help-boun...@r-project.org
> > [mailto:r-help-boun...@r-project.org] Namens rajesh j
> > Verzonden: maandag 6 september 2010 16:57
> > Aan: r-help@r-project.org
> > Onderwerp: [R] rbind() overwriting data.frame()
> >
> > Hi,
> >
> > first off, I wanna ask how do I declare a data.frame of 0
> > rows and n columns?
> >
> > Coming to my problem,
> >
> > I have a data.frame of 22 columns by dynamic rows which I
> > insert using rbind. The total number of rows could go upto
> > 2,00,000. The problem is that after about 800 or 900 get
> > inserted rbind starts overwriting the data.frame and I end up
> > with a total of 800-900 rows. What is up with that?
> > The 22 columns are all strings each having about 10 characters
> > --
> > Rajesh.J
> >
> >   [[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.
> >
>
> Druk dit bericht a.u.b. niet onnodig af.
> Please do not print this message unnecessarily.
>
> Dit bericht en eventuele bijlagen geven enkel de visie van de schrijver
> weer
> en binden het INBO onder geen enkel beding, zolang dit bericht niet
> bevestigd is
> door een geldig ondertekend document. The views expressed in  this message
> and any annex are purely those of the writer and may not be regarded as
> stating
> an official position of INBO, as long as the message is not confirmed by a
> duly
> signed document.
>



-- 
Rajesh.J

[[alternative HTML version deleted]]

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Re: [R] Time Series

2010-09-06 Thread trb1

Thank you very much for your post. 
Your answer has been very helpful.
Is it possible to merge >2 time series?

-- 
View this message in context: 
http://r.789695.n4.nabble.com/Time-Series-tp2528444p2528584.html
Sent from the R help mailing list archive at Nabble.com.

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Re: [R] rbind() overwriting data.frame()

2010-09-06 Thread rajesh j
Also, when I create the data.frame with matrix and try to rbind, I get
warnings..
Warning messages:
1: In `[<-.factor`(`*tmp*`, ri, value = "4") :
  invalid factor level, NAs generated
2: In `[<-.factor`(`*tmp*`, ri, value = "5") :
  invalid factor level, NAs generated
3: In `[<-.factor`(`*tmp*`, ri, value = "6") :
  invalid factor level, NAs generated


On Mon, Sep 6, 2010 at 9:17 PM, rajesh j  wrote:

> But If I do that how will I resize later?
>
>
> On Mon, Sep 6, 2010 at 8:54 PM, ONKELINX, Thierry <
> thierry.onkel...@inbo.be> wrote:
>
>> This will give a matrix with 0 rows.
>>
>> data.frame(matrix(nrow = 0,  ncol = 22, dimnames = list(NULL,
>> LETTERS[1:22])))
>>
>> But you should avoid growing dataframes is the final dataframe is going
>> to be large. You are very likely to get memory problems. It is much to
>> better to create a large enough dataframe and then overwrite the rows.
>> And it is faster too...
>>
>> > nrows <- 2000
>> > ncols <- 22
>> > system.time({
>> + tmp <- data.frame(matrix(nrow = 0,  ncol = ncols))
>> + for(i in seq_len(nrows)){
>> + tmp <- rbind(tmp, rnorm(ncols))
>> + }
>> + })
>>   user  system elapsed
>>   7.830.027.86
>> > system.time({
>> + tmp <- data.frame(matrix(nrow = nrows,  ncol = ncols))
>> + for(i in seq_len(nrows)){
>> + tmp[i, ] <- rnorm(ncols)
>> + }
>> + })
>>   user  system elapsed
>>   3.750.003.76
>>
>> #In this case an apply construction was even faster
>>
>> > system.time({
>> + tmp <- t(sapply(seq_len(nrows), function(i){
>> + rnorm(ncols)
>> + }))
>> + })
>>   user  system elapsed
>>   0.020.000.02
>>
>>
>>
>>
>> 
>> 
>> ir. Thierry Onkelinx
>> Instituut voor natuur- en bosonderzoek
>> team Biometrie & Kwaliteitszorg
>> Gaverstraat 4
>> 9500 Geraardsbergen
>> Belgium
>>
>> Research Institute for Nature and Forest
>> team Biometrics & Quality Assurance
>> Gaverstraat 4
>> 9500 Geraardsbergen
>> Belgium
>>
>> tel. + 32 54/436 185
>> thierry.onkel...@inbo.be
>> www.inbo.be
>>
>> To call in the statistician after the experiment is done may be no more
>> than asking him to perform a post-mortem examination: he may be able to
>> say what the experiment died of.
>> ~ Sir Ronald Aylmer Fisher
>>
>> The plural of anecdote is not data.
>> ~ Roger Brinner
>>
>> The combination of some data and an aching desire for an answer does not
>> ensure that a reasonable answer can be extracted from a given body of
>> data.
>> ~ John Tukey
>>
>>
>> > -Oorspronkelijk bericht-
>> > Van: r-help-boun...@r-project.org
>> > [mailto:r-help-boun...@r-project.org] Namens rajesh j
>> > Verzonden: maandag 6 september 2010 16:57
>> > Aan: r-help@r-project.org
>> > Onderwerp: [R] rbind() overwriting data.frame()
>> >
>> > Hi,
>> >
>> > first off, I wanna ask how do I declare a data.frame of 0
>> > rows and n columns?
>> >
>> > Coming to my problem,
>> >
>> > I have a data.frame of 22 columns by dynamic rows which I
>> > insert using rbind. The total number of rows could go upto
>> > 2,00,000. The problem is that after about 800 or 900 get
>> > inserted rbind starts overwriting the data.frame and I end up
>> > with a total of 800-900 rows. What is up with that?
>> > The 22 columns are all strings each having about 10 characters
>> > --
>> > Rajesh.J
>> >
>> >   [[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.
>> >
>>
>> Druk dit bericht a.u.b. niet onnodig af.
>> Please do not print this message unnecessarily.
>>
>> Dit bericht en eventuele bijlagen geven enkel de visie van de schrijver
>> weer
>> en binden het INBO onder geen enkel beding, zolang dit bericht niet
>> bevestigd is
>> door een geldig ondertekend document. The views expressed in  this message
>> and any annex are purely those of the writer and may not be regarded as
>> stating
>> an official position of INBO, as long as the message is not confirmed by a
>> duly
>> signed document.
>>
>
>
>
> --
> Rajesh.J
>
>
>


