Re: [R] write.xls dont find the object in function
Em 20-12-2011 14:09, MacQueen, Don escreveu:
Or:
require(xlsx)
test- function(x){
+a- data.frame(A=c(1,2),B=c(10,11))
+write.xlsx(a,file=a.xlsx)
+ }
test()
list.files(patt='xlsx')
[1] a.xlsx
Thanks,
with the write.xlsx all work.
Inte
Ronaldo
--
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certo, na maior parte do tempo.
--Herman, I. P. 2007. Following the law. NATURE, Vol 445, p. 228.
Prof. Ronaldo Reis Júnior
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[R] simple ggplot2 question
Hello, I am trying to make a plot using the code below. The plot is divided into two parts, using facet_grid. I would like the vertical axis (labelled 'place') to be different for each location (=part). So in the upper part, only places 'n' through 'z' are shown, while in the lower part, only places 'a' through 'm' are shown. I thought 'free_y' would do the trick. I also tried converting variable place into class 'factor'. require(ggplot2) DF - data.frame(place=letters, value=runif(26), location=c(rep(1, 13), rep(0, 13))) qplot(data=DF, x=place, y=value, geom=bar, stat=identity) + coord_flip() + geom_abline(intercept=35, slope=0, colour=red) + facet_grid(location ~ ., scales=free_y) R.version.string # R version 2.10.1 (2009-12-14) Thank you in advance merry xmas! Cheers!! Albert-Jan ~~ All right, but apart from the sanitation, the medicine, education, wine, public order, irrigation, roads, a fresh water system, and public health, what have the Romans ever done for us? ~~ [[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] GIS operations
Hello, Maybe this will help: There is rgeos package with gContains, gCovers, etc. It is a full implementation of Java Topology Suite http://cran.r-project.org/web/packages/rgeos/rgeos.pdf -- Kind regards, Antonio Rodriges __ 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 Source Code Request Office For National Statistics UK
On Wed, Dec 21, 2011 at 1:06 PM, David Samuel david.sam...@ons.gsi.gov.uk wrote: To R Support Team, Could you please let me know if ONS would be allowed to view your software's source code? If proof was ever needed that people don't read software license agreements, this is it. Is there a license that forces you to read the source code before using the software? I'd license my stuff under that: THE BAZ PUBLIC LICENSE (hereinafter the BPL) Under the terms of the BPL you MUST first read all source code and get a pretty good understanding of what it all does before using THE SOFTWARE. You MUST report bugs or problems to THE AUTHOR but you MUST NOT expect them to be fixed. You MUST fix them YOURSELF if you think you may be technically capable or if YOU think YOU are a better programmer than THE AUTHOR. Otherwise YOU must SHUT UP. Barry (hereinbefore Baz) __ 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] Help transforming a dist
Perfect! combn was the trick I needed. Although I'll probably rbind the stuff instead of building a new frame. :) On Thu, Dec 22, 2011 at 6:48 PM, Sarah Goslee sarah.gos...@gmail.com wrote: I'm not at all sure what you mean with your matrix: is that supposed to be three columns? What about: fakedata - data.frame(X=runif(3), Y = runif(3)) rownames(fakedata) - c(A, B, C) dist(fakedata) A B B 0.8617733 C 0.3813032 0.5124284 data.frame(t(combn(rownames(fakedata), 2)), dist=as.vector(dist(fakedata))) X1 X2 dist 1 A B 0.8617733 2 A C 0.3813032 3 B C 0.5124284 Sarah On Thu, Dec 22, 2011 at 7:03 PM, Taral tar...@gmail.com wrote: I'd like to convert a dist into a table/matrix/thingy of the form: A B value A C value B C value (assuming the dist was over 3 names). Is there a way to do this without using a for loop? -- Taral tar...@gmail.com Please let me know if there's any further trouble I can give you. -- Unknown -- Sarah Goslee http://www.functionaldiversity.org -- Taral tar...@gmail.com Please let me know if there's any further trouble I can give you. -- Unknown __ 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] GIS operations
