Re: [R] Rscript -e, Sweave and tempdir()
My first thought was that this is R 2.7.2 and you have encountered
(from the CHANGES file)
o Rscript -e (and Rterm -e) failed on Vista because the MSVCRT
function 'tmpfile' is broken on that platform.
But that couldn't possibly be the case, as the posting guide required you
to update your R before posting (and tell us lots of information I don't
see, including your OS).
On Thu, 16 Oct 2008, [EMAIL PROTECTED] wrote:
Hello R-Help
I have a question about the behind the scenes behaviour of the Rscript
-e command and particularly its interaction with Sweave and tempdir().
We are trying to deploy R as a web service to do water quality analyses
and have been writing software to call Sweave via Rscript eg:
cmd Rscript -e Sweave('sweavefile.Rnw')
One problem we have been having is that Sweave tries to write temporary
files in a temporary directory given by the tempdir() command (as per an
email from F. Leitsch appended below). This is where our problem seems
to lie.
As a simplified example, if we call
cmd Rscript -e tempdir()
from our web service we get a win32(Cannot read memory:0x0037) error
message unless we grant Read+Write permissions to the root directory,
which we are trying to avoid doing for security reasons. This suggests
Sweave(?) is trying to write to somewhere else OTHER than tempdir()
perhaps.
However if we run:
Rscirpt testdir.r (ie. sourcing a file without the -e option)
where testdir.r contains:
outpath - tempdir()
write.table(outpath,D:\\Internet\\RWebService\\Data\\dir.txt)
there appears to be no problem with the call to the tempdir() command.
Also if we put the call to Sweave within a call2sweave.r file and do
cmdRscript call2sweave.r (ie. run sweave via a file that is sourced,
rather than using the -e option)
then things go OK.
In short, something seems to be happening with the -e version of
Rscript, the tempdir() command and its invocation by Sweave.
Any ideas?
Thanks
Paul.
EARLIER EMAIL FROM F. LEISCH (AUTHOR OF SWEAVE)
___
It's not figures but text output which is collected via temporary files. The
corresponding code in Sweave is:
tmpcon - file()
sink(file=tmpcon)
err - NULL
if(options$eval) err - evalFunc(ce, options)
cat(\n) # make sure final line is complete
sink()
output - readLines(tmpcon)
close(tmpcon)
So you need to figure out where file() opens the anoymous temporary file on
your operating system. His is the same place as a call to
tempdir() shows, e.g. on my Linux box:
R tempdir()
[1] /tmp/RtmpIzi6Yv
so it tries to write under /tmp.
_
Paul Rustomji
Rivers and Estuaries
CSIRO Land and Water
GPO Box 1666
Canberra ACT 2601
ph +61 2 6246 5810
mobile 0406 375 739
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--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
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Re: [R] how to count unique observations by variables
On Oct 16, 2008, at 1:27 AM, Lijiang Guo wrote: Dear R-helpers, I have a data frame with 3 variables, each record is a unique combination of the three variables. I would like to count the number of unique values of v3 in each v1, and save it as a new variable v4 in the same data frame. e.g. df1 [v1] [v2] [v3] [1,] a C 1 [2,] b C 2 [3,] c B 3 [4,] a B 3 [5,] b A 2 [6,] c A 1 In this case, the 4th column would become (2, 1, 2, 2, 1, 2). txt - ' v1 v2 v3 + a C 1 + b C 2 + c B 3 + a B 3 + b A 2 + c A 1' df1 - read.table(textConnection(txt), header=TRUE) grps - tapply(df1$v3, df1$v1,FUN=table) # sapply(grps,length) # a b c # 2 1 2 df1$v4 - sapply(grps,length)[df1$v1] df1 v1 v2 v3 v4 1 a C 1 2 2 b C 2 1 3 c B 3 2 4 a B 3 2 5 b A 2 1 6 c A 1 2 -- David Winsemius, MD Heritage Labs Could someone tell me how to do this? regards, Lijiang -- [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Using source()
Hello, I have never used source and I am a R beginner. If I have text file that contains 1,000's of lines of code. Can I use source to bring this code into R and execute the code? Does it run the code one line at a time? Is there a best way to setup source() for maximum effieciency? After reading ?source() would I just do: source(my_Rcode.txt) Thanks, Michael [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to count unique observations by variables
How about this: df1=data.frame(v1=c(1,1,2,3,2,4,1)) df1$v2 - ave(df1$v1,df1$v1,FUN=length) df1 v1 v2 1 1 3 2 1 3 3 2 2 4 3 1 5 2 2 6 4 1 7 1 3 On Thu, Oct 16, 2008 at 1:27 PM, Lijiang Guo [EMAIL PROTECTED] wrote: Dear R-helpers, I have a data frame with 3 variables, each record is a unique combination of the three variables. I would like to count the number of unique values of v3 in each v1, and save it as a new variable v4 in the same data frame. e.g. df1 [v1] [v2] [v3] [1,] a C 1 [2,] b C 2 [3,] c B 3 [4,] a B 3 [5,] b A 2 [6,] c A 1 In this case, the 4th column would become (2, 1, 2, 2, 1, 2). Could someone tell me how to do this? regards, Lijiang -- [[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. -- HUANG Ronggui, Wincent Tel: (00852) 3442 3832 Ph.D. Candidate, CityU of HK Master of sociology, Fudan University, China Bachelor of Social Work, Fudan University, China Personal homepage: http://ronggui.huang.googlepages.com/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] qt with df1 (repost)
On Sun, 12 Oct 2008, Peter Dalgaard wrote: Prof Brian Ripley wrote: Please do RTFM, for the help says df: degrees of freedom ( 0, maybe non-integer). 'df = Inf' is allowed. For 'qt' only values of at least one are currently supported. On Sun, 12 Oct 2008, Enrico Rossi wrote: Sorry about the html-formatted message. Here it is again in plain text. Hello, The function qt returns NaN for degrees of freedom 1. For example: qt(0.5,0.5) [1] NaN Warning message: In qt(p, df, lower.tail, log.p) : NaNs produced But qt(0.5,0.5) should be 0, since the distribution is symmetric. pt(0,0.5) [1] 0.5 It actually fails with any value, as long as df1. Is this a bug, or is there some fundamental reason why this cannot be computed? Neither Well, presumably, it is using an algorithm that only works for df =1. However, qf() works fine with df1 1, so there's a fairly easy workaround. Are you referring to If X ~ t(df) then X^2 ~ F(1, df) ? That needs df2 1. So the workaround is fake_qt - function(p, df) sign(p-0.5) * sqrt(qf(2*min(p, 1-p), 1, df, lower=FALSE)) Another workaround is qt(x, df, ncp=1e-10) -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Using source()
On Thu, 16 Oct 2008, Michael Just wrote: Hello, I have never used source and I am a R beginner. Not a 'beginner' R-help poster, though. (32 posts this month so far under this name.) If I have text file that contains 1,000's of lines of code. Can I use source to bring this code into R and execute the code? Yes. But surely after you have written (or even understood) a file with 1,000's of lines of code you are no longer a 'beginner'. Does it run the code one line at a time? No. It parses the file and runs the parsed code one expression at a time. Is there a best way to setup source() for maximum effieciency? Worry about efficiency when you have to (when you are no longer a 'beginner'). Remember Jackson's Rules of Optimization (in the context of programming): 1) Don't do it. 2) (For experts only) Don't do it yet. But you can use other means to run large collections of code, e.g. make a package or use Rscript. After reading ?source() would I just do: source(my_Rcode.txt) Yes (most people use extension 'R' for R source, though). Note that you have to explicitly print() in code run this way: auto-printing is not done. Thanks, Michael [[alternative HTML version deleted]] After so many postings, please do follow the posting guide. (No HTML.) PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ 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] Statistical courses
Hi, Can anyone suggest some short term statistical courses using R that one could take? Name or Links will be much appreciated. many thanks, 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.
Re: [R] Using source()
On Thu, Oct 16, 2008 at 1:42 AM, Prof Brian Ripley [EMAIL PROTECTED] wrote: On Thu, 16 Oct 2008, Michael Just wrote: Hello, I have never used source and I am a R beginner. Not a 'beginner' R-help poster, though. (32 posts this month so far under this name.) I hope this isn't bad form on my part. My thought was beginners may need the most help and therefore post the most. I hope my name or posts won't cause fatigue. If I have text file that contains 1,000's of lines of code. Can I use source to bring this code into R and execute the code? Yes. But surely after you have written (or even understood) a file with 1,000's of lines of code you are no longer a 'beginner'. I think I am still a beginner, hence all the questions. I am pretty sure my code has redundancies and I use excel as an text editor, its the fastest way I know how (lots of dragging formulas). I have 1,000's because I have 1000's of locations in my dataset and I am running a few regressions on each. I think that this 1,000's of lines R code should not discredit my 'beginner' status. I am still just using univariate statistics and learning how to do EDA in R. Does it run the code one line at a time? No. It parses the file and runs the parsed code one expression at a time. Thank you. Is there a best way to setup source() for maximum effieciency? Worry about efficiency when you have to (when you are no longer a 'beginner'). Remember Jackson's Rules of Optimization (in the context of programming): 1) Don't do it. 2) (For experts only) Don't do it yet. Thanks, I had to look up Jackson's Rules of Optimization. I asked about efficiency to make sure that I wasn't suggesting to do something totally ridiculous. The only 'programming' I do is for R and this is partly why I have so many questions and furthermore posts. I came to R after using Excel or Systat for statistical analysis. But you can use other means to run large collections of code, e.g. make a package or use Rscript. After reading ?source() would I just do: source(my_Rcode.txt) Yes (most people use extension 'R' for R source, though). Note that you have to explicitly print() in code run this way: auto-printing is not done. Thanks for .R and print() tips. I would not have done those. Thanks, Michael [[alternative HTML version deleted]] After so many postings, please do follow the posting guide. (No HTML.) Thanks for bringing this to my attention. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Using source()
2008/10/16 Michael Just [EMAIL PROTECTED]:
On Thu, 16 Oct 2008, Michael Just wrote:
I think I am still a beginner, hence all the questions. I am pretty
sure my code has redundancies and I use excel as an text editor, its
the fastest way I know how (lots of dragging formulas). I have
1,000's because I have 1000's of locations in my dataset and I am
running a few regressions on each. I think that this 1,000's of lines
R code should not discredit my 'beginner' status. I am still just
using univariate statistics and learning how to do EDA in R.
Having 1,000s of lines of R code that do almost the same thing 1000
times is not beginner R status, it's beginner programmer status. You
should never have:
doThing(1)
doThing(2)
doThing(3)
...
for 1000 lines. Can you see what would be better?
for(i in 1:1000){doThing(i)}
Why is it better? Because I can make it go to 10,000 by adding one
character. Because I can see the whole thing on one line. Because it
just IS.
In the most beautiful programs, everything unique only happens
exactly once[1]. For example if your dataset is of size 143, you don't
have all your loops going for(i in 1:143) - you do size=143 and then
use for(i in 1:size). It makes things much more easy to maintain,
modify, and reuse.
Maybe you could post an extract from this big file so we can get the
gist of what you are doing.
Barry
[1] This may be a bit too zen for a Thursday
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[R] Request: How to draw a tree
Dear friends As a result I get an array containing certain no of rows and columns. In the resultant array first row represents first node of a tree starting from left side, second row represents second node of that tree and so on. In the example below, the resultant tree contains 6 nodes. We get the first node on L.H.S by splitting top node using variable 10, node 2 by taking start from the R.H.S node then splitting it using variables 7 and 11 respectively [ Note: 0's are not the variables ]. All the nodes are connected on this way to get the whole tree. We can easily draw it manually. Now the question is, is it possible to draw it using R. Any help in this regard is needed. Thanks [,1] [,2] [,3] [,4] [1,] 10000 [2,] 107 110 [3,] 107 110 [4,] 107 13 11 [5,] 107 13 11 [6,] 107 130 regards M. Azam [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] qt with df1 (repost)
Prof Brian Ripley wrote: On Sun, 12 Oct 2008, Peter Dalgaard wrote: Prof Brian Ripley wrote: Please do RTFM, for the help says df: degrees of freedom ( 0, maybe non-integer). 'df = Inf' is allowed. For 'qt' only values of at least one are currently supported. On Sun, 12 Oct 2008, Enrico Rossi wrote: Sorry about the html-formatted message. Here it is again in plain text. Hello, The function qt returns NaN for degrees of freedom 1. For example: qt(0.5,0.5) [1] NaN Warning message: In qt(p, df, lower.tail, log.p) : NaNs produced But qt(0.5,0.5) should be 0, since the distribution is symmetric. pt(0,0.5) [1] 0.5 It actually fails with any value, as long as df1. Is this a bug, or is there some fundamental reason why this cannot be computed? Neither Well, presumably, it is using an algorithm that only works for df =1. However, qf() works fine with df1 1, so there's a fairly easy workaround. Are you referring to If X ~ t(df) then X^2 ~ F(1, df) ? That needs df2 1. So the workaround is fake_qt - function(p, df) sign(p-0.5) * sqrt(qf(2*min(p, 1-p), 1, df, lower=FALSE)) Yes, that's what I meant. For some reason I couldn't come up with the compact expression at the time. (df1 was a typo, but not actually wrong since 1/F(1,df) ~ F(df,1)...) Another workaround is qt(x, df, ncp=1e-10) Somewhat easier, yes! -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to count unique observations by variables
On 10/16/2008 1:27 AM, Lijiang Guo wrote:
Dear R-helpers,
I have a data frame with 3 variables, each record is a unique combination of
the three variables. I would like to count the number of unique values of v3
in each v1, and save it as a new variable v4 in the same data frame.
e.g.
df1
[v1] [v2] [v3]
[1,] a C 1
[2,] b C 2
[3,] c B 3
[4,] a B 3
[5,] b A 2
[6,] c A 1
In this case, the 4th column would become (2, 1, 2, 2, 1, 2).
Could someone tell me how to do this?
df1 - data.frame(V1=rep(c('a','b','c'),2),
V2=rep(c('C','B','A'),each=2),
V3=c(1,2,3,3,2,1))
df1$V4 - with(df1, ave(V3, V1, FUN = function(x){length(unique(x))}))
df1
V1 V2 V3 V4
1 a C 1 2
2 b C 2 1
3 c B 3 2
4 a B 3 2
5 b A 2 1
6 c A 1 2
?ave
?unique
?length
regards,
Lijiang
--
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Re: [R] dynlm and lm: should they give same estimates?
Hello Werner,
this is easily clarified. The code in my book contains an error: please
replace the line:
error.lagged - error[-c(99, 100)]
with
error.lagged - error[-c(1, 100)]
I will file this in the errata section on my web-site and will correct
the relevant example in the urca and vars packages for their next
releases.
Best,
Bernhard
Hi,
I was wondering why the results from lm and dynlm are not the
same for what I think is the same model.
I have just modified example 4.2 from the Pfaff book, please
see below for the code and results.
Can anyone tell my what I am doing wrongly?
Many thanks,
Werner
set.seed(123456)
e1 - rnorm(100)
e2 - rnorm(100)
y1 - ts(cumsum(e1))
y2 - ts(0.6*y1 + e2)
lr.reg - lm(y2 ~ y1)
error - ts(residuals(lr.reg))
error.lagged - error[-c(99, 100)]
dy1 - diff(y1)
dy2 - diff(y2)
diff.dat - data.frame(embed(cbind(dy1, dy2), 2))
colnames(diff.dat) - c('dy1', 'dy2', 'dy1.1', 'dy2.1')
ecm.reg - lm(dy2 ~ error.lagged + dy1.1 + dy2.1,
data=diff.dat)
ecm.dynreg - dynlm(d(y2) ~ L(error) + L(d(y1),1) + L(d(y2),1))
summary(ecm.reg)
summary(ecm.dynreg)
summary(ecm.reg)
Call:
lm(formula = dy2 ~ error.lagged + dy1.1 + dy2.1, data = diff.dat)
Residuals:
Min 1Q Median 3Q Max
-2.9588 -0.5439 0.1370 0.7114 2.3065
Coefficients:
Estimate Std. Error t value Pr(|t|)
(Intercept) 0.003398 0.103611 0.0330.974
error.lagged -0.968796 0.158554 -6.110 2.24e-08 ***
dy1.1 0.808633 0.112042 7.217 1.35e-10 ***
dy2.1-1.058913 0.108375 -9.771 5.64e-16 ***
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 1.026 on 94 degrees of freedom
Multiple R-Squared: 0.5464, Adjusted R-squared: 0.5319
F-statistic: 37.74 on 3 and 94 DF, p-value: 4.243e-16
summary(ecm.dynreg)
Time series regression with ts data:
Start = 3, End = 100
Call:
dynlm(formula = d(y2) ~ L(error) + L(d(y1), 1) + L(d(y2), 1))
Residuals:
Min 1Q Median 3Q Max
-2.9588 -0.5439 0.1370 0.7114 2.3065
Coefficients:
Estimate Std. Error t value Pr(|t|)
(Intercept) 0.003398 0.103611 0.033 0.9739
L(error)-0.968796 0.158554 -6.110 2.24e-08 ***
L(d(y1), 1) 0.245649 0.126996 1.934 0.0561 .