-- 
Rajesh.J

[[alternative HTML version deleted]]

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[R] WriteXLS problem

2010-09-06 Thread Kenneth Roy Cabrera Torres
Hi R users:

I don't know if you have had the following problem trying to
export to an "xls" format file in a non windows platform.

I try to use the following packages:
1. dataframes2xls (version 0.4.4) (with phyton 2.7 and 3.1)
2. WriteXLS (version 1.9.0) (with perl and testPerl working)

Even "xlsx" package that take too long and do not finish.

The data frame I try to export has 269363 row and 116 columns.
In the first one (dataframe2xls) I get this message:

Traceback (most recent call last):
 File
"C:/PROGRA~2/R/R-211~1.1PA/library/dataframes2xls/python/csv2xls.py",
line 18, in 
import pyexcelerator
File
"C:\PROGRA~2\R\R-211~1.1PA\library\dataframes2xls\python\pyexcelerator
\__init__.py",
line 12, in  from Workbook import Workbook
File
"C:\PROGRA~2\R\R-211~1.1PA\library\dataframes2xls\python\pyexcelerator
\Workbook.py",
line 526 boundsheets_len += len(BIFFRecords.BoundSheetRecord(0x00L,
sheet.hidden, sheet.name).get())
^
SyntaxError: invalid syntax

Using the second option I get this message:

Error en get(as.character(i)),envr=envir) :
  objeto '089' no encontrado

Object '089' not found.

Im using this R platform:
sessionInfo()
R version 2.11.1 Patched (2010-08-30 r52848)
Platform: x86_64-unknown-linux-gnu (64-bit)

Locale:
LC_CTYPE=es_CO.UTF-8 

Is the only solution to export to ".csv" and then
to ".xls" format with other program like openoffice?

Thank you for your help and advice.

Kenneth

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Re: [R] WriteXLS problem

2010-09-06 Thread Ivan Calandra
  Hi,

Are you sure you used the correct syntax and object names? It might just 
be because of that...(reading the error messages)
There is another function, xlsReadWrite::write.xls(), that I like a lot: 
it is really easy to use and does not require Perl or Python.

HTH,
Ivan

Le 9/6/2010 18:03, Kenneth Roy Cabrera Torres a écrit :
> Hi R users:
>
> I don't know if you have had the following problem trying to
> export to an "xls" format file in a non windows platform.
>
> I try to use the following packages:
> 1. dataframes2xls (version 0.4.4) (with phyton 2.7 and 3.1)
> 2. WriteXLS (version 1.9.0) (with perl and testPerl working)
>
> Even "xlsx" package that take too long and do not finish.
>
> The data frame I try to export has 269363 row and 116 columns.
> In the first one (dataframe2xls) I get this message:
>
> Traceback (most recent call last):
>   File
> "C:/PROGRA~2/R/R-211~1.1PA/library/dataframes2xls/python/csv2xls.py",
> line 18, in
> import pyexcelerator
> File
> "C:\PROGRA~2\R\R-211~1.1PA\library\dataframes2xls\python\pyexcelerator
> \__init__.py",
> line 12, in  from Workbook import Workbook
> File
> "C:\PROGRA~2\R\R-211~1.1PA\library\dataframes2xls\python\pyexcelerator
> \Workbook.py",
> line 526 boundsheets_len += len(BIFFRecords.BoundSheetRecord(0x00L,
> sheet.hidden, sheet.name).get())
>  ^
> SyntaxError: invalid syntax
>
> Using the second option I get this message:
>
> Error en get(as.character(i)),envr=envir) :
>objeto '089' no encontrado
>
> Object '089' not found.
>
> Im using this R platform:
> sessionInfo()
> R version 2.11.1 Patched (2010-08-30 r52848)
> Platform: x86_64-unknown-linux-gnu (64-bit)
>
> Locale:
> LC_CTYPE=es_CO.UTF-8
>
> Is the only solution to export to ".csv" and then
> to ".xls" format with other program like openoffice?
>
> Thank you for your help and advice.
>
> Kenneth
>
> __
> 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

**
http://www.for771.uni-bonn.de
http://webapp5.rrz.uni-hamburg.de/mammals/eng/mitarbeiter.php


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Re: [R] path analysis

2010-09-06 Thread Sarah Goslee
There are lots of options for path analysis in R.

If you go to http://www.rseek.org and type path analysis into the search box,
you will get lots of information on functions/packages, and more general
info as well.

Beyond that, we'd need more specifics about your task.