On 12/20/2011 12:54 PM, Antonio Rodriges wrote: Hello, Is there a way to find points in SpatialPoints which lie inside a given Polygon? Hi, Take a look at the over function from the sp-package. Paul -- Paul Hiemstra, Ph.D. Global Climate Division Royal Netherlands Meteorological Institute (KNMI) Wilhelminalaan 10 | 3732 GK | De Bilt | Kamer B 3.39 P.O. Box 201 | 3730 AE | De Bilt tel: +31 30 2206 494 http://intamap.geo.uu.nl/~paul http://nl.linkedin.com/pub/paul-hiemstra/20/30b/770 __ 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] Axis manipulation in Stackpoly (Plotrix)
On 12/23/2011 07:52 AM, Ben Neal wrote: Have made a stacked area plot, but now want to manipulate the x axis: make in even increments (50) in order to suppress the tick marks (forming solid bar under plot). However, the plot functions do not seem to work, and I cannot find documentation for use of xat in stackpoly. This is a time series of data covering 579 years, and has been successfully converted from zoo to matrix. Data plotting fine, jsut want to change axis - CaribMatrix-as.matrix(ts4Ex) stackpoly(CaribMatrix, stack=TRUE, xlab=Year,ylab=Catch in tons, xaxlab=(1250:2008), main=Historical Florida reef fisheries catch by sector, axis4=FALSE) Any assistance appreciated! Thanks very much, Benjamin P Neal, Scripps Inst. of Oceanogrpahy Hi Ben, In the stackpoly function, you can specify the positions of the axis ticks (xat=...) and the labels for those positions (xaxlab=...). You don't seem to have supplied the x values, so the values in CaribMatrix will be placed display at 1:dim(CaribMatrix)[1]. I'll have to make up CaribMatrix... CaribMatrix-matrix(order(sample(200:500,579,TRUE),nrow=579,ncol=3) stackpoly(CaribMatrix,stack=TRUE,ylim=c(0,1500),xat=seq(1,600,by=50), xaxlab=seq(1450,2000,by=50),staxx=TRUE) This seems to give a reasonable plot for me. 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] simple ggplot2 question
On 12/23/2011 08:26 AM, Albert-Jan Roskam wrote: Hello, I am trying to make a plot using the code below. The plot is divided into two parts, using facet_grid. I would like the vertical axis (labelled 'place') to be different for each location (=part). So in the upper part, only places 'n' through 'z' are shown, while in the lower part, only places 'a' through 'm' are shown. I thought 'free_y' would do the trick. I also tried converting variable place into class 'factor'. require(ggplot2) DF - data.frame(place=letters, value=runif(26), location=c(rep(1, 13), rep(0, 13))) qplot(data=DF, x=place, y=value, geom=bar, stat=identity) + coord_flip() + geom_abline(intercept=35, slope=0, colour=red) + facet_grid(location ~ ., scales=free_y) R.version.string # R version 2.10.1 (2009-12-14) Hi, This code using facet_wrap works: require(ggplot2) DF - data.frame(place=letters, value=runif(26), location=c(rep(1, 13), rep(0, 13))) qplot(data=DF, x=place, y=value, geom=bar, stat=identity) + coord_flip() + facet_wrap(~ location, scales = free, ncol = 1) And using facet_grid: DF - data.frame(place=letters, value=runif(26), location=c(rep(1, 13), rep(0, 13))) qplot(data=DF, x=place, y=value, geom=bar, stat=identity) + coord_flip() + facet_grid(. ~ location, scales = free) Note that I switched the formula in facet_grid. However, I would expect your code to also work. cheers and hope this helps, Paul Thank you in advance merry xmas! Cheers!! Albert-Jan ~~ All right, but apart from the sanitation, the medicine, education, wine, public order, irrigation, roads, a fresh water system, and public health, what have the Romans ever done for us? ~~ [[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. -- Paul Hiemstra, Ph.D. Global Climate Division Royal Netherlands Meteorological Institute (KNMI) Wilhelminalaan 10 | 3732 GK | De Bilt | Kamer B 3.39 P.O. Box 201 | 3730 AE | De Bilt tel: +31 30 2206 494 http://intamap.geo.uu.nl/~paul http://nl.linkedin.com/pub/paul-hiemstra/20/30b/770 [[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] Axis manipulation in Stackpoly (Plotrix)
On 12/23/2011 09:53 PM, Jim Lemon wrote: CaribMatrix-matrix(order(sample(200:500,579,TRUE),nrow=579,ncol=3) Oops, that should be CaribMatrix-matrix(sort(sample(200:500,579,TRUE),nrow=579,ncol=3) 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.