L(d(y2), 1) -0.090117 0.105938 -0.851 0.3971
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 1.026 on 94 degrees of freedom
Multiple R-Squared: 0.5464, Adjusted R-squared: 0.5319
F-statistic: 37.74 on 3 and 94 DF, p-value: 4.243e-16
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[R] about granova library
Dear all Recently the granova package was launched. I installed but after when I invoked it in R it requested for other libraries. They were downloaded and install automatically. I tried to run the example syntax of granova.1w and granova.2w but two things happened: i) either a file called granova.rdb wasnt existent or ii) the GUI clashed and R shut down. Has anyone else experience this? Do the developers have an answer for this troubleshot? Im using a Windows Vista system and I have the R version 2.7.2. Cheers, Fer __ 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] One-sample test for p
Hi, i am doing a statistics course and am having trouble with an exercise where i need to determine whether my success rate at something is higher than 80%. I was successful in 29 out of 60 trials, so these were the commands i entered into R: n=60 p.hat=29/n p.0=0.8 se.0=sqrt(p.0*(1-p.0)/n) z=(p.hat-p.0)/se.0 print(z) Which returned: [1] -6.132224 1-pnorm(z) Which returned [1] 1 My problem is that i am meant to state a null and alternative hypothesis which at the moment i have stated as p0.8 (null) and p≤0.8 (alternative). As things stand, though, a p-value of 1 suggests i should reject the null hypothesis which can't be right since i am obviously successful less than 80% of the time. I am not sure where i am getting muddled. Any advice would be greatly appreciated. Thanks! -- View this message in context: http://www.nabble.com/One-sample-test-for-p-tp20010677p20010677.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] One-sample test for p
rr400 wrote: Hi, i am doing a statistics course and am having trouble with an exercise where i need to determine whether my success rate at something is higher than 80%. I was successful in 29 out of 60 trials, so these were the commands i entered into R: n=60 p.hat=29/n p.0=0.8 se.0=sqrt(p.0*(1-p.0)/n) z=(p.hat-p.0)/se.0 print(z) Which returned: [1] -6.132224 1-pnorm(z) Which returned [1] 1 My problem is that i am meant to state a null and alternative hypothesis which at the moment i have stated as p0.8 (null) and p≤0.8 (alternative). As things stand, though, a p-value of 1 suggests i should reject the null hypothesis which can't be right since i am obviously successful less than 80% of the time. I am not sure where i am getting muddled. Any advice would be greatly appreciated. Thanks! This isn't really about R, and maybe it is homework, but now that we got you in the appropriate frame of mind: (a) p values should look at this or more unfavourable events. You have arranged things so that that translates to -6.13 or _lower_. I.e. you're looking at the wrong tail. (b) Make sure you get your accept/reject logic right. You _reject_ the null when data would be _un_likely if the null hypothesis were true. (c) You might also want to play with binom.test and prop.test -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] merge/combine data
try this:
x - read.table(textConnection( user_id site_id name
+ 11 11 februari
+ 21 11 redbook
+ 31 11 tips
+ 73 6 sleep
+ 83 6monitoring
+ 93 6 alarm), header=TRUE)
closeAllConnections()
y - lapply(split(x, list(x$user_id, x$site_id), drop=TRUE), function(.data){
+ data.frame(user_id=.data$user_id[1], site_id=.data$site_id[1],
+ name=paste(as.character(.data$name), collapse=','))
+ })
do.call(rbind, y)
user_id site_id name
3.63 6 sleep,monitoring,alarm
1.11 1 11 februari,redbook,tips
On Thu, Oct 16, 2008 at 4:43 AM, Dirkheld [EMAIL PROTECTED] wrote:
Hi,
I have the following data imported from a csv file
user_id site_id name
11 11 februari
21 11 redbook
31 11 tips
73 6 sleep
83 6monitoring
93 6 alarm
Which I would like to merge/combine into
user_id site_id name
11 11 februari, redbook, tips
23 6 sleep, monitoring, alarm
3 .
So I would like to combine data from similar user_id and site_id in one line
with a merge of the 'name'.
--
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--
Jim Holtman
Cincinnati, OH
+1 513 646 9390
What is the problem that you are trying to solve?
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Re: [R] Fw: Logistic regresion - Interpreting (SENS) and (SPEC)
Gad Abraham wrote: Frank E Harrell Jr wrote: Gad Abraham wrote: This approach leaves much to be desired. I hope that its practitioners start gauging it by the mean squared error of predicted probabilities. Is the logic here is that low MSE of predicted probabilities equals a better calibrated model? What about discrimination? Perfect calibration Almost. I was addressed more the wish for the use of strategies that maximize precision while keeping bias to a minimim. implies perfect discrimination, but I often find that you can have two That doesn't follow. You can have perfect calibration in the large with no discrimination. I'm not sure I understand: if you have perfect calibration, so that you correctly assign the probability Pr(y=1|x) to each x, doesn't it follow that the x will also be ranked in correct order of probability, which is what the AUC is measuring? You can have a prediction model that assigns Pr(y=1|x) to a range of 0.45 to 0.55 such that the probabilities are perfectly accurate, but the ROC area is 0.6. competing models, the first with higher discrimination (AUC) and worse calibration, and the the second the other way round. Which one is the better model? I judge models on the basis of both discrimination (best measured with log likelihood measures, 2nd best AUC) and calibration. It's a two-dimensional issue and we don't always know how to weigh the two. For many purposes calibration is a must. In those we don't look at discrimination until calibration-in-the-small is verified at high resolution. By log likelihood measures do you mean likelihood-ratio tests? I mean generalized R^2, log-likelihood, or the adequacy index in my book. Frank -- Frank E Harrell Jr Professor and Chair School of Medicine Department of Biostatistics Vanderbilt University __ 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] apply, t-test and p-values
R 2.7.2
Windows XP
I am using apply to compute a series of Student's t-test from two matrices,
sample1 and sample2.
boo-apply(sample1,1,t.test,sample2)
I want to pick of the p-values from the tests, but can't seem to get it to
work. I have tried several methods to get the values including:
boo-apply(sample1,1,t.test$t.test,sample2)
boo-apply(sample1,1,t.test,sample2)$t.test
any suggestions?
Thanks
John
John Sorkin M.D., Ph.D.
Chief, Biostatistics and Informatics
Baltimore VA Medical Center GRECC,
University of Maryland School of Medicine Claude D. Pepper OAIC,
University of Maryland Clinical Nutrition Research Unit, and
Baltimore VA Center Stroke of Excellence
University of Maryland School of Medicine
Division of Gerontology
Baltimore VA Medical Center
10 North Greene Street
GRECC (BT/18/GR)
Baltimore, MD 21201-1524
(Phone) 410-605-7119
(Fax) 410-605-7913 (Please call phone number above prior to faxing)
[EMAIL PROTECTED]
Confidentiality Statement:
This email message, including any attachments, is for th...{{dropped:6}}
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Re: [R] apply, t-test and p-values
Dear John,
Yes. Assuming that 'sample2' is fixed, something like this should do what
you want:
# Data sets
set.seed(123)
sample1=matrix(rnorm(100),ncol=10)
sample2=rnorm(10,5,4)
# t-test p-values
apply(X,1,function(x) t.test(x,sample2)$p.value)
[1] 0.002348970 0.004733230 0.004810952 0.004907549 0.002342979 0.018748229
0.003546655 0.006084410
[9] 0.003100983 0.007007980
HTH,
Jorge
On Thu, Oct 16, 2008 at 8:01 AM, John Sorkin [EMAIL PROTECTED]wrote:
R 2.7.2
Windows XP
I am using apply to compute a series of Student's t-test from two matrices,
sample1 and sample2.
boo-apply(sample1,1,t.test,sample2)
I want to pick of the p-values from the tests, but can't seem to get it to
work. I have tried several methods to get the values including:
boo-apply(sample1,1,t.test$t.test,sample2)
boo-apply(sample1,1,t.test,sample2)$t.test
any suggestions?
Thanks
John
John Sorkin M.D., Ph.D.
Chief, Biostatistics and Informatics
Baltimore VA Medical Center GRECC,
University of Maryland School of Medicine Claude D. Pepper OAIC,
University of Maryland Clinical Nutrition Research Unit, and
Baltimore VA Center Stroke of Excellence
University of Maryland School of Medicine
Division of Gerontology
Baltimore VA Medical Center
10 North Greene Street
GRECC (BT/18/GR)
Baltimore, MD 21201-1524
(Phone) 410-605-7119
(Fax) 410-605-7913 (Please call phone number above prior to faxing)
[EMAIL PROTECTED]
Confidentiality Statement:
This email message, including any attachments, is for ...{{dropped:13}}
__
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] One-sample test for p
On 10/16/2008 7:35 AM, Peter Dalgaard wrote: rr400 wrote: Hi, i am doing a statistics course and am having trouble with an exercise where i need to determine whether my success rate at something is higher than 80%. I was successful in 29 out of 60 trials, so these were the commands i entered into R: n=60 p.hat=29/n p.0=0.8 se.0=sqrt(p.0*(1-p.0)/n) z=(p.hat-p.0)/se.0 print(z) Which returned: [1] -6.132224 1-pnorm(z) Which returned [1] 1 My problem is that i am meant to state a null and alternative hypothesis which at the moment i have stated as p0.8 (null) and p≤0.8 (alternative). As things stand, though, a p-value of 1 suggests i should reject the null hypothesis which can't be right since i am obviously successful less than 80% of the time. I am not sure where i am getting muddled. Any advice would be greatly appreciated. Thanks! This isn't really about R, and maybe it is homework, but now that we got you in the appropriate frame of mind: (a) p values should look at this or more unfavourable events. You have arranged things so that that translates to -6.13 or _lower_. I.e. you're looking at the wrong tail. I think he was looking at the right tail, since his p value was 1. Your comment (b) is the important one; comment (c) might not be allowed by his instructor, which is one reason I'm always reluctant to give advice on other people's homework problems. Duncan Murdoch (b) Make sure you get your accept/reject logic right. You _reject_ the null when data would be _un_likely if the null hypothesis were true. (c) You might also want to play with binom.test and prop.test __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] R package: autocorrelation in gamm
Dear users I am fitting a Generalized Additive Mixed Models (gamm) model to establish possible relationship between explanatory variables (water temperature, dissolved oxygen and chlorophyll) and zooplankton data collected in the inner and outer estuarine waters. I am using monthly time-series which are auto-correlated. In the case of the inner waters, I have applied satisfactoryly (by ACF and PACF) an corAR1() as follows: gamm.zoo - gamm(zoo~s(water.temperature) + s(dissolved.oxygen) + s(chlorophyll), data=datos,family=gaussian,corr=corAR1() ) But in the case of the outer estuarine waters, our data has a seasonal auto-correlation. I have tried to applied corARMA(p,q), but it is not completely satisfactory because I have an autocorrelation of order 1 and 12 as an seasonal ARIMA=(1,0,0)(1,0,0) I have tried to model it with corr=corARMA(p=12, fixed=c(NA,0,0,0,0,0,0,0,0,0,0,NA,NA))) However, I am not completely sure I am modelling what I should. I suspect the autocorrelation structure in the error term has been taken into account with corAR1/corARMA, but I cannot work it out how it is the random structure. I do not need to incorporate a random structure in my model, only need to take into account the autocorrelation term structure in the error term, but I am not sure that the model aforementioned is doing what I am trying to do. Any help would be appreciated. Thanks in advance Guille __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] One-sample test for p
Duncan Murdoch wrote: This isn't really about R, and maybe it is homework, but now that we got you in the appropriate frame of mind: (a) p values should look at this or more unfavourable events. You have arranged things so that that translates to -6.13 or _lower_. I.e. you're looking at the wrong tail. I think he was looking at the right tail, since his p value was 1. Yes, clearly, but then he needs to rearrange his definitions so that deviations go that way when data move away from the null (i.e., to look at p.0 - p.hat). Your comment (b) is the important one; comment (c) might not be allowed by his instructor, which is one reason I'm always reluctant to give advice on other people's homework problems. Not allowed to _play_? Surely not, I hope. Not allowed as the _solution_ is of course quite possible. -p Duncan Murdoch (b) Make sure you get your accept/reject logic right. You _reject_ the null when data would be _un_likely if the null hypothesis were true. (c) You might also want to play with binom.test and prop.test -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ 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] correlation
What test do I use to determine if there is a correlation between a discrete variable and a continuous variable? For example - I have water quality ratings for streams (excellent, good, fair, poor) and a corresponding nitrogen concentration for each rating. I want to know if the the ratings correlate with the concentration of nitrogen in the stream. Help? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Network meta-analysis, varConstPower in nlme - Thank you!
Excellent - now it works. Thank you, Christian! (I still seem to be unable to replicate the results as reported in the paper, but that is unrelated to R-help. Thomas, do you have any idea what I may have entered wrongly?) Best, Christian Christian Ritz wrote: Hi Christian, I believe that the argument var has changed name to weights. The following lines work for me: ... sigma - rep(sqrt(.5), nrow(lumley1)) # not nrow= lme1 - lme(Y1 ~ trt.B + trt.C + trt.D + trt.E, random = ~ 1 | trtpair, data=lumley1, weights = varConstPower(form=~sigma, fixed=list(power=1))) lme2 - lme(Y2 ~ trt.B + trt.C + trt.D + trt.E, random = ~ 1 | trtpair, data=lumley1), weights = varConstPower(form=~sigma, fixed=list(power=1))) I hope you can now make it work!? Christian __ 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] correlation between a discrete variable and a continuous variable
What test would you use to determine if a correlation exists between a discrete variable and a continuous variable? For example, I have a rating for stream water quality (excellent, good, fair, poor) and a corresponding nitrogen concentration. I want to see if there is a correlation between the water quality rating and the concentration of nitrogen in the stream. Help? __ 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] saving result of a for loop
Hi everyone,
I have dataset which I take random samples of it 5 times. each time I get
the mean for rows for each sample.
at the end I need to calculate the Average of all means for each sample and
each row. to clear it up I give an example:
say this is my dataset.
X8 X9X10X12 X13 X14 X15 X16X17X18X19 X20 X21 X22
s1 0 1 000100000000
s2 0 0 001000000010
s3 0 1 000000000010
s4 1 0 001000010000
I get a random sample and mean of row for that sample(5 times , but I just
put 2 of them here to give you an idea)
X12 X9 X10
s10 1 0
s20 0 0
s30 1 0
s40 0 0
s1s2s3s4
0.333 0.000 0.333 0.000
X10 X18 X8
s100 0
s200 0
s300 0
s401 1
s1s2s3s4
0.000 0.000 0.000 0.667
This is the code I used:
for(i in 1:5)
{
temp-sample(A3,3, replace=F)
Avg=rowMeans(temp)
show(temp)
show(Avg)
}
Now, the problem is how can I save the result for each row(s1,s2,s3,s4) so
that I can get the grand average from 5 runs?I thought about using a vector
in the for loop but it's no good, it over right so basically I only get
the means for last sample. any idea how to do it?
Thanks a lot
--
View this message in context:
http://www.nabble.com/saving-result-of-a-%22for%22-loop-tp20013519p20013519.html
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__
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] correlation
This may be incorrect, but couldn't you just assign a number to the ratings 1:4 and then scatterplot them agianst N (x-axis). Or you may want to plot boxplots of N by water quality ratings. If these are water quality ratings derived from aquatic macroinvertebrates and you have access to the actual count information (reguardless of taxanomic level) you may want to consider some ordination techniques to see if there is an ecological gradient underlying your taxa information. hope this helps and good luck Stephen On Thu, Oct 16, 2008 at 9:05 AM, kdebusk [EMAIL PROTECTED] wrote: What test do I use to determine if there is a correlation between a discrete variable and a continuous variable? For example - I have water quality ratings for streams (excellent, good, fair, poor) and a corresponding nitrogen concentration for each rating. I want to know if the the ratings correlate with the concentration of nitrogen in the stream. Help? __ 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. -- Stephen Sefick Research Scientist Southeastern Natural Sciences Academy Let's not spend our time and resources thinking about things that are so little or so large that all they really do for us is puff us up and make us feel like gods. We are mammals, and have not exhausted the annoying little problems of being mammals. -K. Mullis __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] saving result of a for loop
a - c(1:10)
b - c(.5, .6, .9, 10, .4, 3, 4, 9, 0, 11)
d - c(21:30)
z - data.frame(a,b,d)
library(fields)
results - c()
for(i in 1:(length(rownames(z))-1)){
results[i] - rdist(z[i,], z[(i+1),])
}
results.1 - data.frame(results)
f - rownames(z)
r - f[-1]
rownames(results.1) - r
colnames(results.1) - f[1]
This is a for loop that I used not too long ago defining the results
outside of the loop worked for me.
hope this helps
stephen
On Thu, Oct 16, 2008 at 9:10 AM, Alex99 [EMAIL PROTECTED] wrote:
Hi everyone,
I have dataset which I take random samples of it 5 times. each time I get
the mean for rows for each sample.
at the end I need to calculate the Average of all means for each sample and
each row. to clear it up I give an example:
say this is my dataset.
X8 X9X10X12 X13 X14 X15 X16X17X18X19 X20 X21 X22
s1 0 1 000100000000
s2 0 0 001000000010
s3 0 1 000000000010
s4 1 0 001000010000
I get a random sample and mean of row for that sample(5 times , but I just
put 2 of them here to give you an idea)
X12 X9 X10
s10 1 0
s20 0 0
s30 1 0
s40 0 0
s1s2s3s4
0.333 0.000 0.333 0.000
X10 X18 X8
s100 0
s200 0
s300 0
s401 1
s1s2s3s4
0.000 0.000 0.000 0.667
This is the code I used:
for(i in 1:5)
{
temp-sample(A3,3, replace=F)
Avg=rowMeans(temp)
show(temp)
show(Avg)
}
Now, the problem is how can I save the result for each row(s1,s2,s3,s4) so
that I can get the grand average from 5 runs?I thought about using a vector
in the for loop but it's no good, it over right so basically I only get
the means for last sample. any idea how to do it?
Thanks a lot
--
View this message in context:
http://www.nabble.com/saving-result-of-a-%22for%22-loop-tp20013519p20013519.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.
--
Stephen Sefick
Research Scientist
Southeastern Natural Sciences Academy
Let's not spend our time and resources thinking about things that are
so little or so large that all they really do for us is puff us up and
make us feel like gods. We are mammals, and have not exhausted the
annoying little problems of being mammals.
-K. Mullis
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] tune function using sigmoid and polynomial kernels
Hello, When i use tune svm with sigmoid oy polynomial kernels to optimize Cost/gamma/coef0 (sigmoid) or cost/gamma/coef0/degree (polynomial) parameters, i can't get a summary of the best parameters. instead I got this result: dummyparameter 1 0 I use the following command line (7 fold cross validation): tunesv - tune(svm, x, y, kernel=sigmoid, coef0= -1:1, gamma=10^(-6:-2), cost=10^(0:3), tunecontrol=tune.control(sampling=cross, cross=7)) When i use a similar function for radial kernels, I get the best gamma and cost parameters. Thanks for your help. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] saving result of a for loop
Dear Alex,
Is this what you want?
my=read.table(textConnection(
X8 X9 X10 X102 X110 X177 X283 X284 X286 X292 X297 X306 X308 X314
0 1 000100000000
0 0 001000000010
0 1 000000000010
1 0 00100001000
0),header=TRUE)
closeAllConnections()
rownames(my)=paste('s',1:4,sep=)
# Five samples
res=replicate(5,my[,sample(colnames(my),5)],simplify=FALSE)
# mean for each sample
Means=do.call(rbind,lapply(res,function(x) rowMeans(x)))
rownames(Means)=paste('sample',1:5,sep=)
Means
# Global mean for sample
colMeans(Means)
HTH,
Jorge
On Thu, Oct 16, 2008 at 9:10 AM, Alex99 [EMAIL PROTECTED] wrote:
Hi everyone,
I have dataset which I take random samples of it 5 times. each time I get
the mean for rows for each sample.
at the end I need to calculate the Average of all means for each sample
and
each row. to clear it up I give an example:
say this is my dataset.