Sarah

On Mon, Sep 6, 2010 at 10:37 AM, Guy rotem  wrote:
> Hi.
>
> which package i need to install to be able to run "Path analysis" using r?
>
> many thanks, Guy
>


-- 
Sarah Goslee
http://www.functionaldiversity.org

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Re: [R] boxplot knowing Q1, Q3, median, upper and lower whisker value

2010-09-06 Thread Joshua Wiley
Hi Dave,

You can look at the function ?bxp it might work for you.  Alternately,
create a meaningless boxplot object, and then just edit that data, in
which case I know it will work with bxp().

# Create a boxplot, the data does not matter
x <- boxplot(1:10)
x # view the data for the boxplot

x$stats <- c() # put the stats here, min Q1, median, Q3, max
# or hinges or whatever you like

x$n <- c() # the number of observations, though you do not need to change this

bxp(x) # plot boxplot with updated info

HTH,

Josh


On Mon, Sep 6, 2010 at 8:46 AM, David A.  wrote:
>
> Dear list,
>
> I am using a external program that outputs Q1, Q3, median, upper and lower 
> whisker values for various datasets simultaneously in a tab delimited format. 
> After importing this text file into R, I would like to plot a boxplot using 
> these given values and not the original series of data points, i.e. not using 
> something like boxplot(mydata).
>
> Is there an easy way for doing this? If I am not wrong, boxplot() does not 
> accept these values as parameters.
>
> Cheers,
>
> Dave
>
>        [[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.
>



-- 
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/

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[R] How to run R on Emacs+ESS

2010-09-06 Thread Stephen Liu
Hi folks,

Debian 504 64-bit

I found following document;
http://www.biostat.wisc.edu/~kbroman/Rintro/

Whether it is the right document for installing Emacs+ESS and R so that R can 
run on Emacs?

TIA

B.R.
Stephen L




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Re: [R] WriteXLS problem

2010-09-06 Thread Kenneth Roy Cabrera Torres
Thank you Ivan for you answer:
El lun, 06-09-2010 a las 18:11 +0200, Ivan Calandra escribió:
> Hi,
> 
> Are you sure you used the correct syntax and object names? It might just 
> be because of that...(reading the error messages)
Im sure, because it works with write.csv or write.table.
> There is another function, xlsReadWrite::write.xls(), that I like a lot: 
> it is really easy to use and does not require Perl or Python.
Unfortunately it works on windows, and I am in a non windows platform
(ubuntu).

Thank you for you advice and help.

Kenneth
> 
> HTH,
> Ivan
> 
> Le 9/6/2010 18:03, Kenneth Roy Cabrera Torres a crit :
> > Hi R users:
> >
> > I don't know if you have had the following problem trying to
> > export to an "xls" format file in a non windows platform.
> >
> > I try to use the following packages:
> > 1. dataframes2xls (version 0.4.4) (with phyton 2.7 and 3.1)
> > 2. WriteXLS (version 1.9.0) (with perl and testPerl working)
> >
> > Even "xlsx" package that take too long and do not finish.
> >
> > The data frame I try to export has 269363 row and 116 columns.
> > In the first one (dataframe2xls) I get this message:
> >
> > Traceback (most recent call last):
> >   File
> > "C:/PROGRA~2/R/R-211~1.1PA/library/dataframes2xls/python/csv2xls.py",
> > line 18, in
> > import pyexcelerator
> > File
> > "C:\PROGRA~2\R\R-211~1.1PA\library\dataframes2xls\python\pyexcelerator
> > \__init__.py",
> > line 12, in  from Workbook import Workbook
> > File
> > "C:\PROGRA~2\R\R-211~1.1PA\library\dataframes2xls\python\pyexcelerator
> > \Workbook.py",
> > line 526 boundsheets_len += len(BIFFRecords.BoundSheetRecord(0x00L,
> > sheet.hidden, sheet.name).get())
> >  ^
> > SyntaxError: invalid syntax
> >
> > Using the second option I get this message:
> >
> > Error en get(as.character(i)),envr=envir) :
> >objeto '089' no encontrado
> >
> > Object '089' not found.
> >
> > Im using this R platform:
> > sessionInfo()
> > R version 2.11.1 Patched (2010-08-30 r52848)
> > Platform: x86_64-unknown-linux-gnu (64-bit)
> >
> > Locale:
> > LC_CTYPE=es_CO.UTF-8
> >
> > Is the only solution to export to ".csv" and then
> > to ".xls" format with other program like openoffice?
> >
> > Thank you for your help and advice.
> >
> > Kenneth
> >
> > __
> > 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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and provide commented, minimal, self-contained, reproducible code.


Re: [R] How to run R on Emacs+ESS

2010-09-06 Thread Dirk Eddelbuettel

On 6 September 2010 at 09:18, Stephen Liu wrote:
| Hi folks,
| 
| Debian 504 64-bit

Good. All you need is 

   sudo apt-get install ess

| I found following document;
| http://www.biostat.wisc.edu/~kbroman/Rintro/
| 
| Whether it is the right document for installing Emacs+ESS and R so that R can 
| run on Emacs?

There is nothing else to do.  Restart (X)Emacs, whichever variant you use on
Debian, and type M-x R. You now run R inside Emacs.

After that, see http://ess.r-project.org, esp the Documentation tab.

Dirk

-- 
Dirk Eddelbuettel | e...@debian.org | http://dirk.eddelbuettel.com

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Re: [R] WriteXLS problem

2010-09-06 Thread David Winsemius


On Sep 6, 2010, at 12:25 PM, Kenneth Roy Cabrera Torres wrote:


Thank you Ivan for you answer:
El lun, 06-09-2010 a las 18:11 +0200, Ivan Calandra escribió:

Hi,

Are you sure you used the correct syntax and object names? It might  
just

be because of that...(reading the error messages)

Im sure, because it works with write.csv or write.table.