[R] missing value where TRUE/FALSE needed
Merry Xmas to all,
I am writing a function and curiously this runs sometimes on one data set
and fails on another and i cannot figure out why.
Any help much appreciated.
If i run the code below with
data - iris[ ,1:4]
The code runs fine, but if i run on a large dataset i get the following
error (showing data structures as matrix is large)
str(cluster.data)
num [1:9985, 1:811] 0 0 0 0 0 0 0 0 0 0 ...
- attr(*, dimnames)=List of 2
..$ : NULL
..$ : chr [1:811] 1073949105 1073930585 1073843224 1073792624 ...
#(This is intended to be chr)
str(iris)
'data.frame': 150 obs. of 5 variables:
$ Sepal.Length: num 5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ...
$ Sepal.Width : num 3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ...
$ Petal.Length: num 1.4 1.4 1.3 1.5 1.4 1.7 1.4 1.5 1.4 1.5 ...
$ Petal.Width : num 0.2 0.2 0.2 0.2 0.2 0.4 0.3 0.2 0.2 0.1 ...
$ Species : Factor w/ 3 levels setosa,versicolor,..: 1 1 1 1 1 1 1
1 1 1 ...
str(as.matrix(iris[,1:4]))
num [1:150, 1:4] 5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ...
- attr(*, dimnames)=List of 2
..$ : NULL
..$ : chr [1:4] Sepal.Length Sepal.Width Petal.Length Petal.Width
n.cols - ncol(data)
n.rows - nrow(data)
X - as.matrix(data)
stepsize - 0.05
c1 - (2 * pi) ** (n.cols / 2)
c2 - n.rows * (smoothing ** (n.cols + 2))
c3 - n.rows * (smoothing ** n.cols)
Kexp - function(sqs){
return (exp((-1 * sqs) / (2 * smoothing ** 2)))
}
FindGradient - function(x){
XmY - t(x - t(X))
sqsum - rowSums(XmY * XmY)
K - sapply(sqsum, Kexp)
dens - ((c1 * c3) ** -1) * sum(K)
grad - -1 * ((c1 * c2) ** -1) * colSums(K * XmY)
return (list(gradient = grad,
density = dens))
}
attractors - matrix(0, n.rows, n.cols)
densities - matrix(0, n.rows)