X8 X9X10X12 X13 X14 X15 X16X17X18X19 X20 X21 X22
s1 0 1 000100000000
s2 0 0 001000000010
s3 0 1 000000000010
s4 1 0 001000010000
I get a random sample and mean of row for that sample(5 times , but I just
put 2 of them here to give you an idea)
X12 X9 X10
s10 1 0
s20 0 0
s30 1 0
s40 0 0
s1s2s3s4
0.333 0.000 0.333 0.000
X10 X18 X8
s100 0
s200 0
s300 0
s401 1
s1s2s3s4
0.000 0.000 0.000 0.667
This is the code I used:
for(i in 1:5)
{
temp-sample(A3,3, replace=F)
Avg=rowMeans(temp)
show(temp)
show(Avg)
}
Now, the problem is how can I save the result for each row(s1,s2,s3,s4) so
that I can get the grand average from 5 runs?I thought about using a vector
in the for loop but it's no good, it over right so basically I only get
the means for last sample. any idea how to do it?
Thanks a lot
--
View this message in context:
http://www.nabble.com/saving-result-of-a-%22for%22-loop-tp20013519p20013519.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
__
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Re: [R] apply, t-test and p-values
On Oct 16, 2008, at 8:11 AM, Jorge Ivan Velez wrote:
Dear John,
Yes. Assuming that 'sample2' is fixed, something like this should do
what
you want:
# Data sets
set.seed(123)
sample1=matrix(rnorm(100),ncol=10)
sample2=rnorm(10,5,4)
# t-test p-values
apply(X,1,function(x) t.test(x,sample2)$p.value)
It seems clear that you meant to type:
apply(sample1,1,function(x) t.test(x,sample2)$p.value)
[1] 0.002348970 0.004733230 0.004810952 0.004907549 0.002342979
0.018748229
0.003546655 0.006084410
[9] 0.003100983 0.007007980
HTH,
Jorge
On Thu, Oct 16, 2008 at 8:01 AM, John Sorkin [EMAIL PROTECTED]
wrote:
R 2.7.2
Windows XP
I am using apply to compute a series of Student's t-test from two
matrices,
sample1 and sample2.
boo-apply(sample1,1,t.test,sample2)
I want to pick of the p-values from the tests, but can't seem to
get it to
work. I have tried several methods to get the values including:
boo-apply(sample1,1,t.test$t.test,sample2)
boo-apply(sample1,1,t.test,sample2)$t.test
any suggestions?
Thanks
John
John Sorkin M.D., Ph.D.
Chief, Biostatistics and Informatics
Baltimore VA Medical Center GRECC,
University of Maryland School of Medicine Claude D. Pepper OAIC,
University of Maryland Clinical Nutrition Research Unit, and
Baltimore VA Center Stroke of Excellence
University of Maryland School of Medicine
Division of Gerontology
Baltimore VA Medical Center
10 North Greene Street
GRECC (BT/18/GR)
Baltimore, MD 21201-1524
(Phone) 410-605-7119
(Fax) 410-605-7913 (Please call phone number above prior to faxing)
[EMAIL PROTECTED]
Confidentiality Statement:
This email message, including any attachments, is for ...{{dropped:
13}}
__
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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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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] One-sample test for p
Thanks for your responses. I know it is bad form asking about these things but i was really having trouble getting my head around it, and i wanted to make sure that the cause wasn't due to the commands i was entering into R. At least now i know it's a conceptual error i am making rather than a technical one. Thanks again for your help. R. Duncan Murdoch-2 wrote: On 10/16/2008 7:35 AM, Peter Dalgaard wrote: rr400 wrote: Hi, i am doing a statistics course and am having trouble with an exercise where i need to determine whether my success rate at something is higher than 80%. I was successful in 29 out of 60 trials, so these were the commands i entered into R: n=60 p.hat=29/n p.0=0.8 se.0=sqrt(p.0*(1-p.0)/n) z=(p.hat-p.0)/se.0 print(z) Which returned: [1] -6.132224 1-pnorm(z) Which returned [1] 1 My problem is that i am meant to state a null and alternative hypothesis which at the moment i have stated as p0.8 (null) and p≤0.8 (alternative). As things stand, though, a p-value of 1 suggests i should reject the null hypothesis which can't be right since i am obviously successful less than 80% of the time. I am not sure where i am getting muddled. Any advice would be greatly appreciated. Thanks! This isn't really about R, and maybe it is homework, but now that we got you in the appropriate frame of mind: (a) p values should look at this or more unfavourable events. You have arranged things so that that translates to -6.13 or _lower_. I.e. you're looking at the wrong tail. I think he was looking at the right tail, since his p value was 1. Your comment (b) is the important one; comment (c) might not be allowed by his instructor, which is one reason I'm always reluctant to give advice on other people's homework problems. Duncan Murdoch (b) Make sure you get your accept/reject logic right. You _reject_ the null when data would be _un_likely if the null hypothesis were true. (c) You might also want to play with binom.test and prop.test __ 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. -- View this message in context: http://www.nabble.com/One-sample-test-for-p-tp20010677p20012878.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] apply, t-test and p-values
Thanks for the correction David! You're absolutely right! A typo :- )
...
Cheers,
Jorge
On Thu, Oct 16, 2008 at 10:00 AM, David Winsemius [EMAIL PROTECTED]wrote:
On Oct 16, 2008, at 8:11 AM, Jorge Ivan Velez wrote:
Dear John,
Yes. Assuming that 'sample2' is fixed, something like this should do what
you want:
# Data sets
set.seed(123)
sample1=matrix(rnorm(100),ncol=10)
sample2=rnorm(10,5,4)
# t-test p-values
apply(X,1,function(x) t.test(x,sample2)$p.value)
It seems clear that you meant to type:
apply(sample1,1,function(x) t.test(x,sample2)$p.value)
[1] 0.002348970 0.004733230 0.004810952 0.004907549 0.002342979
0.018748229
0.003546655 0.006084410
[9] 0.003100983 0.007007980
HTH,
Jorge
On Thu, Oct 16, 2008 at 8:01 AM, John Sorkin [EMAIL PROTECTED]
wrote:
R 2.7.2
Windows XP
I am using apply to compute a series of Student's t-test from two
matrices,
sample1 and sample2.
boo-apply(sample1,1,t.test,sample2)
I want to pick of the p-values from the tests, but can't seem to get it
to
work. I have tried several methods to get the values including:
boo-apply(sample1,1,t.test$t.test,sample2)
boo-apply(sample1,1,t.test,sample2)$t.test
any suggestions?
Thanks
John
John Sorkin M.D., Ph.D.
Chief, Biostatistics and Informatics
Baltimore VA Medical Center GRECC,
University of Maryland School of Medicine Claude D. Pepper OAIC,
University of Maryland Clinical Nutrition Research Unit, and
Baltimore VA Center Stroke of Excellence
University of Maryland School of Medicine
Division of Gerontology
Baltimore VA Medical Center
10 North Greene Street
GRECC (BT/18/GR)
Baltimore, MD 21201-1524
(Phone) 410-605-7119
(Fax) 410-605-7913 (Please call phone number above prior to faxing)
[EMAIL PROTECTED]
Confidentiality Statement:
This email message, including any attachments, is for ...{{dropped:13}}
__
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] NAs in color2D.matplot
bg708 wrote: Hi, I am trying to plot a large matrix by using color2D.matplot. A substantial part of the matrix has no data (NAs). However, when trying to plot it, I get an error. It is only when I change all NAs to zeros that it works, but that in some way introduces 'wrong' data. Is there any workaround this? Thank you in advance!!! Hi Bernardo, The problem is really in color.scale, and I can't really fix it properly this evening. I will try to rewrite both functions to handle this problem. With any luck, I may have it done this weekend, and it will appear sometime next week in version 2.4-8. The intended solution is to have NAs preserved in the output of color.scale. Jim __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Two last questions: about output
Please read the last line of every message to r-help and provide
reproducible code.
On Thu, Oct 16, 2008 at 12:44 AM, Ted Byers [EMAIL PROTECTED] wrote:
Thanks Gabor,
To be clear, would something like testframe$est[[i]] - fp$estimate be
valid within my loop, as in (assuming I created testframe before the
loop):
for (i in 1:length(V4) ) {
x = read.csv(as.character(V4[[i]]), header = FALSE, na.strings=);
y = x[,1];
fp = fitdistr(y,exponential);
print(c(V1[[i]],V2[[i]],V3[[i]],fp$estimate,fp$sd))
testframe$est[[i]] - fp$estimate
testframe$sd[[i]] - fp$sd
}
Thanks
Ted
On Thu, Oct 16, 2008 at 12:08 AM, Gabor Grothendieck
[EMAIL PROTECTED] wrote:
testframe$newvar - ...whatever...
(or see ?transform for another way)
adds a new column to the data frame. The table does not
have to pre-exist in your MySQL database and you don't need
a create statement; however, if the table does pre-exist the columns
of your data frame and those of the database table should have the
same names in the same order and use dbWriteTable(..., append = TRUE)
On Wed, Oct 15, 2008 at 11:54 PM, Ted Byers [EMAIL PROTECTED] wrote:
Thanks Gabor,
I get how to make a frame using existing vectors. In my example, the
following puts my first three columns into a frame (and displays it:
testframe - data.frame(mid=V1,year=V2,week=V3)
testframe
mid year week
1 251 2008 18
2 251 2008 19
3 251 2008 20
4 251 2008 22
5 251 2008 23
6 251 2008 24
7 251 2008 25
I show the first of about 60 rows, and I am pleased that these values
appear as integers.
But what I don't see is how to add the fp$estimate,fp$sd values
obtained from my analyses to vectors to form the last two columns in
the data frame. Is there something like a vector type, analogous to
the vector class std::vector from C++, that has a push_back function
allowing a vector to grow as new values are generated?
And suppose I have the following table in MySQL (ignoring for the
moment keys and indeces):
CREATE TABLE (
id INTEGER UNSIGNED NOT NULL auto_increment,
mid INTEGER NOT NULL,
y INTEGER NOT NULL,
w INTEGER NOT NULL,
rate DOUBLE NOT NULL,
sd DOUBLE NOT NULL
process_date DATETIME NOT NULL DEFAULT CURRENT_TIMESTAMP
) ENGINE=InnoDB;
How would I tell dbWriteTable() that my frame's five columns
correspond to mid,y,w,rate and sd in that order, and that the fields
id and process_date will take the appropriate default values? Or do I
need a temporary table, in memory, that has only the five columns, and
use a stored procedure to move the data to its final home?
Thanks again,
Ted
On Wed, Oct 15, 2008 at 9:57 PM, Gabor Grothendieck
[EMAIL PROTECTED] wrote:
Put the data in an R data frame and use dbWriteTable() to
write it to your MySQL database directly.
On Wed, Oct 15, 2008 at 9:34 PM, Ted Byers [EMAIL PROTECTED] wrote:
Here is my little scriptlet:
optdata =
read.csv(K:\\MerchantData\\RiskModel\\AutomatedRiskModel\\soptions.dat,
header = FALSE, na.strings=)
attach(optdata)
library(MASS)
setwd(K:\\MerchantData\\RiskModel\\AutomatedRiskModel)
for (i in 1:length(V4) ) {
x = read.csv(as.character(V4[[i]]), header = FALSE, na.strings=);
y = x[,1];
fp = fitdistr(y,exponential);
print(c(V1[[i]],V2[[i]],V3[[i]],fp$estimate,fp$sd))
}
And here are the first few lines of output:
rate rate
2.51e+02 2.008000e+03 1.80e+01 6.869301e-02 6.462095e-03
rate rate
2.51e+02 2.008000e+03 1.90e+01 5.958023e-02 4.491029e-03
rate rate
2.51e+02 2.008000e+03 2.00e+01 8.631714e-02 7.428996e-03
rate rate
2.51e+02 2.008000e+03 2.20e+01 1.261538e-01 1.137491e-02
rate rate
2.51e+02 2.008000e+03 2.30e+01 1.339523e-01 1.332875e-02
rate rate
2.51e+02 2.008000e+03 2.40e+01 8.916084e-02 1.248501e-02
There are only two things wrong, here.
1) the first three columns are integers, and are output variously as
integers, floating point numbers and, as shown here, in scientific
notation.
2) this output isn't going to a file or to my DB. This second issue isn't
much of a problem, as I think I know now how to deal with it.
This output data is, in one sense, perfectly organized, and there is a
table
with a nearly identical structure (these five columns, plus one to hold
the
date on which the analysis is performed (and of course, therefore, it has
a
default value of the current timestamp - handled in MySQL). If I can get
the data written to a CSV file, with the first three columns provided as
integers, I can use the DB's bulk load utility to get the data into the
DB,
and this may be faster
[R] package Utils Sweave Example Error
Hi: I'm still trying to figure out how use Sweave. Trying the example below I get the error message when texi2dvi is executed. Any ideas about how to make texi2dvi work? library(tools) testfile - system.file(Sweave, Sweave-test-1.Rnw, package = utils) options(device.ask.default=FALSE) Sweave(testfile) Writing to file Sweave-test-1.tex Processing code chunks ... 1 : print term verbatim 2 : term hide 3 : echo print term verbatim 4 : term verbatim 5 : echo term verbatim 6 : echo term verbatim eps pdf 7 : echo term verbatim eps pdf You can now run LaTeX on 'Sweave-test-1.tex' texi2dvi(Sweave-test-1.tex, pdf=TRUE) # Error here C:/Documents and Settings/fcarrillo/Desktop/Sweave-test-1.tex:11: Missing C:/Documents and Settings/fcarrillo/Desktop/Sweave-test-1.tex:11: C:/Documents and Settings/fcarrillo/Desktop/Sweave-test-1.tex:11: Extra C:/Documents and Settings/fcarrillo/Desktop/Sweave-test-1.tex:11: Missing C:/Documents and Settings/fcarrillo/Desktop/Sweave-test-1.tex:11: Extra C:/Documents and Settings/fcarrillo/Desktop/Sweave-test-1.tex:11: Missing C:/Documents and Settings/fcarrillo/Desktop/Sweave-test-1.tex:11: Extra C:/Documents and Settings/fcarrillo/Desktop/Sweave-test-1.tex:11: Missing C:/Documents and Settings/fcarrillo/Desktop/Sweave-test-1.tex:11: Missing C:/Documents and Settings/fcarrillo/Desktop/Sweave-test-1.tex:11: Extra C:/Documents and Settings/fcarrillo/Desktop/Sweave-test-1.tex:11: Missing C:/Documents and Settings/fcarrillo/Desktop/Sweave-test-1.tex:11: Extra Error in texi2dvi(Sweave-test-1.tex, pdf = TRUE) : running 'texi2dvi' on 'Sweave-test-1.tex' failed Felipe D. Carrillo Supervisory Fishery Biologist Department of the Interior US Fish Wildlife Service California, USA __ 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] lmer for two models followed by anova to compare the two models
Dear Colleagues, I run this model: mod1 - lmer(x~category+subcomp+category*subcomp+(1|id),data=impchiefsrm) obtain this summary result: Linear mixed-effects model fit by REML Formula: x ~ category + subcomp + category * subcomp + (1 | id) Data: impchiefsrm AIC BIC logLik MLdeviance REMLdeviance 4102 4670 -1954 3665 3908 Random effects: Groups NameVariance Std.Dev. id (Intercept) 0.11562 0.34003 Residual 0.22765 0.47713 number of obs: 2568, groups: id, 107 run this model (only difference is I've removed the interaction term): mod2 - lmer(x~category+subcomp+(1|id),data=impchiefsrm) obtain this summary result: Linear mixed-effects model fit by REML Formula: x ~ category + subcomp + (1 | id) Data: impchiefsrm AIC BIC logLik MLdeviance REMLdeviance 3987 4151 -1966 3823 3931 Random effects: Groups NameVariance Std.Dev. id (Intercept) 0.11528 0.33953 Residual 0.23584 0.48564 number of obs: 2568, groups: id, 107 Note that the loglik from the first model is -1954 and from the second model loglik is -1966. Next, to test the difference between the two models I run: anova(mod1, mod2) and obtain this result: Data: impchiefsrm Models: mod2: x ~ category + subcomp + (1 | id) mod1: x ~ category + subcomp + category * subcomp + (1 | id) Df AIC BIC logLik Chisq Chi Df Pr(Chisq) mod2 28 3879.1 4042.9 -1911.5 mod1 97 3859.3 4426.9 -1832.7 157.72 69 6.71e-09 *** --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Note that in this result the logLiks are reported as -1832.7 and -1911.5 respectively for models 1 and 2. Now my question: Why are the logLiks from the anova command that compares the two models different from what was reported in the separate model results? Thanks, Larry __ 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] merge/combine data
Hi, I have the following data imported from a csv file user_id site_id name 11 11 februari 21 11 redbook 31 11 tips 73 6 sleep 83 6monitoring 93 6 alarm Which I would like to merge/combine into user_id site_id name 11 11 februari, redbook, tips 23 6 sleep, monitoring, alarm 3 . So I would like to combine data from similar user_id and site_id in one line with a merge of the 'name'. -- View this message in context: http://www.nabble.com/merge-combine-data-tp20009538p20009538.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R: in-place appending to a matrix.
Hello Gabor,
First of all, thanks for your reply.
Indeed, your strategy solves the problem of speed=). My question was,
however, more of an R design question.
Your suggestion of pre-allocating a big and continuous portion of memory is
probably the first manual work around to try. And it will work.
Now, modern high level languages like the continually improving R, Python
and also Matlab and Perl come with built-in goodies that run at C/C++ speed.
Those built-in functions are high level versions of functionalities that we
would code manually in low level languages.
Being relatively new to R, my expectations are that in a high level language
like this, I do not need to manually program a simple operation such as
in-place appending. Manipulation of rectangular data should be extremely
efficient. (I use it to make computation of DNA properties in organisms with
huge genomes like human and mouse.)
So, do you know whether or not such functions exists or is planned to be
incorporated in the near future? An optional module perhaps?
Again, thank you,
culpritNr1
Gabor Grothendieck wrote:
Create an empty matrix first and then fill it in. That
will avoid the overhead in repeatedly expanding it. If
you don't know how many rows then make it 1000 rows
and remove the unused ones once finished.
On Wed, Oct 15, 2008 at 4:05 PM, culpritNr1 [EMAIL PROTECTED]
wrote:
Hello fellow R sufferers,
Is there a way to perform an appending operation in place?
Currently, the way my pseudo-code goes is like this
for (i in 1:1000) {
if (some condition) {
newRow - myFunction(myArguments)
X - rbind(X, newRow) # - this is the bottleneck!!
}
}
As you can see, it works but as the matrix X gets the size of a few
million
rows, the code runs very slow.
I am looking for something like the natively in place appending python
function called append() or the perl function push. In-place
operations would allow me to do (in pseudocode)
for (i in 1:1000) {
if (some condition) {
newRow - myFunction(myArguments)
append(X, newRow)
}
}
You see? I do not have to call and re-assign the giant X matrix every
loop
cycle.