Sure? You are making the incorrect assumption that those write  
functions have the same syntax. At least for WriteXLS that assumption  
is false. The help page clearly states that the objects need to be  
quoted rather than being referred to by their "naked" names. The error  
you are getting with your "second option" suggests to me that you  
offered an unquoted name of an object.


You can offer a vector of quoted names of dataframes to WriteXLS and  
each named dataframe will be converted to a worksheet within the  
workbook.


--
David.


There is another function, xlsReadWrite::write.xls(), that I like a  
lot:

it is really easy to use and does not require Perl or Python.

Unfortunately it works on windows, and I am in a non windows platform
(ubuntu).

Thank you for you advice and help.

Kenneth


HTH,
Ivan

Le 9/6/2010 18:03, Kenneth Roy Cabrera Torres a crit :

Hi R users:

I don't know if you have had the following problem trying to
export to an "xls" format file in a non windows platform.

I try to use the following packages:
1. dataframes2xls (version 0.4.4) (with phyton 2.7 and 3.1)
2. WriteXLS (version 1.9.0) (with perl and testPerl working)

Even "xlsx" package that take too long and do not finish.

The data frame I try to export has 269363 row and 116 columns.
In the first one (dataframe2xls) I get this message:

Traceback (most recent call last):
 File
"C:/PROGRA~2/R/R-211~1.1PA/library/dataframes2xls/python/ 
csv2xls.py",

line 18, in
import pyexcelerator
File
"C:\PROGRA~2\R\R-211~1.1PA\library\dataframes2xls\python 
\pyexcelerator

\__init__.py",
line 12, in  from Workbook import Workbook
File
"C:\PROGRA~2\R\R-211~1.1PA\library\dataframes2xls\python 
\pyexcelerator

\Workbook.py",
line 526 boundsheets_len +=  
len(BIFFRecords.BoundSheetRecord(0x00L,

sheet.hidden, sheet.name).get())
^
SyntaxError: invalid syntax

Using the second option I get this message:

Error en get(as.character(i)),envr=envir) :
  objeto '089' no encontrado

Object '089' not found.

Im using this R platform:
sessionInfo()
R version 2.11.1 Patched (2010-08-30 r52848)
Platform: x86_64-unknown-linux-gnu (64-bit)

Locale:
LC_CTYPE=es_CO.UTF-8

Is the only solution to export to ".csv" and then
to ".xls" format with other program like openoffice?

Thank you for your help and advice.

Kenneth

__
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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.



__
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David Winsemius, MD
West Hartford, CT

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Re: [R] WriteXLS problem

2010-09-06 Thread Kenneth Roy Cabrera Torres
I use the following sintaxis for the packages:

For WriteXLS I use:

writeXLS(todo2009,"todo2009.xls")

And for dataframes2xls I use:
dataframe2xls::write.xls(todo2009,"todo2009.xls")



El lun, 06-09-2010 a las 12:34 -0400, David Winsemius escribió:
> On Sep 6, 2010, at 12:25 PM, Kenneth Roy Cabrera Torres wrote:
> 
> > Thank you Ivan for you answer:
> > El lun, 06-09-2010 a las 18:11 +0200, Ivan Calandra escribió:
> >> Hi,
> >>
> >> Are you sure you used the correct syntax and object names? It might  
> >> just
> >> be because of that...(reading the error messages)
> > Im sure, because it works with write.csv or write.table.
> 
> Sure? You are making the incorrect assumption that those write  
> functions have the same syntax. At least for WriteXLS that assumption  
> is false. The help page clearly states that the objects need to be  
> quoted rather than being referred to by their "naked" names. The error  
> you are getting with your "second option" suggests to me that you  
> offered an unquoted name of an object.
> 
> You can offer a vector of quoted names of dataframes to WriteXLS and  
> each named dataframe will be converted to a worksheet within the  
> workbook.
>

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Re: [R] How can I fixe convergence=1 in optim

2010-09-06 Thread Ben Bolker
  [forwarding back to r-help for archiving/further discussion]

On 10-09-05 08:48 PM, Sally Luo wrote:
> Prof. Bolker,
>  
> Thanks for your reply and the helpful info.
>  
> I still have a few questions. 
>  
> 1. I also tried to use different methods other than "BFGS" without
> changing their default maxit values, and got very different results. 
> Since I am not that experienced with the optim funciton, I am not sure
> which one is correct or trustworthy?
>  
> *If I use method=SANN, that is,*
>  
> > p<-optim(c(-0.2392925,0.4653128,-0.8332286, 0.0657, -0.0031,
> -0.00245, 3.366, 0.5885, -0.8,
> +0.0786,-0.00292,-0.00081, 3.266, -0.3632, -0.49,   
> 0.1856, 0.00394, -0.00193, -0.889, 0.5379, -0.63,
> +0.213, 0.00338, -0.00026, -0.8912, -0.3023, -0.56),
> f, method ="SANN", y=y,X=X,W=W)
>  
> I get:
>
> There were 50 or more warnings (use warnings() to see the first 50)
> > p
> $par
>  [1] -0.2392925  0.4653128 -0.8332286  0.0657000 -0.0031000
> -0.0024500  3.366  0.5885000 -0.800  0.0786000 -0.0029200
> -0.0008100
> [13]  3.266 -0.3632000 -0.490  0.1856000  0.0039400 -0.0019300
> -0.889  0.5379000 -0.630  0.213  0.0033800 -0.0002600
> [25] -0.8912000 -0.3023000 -0.560
>  
> $value
> [1] -772.3262
>  
> $counts
> function gradient
>1   NA
>  
> $convergence
> [1] 0
>  
> $message
> NULL
> > warnings()
> Warning messages:
> 1: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
> 2: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
> .
> .
>  
> 49: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
> 50: In log(det(I_N - pd * wd - po * wo - pw * ww)) : NaNs produced
>  
>  


for "SANN", the convergence criterion is not meaningful, because
SANN does not use a tolerance-based stopping criterion.  As ?optim says,
"‘0’ indicates successful completion (which is always the case for
‘"SANN"’)."

> *If I change method to "Nelder-Mead", that is:*
>  
> > p<-optim(c(-0.2392925,0.4653128,-0.8332286, 0.0657, -0.0031,
> -0.00245, 3.366, 0.5885, -0.8,
> +0.0786,-0.00292,-0.00081, 3.266, -0.3632, -0.49,   
> 0.1856, 0.00394, -0.00193, -0.889, 0.5379, -0.63,
> +0.213, 0.00338, -0.00026, -0.8912, -0.3023, -0.56),
> f, method ="Nelder-Mead", y=y,X=X,W=W)
>  
> Then I get:
>
> There were 21 warnings (use warnings() to see them)
> > p
> $par
>  [1] -0.2392925  0.4653128 -0.8332286  0.0657000 -0.0031000
> -0.0024500  3.366  0.5885000 -0.800  0.0786000 -0.0029200
> -0.0008100
> [13]  3.5184500 -0.3632000 -0.490  0.1856000  0.0039400 -0.0019300
> -0.889  0.5379000 -0.630  0.213  0.0033800 -0.0002600
> [25] -0.8912000 -0.3023000 -0.560
>  
> $value
> [1] -772.3568
>  
> $counts
> function gradient
>  192   NA
>  
> $convergence
> [1] 10
>  
> ACCORDING TO the R manual, convergence=10 indicates degeneracy of the
> Nelder–Mead simplex.  Could you explain to me what this degeneracy
> means?  Does it mean the optimization gets stuck with a local minimal?


  This means that the Nelder-Mead simplex has shrunk in such a way that
in a least one dimension the extent
of the simplex has shrunk to a point.  It doesn't have anything to do
with local minima (there's not really any
way that a single run of a local optimizer can detect a local minimum). 
You could try re-starting the optimization
from the point at which the previous run stopped.
> 2.  I also tried to change the maxit value of BFGS to 1, and got
> the following results. It seems this time the algorithm coverges, but
> the estimation results are quite different from what I got by using
> the method "SANN".  In this case, which method should I use? 
>  
> > p<-optim(c(-0.2392925,0.4653128,-0.8332286, 0.0657, -0.0031,
> -0.00245, 3.366, 0.5885, -0.8,
> +0.0786,-0.00292,-0.00081, 3.266, -0.3632, -0.49,   
> 0.1856, 0.00394, -0.00193, -0.889, 0.5379, -0.63,
> +0.213, 0.00338, -0.00026, -0.8912, -0.3023, -0.56),
> f, method ="BFGS", hessian =TRUE, control=list(maxit=1),y=y,X=X,W=W)
>
> There were 50 or more warnings (use warnings() to see the first 50)
> > p
> $par
>  [1]  1.113491e-01  6.347504e-02 -1.570647e-01  7.793766e-02 
> 7.011026e-02 -3.075866e-03  3.365178e+00  8.123945e-02 -2.670111e-04
> [10]  7.941502e-02 -2.249492e-04 -1.388776e-03  3.266022e+00
> -4.023881e-01 -6.195116e-03  1.829491e-01 -1.116388e-02 -3.088426e-03
> [19] -8.888543e-01  6.394912e-01  3.425666e-03  2.193541e-01 
> 3.743851e-02  8.376799e-05 -8.915029e-01 -5.596738e-01 -1.845092e-03
>  
> $value
> [1] -950.553
>  
> $counts
> function gradient
>31321 1741
>  
> $convergence
> [1] 0

  The BFGS result (-950) is much better than the SANN or Nelder-Mead
results (-722).

>
> 3. In Peng's email, he pointed out the importance of choosing good
> initial values in order to get sensible estimates by using optim. 
> Since I am not confident 

[R] poisson distribution

2010-09-06 Thread tamas barjak
Hello!

I need some help.
How I know it to draw the formula of the poisson distribution?

expr<-expression(P(xi == k) == frac(lambda^k, factorial(k))*e^-lambda) --->
not good

on the screen the " k! " not the Poisson Formula, but "factorial(k)"

Thanx!

[[alternative HTML version deleted]]

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[R] Failure to aggregate

2010-09-06 Thread Dimitri Shvorob

I have a (very big - 1.5 rows) dataframe with a (POSIXt"  "POSIXlt") column h
(hour). Surprisingly, I cannot calculate a simple aggregate over the
dataframe.

> n.h1 = sqldf("select distinct h, count(*) from x group by h")
Error in sqliteExecStatement(con, statement, bind.data) : 
  RS-DBI driver: (error in statement: no such table: x)
In addition: Warning message:
In value[[3L]](cond) : RAW() can only be applied to a 'raw', not a 'double'

> n.h2 = aggregate(x$price, by = x$h, FUN = nrow)
Error in names(y) <- c(names(by), names(x)) : 
  'names' attribute [10] must be the same length as the vector [2]

Arrgh...
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[R] sample a matrix with one element to be 1 from wishart distribution

2010-09-06 Thread mou sonia
Hi,

I am not sure if this make sense at all. I'd like to sample a matrix, which
follows a wishart / inverted wishart distribution. However, the (1,1)
element of this matrix should always be equal to 1. How can I handle it in
R? Any suggestion is greatly appreciated. Thanks a lot.