Any help?
Thank you,
Your culpritNr1
--
View this message in context:
http://www.nabble.com/R%3A-%22in-place%22-appending-to-a-matrix.-tp20001258p20001258.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-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.
--
View this message in context:
http://www.nabble.com/R%3A-%22in-place%22-appending-to-a-matrix.-tp20001258p20014954.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] R: in-place appending to a matrix.
The real problem is that running a loop to build up an object bit by bit
is not the preferred way to write an R program. The preferred way
is to use lapply or other apply command in which case the object
gets built for you.
On Thu, Oct 16, 2008 at 10:50 AM, culpritNr1 [EMAIL PROTECTED] wrote:
Hello Gabor,
First of all, thanks for your reply.
Indeed, your strategy solves the problem of speed=). My question was,
however, more of an R design question.
Your suggestion of pre-allocating a big and continuous portion of memory is
probably the first manual work around to try. And it will work.
Now, modern high level languages like the continually improving R, Python
and also Matlab and Perl come with built-in goodies that run at C/C++ speed.
Those built-in functions are high level versions of functionalities that we
would code manually in low level languages.
Being relatively new to R, my expectations are that in a high level language
like this, I do not need to manually program a simple operation such as
in-place appending. Manipulation of rectangular data should be extremely
efficient. (I use it to make computation of DNA properties in organisms with
huge genomes like human and mouse.)
So, do you know whether or not such functions exists or is planned to be
incorporated in the near future? An optional module perhaps?
Again, thank you,
culpritNr1
Gabor Grothendieck wrote:
Create an empty matrix first and then fill it in. That
will avoid the overhead in repeatedly expanding it. If
you don't know how many rows then make it 1000 rows
and remove the unused ones once finished.
On Wed, Oct 15, 2008 at 4:05 PM, culpritNr1 [EMAIL PROTECTED]
wrote:
Hello fellow R sufferers,
Is there a way to perform an appending operation in place?
Currently, the way my pseudo-code goes is like this
for (i in 1:1000) {
if (some condition) {
newRow - myFunction(myArguments)
X - rbind(X, newRow) # - this is the bottleneck!!
}
}
As you can see, it works but as the matrix X gets the size of a few
million
rows, the code runs very slow.
I am looking for something like the natively in place appending python
function called append() or the perl function push. In-place
operations would allow me to do (in pseudocode)
for (i in 1:1000) {
if (some condition) {
newRow - myFunction(myArguments)
append(X, newRow)
}
}
You see? I do not have to call and re-assign the giant X matrix every
loop
cycle.
Any help?
Thank you,
Your culpritNr1
--
View this message in context:
http://www.nabble.com/R%3A-%22in-place%22-appending-to-a-matrix.-tp20001258p20001258.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-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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http://www.nabble.com/R%3A-%22in-place%22-appending-to-a-matrix.-tp20001258p20014954.html
Sent from the R help mailing list archive at Nabble.com.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] NAs in color2D.matplot
Hi, I am trying to plot a large matrix by using color2D.matplot. A substantial part of the matrix has no data (NAs). However, when trying to plot it, I get an error. It is only when I change all NAs to zeros that it works, but that in some way introduces 'wrong' data. Is there any workaround this? Thank you in advance!!! Bernardo -- View this message in context: http://www.nabble.com/NAs-in-color2D.matplot-tp20009959p20009959.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] lmer for two models followed by anova to compare the two models
Lawrence Hanser wrote: Dear Colleagues, I run this model: mod1 - lmer(x~category+subcomp+category*subcomp+(1|id),data=impchiefsrm) obtain this summary result: Linear mixed-effects model fit by REML Formula: x ~ category + subcomp + category * subcomp + (1 | id) Data: impchiefsrm AIC BIC logLik MLdeviance REMLdeviance 4102 4670 -1954 3665 3908 Random effects: Groups NameVariance Std.Dev. id (Intercept) 0.11562 0.34003 Residual 0.22765 0.47713 number of obs: 2568, groups: id, 107 run this model (only difference is I've removed the interaction term): mod2 - lmer(x~category+subcomp+(1|id),data=impchiefsrm) obtain this summary result: Linear mixed-effects model fit by REML Formula: x ~ category + subcomp + (1 | id) Data: impchiefsrm AIC BIC logLik MLdeviance REMLdeviance 3987 4151 -1966 3823 3931 Random effects: Groups NameVariance Std.Dev. id (Intercept) 0.11528 0.33953 Residual 0.23584 0.48564 number of obs: 2568, groups: id, 107 Note that the loglik from the first model is -1954 and from the second model loglik is -1966. Next, to test the difference between the two models I run: anova(mod1, mod2) and obtain this result: Data: impchiefsrm Models: mod2: x ~ category + subcomp + (1 | id) mod1: x ~ category + subcomp + category * subcomp + (1 | id) Df AIC BIC logLik Chisq Chi Df Pr(Chisq) mod2 28 3879.1 4042.9 -1911.5 mod1 97 3859.3 4426.9 -1832.7 157.72 69 6.71e-09 *** --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Note that in this result the logLiks are reported as -1832.7 and -1911.5 respectively for models 1 and 2. Now my question: Why are the logLiks from the anova command that compares the two models different from what was reported in the separate model results? REML likelihoods cannot be compared between models with different mean value structure, so it is switching to ordinary likelihood. This is arguably not all that smart, but at least it makes some sense. Compare the anova table with half the MLdeviances and you'll see the light. -- O__ Peter Dalgaard Øster Farimagsgade 5, Entr.B c/ /'_ --- Dept. of Biostatistics PO Box 2099, 1014 Cph. K (*) \(*) -- University of Copenhagen Denmark Ph: (+45) 35327918 ~~ - ([EMAIL PROTECTED]) FAX: (+45) 35327907 __ 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] histogram without bars but with density line
Hi, I know how to plot an histogram and how to add a density line. But how can I plot only the density line without the bars? Is there a way to say the hist() function not to plot bars, but a density line instead? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] saving result of a for loop
Jorge you are awesome,
This code is very efficient. it's exactly what I wanted to do and it's very
short and not complicated.
Thanks a lot
Jorge Ivan Velez wrote:
Dear Alex,
Is this what you want?
my=read.table(textConnection(
X8 X9 X10 X102 X110 X177 X283 X284 X286 X292 X297 X306 X308 X314
0 1 000100000000
0 0 001000000010
0 1 000000000010
1 0 00100001000
0),header=TRUE)
closeAllConnections()
rownames(my)=paste('s',1:4,sep=)
# Five samples
res=replicate(5,my[,sample(colnames(my),5)],simplify=FALSE)
# mean for each sample
Means=do.call(rbind,lapply(res,function(x) rowMeans(x)))
rownames(Means)=paste('sample',1:5,sep=)
Means
# Global mean for sample
colMeans(Means)
HTH,
Jorge
On Thu, Oct 16, 2008 at 9:10 AM, Alex99 [EMAIL PROTECTED] wrote:
Hi everyone,
I have dataset which I take random samples of it 5 times. each time I get
the mean for rows for each sample.
at the end I need to calculate the Average of all means for each sample
and
each row. to clear it up I give an example:
say this is my dataset.
X8 X9X10X12 X13 X14 X15 X16X17X18X19 X20 X21 X22
s1 0 1 000100000000
s2 0 0 001000000010
s3 0 1 000000000010
s4 1 0 001000010000
I get a random sample and mean of row for that sample(5 times , but I
just
put 2 of them here to give you an idea)
X12 X9 X10
s10 1 0
s20 0 0
s30 1 0
s40 0 0
s1s2s3s4
0.333 0.000 0.333 0.000
X10 X18 X8
s100 0
s200 0
s300 0
s401 1
s1s2s3s4
0.000 0.000 0.000 0.667
This is the code I used:
for(i in 1:5)
{
temp-sample(A3,3, replace=F)
Avg=rowMeans(temp)
show(temp)
show(Avg)
}
Now, the problem is how can I save the result for each row(s1,s2,s3,s4)
so
that I can get the grand average from 5 runs?I thought about using a
vector
in the for loop but it's no good, it over right so basically I only get
the means for last sample. any idea how to do it?
Thanks a lot
--
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[[alternative HTML version deleted]]
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Re: [R] histogram without bars but with density line
plot(density(x)) Regards, Yihui -- Yihui Xie [EMAIL PROTECTED] Phone: +86-(0)10-82509086 Fax: +86-(0)10-82509086 Mobile: +86-15810805877 Homepage: http://www.yihui.name School of Statistics, Room 1037, Mingde Main Building, Renmin University of China, Beijing, 100872, China On Thu, Oct 16, 2008 at 10:54 PM, Jörg Groß [EMAIL PROTECTED] wrote: Hi, I know how to plot an histogram and how to add a density line. But how can I plot only the density line without the bars? Is there a way to say the hist() function not to plot bars, but a density line instead? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] saving result of a for loop
Thanks a lot Stephen. It worked for me. also Jorge had a solution, it's a
short and easy code. U could have a look in case U want to use it in future.
Thanks again for all your help
stephen sefick wrote:
a - c(1:10)
b - c(.5, .6, .9, 10, .4, 3, 4, 9, 0, 11)
d - c(21:30)
z - data.frame(a,b,d)
library(fields)
results - c()
for(i in 1:(length(rownames(z))-1)){
results[i] - rdist(z[i,], z[(i+1),])
}
results.1 - data.frame(results)
f - rownames(z)
r - f[-1]
rownames(results.1) - r
colnames(results.1) - f[1]
This is a for loop that I used not too long ago defining the results
outside of the loop worked for me.
hope this helps
stephen
On Thu, Oct 16, 2008 at 9:10 AM, Alex99 [EMAIL PROTECTED] wrote:
Hi everyone,
I have dataset which I take random samples of it 5 times. each time I get
the mean for rows for each sample.
at the end I need to calculate the Average of all means for each sample
and
each row. to clear it up I give an example:
say this is my dataset.
X8 X9X10X12 X13 X14 X15 X16X17X18X19 X20 X21 X22
s1 0 1 000100000000
s2 0 0 001000000010
s3 0 1 000000000010
s4 1 0 001000010000
I get a random sample and mean of row for that sample(5 times , but I
just
put 2 of them here to give you an idea)
X12 X9 X10
s10 1 0
s20 0 0
s30 1 0
s40 0 0
s1s2s3s4
0.333 0.000 0.333 0.000
X10 X18 X8
s100 0
s200 0
s300 0
s401 1
s1s2s3s4
0.000 0.000 0.000 0.667
This is the code I used:
for(i in 1:5)
{
temp-sample(A3,3, replace=F)
Avg=rowMeans(temp)
show(temp)
show(Avg)
}
Now, the problem is how can I save the result for each row(s1,s2,s3,s4)
so
that I can get the grand average from 5 runs?I thought about using a
vector
in the for loop but it's no good, it over right so basically I only get
the means for last sample. any idea how to do it?
Thanks a lot
--
View this message in context:
http://www.nabble.com/saving-result-of-a-%22for%22-loop-tp20013519p20013519.html
Sent from the R help mailing list archive at Nabble.com.
__
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PLEASE do read the posting guide
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and provide commented, minimal, self-contained, reproducible code.
--
Stephen Sefick
Research Scientist
Southeastern Natural Sciences Academy
Let's not spend our time and resources thinking about things that are
so little or so large that all they really do for us is puff us up and
make us feel like gods. We are mammals, and have not exhausted the
annoying little problems of being mammals.
-K. Mullis
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[R] draw a 5cm x 3cm rectangle
Hi I want to draw sth in a pdf file with a predefined defined size. Say a 5cm x 3cm rectangle (a ruler): pdf(rect.pdf, paper=a4) plot(c(1,5,5,1,1),c(1,1,3,3,1),asp=1,axes=n) dev.off() but how do I fix that one unit is 1cm? Thanks thomas __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] correlation
On Thu, Oct 16, 2008 at 8:05 AM, kdebusk [EMAIL PROTECTED] wrote: What test do I use to determine if there is a correlation between a discrete variable and a continuous variable? For example - I have water quality ratings for streams (excellent, good, fair, poor) and a corresponding nitrogen concentration for each rating. I want to know if the the ratings correlate with the concentration of nitrogen in the stream. What does correlation mean in this example? How can you interpret a linear correlation when one of your variables is discrete? Why not start with some exploratory graphics? For your case, I would suggest drawing a histogram of nitrogen concentration for each water quality. Then look at the histograms - do they look the same? Do they look different? If they are different, what features are different? Hadley -- http://had.co.nz/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R: in-place appending to a matrix.
On 10/16/2008 10:50 AM, culpritNr1 wrote:
Hello Gabor,
First of all, thanks for your reply.
Indeed, your strategy solves the problem of speed=). My question was,
however, more of an R design question.
Your suggestion of pre-allocating a big and continuous portion of memory is
probably the first manual work around to try. And it will work.
Now, modern high level languages like the continually improving R, Python
and also Matlab and Perl come with built-in goodies that run at C/C++ speed.
Those built-in functions are high level versions of functionalities that we
would code manually in low level languages.
Being relatively new to R, my expectations are that in a high level language
like this, I do not need to manually program a simple operation such as
in-place appending. Manipulation of rectangular data should be extremely
efficient. (I use it to make computation of DNA properties in organisms with
huge genomes like human and mouse.)
I think it's unlikely that this will soon be very efficient in R,
because the only way to make it so is to pre-allocate a lot of extra
space (so the appends don't grow the memory footprint). Generally
people run into space limitations that are harder to deal with than the
speed limitations, so inefficient allocations for the sake of speed
aren't going to happen unless the user asks for them (and Gabor told you
how to do that).
Duncan Murdoch
So, do you know whether or not such functions exists or is planned to be
incorporated in the near future? An optional module perhaps?
Again, thank you,
culpritNr1
Gabor Grothendieck wrote:
Create an empty matrix first and then fill it in. That
will avoid the overhead in repeatedly expanding it. If
you don't know how many rows then make it 1000 rows
and remove the unused ones once finished.
On Wed, Oct 15, 2008 at 4:05 PM, culpritNr1 [EMAIL PROTECTED]
wrote:
Hello fellow R sufferers,
Is there a way to perform an appending operation in place?
Currently, the way my pseudo-code goes is like this
for (i in 1:1000) {
if (some condition) {
newRow - myFunction(myArguments)
X - rbind(X, newRow) # - this is the bottleneck!!
}
}
As you can see, it works but as the matrix X gets the size of a few
million
rows, the code runs very slow.
I am looking for something like the natively in place appending python
function called append() or the perl function push. In-place
operations would allow me to do (in pseudocode)
for (i in 1:1000) {
if (some condition) {
newRow - myFunction(myArguments)
append(X, newRow)
}
}
You see? I do not have to call and re-assign the giant X matrix every
loop
cycle.
Any help?
Thank you,
Your culpritNr1
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Re: [R] package Utils Sweave Example Error
On 10/16/2008 10:18 AM, Felipe Carrillo wrote: Hi: I'm still trying to figure out how use Sweave. Trying the example below I get the error message when texi2dvi is executed. Any ideas about how to make texi2dvi work? Try running it as texi2dvi(Sweave-test-1.tex, pdf=TRUE, quiet=FALSE), and post the list of error messages. You should also show us the results of sessionInfo() and Sys.getenv(PATH), and tell us which version of LaTeX you have installed. Duncan Murdoch library(tools) testfile - system.file(Sweave, Sweave-test-1.Rnw, package = utils) options(device.ask.default=FALSE) Sweave(testfile) Writing to file Sweave-test-1.tex Processing code chunks ... 1 : print term verbatim 2 : term hide 3 : echo print term verbatim 4 : term verbatim 5 : echo term verbatim 6 : echo term verbatim eps pdf 7 : echo term verbatim eps pdf You can now run LaTeX on 'Sweave-test-1.tex' texi2dvi(Sweave-test-1.tex, pdf=TRUE) # Error here C:/Documents and Settings/fcarrillo/Desktop/Sweave-test-1.tex:11: Missing C:/Documents and Settings/fcarrillo/Desktop/Sweave-test-1.tex:11: C:/Documents and Settings/fcarrillo/Desktop/Sweave-test-1.tex:11: Extra C:/Documents and Settings/fcarrillo/Desktop/Sweave-test-1.tex:11: Missing C:/Documents and Settings/fcarrillo/Desktop/Sweave-test-1.tex:11: Extra C:/Documents and Settings/fcarrillo/Desktop/Sweave-test-1.tex:11: Missing C:/Documents and Settings/fcarrillo/Desktop/Sweave-test-1.tex:11: Extra C:/Documents and Settings/fcarrillo/Desktop/Sweave-test-1.tex:11: Missing C:/Documents and Settings/fcarrillo/Desktop/Sweave-test-1.tex:11: Missing C:/Documents and Settings/fcarrillo/Desktop/Sweave-test-1.tex:11: Extra C:/Documents and Settings/fcarrillo/Desktop/Sweave-test-1.tex:11: Missing C:/Documents and Settings/fcarrillo/Desktop/Sweave-test-1.tex:11: Extra Error in texi2dvi(Sweave-test-1.tex, pdf = TRUE) : running 'texi2dvi' on 'Sweave-test-1.tex' failed Felipe D. Carrillo Supervisory Fishery Biologist Department of the Interior US Fish Wildlife Service California, USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] apply, t-test and p-values
Jorge, David,
Thank you for a brilliant solution to my problem!
Thanks
John
John David Sorkin M.D., Ph.D.
Chief, Biostatistics and Informatics
University of Maryland School of Medicine Division of Gerontology
Baltimore VA Medical Center
10 North Greene Street
GRECC (BT/18/GR)
Baltimore, MD 21201-1524
(Phone) 410-605-7119
(Fax) 410-605-7913 (Please call phone number above prior to faxing)
Jorge Ivan Velez [EMAIL PROTECTED] 10/16/2008 10:04 AM
Thanks for the correction David! You're absolutely right! A typo :- )
...
Cheers,
Jorge
On Thu, Oct 16, 2008 at 10:00 AM, David Winsemius [EMAIL PROTECTED]wrote:
On Oct 16, 2008, at 8:11 AM, Jorge Ivan Velez wrote:
Dear John,
Yes. Assuming that 'sample2' is fixed, something like this should do what
you want:
# Data sets
set.seed(123)
sample1=matrix(rnorm(100),ncol=10)
sample2=rnorm(10,5,4)
# t-test p-values
apply(X,1,function(x) t.test(x,sample2)$p.value)
It seems clear that you meant to type:
apply(sample1,1,function(x) t.test(x,sample2)$p.value)
[1] 0.002348970 0.004733230 0.004810952 0.004907549 0.002342979
0.018748229
0.003546655 0.006084410
[9] 0.003100983 0.007007980
HTH,
Jorge
On Thu, Oct 16, 2008 at 8:01 AM, John Sorkin [EMAIL PROTECTED]
wrote:
R 2.7.2
Windows XP
I am using apply to compute a series of Student's t-test from two
matrices,
sample1 and sample2.
boo-apply(sample1,1,t.test,sample2)
I want to pick of the p-values from the tests, but can't seem to get it
to
work. I have tried several methods to get the values including:
boo-apply(sample1,1,t.test$t.test,sample2)
boo-apply(sample1,1,t.test,sample2)$t.test
any suggestions?