Sonia

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Re: [R] Creating named.list from two matrix columns

2010-09-06 Thread Viki S

Hi Jim,
Thanks,

That´s right. But the problem is that it
introduces unnecessary quotes, perhaps due to the format of first
column data in this case :

x<-cbind(c("row:1", "row:2", "row:3"), c("4889", "9987", "494"))
x1<-as.list(x[,2])
names(x1)<-x[,1]
> x1
$`row:1`
[1] "4889"

$`row:2`
[1] "9987"

$`row:3`
[1] "494"

How can I avoid unnecessary ` `quotes around the names ?

V

> Date: Mon, 6 Sep 2010 09:14:23 -0400
> Subject: Re: [R] Creating named.list from two matrix columns
> From: jholt...@gmail.com
> To: is...@live.com
> CC: r-help@r-project.org
> 
> Is this what you want:
> 
> > x
> V1   V2
> 1 row1 2334
> 2 row2  347
> 3 row3  379
> > x.list <- as.list(x$V2)
> > names(x.list) <- x$V1
> > x.list
> $row1
> [1] 2334
> 
> $row2
> [1] 347
> 
> $row3
> [1] 379
> 
> 
> 
> On Mon, Sep 6, 2010 at 7:55 AM, Viki S  wrote:
> >
> > Hi Friends,
> > I am new to R.
> >
> > On R utility class pages, creating "named.list" is described with this 
> > command :
> > new("named.list",a=1,b=2)
> >
> >
> > For large matrix having two columns, such as :
> >
> > "row1"   2334
> > "row2"   347
> > "row3"   379
> > ...
> >
> > I want to create a named.list like :
> > $row1
> > [1] 2334
> >
> > $row2
> > [1] 347
> >
> > ...
> >
> > Can anyone explain how "named.list" variable can be created by using two 
> > specified columns of a dataframe or matrix object, where one of the two 
> > columns is assigned as a name (string) and
> > other as its corresponding value ?
> >
> > Thanks
> >
> >[[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.
> >
> 
> 
> 
> -- 
> Jim Holtman
> Cincinnati, OH
> +1 513 646 9390
> 
> What is the problem that you are trying to solve?
  
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[R] c++ equivalent switch statement?

2010-09-06 Thread rajesh j
Is there a c++ equivalent switch statement in R?

-- 
Rajesh.J

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[R] Over lay 2 scale in same plot

2010-09-06 Thread mamunbabu2001

Hi Everyone,
I have two different data set in 2 different scale. 
I want to plot these two data in the same plot
in their respective scale. So the plot will have 2 different scale.
I have added an image below to show how it should look.
does any bode has any idea how this can be done. 

2 different y scale in same plot..??

http://r.789695.n4.nabble.com/file/n2528661/2scale_ovelay.jpg 

Thanks in advance.

Mamun
-- 
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[R] sample a matrix with one element to be 1 from wishart distribution

2010-09-06 Thread mou sonia
Hi,

I am not sure if this make sense at all. I'd like to sample a matrix, which
follows a wishart / inverted wishart distribution. However, the (1,1)
element of this matrix should always be equal to 1. How can I handle it in
R? Any suggestion is greatly appreciated. Thanks a lot.

Sonia

[[alternative HTML version deleted]]

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Re: [R] c++ equivalent switch statement?

2010-09-06 Thread romain
 Le 06/09/10 19:17, rajesh j a écrit :
> Is there a c++ equivalent switch statement in R?

yes, with the same name. See the "R Language definition" :
http://cran.r-project.org/doc/manuals/R-lang.html#switch

If this is not what you want, maybe you could at least share some C++ code 
snippet of what you want to reproduce in R. 

Romain

-- 
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Professional R Enthusiast
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|- http://bit.ly/bzoWrs : Rcpp svn revision 2000
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Re: [R] poisson distribution

2010-09-06 Thread David Winsemius


On Sep 6, 2010, at 1:13 PM, tamas barjak wrote:


Hello!

I need some help.
How I know it to draw the formula of the poisson distribution?

expr<-expression(P(xi == k) == frac(lambda^k, factorial(k))*e^- 
lambda) --->

not good


?plotmath

(Do not see factorial as a plotmath "function"

Try:

expr<-expression(P(xi == k) == frac(lambda^k, k*"!")*e^-lambda)



on the screen the " k! " not the Poisson Formula, but "factorial(k)"

Thanx!


-- David.

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Re: [R] poisson distribution

2010-09-06 Thread tamas barjak
Successful!

Thank you!



2010/9/6 David Winsemius 

>
> On Sep 6, 2010, at 1:13 PM, tamas barjak wrote:
>
>  Hello!
>>
>> I need some help.
>> How I know it to draw the formula of the poisson distribution?
>>
>> expr<-expression(P(xi == k) == frac(lambda^k, factorial(k))*e^-lambda)
>> --->
>> not good
>>
>
> ?plotmath
>
> (Do not see factorial as a plotmath "function"
>
> Try:
>
> expr<-expression(P(xi == k) == frac(lambda^k, k*"!")*e^-lambda)
>
>
>
>> on the screen the " k! " not the Poisson Formula, but "factorial(k)"
>>
>> Thanx!
>>
>
> -- David.
>

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[R] ERCIM'10: Submission of abstracts

2010-09-06 Thread Uwe Ligges

Dear useRs,

the deadline for submission of abstracts is approaching for ERCIM'10. 
Please upload your abstract until 2010-09-08  if you would like to give 
a presentation at our track on "Statistical Algorithms and Software" at the


3rd International Conference of the ERCIM WG on
COMPUTING & STATISTICS (ERCIM'10)
10-12 December 2010, Senate House, University of London, UK
http://www.cfe-csda.org/ercim10

Important Dates are:

Submission of abstracts:  2010-09-08
Standard registration:2010-09-14
Tutorial: 2010-12-09
Conference:   2010-12-10 to 2010-12-12

Hope to see you there,
Achim, Bettina and Uwe

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Re: [R] Failure to aggregate

2010-09-06 Thread David Winsemius


On Sep 6, 2010, at 12:15 PM, Dimitri Shvorob wrote:



I have a (very big - 1.5 rows) dataframe with a (POSIXt"  "POSIXlt")  
column h

(hour). Surprisingly, I cannot calculate a simple aggregate over the
dataframe.


n.h1 = sqldf("select distinct h, count(*) from x group by h")

Error in sqliteExecStatement(con, statement, bind.data) :
 RS-DBI driver: (error in statement: no such table: x)
In addition: Warning message:
In value[[3L]](cond) : RAW() can only be applied to a 'raw', not a  
'double'



n.h2 = aggregate(x$price, by = x$h, FUN = nrow)


A vector argument (x$price) would only have one row (at most).

nrow(c(1,2)
NULL


Error in names(y) <- c(names(by), names(x)) :
 'names' attribute [10] must be the same length as the vector [2]


Try:

tapply(x$price, by = x$h, FUN = length)

--
David.

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Re: [R] Over lay 2 scale in same plot

2010-09-06 Thread Joshua Wiley
Hi,

Looking at the picture, I think you are just talking about plotting
two datasets.  Here is an example I made up, that looks sort of like
your picture:

# make a barplot
barplot(-50:50)
# add points into the existing plot at the coordinates set by x and y
# and use a line to connect them
points(x = 1:101, y = seq(from = 30, to = -20, length.out = 101), type = "l")

Do you have some sample data you could send us of what you are trying
to plot?  We can give more specific feedback if we have some actual
data to work with.

Hope that helps,

Josh

On Mon, Sep 6, 2010 at 9:57 AM, mamunbabu2001  wrote:
>
> Hi Everyone,
> I have two different data set in 2 different scale.
> I want to plot these two data in the same plot
> in their respective scale. So the plot will have 2 different scale.
> I have added an image below to show how it should look.
> does any bode has any idea how this can be done.
>
> 2 different y scale in same plot..??
>
> http://r.789695.n4.nabble.com/file/n2528661/2scale_ovelay.jpg
>
> Thanks in advance.
>
> Mamun
> --
> View this message in context: 
> http://r.789695.n4.nabble.com/Over-lay-2-scale-in-same-plot-tp2528661p2528661.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.
>



-- 
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/

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Re: [R] Time Series

2010-09-06 Thread Gabor Grothendieck
On Mon, Sep 6, 2010 at 11:56 AM, trb1  wrote:
>
> Thank you very much for your post.
> Your answer has been very helpful.
> Is it possible to merge >2 time series?
>

zz is my posted code was formed by merging two univariate and one
multivariate series.

-- 
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[R] combining collumns for data.frames

2010-09-06 Thread Martin Hughes

Hi

This question is far less simple than the title suggests, please read 
carefully, thanks.

I have 2 sets of data, both read into R

>data1<-read.table ("1.txt", header=T, sep="\t")
>data2<-read.table ("2.txt", header=T, sep="\t")

>data1

Taxon   stage1   stage2   stage3   stage4
T1  0  0  1  1
T2  0  1  1  0
T3  0  0  0  1
T4  1  0  0  0


>data2 # this is a library file, it contains all possible values of stage 
>(Col_1) that may be contained in the data1 file (headers of each column), and 
>what they correspond to  
   # in the Col_2 ie stages 1:2 == Group1

Col_1Col_2
Stage1  Group1
Stage2  Group1
Stage3  Group2
Stage4  Group2

 I want to get R to combine the columns in data1 based on the information in 
data2 (Col_2), eg in this instance reduce the columns in data1 from 4 to 2, 
summing up the 
 values within each column of data1 to get the result below

Taxon   group1   group2

T1  0  1

T2  1  1

T3  0  1

T4  1  0

i have many datasets which have different numbers of stage eg one dataset will 
have stage1-10, another will have stage15-35 (data2, Col_2 has all possilbe 
stage values so will say what group they correspond to)

so far i can isolate the rows of data2 which contains the stages in data1 with 
this:

> data1.names<-names(data1[,-1])#take the header names 
> from data1 minus the 1st column (this is not found in the data2 library file)
> row.numbers<-match(data1.names, data2[,1]) #match the vector containing 
> the data1 column header names to those found in the library file of data2
> data2.small<-data2[row.numbers]   #reduce the data2 to 
> only include the same stages as found in the data1 file 

 from here on i dont know what to, really i wanted to just be able to change 
the header names of data1 to their corresponding name that is found in Col_2 
and then use some statement that could merge columns in data1 which were the 
same (and also sum the values at each row and dividing by their value if they 
were greater than 1 (so i only have 0 or 1 again) but i dont know how to do 
that.

Can someone help me to get the desired result  (as in the example above) that 
doe not require me to manually merge columns? ie get the example output in an 
automated way that could take any version of the data1 file (ie with different 
stage values) and using the data2 file (library file - same in each instance) 
get the output similar as in the example above?


Thanks

Martin








  
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[R] how to change the xlab name?