Thanks
John
John Sorkin M.D., Ph.D.
Chief, Biostatistics and Informatics
Baltimore VA Medical Center GRECC,
University of Maryland School of Medicine Claude D. Pepper OAIC,
University of Maryland Clinical Nutrition Research Unit, and
Baltimore VA Center Stroke of Excellence
University of Maryland School of Medicine
Division of Gerontology
Baltimore VA Medical Center
10 North Greene Street
GRECC (BT/18/GR)
Baltimore, MD 21201-1524
(Phone) 410-605-7119
(Fax) 410-605-7913 (Please call phone number above prior to faxing)
[EMAIL PROTECTED]
Confidentiality Statement:
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Re: [R] plot - central limit theorem
I wonder if including the p-values for the normality test is the best approach in you animation? The clt does not say that the distribution of the means will be normal, just that it approaches normality (and therefore may be a decent approximation). The normality test can just reject the null that the data (simulated means) comes from a normal distribution. Since the true distribution of the means is not normal (unless you use a sample size of Inf, and I for one have better things to than wait for a computer to simulate several samples of size Inf) the null for the normality test is always false and therefore the test will always result in either saying it is not normal or a type II error. The real goal is not to show normality, but to show that using the normal gives a good enough approximation. I would prefer the bottom plot to show either the proportion of p-values from a normal based test on the simulated data that is less than alpha, or the proportion of confidence intervals based on the normal based test that include the true parameter. Then the user can see when those values become close enough an approximation. What is your target audience for this demo? In my opinion, anyone who could understand the bottom plot should already understand the clt enough not to need the demo, those that I would aim the demo at would just be confused by the current bottom plot. -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare [EMAIL PROTECTED] 801.408.8111 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] project.org] On Behalf Of Yihui Xie Sent: Wednesday, October 15, 2008 10:51 PM To: roger koenker Cc: r-help Subject: Re: [R] plot - central limit theorem Thanks, Roger, your demo is interesting. I'm thinking about improving it later. I've also made a demo for the CLT in my package 'animation', in which there's also normality testing for the sample means, because I don't think bell-shaped alone means normality - so I performed the Shapiro-Wilk test and plotted the P-values under the demo. See the function clt.ani() in the package 'animation', or http://animation.yihui.name/prob:central_limit_theorem You can use any function to denote the population (specify the argument 'FUN') in clt.ani(). Regards, Yihui -- Yihui Xie [EMAIL PROTECTED] Phone: +86-(0)10-82509086 Fax: +86-(0)10-82509086 Mobile: +86-15810805877 Homepage: http://www.yihui.name School of Statistics, Room 1037, Mingde Main Building, Renmin University of China, Beijing, 100872, China On Thu, Oct 16, 2008 at 4:22 AM, roger koenker [EMAIL PROTECTED] wrote: Galton's 19th century mechanical version of this is the quincunx. I have a (very primitive) version of this for R at: http://www.econ.uiuc.edu/~roger/courses/476/routines/quincunx.R url:www.econ.uiuc.edu/~rogerRoger Koenker email[EMAIL PROTECTED]Department of Economics vox: 217-333-4558University of Illinois fax: 217-244-6678Champaign, IL 61820 Jörg Groß wrote: Hi, Is there a way to simulate a population with R and pull out m samples, each with n values for calculating m means? I need that kind of data to plot a graphic, demonstrating the central limit theorem and I don't know how to begin. So, perhaps someone can give me some tips and hints how to start and which functions to use. thanks for any help, joerg __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] histogram without bars but with density line - frequency-lines?
Am 16.10.2008 um 17:12 schrieb Yihui Xie: plot(density(x)) Regards, Yihui -- Thanks! Is there also a way to plot the frequency-bars (not the density) as lines instead of bars? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] apply, t-test and p-values PREVIOUS SUGGESTION DID NOT WORK
Although I am grateful to Jorge and David for their suggestions, their solution
will not solve my problem as sample2 is not fixed. Let me demonstrate what I
want to do:
# Define two matricies
sample1-matrix(data=c(1:20),byrow=TRUE,ncol=5,nrow=4)
sample2-sample1+rnorm(20)
These are the computations that I would like to perform using apply (or some
similar method):
# Compute t-test comparing first row of sample 1 to first row of sample 2
t.test(sample1[1,],sample2[1,])
# Compute t-test comparing second row of sample 1 to second row of sample 2
t.test(sample1[2,],sample2[2,])
# Compute t-test comparing third row of sample 1 to third row of sample 2
t.test(sample1[3,],sample2[3,])
# Compute t-test comparing fourth row of sample 1 to fourth row of sample 2
t.test(sample1[4,],sample2[4,])
As you can see, the code below does not return the same values! It does not
work.
apply(sample1,1,function(x) t.test(x,sample1)$p.value)
Sugestions would be appreciated!
John
John David Sorkin M.D., Ph.D.
Chief, Biostatistics and Informatics
University of Maryland School of Medicine Division of Gerontology
Baltimore VA Medical Center
10 North Greene Street
GRECC (BT/18/GR)
Baltimore, MD 21201-1524
(Phone) 410-605-7119
(Fax) 410-605-7913 (Please call phone number above prior to faxing)
Jorge Ivan Velez [EMAIL PROTECTED] 10/16/2008 10:04 AM
Thanks for the correction David! You're absolutely right! A typo :- )
...
Cheers,
Jorge
On Thu, Oct 16, 2008 at 10:00 AM, David Winsemius [EMAIL PROTECTED]wrote:
On Oct 16, 2008, at 8:11 AM, Jorge Ivan Velez wrote:
Dear John,
Yes. Assuming that 'sample2' is fixed, something like this should do what
you want:
# Data sets
set.seed(123)
sample1=matrix(rnorm(100),ncol=10)
sample2=rnorm(10,5,4)
# t-test p-values
apply(X,1,function(x) t.test(x,sample2)$p.value)
It seems clear that you meant to type:
apply(sample1,1,function(x) t.test(x,sample2)$p.value)
[1] 0.002348970 0.004733230 0.004810952 0.004907549 0.002342979
0.018748229
0.003546655 0.006084410
[9] 0.003100983 0.007007980
HTH,
Jorge
On Thu, Oct 16, 2008 at 8:01 AM, John Sorkin [EMAIL PROTECTED]
wrote:
R 2.7.2
Windows XP
I am using apply to compute a series of Student's t-test from two
matrices,
sample1 and sample2.
boo-apply(sample1,1,t.test,sample2)
I want to pick of the p-values from the tests, but can't seem to get it
to
work. I have tried several methods to get the values including:
boo-apply(sample1,1,t.test$t.test,sample2)
boo-apply(sample1,1,t.test,sample2)$t.test
any suggestions?
Thanks
John
John Sorkin M.D., Ph.D.
Chief, Biostatistics and Informatics
Baltimore VA Medical Center GRECC,
University of Maryland School of Medicine Claude D. Pepper OAIC,
University of Maryland Clinical Nutrition Research Unit, and
Baltimore VA Center Stroke of Excellence
University of Maryland School of Medicine
Division of Gerontology
Baltimore VA Medical Center
10 North Greene Street
GRECC (BT/18/GR)
Baltimore, MD 21201-1524
(Phone) 410-605-7119
(Fax) 410-605-7913 (Please call phone number above prior to faxing)
[EMAIL PROTECTED]
Confidentiality Statement:
This email message, including any attachments, is for ...{{dropped:13}}
__
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Re: [R] One-sample test for p
I don't think that your question was bad form, there are some things you can do in the future to make it better form. Your questions was certainly better form than others who just ask for an answer to their homework without even admitting that it is homework. You were confused about some very fundamental concepts, but that in not necessarily a bad thing. Confusion correctly applied is the first step on the road to greater understanding. I would suggest that you go back to your text book and reread the information on testing and work through some more of the examples from the start (you have your hypotheses switched and the understanding of p-values backwards). For future posting on this list, those of us who are current/former teachers are hesitant to give answers to what looks like homework when we don't know what the teachers intent in giving the homework was (do they want you to work something out for yourself, and just being given the answer will not help you understand where it came from; or have you already done that and learning different ways to do the same thing will expand your knowledge). When others have posted that they have a hw problem that they have already answered by hand or using the official software of the class, they just now want to learn how to do the same thing in R, we have been happy to help. For your future posts I would suggest (beyond reading the posting guide) that you give detail on what you have already done, what the teacher has said about using software, and what your focus is in asking the question (clarifying if an unexpected result is due to misunderstanding or wrong syntax, vs. finding if there is a better way to do something, etc.) I have had many students who just get an answer, write it down, and go on without ever stopping to ask if the answer makes sense. You are ahead of them already in the quest for understanding. We don't want students to take shortcuts that bypass their understanding of something, but I for one would think it was a good thing if any students, after doing the homework the long way, found another way to do it to check their answers and expand their options. Good luck in learning statistics and R, -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare [EMAIL PROTECTED] 801.408.8111 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] project.org] On Behalf Of rr400 Sent: Thursday, October 16, 2008 6:38 AM To: r-help@r-project.org Subject: Re: [R] One-sample test for p Thanks for your responses. I know it is bad form asking about these things but i was really having trouble getting my head around it, and i wanted to make sure that the cause wasn't due to the commands i was entering into R. At least now i know it's a conceptual error i am making rather than a technical one. Thanks again for your help. R. Duncan Murdoch-2 wrote: On 10/16/2008 7:35 AM, Peter Dalgaard wrote: rr400 wrote: Hi, i am doing a statistics course and am having trouble with an exercise where i need to determine whether my success rate at something is higher than 80%. I was successful in 29 out of 60 trials, so these were the commands i entered into R: n=60 p.hat=29/n p.0=0.8 se.0=sqrt(p.0*(1-p.0)/n) z=(p.hat-p.0)/se.0 print(z) Which returned: [1] -6.132224 1-pnorm(z) Which returned [1] 1 My problem is that i am meant to state a null and alternative hypothesis which at the moment i have stated as p0.8 (null) and p≤0.8 (alternative). As things stand, though, a p-value of 1 suggests i should reject the null hypothesis which can't be right since i am obviously successful less than 80% of the time. I am not sure where i am getting muddled. Any advice would be greatly appreciated. Thanks! This isn't really about R, and maybe it is homework, but now that we got you in the appropriate frame of mind: (a) p values should look at this or more unfavourable events. You have arranged things so that that translates to -6.13 or _lower_. I.e. you're looking at the wrong tail. I think he was looking at the right tail, since his p value was 1. Your comment (b) is the important one; comment (c) might not be allowed by his instructor, which is one reason I'm always reluctant to give advice on other people's homework problems. Duncan Murdoch (b) Make sure you get your accept/reject logic right. You _reject_ the null when data would be _un_likely if the null hypothesis were true. (c) You might also want to play with binom.test and prop.test __ 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. -- View this message in context:
Re: [R] package Utils Sweave Example Error
This is my sessionInfo() and Sys.getenv(PATH)
library(tools)
testfile - system.file(Sweave, Sweave-test-1.Rnw, package = utils)
## enforce par(ask=FALSE)
options(device.ask.default=FALSE)
## create a LaTeX file
Sweave(testfile)
Writing to file Sweave-test-1.tex
Processing code chunks ...
1 : print term verbatim
2 : term hide
3 : echo print term verbatim
4 : term verbatim
5 : echo term verbatim
6 : echo term verbatim eps pdf
7 : echo term verbatim eps pdf
You can now run LaTeX on 'Sweave-test-1.tex'
## This can be compiled to PDF by
texi2dvi(Sweave-test-1.tex, pdf=TRUE,quiet=FALSE)
This is pdfTeX, Version 3.1415926-1.40.9 (MiKTeX 2.7)
entering extended mode
(C:/Program Files/R/R-2.7.2/bin/Sweave-test-1.tex
LaTeX2e 2005/12/01
Babel v3.8l and hyphenation patterns for english, dumylang, nohyphenation, ge
rman, ngerman, german-x-2008-06-18, ngerman-x-2008-06-18, french, loaded.
(C:\Program Files\MiKTeX 2.7\tex\latex\base\article.cls
Document Class: article 2005/09/16 v1.4f Standard LaTeX document class
(C:\Program Files\MiKTeX 2.7\tex\latex\base\size10.clo))
(C:\Program Files\MiKTeX 2.7\tex\latex\ltxmisc\a4wide.sty
(C:\Program Files\MiKTeX 2.7\tex\latex\ntgclass\a4.sty))
! Missing \endcsname inserted.
to be read again
\protect
l.11 \begin
{document}
? pdflatex.EXE: Bad file descriptor
texify: pdflatex failed for some reason (see log file).
Error in texi2dvi(Sweave-test-1.tex, pdf = TRUE, quiet = FALSE) :
running 'texi2dvi' on 'Sweave-test-1.tex' failed
sessionInfo()
R version 2.7.2 (2008-08-25)
i386-pc-mingw32
locale:
LC_COLLATE=English_United States.1252;LC_CTYPE=English_United
States.1252;LC_MONETARY=English_United
States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252
attached base packages:
[1] tools stats graphics grDevices utils datasets methods base
Sys.getenv(PATH)
PATH
C:\\Program Files\\MiKTeX
2.7\\miktex\\bin;C:\\GTK\\bin;C:\\WINDOWS\\system32;C:\\WINDOWS;C:\\WINDOWS\\System32\\Wbem;C:\\Program
Files\\SYSTAT 10\\;C:\\Program Files\\SYSTAT 10\\XGRAPH\\;C:\\Program
Files\\Wave Systems Corp\\Dell Preboot Manager\\Access Client\\v5\\;C:\\Program
Files\\ATI Technologies\\ATI.ACE\\Core-Static;c:\\Program Files\\Microsoft SQL
Server\\90\\Tools\\binn\\
Thank you
Felipe D. Carrillo
Supervisory Fishery Biologist
Department of the Interior
US Fish Wildlife Service
California, USA
--- On Thu, 10/16/08, Duncan Murdoch [EMAIL PROTECTED] wrote:
Try running it as texi2dvi(Sweave-test-1.tex,
pdf=TRUE, quiet=FALSE),
and post the list of error messages.
You should also show us the results of sessionInfo() and
Sys.getenv(PATH), and tell us which version of
LaTeX you have installed.
Duncan Murdoch
__
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PLEASE do read the posting guide
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reproducible code.
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] apply, t-test and p-values PREVIOUS SUGGESTION DID NOT WORK
How about the following? sapply(1:nrow(sample1), function(x) t.test(sample1[x,], sample2[x,])$p.value) Gabor On Thu, Oct 16, 2008 at 6:07 PM, John Sorkin [EMAIL PROTECTED] wrote: Although I am grateful to Jorge and David for their suggestions, their solution will not solve my problem as sample2 is not fixed. Let me demonstrate what I want to do: # Define two matricies sample1-matrix(data=c(1:20),byrow=TRUE,ncol=5,nrow=4) sample2-sample1+rnorm(20) These are the computations that I would like to perform using apply (or some similar method): # Compute t-test comparing first row of sample 1 to first row of sample 2 t.test(sample1[1,],sample2[1,]) # Compute t-test comparing second row of sample 1 to second row of sample 2 t.test(sample1[2,],sample2[2,]) # Compute t-test comparing third row of sample 1 to third row of sample 2 t.test(sample1[3,],sample2[3,]) # Compute t-test comparing fourth row of sample 1 to fourth row of sample 2 t.test(sample1[4,],sample2[4,]) As you can see, the code below does not return the same values! It does not work. apply(sample1,1,function(x) t.test(x,sample1)$p.value) Sugestions would be appreciated! John [...] -- Gabor Csardi [EMAIL PROTECTED] UNIL DGM __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] apply, t-test and p-values PREVIOUS SUGGESTION DID NOT WORK
Dear John,
Perhaps is possible to do the job using apply or any of its friends. For
now, here is a solution using for() :
# Data sets
set.seed(123)
sample1-matrix(data=c(1:20),byrow=TRUE,ncol=5,nrow=4)
sample2-sample1+rnorm(20)
# t-test p-values
k=nrow(sample1)
res=NULL
for(i in 1:k) res[i]=t.test(sample1[i,],sample2[i,])$p.value
res
Another approach could be pag 5 in [1].
HTH,
Jorge
[1]
http://www.amstat-online.org/sections/graphics/newsletter/Volumes/v181.pdf
On Thu, Oct 16, 2008 at 12:07 PM, John Sorkin
[EMAIL PROTECTED]wrote:
Although I am grateful to Jorge and David for their suggestions, their
solution will not solve my problem as sample2 is not fixed. Let me
demonstrate what I want to do:
# Define two matricies
sample1-matrix(data=c(1:20),byrow=TRUE,ncol=5,nrow=4)
sample2-sample1+rnorm(20)
These are the computations that I would like to perform using apply (or
some similar method):
# Compute t-test comparing first row of sample 1 to first row of sample 2
t.test(sample1[1,],sample2[1,])
# Compute t-test comparing second row of sample 1 to second row of sample 2
t.test(sample1[2,],sample2[2,])
# Compute t-test comparing third row of sample 1 to third row of sample 2
t.test(sample1[3,],sample2[3,])
# Compute t-test comparing fourth row of sample 1 to fourth row of sample 2
t.test(sample1[4,],sample2[4,])
As you can see, the code below does not return the same values! It does not
work.
apply(sample1,1,function(x) t.test(x,sample1)$p.value)
Sugestions would be appreciated!
John
John David Sorkin M.D., Ph.D.
Chief, Biostatistics and Informatics
University of Maryland School of Medicine Division of Gerontology
Baltimore VA Medical Center
10 North Greene Street
GRECC (BT/18/GR)
Baltimore, MD 21201-1524
(Phone) 410-605-7119
(Fax) 410-605-7913 (Please call phone number above prior to faxing)
Jorge Ivan Velez [EMAIL PROTECTED] 10/16/2008 10:04 AM
Thanks for the correction David! You're absolutely right! A typo :- )
...