2010-09-06 Thread tooblue

I simply put,  plot(density(), main="", + xlab = "XXX"), it says that
I have an unexpected "=" in it.
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Re: [R] Failure to aggregate

2010-09-06 Thread Gabor Grothendieck
On Mon, Sep 6, 2010 at 12:15 PM, Dimitri Shvorob
 wrote:
>
> I have a (very big - 1.5 rows) dataframe with a (POSIXt"  "POSIXlt") column h
> (hour). Surprisingly, I cannot calculate a simple aggregate over the
> dataframe.
>
>> n.h1 = sqldf("select distinct h, count(*) from x group by h")
> Error in sqliteExecStatement(con, statement, bind.data) :
>  RS-DBI driver: (error in statement: no such table: x)
> In addition: Warning message:
> In value[[3L]](cond) : RAW() can only be applied to a 'raw', not a 'double'
>
>> n.h2 = aggregate(x$price, by = x$h, FUN = nrow)
> Error in names(y) <- c(names(by), names(x)) :
>  'names' attribute [10] must be the same length as the vector [2]


Since you are using group by you don't want "distinct".

In aggregate use x["price"] and x["h"] rather than
x$price and x$h or use a formula.

Also use nrow or length in place of NROW.

library(sqldf)
x <- data.frame(price = 1:4, h = c(1, 1, 2, 3))
sqldf("select h, count(*) from x group by h")

aggregate(x["price"], by = x["h"], FUN = NROW)
aggregate(x["price"], by = x["h"], FUN = length)
aggregate(price ~ h, x, FUN = length)







-- 
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Re: [R] how to change the xlab name?

2010-09-06 Thread Joshua Wiley
On Mon, Sep 6, 2010 at 11:07 AM, tooblue  wrote:
>
> I simply put,  plot(density(), main="", + xlab = "XXX"), it says that
> I have an unexpected "=" in it.

You just have an extra ' + ' before the xlab argument:

plot(density(rnorm(100)), main = "", xlab = "XXX")

ought to do it.

Cheers,
Josh

> --
> View this message in context: 
> http://r.789695.n4.nabble.com/how-to-change-the-xlab-name-tp2528733p2528733.html
> Sent from the R help mailing list archive at Nabble.com.
>
> __
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> and provide commented, minimal, self-contained, reproducible code.
>



-- 
Joshua Wiley
Ph.D. Student, Health Psychology
University of California, Los Angeles
http://www.joshuawiley.com/

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Re: [R] how to change the xlab name?

2010-09-06 Thread David Winsemius


On Sep 6, 2010, at 2:07 PM, tooblue wrote:



I simply put,  plot(density(), main="", + xlab = "XXX"), it  
says that

I have an unexpected "=" in it.


It may be a case of a confused parser. You have an extraneous "+" in  
there:


>  = rnorm(100)
> plot(density(), main="",  xlab = "XXX")  # "works"

If on the other hand you wanted to construct a more complex title then  
you will probably need to read the expression and bquote help pages  
and submit a more descriptive problem statement.


--
David.

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[R] likelyhood maximization problem with polr

2010-09-06 Thread blackscorpio

Dear community,

I am currently trying to fit an ordinal logistic regression model with the
polr function. I often get the same error message :

"attempt to find suitable starting values failed", for example with :

require(MASS) 
data(iris) 
polr(Species~Sepal.Length+Sepal.Width+Petal.Length+Petal.Width,iris) 

(I know the response variable Species should be nominal but I do as levels
were ordered for the example). 
I think this is a likelyhood maximization problem ; I tried to solve this by
setting the "start" option of polr to a null or a random vector by it
doesn't garantee to find "a good" solution at the end. 

Does anyone have a clue ?

Thanks a lot ! 
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[R] two questions

2010-09-06 Thread Iasonas Lamprianou

Dear friends, two questions

(1) does anyone know if there are any non-parametric equivalents of the two-way 
ANOVA in R? I have an ordinal non-normally distributed dependent variable and 
two factors (gender and city of birth). Normally, one would try a two-way 
anova, but if R has any non-parametric equivalents, that might be great. 
(2) Also, if the interaction of gender and city of birth is statistically 
significant, which post-hoc tests should I run?

Thanks

Jason 


Dr. Iasonas Lamprianou


Assistant Professor (Educational Research and Evaluation)
Department of Education Sciences
European University-Cyprus
P.O. Box 22006
1516 Nicosia
Cyprus 
Tel.: +357-22-713178
Fax: +357-22-590539


Honorary Research Fellow
Department of Education
The University of Manchester
Oxford Road, Manchester M13 9PL, UK
Tel. 0044  161 275 3485
iasonas.lampria...@manchester.ac.uk




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[R] Help with unexpected symbol errors

2010-09-06 Thread Amit Patel
Hi

I have got a long script which will not run for me as i keep getting errors :

> source("clusterfixV1_4.r")
Error in source("clusterfixV1_4.r") : 
clusterfixV1_4.r: unexpected symbol at
158: eck[k,2] <- as.numeric(1)
159:   #ClusterInfo[k,2] <- "Clustered

I have sorted all the ones i can but i am having a problem here
Can anyone tell me the cause of these problems. Its not a very short or 
straightforward script so i dont expect you to go through the whole thing but 
it 
would be great if you can give me an indication as to what I may be doing wrong.
I havent attached the data that the script uses because I'm pretty sure the 
principles i have used are right
when running the script i get the above error on line 158 and 159
I am more than happy to provide further information(e.g the dataset) if it helps

Many thanks in advance

Amit Patel


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