Cheers,
Jorge
On Thu, Oct 16, 2008 at 10:00 AM, David Winsemius [EMAIL PROTECTED]
wrote:
On Oct 16, 2008, at 8:11 AM, Jorge Ivan Velez wrote:
Dear John,
Yes. Assuming that 'sample2' is fixed, something like this should do
what
you want:
# Data sets
set.seed(123)
sample1=matrix(rnorm(100),ncol=10)
sample2=rnorm(10,5,4)
# t-test p-values
apply(X,1,function(x) t.test(x,sample2)$p.value)
It seems clear that you meant to type:
apply(sample1,1,function(x) t.test(x,sample2)$p.value)
[1] 0.002348970 0.004733230 0.004810952 0.004907549 0.002342979
0.018748229
0.003546655 0.006084410
[9] 0.003100983 0.007007980
HTH,
Jorge
On Thu, Oct 16, 2008 at 8:01 AM, John Sorkin
[EMAIL PROTECTED]
wrote:
R 2.7.2
Windows XP
I am using apply to compute a series of Student's t-test from two
matrices,
sample1 and sample2.
boo-apply(sample1,1,t.test,sample2)
I want to pick of the p-values from the tests, but can't seem to get it
to
work. I have tried several methods to get the values including:
boo-apply(sample1,1,t.test$t.test,sample2)
boo-apply(sample1,1,t.test,sample2)$t.test
any suggestions?
Thanks
John
John Sorkin M.D., Ph.D.
Chief, Biostatistics and Informatics
Baltimore VA Medical Center GRECC,
University of Maryland School of Medicine Claude D. Pepper OAIC,
University of Maryland Clinical Nutrition Research Unit, and
Baltimore VA Center Stroke of Excellence
University of Maryland School of Medicine
Division of Gerontology
Baltimore VA Medical Center
10 North Greene Street
GRECC (BT/18/GR)
Baltimore, MD 21201-1524
(Phone) 410-605-7119
(Fax) 410-605-7913 (Please call phone number above prior to faxing)
[EMAIL PROTECTED]
Confidentiality Statement:
This email message, including any attachments, is for ...{{dropped:13}}
__
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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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Confidentiality Statement:
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] draw a 5cm x 3cm rectangle
2008/10/16 Thomas Steiner [EMAIL PROTECTED]: Hi I want to draw sth in a pdf file with a predefined defined size. Say a 5cm x 3cm rectangle (a ruler): pdf(rect.pdf, paper=a4) plot(c(1,5,5,1,1),c(1,1,3,3,1),asp=1,axes=n) dev.off() but how do I fix that one unit is 1cm? I just did help.search(inch) and found xinch: 'xinch' and 'yinch' convert the specified number of inches given as their arguments into the correct units for plotting with graphics functions. Usually, this only makes sense when normal coordinates are used, i.e., _no_ 'log' scale (see the 'log' argument to 'par'). Then help.search(cm) gave me cm: Description: Translates from inches to cm (centimeters). You just need the inverse of this. Not too hard. Also, the symbol() function takes an inches=TRUE parameter that lets you specify inches as the object size. The grid graphics package has a units system as well, but I've not played with that. See help(unit) for info. Barry __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] correlation between a discrete variable and a continuous variable
Since water quality is ordered, you could use cor.test with method of kendall or spearman. This will basically give a test of whether the rank based correlation is significantly different from 0. Often more interesting information can come from the data if you do some more work. Start by graphing your data to see what is there and if anything stands out. Then decide what you really want to learn from the data to decide what other tests to do. If you want to predict nitrogen given water quality, then you can use lm or aov (or other related methods), if you want to predict quality given nitrogen, then possibilities include proportional odds logistic regression (polr in MASS, lrm in Design) or recursive partitioning (rpart and other packages). Hope this helps, -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare [EMAIL PROTECTED] 801.408.8111 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] project.org] On Behalf Of kdebusk Sent: Thursday, October 16, 2008 6:54 AM To: r-help@r-project.org Subject: [R] correlation between a discrete variable and a continuous variable What test would you use to determine if a correlation exists between a discrete variable and a continuous variable? For example, I have a rating for stream water quality (excellent, good, fair, poor) and a corresponding nitrogen concentration. I want to see if there is a correlation between the water quality rating and the concentration of nitrogen in the stream. Help? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] counting the frequencies of a vector
Hi, Is there a function which counts the frequencies of the occurence of a number within an interval? for example I have this vector: x - c(1, 3, 1.2, 5, 5.9) and I want a vector that gives me the frequencies within an interval of 2, beginning at 0 (so the intervals are 0-2, 2-4, 4-6 and so on) so I get these frequencies: 2, 1, 2 Which functions do I have to use for this purpose? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] draw a 5cm x 3cm rectangle
Add to that list the functions grconvertX and grconvertY which can take coordinates in many different units and convert them to others (including inches, then just multiply/divide by 2.54 for the conversion from/to cm). -- Gregory (Greg) L. Snow Ph.D. Statistical Data Center Intermountain Healthcare [EMAIL PROTECTED] 801.408.8111 -Original Message- From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] project.org] On Behalf Of Barry Rowlingson Sent: Thursday, October 16, 2008 10:34 AM To: Thomas Steiner Cc: [EMAIL PROTECTED] Subject: Re: [R] draw a 5cm x 3cm rectangle 2008/10/16 Thomas Steiner [EMAIL PROTECTED]: Hi I want to draw sth in a pdf file with a predefined defined size. Say a 5cm x 3cm rectangle (a ruler): pdf(rect.pdf, paper=a4) plot(c(1,5,5,1,1),c(1,1,3,3,1),asp=1,axes=n) dev.off() but how do I fix that one unit is 1cm? I just did help.search(inch) and found xinch: 'xinch' and 'yinch' convert the specified number of inches given as their arguments into the correct units for plotting with graphics functions. Usually, this only makes sense when normal coordinates are used, i.e., _no_ 'log' scale (see the 'log' argument to 'par'). Then help.search(cm) gave me cm: Description: Translates from inches to cm (centimeters). You just need the inverse of this. Not too hard. Also, the symbol() function takes an inches=TRUE parameter that lets you specify inches as the object size. The grid graphics package has a units system as well, but I've not played with that. See help(unit) for info. Barry __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] counting the frequencies of a vector
2008/10/16 Jörg Groß [EMAIL PROTECTED]: Hi, Is there a function which counts the frequencies of the occurence of a number within an interval? for example I have this vector: x - c(1, 3, 1.2, 5, 5.9) and I want a vector that gives me the frequencies within an interval of 2, beginning at 0 (so the intervals are 0-2, 2-4, 4-6 and so on) so I get these frequencies: 2, 1, 2 Which functions do I have to use for this purpose? 'cut' cuts your vector into chunks, and 'table' counts the number in each chunk: x [1] 1.0 3.0 1.2 5.0 5.9 table(cut(x,c(0,2,4,6))) (0,2] (2,4] (4,6] 2 1 2 Barry __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Grouped Histogram (colored)
Hi all, I'm trying to create a histogram which shows the frequency of variables within a certain timeframe. I've been using SPSS before, but I didn't quite like it... To describe my problem further here are some example variables: the event is actually a string which I recoded using: [code] data$event_class = as.numeric(as.factor(data$event)) [/code] I've recoded them into numerics csv: [code] time,event,event_class 01,cookies,1 05,cookies,1 06,pie,2 07,coffee,3 08,cookies,1 30,pie,2 31,coffee,3 [/code] and so on... Now I'd like to create a histogram where X is the time, the color of the area is the event_class and Y is defined by the frequency of event_class around some accumulated time In SPSS I used this: Graphs - Chart Builder Gallery-Histogram use some horizontal histogram put time on the x-axis select grouping/stacking variables in the groups/point id tab and then set Stack: set-color to event_class the y-axis will be automatically set to histogram thanks a lot in advance! Regards, - x0rr0x -- View this message in context: http://www.nabble.com/Grouped-Histogram-%28colored%29-tp20015941p20015941.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] draw a 5cm x 3cm rectangle
I would personally use grid for this. library(grid) grid.rect(width = 5, height = 3, default.units = cm) Barry Rowlingson wrote: 2008/10/16 Thomas Steiner [EMAIL PROTECTED]: Hi I want to draw sth in a pdf file with a predefined defined size. Say a 5cm x 3cm rectangle (a ruler): pdf(rect.pdf, paper=a4) plot(c(1,5,5,1,1),c(1,1,3,3,1),asp=1,axes=n) dev.off() but how do I fix that one unit is 1cm? I just did help.search(inch) and found xinch: 'xinch' and 'yinch' convert the specified number of inches given as their arguments into the correct units for plotting with graphics functions. Usually, this only makes sense when normal coordinates are used, i.e., _no_ 'log' scale (see the 'log' argument to 'par'). Then help.search(cm) gave me cm: Description: Translates from inches to cm (centimeters). You just need the inverse of this. Not too hard. Also, the symbol() function takes an inches=TRUE parameter that lets you specify inches as the object size. The grid graphics package has a units system as well, but I've not played with that. See help(unit) for info. Barry __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Grouped Histogram (colored)
On Thu, Oct 16, 2008 at 11:42 AM, x0rr0x [EMAIL PROTECTED] wrote: Hi all, I'm trying to create a histogram which shows the frequency of variables within a certain timeframe. I've been using SPSS before, but I didn't quite like it... To describe my problem further here are some example variables: the event is actually a string which I recoded using: [code] data$event_class = as.numeric(as.factor(data$event)) [/code] I've recoded them into numerics csv: [code] time,event,event_class 01,cookies,1 05,cookies,1 06,pie,2 07,coffee,3 08,cookies,1 30,pie,2 31,coffee,3 [/code] and so on... Now I'd like to create a histogram where X is the time, the color of the area is the event_class and Y is defined by the frequency of event_class around some accumulated time install.packages(ggplot2) library(ggplot2) qplot(time, fill = event, data = mydata, geom = histogram) You can find out more about ggplot2 at http://had.co.nz/ggplot2 - it's inspired by the Grammar of Graphics, which is also the theory that underlies SPSS's plotting systems. Hadley -- http://had.co.nz/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Grouped Histogram (colored)
Does this do what you want?
colhist - function(x,col){
tmp - hist(x,plot=F)
br - tmp$breaks
w - as.numeric(cut(x,br,include.lowest=TRUE))
sy - unlist(lapply(tmp$counts,function(x)seq(length=x)))
sy - sy[order(order(x))]
plot( range(br), range( 0, sy ), xlab=deparse(substitute(x)),
ylab='Frequency', type='n')
rect(br[w], sy-1, br[w+1], sy,
col=col,
border=NA)
rect(br[-length(br)], 0, br[-1], tmp$counts)
}
x - rnorm(75, rep( c(90,100,110), each=25), 5 )
g - rep( c('red','green','blue'), each=25 )
colhist(x,g)
note: this colhist function is a modified version of the one from the help file
for the tkBrush function in the TeachingDemos package.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
801.408.8111
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
project.org] On Behalf Of x0rr0x
Sent: Thursday, October 16, 2008 10:42 AM
To: r-help@r-project.org
Subject: [R] Grouped Histogram (colored)
Hi all,
I'm trying to create a histogram which shows the frequency of variables
within a certain timeframe.
I've been using SPSS before, but I didn't quite like it...
To describe my problem further here are some example variables:
the event is actually a string which I recoded using:
[code]
data$event_class = as.numeric(as.factor(data$event))
[/code]
I've recoded them into numerics
csv:
[code]
time,event,event_class
01,cookies,1
05,cookies,1
06,pie,2
07,coffee,3
08,cookies,1
30,pie,2
31,coffee,3
[/code]
and so on...
Now I'd like to create a histogram where X is the time, the color of
the
area is the event_class
and Y is defined by the frequency of event_class around some
accumulated
time
In SPSS I used this:
Graphs - Chart Builder
Gallery-Histogram
use some horizontal histogram
put time on the x-axis
select grouping/stacking variables in the groups/point id tab
and then set Stack: set-color to event_class
the y-axis will be automatically set to histogram
thanks a lot in advance!
Regards,
- x0rr0x
--
View this message in context: http://www.nabble.com/Grouped-Histogram-
%28colored%29-tp20015941p20015941.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-
guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] increase memory limit in R
AFAICS, you have a 2GB address space: please do read the rw-FAQ and fix that, in which case R will automatically make use of the increased space. Hint: what is /3GB for? On Thu, 16 Oct 2008, liujb wrote: Dear R users, I am running lmer() and having memory problem: reached total allocation of 1535Mb. The lmer() function was successful (no errors or warnings). However when I do summary(fit.lmer), I got this Error: cannot allocate vector of size 19.5 Mb). I used memory.size() and got 1880. I used memory.limit(2500) to increase the memory size to 2500Mb, however I still get the warning messages as above. I entered memory.limit() to check whether the size has been increased or not, and R returns 2500. I am running the program on Windows XP Professional, CPU 2.00GHz and 3.3 GB of RAM. I appreciate your advices. Thank you. Julia -- View this message in context: http://www.nabble.com/increase-memory-limit-in-R-tp20015931p20015931.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. -- Brian D. Ripley, [EMAIL PROTECTED] Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ 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] rpart display
Hi, is there a way to replace long splits with some short naming conventions? For example if my resulting tree from rpart has 50-100 values can I display some summary text like group1 instead of group num = 1 or 2 or 3100? is there a way to specify different display text for nodes instead of the actual split conditions? thanks Dhruv [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to save/load RWeka models into/from a file?
Hi, I want to save a RWeka model into a file, in order to retrive it latter with a load function. See this example: library(RWeka) NB - make_Weka_classifier(weka/classifiers/bayes/NaiveBayes) model-NB(formula,data=data,...) # does not run but you get the idea save(model,file=model.dat) # simple save R command # ... load(model.dat) # load the model from the previously saved file... model # should work but I get instead this error: Error in .jcall(x$classifier, S, toString) : RcallMethod: attempt to call a method of a NULL object. What is wrong and how can I solve this problem? Regards, -- Paulo Alexandre Ribeiro Cortez (PhD, MSc) Lecturer (Prof. Auxiliar) at the Department of Information Systems (DSI) University of Minho, Campus de AzurÈm, 4800-058 Guimaraes, Portugal http://www.dsi.uminho.pt/~pcortez +351253510313 Fax:+351253510300 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to save/load RWeka models into/from a file?
On Thu, 16 Oct 2008, Paulo Cortez wrote: Hi, I want to save a RWeka model into a file, in order to retrive it latter with a load function. See this example: library(RWeka) NB - make_Weka_classifier(weka/classifiers/bayes/NaiveBayes) model-NB(formula,data=data,...) # does not run but you get the idea save(model,file=model.dat) # simple save R command # ... load(model.dat) # load the model from the previously saved file... model # should work but I get instead this error: Error in .jcall(x$classifier, S, toString) : RcallMethod: attempt to call a method of a NULL object. What is wrong and how can I solve this problem? The R object is just a reference to the corresponding object on the Java side (in Weka). When you close R, the Java/Weka session is also closed and the model is gone. Thus, when you load the object again in a new session, you only have a reference to a Java object that does not live anymore. Z Regards, -- Paulo Alexandre Ribeiro Cortez (PhD, MSc) Lecturer (Prof. Auxiliar) at the Department of Information Systems (DSI) University of Minho, Campus de AzurÈm, 4800-058 Guimaraes, Portugal http://www.dsi.uminho.pt/~pcortez +351253510313 Fax:+351253510300 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] plot - central limit theorem *solved*
Thanks for your great help!
I have now what I wanted.
For sure it's not well written and you can realize it with much less
lines.
But it works and it's like how I wanted it to look like;
z1 - rnorm(4500, mean=20, sd=5)
z2 - rnorm(3600, mean=28, sd=5)
z3 - rnorm(1300, mean=40, sd=7)
z4 - rnorm(1200, mean=10, sd=7)
z5 - rnorm(1500, mean=50, sd=5)
z6 - rnorm(1500, mean=10, sd=5)
z7 - rnorm(1400, mean=37, sd=3)
pop - c(z1, z2, z3, z4, z5, z6, z7)
meansN2 - numeric()
meansN10 - numeric()
meansN40 - numeric()
for (i in 1:1000) {
x - runif(2, min=1, max=15000)
x - round(x)
sample - c(pop[x])
meansN2 - append(meansN2, mean(sample))
}
for (i in 1:1000) {
x - runif(10, min=1, max=15000)
x - round(x)
sample - c(pop[x])
meansN10 - append(meansN10, mean(sample))
}
for (i in 1:1000) {
x - runif(40, min=1, max=15000)
x - round(x)
sample - c(pop[x])
meansN40 - append(meansN40, mean(sample))
}
pop - table(cut(pop,breaks=seq(from=-10, to=70, length.out=80)))
meansN2 - table(cut(meansN2,breaks=seq(from=-10, to=70,
length.out=80)))
meansN10 - table(cut(meansN10,breaks=seq(from=-10, to=70,
length.out=80)))
meansN40 - table(cut(meansN40,breaks=seq(from=-10, to=70,
length.out=80)))
pop - smooth(pop)
meansN2 - smooth(meansN2)
meansN10 - smooth(meansN10)
meansN40 - smooth(meansN40)
plot(pop, type=l, lwd=1.5)
lines(meansN2, lty=dotted, lwd=1.5)
lines(meansN10, lty=dashed, lwd=1.5)
lines(meansN40, lty=twodash, lwd=1.5)
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] package Utils Sweave Example Error
On 10/16/2008 12:19 PM, Felipe Carrillo wrote:
This is my sessionInfo() and Sys.getenv(PATH)
library(tools)
testfile - system.file(Sweave, Sweave-test-1.Rnw, package = utils)
## enforce par(ask=FALSE)
options(device.ask.default=FALSE)
## create a LaTeX file
Sweave(testfile)
Writing to file Sweave-test-1.tex
Processing code chunks ...
1 : print term verbatim
2 : term hide
3 : echo print term verbatim
4 : term verbatim
5 : echo term verbatim
6 : echo term verbatim eps pdf
7 : echo term verbatim eps pdf
You can now run LaTeX on 'Sweave-test-1.tex'
## This can be compiled to PDF by
texi2dvi(Sweave-test-1.tex, pdf=TRUE,quiet=FALSE)
This is pdfTeX, Version 3.1415926-1.40.9 (MiKTeX 2.7)
entering extended mode
(C:/Program Files/R/R-2.7.2/bin/Sweave-test-1.tex
LaTeX2e 2005/12/01
Babel v3.8l and hyphenation patterns for english, dumylang, nohyphenation, ge
rman, ngerman, german-x-2008-06-18, ngerman-x-2008-06-18, french, loaded.
(C:\Program Files\MiKTeX 2.7\tex\latex\base\article.cls
Document Class: article 2005/09/16 v1.4f Standard LaTeX document class
(C:\Program Files\MiKTeX 2.7\tex\latex\base\size10.clo))
(C:\Program Files\MiKTeX 2.7\tex\latex\ltxmisc\a4wide.sty
(C:\Program Files\MiKTeX 2.7\tex\latex\ntgclass\a4.sty))
! Missing \endcsname inserted.
to be read again
\protect
l.11 \begin
{document}
? pdflatex.EXE: Bad file descriptor
This appears to be the problem. I'm not certain, but I would guess if
you look in the C:/Program Files/R/R-2.7.2/bin/Sweave-test-1.tex file
you'll see a line something like this:
\usepackage{C:/Program Files/R/R-2.7.2/share/texmf/Sweave}
That line (which was added by Sweave) will not work, because LaTeX does
not understand blanks in file paths. There are a couple of workarounds:
1. Install R in a directory that doesn't contain blanks.
2. Include \usepackage{Sweave} in the Rnw file you're processing; this
will stop Sweave from adding anything. I believe we set things up so
that MikTeX will find it in the right place (though I don't normally use
texi2dvi, so I'm not sure).
To test this, you can just go and edit the Sweave-test-1.tex file now,
replacing the long line with the short version.
Duncan Murdoch
texify: pdflatex failed for some reason (see log file).
Error in texi2dvi(Sweave-test-1.tex, pdf = TRUE, quiet = FALSE) :
running 'texi2dvi' on 'Sweave-test-1.tex' failed
sessionInfo()
R version 2.7.2 (2008-08-25)
i386-pc-mingw32
locale:
LC_COLLATE=English_United States.1252;LC_CTYPE=English_United
States.1252;LC_MONETARY=English_United
States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252
attached base packages:
[1] tools stats graphics grDevices utils datasets methods base
Sys.getenv(PATH)
PATH
C:\\Program Files\\MiKTeX
2.7\\miktex\\bin;C:\\GTK\\bin;C:\\WINDOWS\\system32;C:\\WINDOWS;C:\\WINDOWS\\System32\\Wbem;C:\\Program
Files\\SYSTAT 10\\;C:\\Program Files\\SYSTAT 10\\XGRAPH\\;C:\\Program Files\\Wave
Systems Corp\\Dell Preboot Manager\\Access Client\\v5\\;C:\\Program Files\\ATI
Technologies\\ATI.ACE\\Core-Static;c:\\Program Files\\Microsoft SQL
Server\\90\\Tools\\binn\\
Thank you
Felipe D. Carrillo
Supervisory Fishery Biologist
Department of the Interior
US Fish Wildlife Service
California, USA
--- On Thu, 10/16/08, Duncan Murdoch [EMAIL PROTECTED] wrote:
Try running it as texi2dvi(Sweave-test-1.tex,
pdf=TRUE, quiet=FALSE),
and post the list of error messages.
You should also show us the results of sessionInfo() and
Sys.getenv(PATH), and tell us which version of
LaTeX you have installed.
Duncan Murdoch
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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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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] counting the frequencies of a vector
On Oct 16, 2008, at 12:55 PM, Jorge Ivan Velez wrote: Dear Jörg, See ?cut and ?table. Is this what you want? x - c(1, 3, 1.2, 5, 5.9) table(cut(x,breaks=c(0,2,4,6))) (0,2] (2,4] (4,6] 2 1 2 Perhaps even greater future efficiency could be had by also adding ?seq table(cut(x, breaks=seq(0, 6, by=2))) A couple more keysrokes in this toy example but scales up much better to the requested purpose of evenly divided interval. The OP should also note carefully the open aspect on the left of these intervals. That can be controlled with option in cut: x - c(1, 3, 1.2, 5, 5.9, 6) table(cut(x, breaks=seq(0,6,by=2))) (0,2] (2,4] (4,6] 2 1 3 x - c(1, 3, 1.2, 4, 5, 5.9, 6) table(cut(x, breaks=seq(0,6,by=2))) (0,2] (2,4] (4,6] 2 2 3 table(cut(x, breaks=seq(0,6,by=2), right=FALSE)) [0,2) [2,4) [4,6) 2 1 3 -- David Winsemius Heritage Labs HTH, Jorge On Thu, Oct 16, 2008 at 12:46 PM, Jörg Groß [EMAIL PROTECTED] wrote: Hi, Is there a function which counts the frequencies of the occurence of a number within an interval? for example I have this vector: x - c(1, 3, 1.2, 5, 5.9) and I want a vector that gives me the frequencies within an interval of 2, beginning at 0 (so the intervals are 0-2, 2-4, 4-6 and so on) so I get these frequencies: 2, 1, 2 Which functions do I have to use for this purpose? __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ 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] Sweave working now!!!
Duncan:
I installed R in C:\R\R-2.7.2 and worked beautifully. I hope this will help
others with the same problem. Thanks a lot for your help.
--- On Thu, 10/16/08, Duncan Murdoch [EMAIL PROTECTED] wrote:
On 10/16/2008 12:19 PM, Felipe Carrillo wrote:
This is my sessionInfo() and
Sys.getenv(PATH)
library(tools)
testfile - system.file(Sweave,
Sweave-test-1.Rnw, package = utils)
## enforce par(ask=FALSE)
options(device.ask.default=FALSE)
## create a LaTeX file
Sweave(testfile)
Writing to file Sweave-test-1.tex
Processing code chunks ...
1 : print term verbatim
2 : term hide
3 : echo print term verbatim
4 : term verbatim
5 : echo term verbatim
6 : echo term verbatim eps pdf
7 : echo term verbatim eps pdf
You can now run LaTeX on 'Sweave-test-1.tex'
## This can be compiled to PDF by
texi2dvi(Sweave-test-1.tex,
pdf=TRUE,quiet=FALSE)
This is pdfTeX, Version 3.1415926-1.40.9 (MiKTeX 2.7)
entering extended mode
(C:/Program
Files/R/R-2.7.2/bin/Sweave-test-1.tex
LaTeX2e 2005/12/01
Babel v3.8l and hyphenation patterns for
english, dumylang, nohyphenation, ge
rman, ngerman, german-x-2008-06-18,
ngerman-x-2008-06-18, french, loaded.
(C:\Program Files\MiKTeX
2.7\tex\latex\base\article.cls
Document Class: article 2005/09/16 v1.4f Standard
LaTeX document class
(C:\Program Files\MiKTeX
2.7\tex\latex\base\size10.clo))
(C:\Program Files\MiKTeX
2.7\tex\latex\ltxmisc\a4wide.sty
(C:\Program Files\MiKTeX
2.7\tex\latex\ntgclass\a4.sty))
! Missing \endcsname inserted.
to be read again
\protect
l.11 \begin
{document}
? pdflatex.EXE: Bad file descriptor
This appears to be the problem. I'm not certain, but I
would guess if
you look in the C:/Program
Files/R/R-2.7.2/bin/Sweave-test-1.tex file
you'll see a line something like this:
\usepackage{C:/Program
Files/R/R-2.7.2/share/texmf/Sweave}
That line (which was added by Sweave) will not work,
because LaTeX does
not understand blanks in file paths. There are a couple of
workarounds:
1. Install R in a directory that doesn't contain
blanks.
2. Include \usepackage{Sweave} in the Rnw file
you're processing; this
will stop Sweave from adding anything. I believe we set
things up so
that MikTeX will find it in the right place (though I
don't normally use
texi2dvi, so I'm not sure).
To test this, you can just go and edit the
Sweave-test-1.tex file now,
replacing the long line with the short version.
Duncan Murdoch
texify: pdflatex failed for some reason (see log
file).
Error in texi2dvi(Sweave-test-1.tex, pdf =
TRUE, quiet = FALSE) :
running 'texi2dvi' on
'Sweave-test-1.tex' failed
sessionInfo()
R version 2.7.2 (2008-08-25)
i386-pc-mingw32
locale:
LC_COLLATE=English_United
States.1252;LC_CTYPE=English_United
States.1252;LC_MONETARY=English_United
States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252
attached base packages:
[1] tools stats graphics grDevices utils
datasets methods base
Sys.getenv(PATH)
PATH
C:\\Program Files\\MiKTeX
2.7\\miktex\\bin;C:\\GTK\\bin;C:\\WINDOWS\\system32;C:\\WINDOWS;C:\\WINDOWS\\System32\\Wbem;C:\\Program
Files\\SYSTAT 10\\;C:\\Program
Files\\SYSTAT
10\\XGRAPH\\;C:\\Program
Files\\Wave Systems Corp\\Dell Preboot
Manager\\Access
Client\\v5\\;C:\\Program
Files\\ATI
Technologies\\ATI.ACE\\Core-Static;c:\\Program
Files\\Microsoft SQL
Server\\90\\Tools\\binn\\
Thank you
Felipe D. Carrillo
Supervisory Fishery Biologist
Department of the Interior
US Fish Wildlife Service
California, USA
--- On Thu, 10/16/08, Duncan Murdoch
[EMAIL PROTECTED] wrote:
Try running it as
texi2dvi(Sweave-test-1.tex,
pdf=TRUE, quiet=FALSE),
and post the list of error messages.
You should also show us the results of
sessionInfo() and
Sys.getenv(PATH), and tell us which
version of
LaTeX you have installed.
Duncan Murdoch
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal,
self-contained,
reproducible code.
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Programing and writing function help
Thank- you all for you suggestions and help. I have now resolved my issue. This help forum is one of the things that makes R a great platform for statistics. Stephen Cole Marine Ecology Lab Saint Francis Xavier University On Wed, Oct 8, 2008 at 8:07 AM, Erin Hodgess [EMAIL PROTECTED]wrote: Actually, you will have duplicates with 400 pairs. Here you will have 13^2 pairs with replacement and 13*12 pairs without replacement and with regard to order. How about this: z - unique(x) y - expand.grid(z[1:13],z[1:13]) xx - y[,1] != y[,2] y[xx,] Just another thought. Erin On Wed, Oct 8, 2008 at 5:55 AM, Erin Hodgess [EMAIL PROTECTED] wrote: For all possible pairs, you'll have 20^2 pairs. This is a way to do it: expand.grid(x[1:20],x[1:20]) HTH, Erin On Wed, Oct 8, 2008 at 4:43 AM, Jim Lemon [EMAIL PROTECTED] wrote: Stephen Cole wrote: ... I have a vector of 20 values x - c(20,18, 45, 16, 47, 47, 15, 26, 14,14,12,16,35,27,18,94,16,26,26,30) 1. I want to select random pairs from this data set but do it without replacement exhaustively matrix(x[sample(1:20,20)],nrow=2) then step through the columns of the resulting matrix I know i can select random pairs without replacement using sample(N,n,replace=F) However i am wondering if there is any way to get 10 random pairs from this data set without repeating any of the data points that is to say if i got a (20, 94) for one pair, i would like to get 9 other pairs from the data without again getting 20 or 94? 2. The second thing i would like to do is be able to select all possible pairs of numbers and calculate each pairs variance. I think you want to use the combn function, but you are going to get a lot of pairs... Jim __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Erin Hodgess Associate Professor Department of Computer and Mathematical Sciences University of Houston - Downtown mailto: [EMAIL PROTECTED] -- Erin Hodgess Associate Professor Department of Computer and Mathematical Sciences University of Houston - Downtown mailto: [EMAIL PROTECTED] [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] package Utils Sweave Example Error
On Thu, 16 Oct 2008, Duncan Murdoch wrote:
On 10/16/2008 12:19 PM, Felipe Carrillo wrote:
This is my sessionInfo() and Sys.getenv(PATH)
library(tools)
testfile - system.file(Sweave, Sweave-test-1.Rnw, package = utils)
## enforce par(ask=FALSE)
options(device.ask.default=FALSE)
## create a LaTeX file
Sweave(testfile)
Writing to file Sweave-test-1.tex
Processing code chunks ...
1 : print term verbatim
2 : term hide
3 : echo print term verbatim
4 : term verbatim
5 : echo term verbatim
6 : echo term verbatim eps pdf
7 : echo term verbatim eps pdf
You can now run LaTeX on 'Sweave-test-1.tex'
## This can be compiled to PDF by
texi2dvi(Sweave-test-1.tex, pdf=TRUE,quiet=FALSE)
This is pdfTeX, Version 3.1415926-1.40.9 (MiKTeX 2.7)
entering extended mode
(C:/Program Files/R/R-2.7.2/bin/Sweave-test-1.tex
LaTeX2e 2005/12/01
Babel v3.8l and hyphenation patterns for english, dumylang,
nohyphenation, ge
rman, ngerman, german-x-2008-06-18, ngerman-x-2008-06-18, french, loaded.
(C:\Program Files\MiKTeX 2.7\tex\latex\base\article.cls
Document Class: article 2005/09/16 v1.4f Standard LaTeX document class
(C:\Program Files\MiKTeX 2.7\tex\latex\base\size10.clo))
(C:\Program Files\MiKTeX 2.7\tex\latex\ltxmisc\a4wide.sty
(C:\Program Files\MiKTeX 2.7\tex\latex\ntgclass\a4.sty))
! Missing \endcsname inserted.
to be read again\protect l.11 \begin
{document}
? pdflatex.EXE: Bad file descriptor
This appears to be the problem. I'm not certain, but I would guess if you
look in the C:/Program Files/R/R-2.7.2/bin/Sweave-test-1.tex file you'll
see a line something like this:
\usepackage{C:/Program Files/R/R-2.7.2/share/texmf/Sweave}
That line (which was added by Sweave) will not work, because LaTeX does not
understand blanks in file paths. There are a couple of workarounds:
1. Install R in a directory that doesn't contain blanks.
2. Include \usepackage{Sweave} in the Rnw file you're processing; this will
stop Sweave from adding anything. I believe we set things up so that MikTeX
will find it in the right place (though I don't normally use texi2dvi, so I'm
not sure).
Yes, and 2.8.0 RC does this even better.
To test this, you can just go and edit the Sweave-test-1.tex file now,
replacing the long line with the short version.
Duncan Murdoch
texify: pdflatex failed for some reason (see log file).
Error in texi2dvi(Sweave-test-1.tex, pdf = TRUE, quiet = FALSE) :
running 'texi2dvi' on 'Sweave-test-1.tex' failed
sessionInfo()
R version 2.7.2 (2008-08-25) i386-pc-mingw32
locale:
LC_COLLATE=English_United States.1252;LC_CTYPE=English_United
States.1252;LC_MONETARY=English_United
States.1252;LC_NUMERIC=C;LC_TIME=English_United States.1252
attached base packages:
[1] tools stats graphics grDevices utils datasets methods
base
Sys.getenv(PATH)
PATH
C:\\Program Files\\MiKTeX
2.7\\miktex\\bin;C:\\GTK\\bin;C:\\WINDOWS\\system32;C:\\WINDOWS;C:\\WINDOWS\\System32\\Wbem;C:\\Program
Files\\SYSTAT 10\\;C:\\Program Files\\SYSTAT 10\\XGRAPH\\;C:\\Program
Files\\Wave Systems Corp\\Dell Preboot Manager\\Access
Client\\v5\\;C:\\Program Files\\ATI
Technologies\\ATI.ACE\\Core-Static;c:\\Program Files\\Microsoft SQL
Server\\90\\Tools\\binn\\
Thank you
Felipe D. Carrillo Supervisory Fishery Biologist Department of the
Interior US Fish Wildlife Service California, USA
--- On Thu, 10/16/08, Duncan Murdoch [EMAIL PROTECTED] wrote:
Try running it as texi2dvi(Sweave-test-1.tex,
pdf=TRUE, quiet=FALSE), and post the list of error messages.
You should also show us the results of sessionInfo() and
Sys.getenv(PATH), and tell us which version of
LaTeX you have installed.
Duncan Murdoch
__
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.
--
Brian D. Ripley, [EMAIL PROTECTED]
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Variable shortlisting for the logistic regression
Let's try to bring this discussion back again after Frank made very funny remark! What I'm doing at the moment is: 1. I split dataset in two (development and holdout) 2. I fit single predictor logistic model for every variable and collect following stats: DMaxDeriv=modelD$stats[2] DModelLR=modelD$stats[3] DP=modelD$stats[5] DC=modelD$stats[6] DDxy=modelD$stats[7] DGamma=modelD$stats[8] DTau=modelD$stats[9] DR2=modelD$stats[10] DBier=modelD$stats[11] HMaxDeriv=modelH$stats[2] HModelLR=modelH$stats[3] HP=modelH$stats[5] HC=modelH$stats[6] HDxy=modelH$stats[7] HGamma=modelH$stats[8] HTau=modelH$stats[9] HR2=modelH$stats[10] HBier=modelH$stats[11] where D is prefix for stats on development sample and H is prefix for stats derived from hold out sample 3. Now I screen factor with sommers d grather than 0.3 and relative change on hold out sample is smaller than 5% Any comments are very welcomed On Oct 14, 2:48 pm, John Kane [EMAIL PROTECTED] wrote: --- On Mon, 10/13/08, David Scott [EMAIL PROTECTED] wrote: From: David Scott [EMAIL PROTECTED] Subject: Re: [R] Variable shortlisting for the logistic regression To: Frank E Harrell Jr [EMAIL PROTECTED] Cc: [EMAIL PROTECTED] Received: Monday, October 13, 2008, 6:32 PM On Mon, 13 Oct 2008, Frank E Harrell Jr wrote: useR wrote: Hi R helpers, One rather statistical question? What would be the best startegy to shortlist thousands of continous variables automaticaly using R as the preparation for logistic regression modleing! Thanks The easiest approach is to use a random number generator. Frank Got a laugh from me Frank! Can I nominate it for a fortune? David Seconded. __ [[elided Yahoo spam]] __ [EMAIL PROTECTED] mailing listhttps://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guidehttp://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] defining a function using strings
On 17/10/2008, at 7:40 AM, Chernomoretz Ariel wrote:
Hi All,
I need to evaluate a series expansion using Legendre polynomials.
Using the 'orthopolinom' package I can get a list of the first n
Legendre polynomials as character strings.
library(orthopolynom)
l-legendre.polynomials(4)
l
[[1]]
1
[[2]]
x
[[3]]
-0.5 + 1.5*x^2
[[4]]
-1.5*x + 2.5*x^3
[[5]]
0.375 - 3.75*x^2 + 4.375*x^4
But I can't figure out how to implement functions that could be
evaluated for arbitrary 'x', from this list,
Thanks for your help.
?as.function
methods(as.function)
?as.function.polynomial
foo - as.function(l[[4]])
foo(3)
cheers,
Rolf Turner
##
Attention:\ This e-mail message is privileged and confid...{{dropped:9}}
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Variable shortlisting for the logistic regression
On 17/10/2008, at 8:22 AM, useR wrote:
Let's try to bring this discussion back again after Frank made
very funny remark!
Frank's remark was *serious*. Take it seriously.
cheers,
Rolf Turner
##
Attention:\ This e-mail message is privileged and confid...{{dropped:9}}
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] defining a function using strings
try this: library(orthopolynom) l - legendre.polynomials(4) fun.l - lapply(l, as.function) fun.l[[3]](1:3) I hope it helps. Best, Dimitris Chernomoretz Ariel wrote: Hi All, I need to evaluate a series expansion using Legendre polynomials. Using the 'orthopolinom' package I can get a list of the first n Legendre polynomials as character strings. library(orthopolynom) l-legendre.polynomials(4) l [[1]] 1 [[2]] x [[3]] -0.5 + 1.5*x^2 [[4]] -1.5*x + 2.5*x^3 [[5]] 0.375 - 3.75*x^2 + 4.375*x^4 But I can't figure out how to implement functions that could be evaluated for arbitrary 'x', from this list, Thanks for your help. Ariel./ -- Dimitris Rizopoulos Assistant Professor Department of Biostatistics Erasmus Medical Center Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands Tel: +31/(0)10/7043478 Fax: +31/(0)10/7043014 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] defining a function using strings
Try this (to evaluate them at 10): polynomial.values(l, 10) On Thu, Oct 16, 2008 at 2:40 PM, Chernomoretz Ariel [EMAIL PROTECTED] wrote: Hi All, I need to evaluate a series expansion using Legendre polynomials. Using the 'orthopolinom' package I can get a list of the first n Legendre polynomials as character strings. library(orthopolynom) l-legendre.polynomials(4) l [[1]] 1 [[2]] x [[3]] -0.5 + 1.5*x^2 [[4]] -1.5*x + 2.5*x^3 [[5]] 0.375 - 3.75*x^2 + 4.375*x^4 But I can't figure out how to implement functions that could be evaluated for arbitrary 'x', from this list, Thanks for your help. Ariel./ -- Dr. Ariel Chernomoretz Departamento de Fisica, FCEyN,Universidad de Buenos Aires, (1428) Ciudad Universitaria,Ciudad de Buenos Aires, Argentina. TE +54 11 4576 3390 ext 817 Fax +54 11 4576 3357 email: [EMAIL PROTECTED]Webpage: http://www.df.uba.ar/users/ariel __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Loop avoidance in simulating a vector
All,
I'd like to simulate a vector that is formed from many distinct
distributions and avoid a loop if possible. E.g, consider:
mu = c(1, 2, 3)
sigma = c(1, 2, 3)
n = c(10, 10, 10)
And we simulate a vector of length 30 that consists of N(mu[i], sigma[i])
distributed data, each of length n[i]. Of course for just three groups we
can simply write it out as:
DV = c(rnorm(n[1], mu[1], sigma[1]), rnorm(n[2], mu[2], sigma[2]),
rnorm(n[3], mu[3], sigma[3]) )
For many groups we can use a loop (assuming equal numbers per group):
n = n[1]
DV = numeric(N*n)
for (i in 1:N) {
DV[(n*i - (n-1)): (n*i)] = rnorm(n, mu[i], sigma[i])
}
Is there any way to do the general cas without using a loop?
Cheers,
David
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Loop avoidance in simulating a vector
Have a look at mapply.
-Christos
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of David Afshartous
Sent: Thursday, October 16, 2008 3:47 PM
To: r-help@r-project.org
Subject: [R] Loop avoidance in simulating a vector
All,
I'd like to simulate a vector that is formed from many
distinct distributions and avoid a loop if possible. E.g, consider:
mu = c(1, 2, 3)
sigma = c(1, 2, 3)
n = c(10, 10, 10)
And we simulate a vector of length 30 that consists of
N(mu[i], sigma[i])
distributed data, each of length n[i]. Of course for just
three groups we
can simply write it out as:
DV = c(rnorm(n[1], mu[1], sigma[1]), rnorm(n[2], mu[2],
sigma[2]), rnorm(n[3], mu[3], sigma[3]) )
For many groups we can use a loop (assuming equal numbers per group):
n = n[1]
DV = numeric(N*n)
for (i in 1:N) {
DV[(n*i - (n-1)): (n*i)] = rnorm(n, mu[i], sigma[i])
}
Is there any way to do the general cas without using a loop?
Cheers,
David
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Loop avoidance in simulating a vector
mapply is still a (disguised) loop (at the interpreted level). So other than
improving code readability (always a good thing!), it shouldn't make much of
an efficiency difference.
A longer answer is: if all you're doing is a location-scale family of
distributions, then creating a matrix of standard normal (or whatever)
distributed data for all 1:N at once and then using matrix operations to
multiply and add, say, so each column becomes your different distribution
might be faster. This gets the loops down to C code.
A shorter answer is: it's unlikely that any of this makes enough of a
difference to be worth the effort. Random number generation is so efficient
in R that avoiding loops rarely matters.
Also see ?replicate for a way to perhaps write cleaner code (but still using
hidden interpreted loops).
-- Bert Gunter
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of Christos Hatzis
Sent: Thursday, October 16, 2008 1:06 PM
To: 'David Afshartous'; r-help@r-project.org
Subject: Re: [R] Loop avoidance in simulating a vector
Have a look at mapply.
-Christos
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of David Afshartous
Sent: Thursday, October 16, 2008 3:47 PM
To: r-help@r-project.org
Subject: [R] Loop avoidance in simulating a vector
All,
I'd like to simulate a vector that is formed from many
distinct distributions and avoid a loop if possible. E.g, consider:
mu = c(1, 2, 3)
sigma = c(1, 2, 3)
n = c(10, 10, 10)
And we simulate a vector of length 30 that consists of
N(mu[i], sigma[i])
distributed data, each of length n[i]. Of course for just
three groups we
can simply write it out as:
DV = c(rnorm(n[1], mu[1], sigma[1]), rnorm(n[2], mu[2],
sigma[2]), rnorm(n[3], mu[3], sigma[3]) )
For many groups we can use a loop (assuming equal numbers per group):
n = n[1]
DV = numeric(N*n)
for (i in 1:N) {
DV[(n*i - (n-1)): (n*i)] = rnorm(n, mu[i], sigma[i])
}
Is there any way to do the general cas without using a loop?
Cheers,
David
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Loop avoidance in simulating a vector
Yes, but the difference is that the looping in mapply is done in C.
There are no interpreted loops in mapply,as far as I can see.
-Christos
-Original Message-
From: Bert Gunter [mailto:[EMAIL PROTECTED]
Sent: Thursday, October 16, 2008 4:13 PM
To: [EMAIL PROTECTED]; 'David Afshartous';
r-help@r-project.org
Subject: RE: [R] Loop avoidance in simulating a vector
mapply is still a (disguised) loop (at the interpreted
level). So other than improving code readability (always a
good thing!), it shouldn't make much of an efficiency difference.
A longer answer is: if all you're doing is a location-scale
family of distributions, then creating a matrix of standard
normal (or whatever) distributed data for all 1:N at once and
then using matrix operations to multiply and add, say, so
each column becomes your different distribution might be
faster. This gets the loops down to C code.
A shorter answer is: it's unlikely that any of this makes
enough of a difference to be worth the effort. Random number
generation is so efficient in R that avoiding loops rarely matters.
Also see ?replicate for a way to perhaps write cleaner code
(but still using hidden interpreted loops).
-- Bert Gunter
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Christos Hatzis
Sent: Thursday, October 16, 2008 1:06 PM
To: 'David Afshartous'; r-help@r-project.org
Subject: Re: [R] Loop avoidance in simulating a vector
Have a look at mapply.
-Christos
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of David Afshartous
Sent: Thursday, October 16, 2008 3:47 PM
To: r-help@r-project.org
Subject: [R] Loop avoidance in simulating a vector
All,
I'd like to simulate a vector that is formed from many distinct
distributions and avoid a loop if possible. E.g, consider:
mu = c(1, 2, 3)
sigma = c(1, 2, 3)
n = c(10, 10, 10)
And we simulate a vector of length 30 that consists of N(mu[i],
sigma[i])
distributed data, each of length n[i]. Of course for just
three groups we
can simply write it out as:
DV = c(rnorm(n[1], mu[1], sigma[1]), rnorm(n[2], mu[2], sigma[2]),
rnorm(n[3], mu[3], sigma[3]) )
For many groups we can use a loop (assuming equal numbers
per group):
n = n[1]
DV = numeric(N*n)
for (i in 1:N) {
DV[(n*i - (n-1)): (n*i)] = rnorm(n, mu[i], sigma[i])
}
Is there any way to do the general cas without using a loop?
Cheers,
David
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[R] Saving results of Kruskal Walis test
Hello, I am running Kruskal-Walis test in R. When I try to save results using write.table it gives me the following error : Error in as.data.frame.default(x[[i]], optional = TRUE, stringsAsFactors = stringsAsFactors) : cannot coerce class htest into a data.frame The overall code is as follows : data_file = read.table(~/DATA.dir/data_file.txt, header=T) attach(data_file) data_file.out - krukal.test(data_file) write.table(data_file.out, ~/DATA/results/data_file_out.txt) Error in as.data.frame.default(x[[i]], optional = TRUE, stringsAsFactors = stringsAsFactors) : cannot coerce class htest into a data.frame Results do come in data_file.out after analysis as seen below: data_file.out Kruskal-Wallis rank sum test data: value by pathway Kruskal-Wallis chi-squared = 5.6031, df = 3, p-value = 0.1326 I am wondering if I am making a mistake with using write.table (It works very well saving results from anova analysis) or is there any other way to save results in a file for future use.. Thanks Himanshu \\ [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Loop avoidance in simulating a vector
If you want to avoid using a loop, rather than coding a loop, then someting
like:
DV - rnorm(sum(n))
DV - DV * rep(sigma, times=n) + rep(mu, times=n)
will achieve the same as the loop you specified, but generalised to
non-constant n.
HTH
Ray Brownrigg
On Fri, 17 Oct 2008, Bert Gunter wrote:
mapply is still a (disguised) loop (at the interpreted level). So other
than improving code readability (always a good thing!), it shouldn't make
much of an efficiency difference.
A longer answer is: if all you're doing is a location-scale family of
distributions, then creating a matrix of standard normal (or whatever)
distributed data for all 1:N at once and then using matrix operations to
multiply and add, say, so each column becomes your different distribution
might be faster. This gets the loops down to C code.
A shorter answer is: it's unlikely that any of this makes enough of a
difference to be worth the effort. Random number generation is so efficient
in R that avoiding loops rarely matters.
Also see ?replicate for a way to perhaps write cleaner code (but still
using hidden interpreted loops).
-- Bert Gunter
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of Christos Hatzis
Sent: Thursday, October 16, 2008 1:06 PM
To: 'David Afshartous'; r-help@r-project.org
Subject: Re: [R] Loop avoidance in simulating a vector
Have a look at mapply.
-Christos
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of David Afshartous
Sent: Thursday, October 16, 2008 3:47 PM
To: r-help@r-project.org
Subject: [R] Loop avoidance in simulating a vector
All,
I'd like to simulate a vector that is formed from many
distinct distributions and avoid a loop if possible. E.g, consider:
mu = c(1, 2, 3)
sigma = c(1, 2, 3)
n = c(10, 10, 10)
And we simulate a vector of length 30 that consists of
N(mu[i], sigma[i])
distributed data, each of length n[i]. Of course for just
three groups we
can simply write it out as:
DV = c(rnorm(n[1], mu[1], sigma[1]), rnorm(n[2], mu[2],
sigma[2]), rnorm(n[3], mu[3], sigma[3]) )
For many groups we can use a loop (assuming equal numbers per group):
n = n[1]
DV = numeric(N*n)
for (i in 1:N) {
DV[(n*i - (n-1)): (n*i)] = rnorm(n, mu[i], sigma[i])
}
Is there any way to do the general cas without using a loop?
Cheers,
David
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[R] bivariate uniform density estimation
Dear R users, I have two uniform random variables and I need to estimate the joint density. I like to know whether there is any package which estimates bivariate uniform densities. Thanks, Lavan -- View this message in context: http://www.nabble.com/bivariate-uniform-density-estimation-tp20016805p20016805.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Statistical courses
XLSolutions-corp has 2 day training courses; one for beginners and one for advanced. Here is the website: www.xlsolutions-corp.com Hope this helps! Sincerely, Erin On Thu, Oct 16, 2008 at 1:50 AM, timpanister [EMAIL PROTECTED] wrote: Hi, Can anyone suggest some short term statistical courses using R that one could take? Name or Links will be much appreciated. many thanks, 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. -- Erin Hodgess Associate Professor Department of Computer and Mathematical Sciences University of Houston - Downtown mailto: [EMAIL PROTECTED] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] gWidgets install
On Tue, Oct 14, 2008 at 7:15 PM, john verzani [EMAIL PROTECTED] wrote: Tim Smith tim_smith_666 at yahoo.com writes: Thanks Charlie - I just tried it, but still get the same error: --- install.packages(gWidgets,dependencies=TRUE) ... library(gWidgets) Error in fun(...) : *** gWidgets requires a toolkit implementation to be installed, for instance gWidgetsRGtk2, gWidgetstcltk, or gWidgetsrJava*** Try installing gWidgetstcltk. That will allow you to use gWidgets with the tcltk package. Although the RGtk2 package works better with gWidgets, the tcltk one will require no extra software to install under windows. Perhaps the gWidgets package should prompt the user to install one of these? --John __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] stablefit can fit the parameters of a truncated normal distribution?
This idea is very wrong, Have a look at the function cnormal1 in the VGAM package John frain 2008/10/15 drbn [EMAIL PROTECTED]: I'm using stableFit from the package fBasics to estimate the parameters of a truncated normal distribution (I'm interested in the parameters of the underlying normal distribution). It is correct to generalize this truncated normal distribution as a stable distribution ? Thanks David -- View this message in context: http://www.nabble.com/stablefit-can-fit-the-parameters-of-a-truncated-normal-distribution--tp20002335p20002335.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. -- John C Frain Trinity College Dublin Dublin 2 Ireland www.tcd.ie/Economics/staff/frainj/home.html mailto:[EMAIL PROTECTED] mailto:[EMAIL PROTECTED] __ 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] Roman numeral question
Hi R People! Is there a setup for Roman numerals similar to that of LETTERS and letters, please? I was putting together a randomized block design and thought that it might be nice for factors. Thanks, Erin -- Erin Hodgess Associate Professor Department of Computer and Mathematical Sciences University of Houston - Downtown mailto: [EMAIL PROTECTED] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R: in-place appending to a matrix.
On 10/16/2008 10:50 AM, culpritNr1 wrote: Now, modern high level languages like the continually improving R, References: Becker, R. A., Chambers, J. M. and Wilks, A. R. (1988) _The New S Language_. Wadsworth Brooks/Cole. This is most important reference in many help pages in R. Modern Language, haha. Old crap, not a modern language. \misiek [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R: in-place appending to a matrix.
that's extremely rude , especially to all the people who made and make R what it is. If you don't like R, noone is forcing you to use it. On Thu, Oct 16, 2008 at 5:50 PM, repkakala Gazeta.pl wrote: On 10/16/2008 10:50 AM, culpritNr1 wrote: Now, modern high level languages like the continually improving R, References: Becker, R. A., Chambers, J. M. and Wilks, A. R. (1988) _The New S Language_. Wadsworth Brooks/Cole. This is most important reference in many help pages in R. Modern Language, haha. Old crap, not a modern language. \misiek [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] MLE Constraints
Dears,
Any help?
Thanks,
LFRC
LFRC wrote:
Dears,
I'm trying to find the parameters (a,b, ... l) that optimize the function
(Model)
described below.
1) How can I set some constraints with MLE2 function? I want to set p10,
p20,
p30, p1p3.
2) The code is giving the following warning.
Warning: optimization did not converge (code 1)
How can I solve this problem?
Can someone help me?
M - 14
Y = c(0, 1, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 1)
x1 = c(0.50, 0.50, 0.50, 0.50, 0.50, 0.50, 0.50, 0.50, 0.50, 0.50, 0.25,
0.25, 0.25, 0.25)
x2 = c(-1, -1, -1, -1, -1, 1, 1, 1, 1, 1, 1, 1, 1, 1)
x3 = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0)
states = c(1, 1, 2, 3, 1, 2, 3, 1, 1, 2, 2, 3, 1, 1)
prob_fn = rep(0,M)
Model=function(a, b, c, d, e, f, g, h, i, j, k, l)
{
p1 = exp(-(a g*x1 d*x2 j*x3))
p2 = exp(-(b h*x1 e*x2 k*x3))
p3 = exp(-(c i*x1 f*x2 l*x3))
### Set P
t5 = 0
while(t5M)
{
t5 = t5 1
if(states[t5]==1) {prob_ok = p1[1]}
if(states[t5]==2) {prob_ok = p2[1]}
if(states[t5]==3) {prob_ok = p3[1]}
prob_fn[t5] = c(prob_ok)
}
prob_fn[prob_fn==0] = 0.1
### LL
ll_calc = -(sum(Y*log(prob_fn)))
return(ll_calc)
}
res = mle2(Model, start=list(a=1, b=1, c=1, d=0.15, e=0.15,
f=0.15, g=0.9, h=0.9, i=0.9, j=0.1, k=0.1, l=0.1), method = Nelder-
Mead)
res
Best regards,
LFRC
--
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Re: [R] Roman numeral question
2008/10/16 Erin Hodgess [EMAIL PROTECTED]: Hi R People! Is there a setup for Roman numerals similar to that of LETTERS and letters, please? I was putting together a randomized block design and thought that it might be nice for factors. Thanks, Erin as.roman() in utils: library(utils) as.roman(1:10) [1] III III IV VVI VII VIII IX X Barry __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to constrain the residual variance to be 0 in linear mixed-effects models?
Hi everyone, How to constrain the residual variance to be 0 when using the lmer function? Wen Luo Wen Luo, Ph.D. Assistant Professor Department of Educational Psychology University of Wisconsin -- Milwaukee Enderis Hall 759 P.O. Box 413 Milwaukee, WI 53201-0413 Office: 414 229 4998 Fax: 414 229 4939 [EMAIL PROTECTED] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R: in-place appending to a matrix.
Rigth, sorry. R is great for many usages even when it did not got modernised as many realy modern languages did. Appologies for developers, I were too rude. \misiek 2008/10/16 [EMAIL PROTECTED] that's extremely rude , especially to all the people who made and make R what it is. If you don't like R, noone is forcing you to use it. On Thu, Oct 16, 2008 at 5:50 PM, repkakala Gazeta.pl wrote: On 10/16/2008 10:50 AM, culpritNr1 wrote: Now, modern high level languages like the continually improving R, References: Becker, R. A., Chambers, J. M. and Wilks, A. R. (1988) _The New S Language_. Wadsworth Brooks/Cole. This is most important reference in many help pages in R. Modern Language, haha. Old crap, not a modern language. \misiek [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