density.attractors -
sapply(rep(1:n.rows), function(i) {
notconverged - TRUE
# For each row loop through and find the attractor and density value.
x - (X[i, ])
iters - as.integer(1)
# Run gradient ascent for each point to obtain x*
while(notconverged == TRUE) {
find.gradient - FindGradient(x)
next.x - x + stepsize * find.gradient$gradient
change - sqrt(sum((next.x - x) * (next.x - x)))
notconverged - ifelse(change tol, TRUE, FALSE)
x - next.x
iters - iters + 1
}
# store the attractor and density value
return(c(densities[i, ] - find.gradient$density,
attractors[i, ] - x))
})
Error in while (notconverged == TRUE) { :
missing value where TRUE/FALSE needed
Any help would be great
Mike
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[R] Problem understanding behaviour of mmap package
I am trying to solve the following problem using the mmap package: 1.) Store some data as a memory mapped file that resides on the disk. 2.) Open the data and add some new data (or change it) to that file. 1.) works fine using the following code: cFile -tempfile() dfrData - data.frame(matrix(data=1, ncol=2, nrow=10)) mmapData - as.mmap(dfrData, file=cFile) stcData - mmapData$storage.mode munmap(mmapData) 2.) Now I want to link mmapAfter to the file that should still reside as cFile, or? What is the problem here? The error I get is at the second line of the code: Error in if (!x$bytes) stop(no data to extract) : missing value where TRUE/FALSE needed. mmapAfter - mmap(cFile, mode=stcData, extractFUN=as.data.frame) mmapAfter[]; mmapAfter [1,2] - 2; mmapAfter []; munmap(mmapAfter) I also noted when I execute the following code that the mmapData$filedesc doesn't show the filename I specified in cFile. Is this not how it is supposed to be used? cFile - 'C:/temp/testmapfile' dfrData - data.frame(matrix(data=1, ncol=2, nrow=10)) mmapData - as.mmap(dfrData, file=cFile) mmapData$filedesc Thanks for your help. Regards, Wolfgang Wu __ 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] lasso based selection for mixed model
Dear Dieter There has been some effort examining Lasso-type estimators for mixed models. It is more involved than with other type of models. We have studied Gaussian and generalized linear mixed models including a Lasso-type penalty for the fixed effects and we are now able to provide two packages: lmmlasso and glmmlasso. Both are available from R-Forge (the former even from CRAN). Andreas Groll provides the glmmLasso package, which is available from CRAN. Juerg Schelldorfer -- View this message in context: http://r.789695.n4.nabble.com/lasso-based-selection-for-mixed-model-tp887308p4228495.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] ggplot2: behaviour with empty datasets
For example, prepare like this df.0 - data.frame(x = 0, y = 0, note = 1) df.1 - subset(df.0, note == 1) df.2 - subset(df.0, note == 2) Then a call to ggplot() + aes(x = x, y = y) + geom_point(data = df.1) + geom_point(data = df.2) produces the error Error in eval(expr, envir, enclos) : object 'x' not found And a call to ggplot(df.1, aes(x = x, y = y, shape = 4)) + geom_point() + geom_point(aes(shape = 1), df.2) plots both a cross (shape = 4) and circle (shape = 1) on position (0, 0). However, as expected, only a cross should be plotted because `dt.2' is empty. So with the current behaviour, if someone writes a script to plot datasets automatically and distinguish between subsets, he will need to write different code path for empty and non-empty data subsets. I think ggplot2 can just choose to plot nothing for empty data subsets, and only produce error when the dataset (union of a subsets) to plot is completely empty. This can make scripting easier based on ggplot2. -- Using GPG/PGP? Please get my current public key (ID: 0xAEF6A134, valid from 2010 to 2013) from a key server. signature.asc Description: Digital signature __ 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] missing value where TRUE/FALSE needed
Does this look similar to the error you are getting:
while(NA == TRUE) 1
Error in while (NA == TRUE) 1 : missing value where TRUE/FALSE needed
SO 'notconverged' is probably equal to NA. BTW, what is the value of
'tol'; I do not see it defined. So when computing 'notconverged' you
have generated an NA. You can test it to see if this is true.
You can use the following command:
options(error=utils::recover)
and then learn how to use the 'browser' to examine variables when the
error occurs.
On Fri, Dec 23, 2011 at 5:44 AM, Michael Pearmain
michael.pearm...@gmail.com wrote:
Merry Xmas to all,
I am writing a function and curiously this runs sometimes on one data set
and fails on another and i cannot figure out why.
Any help much appreciated.
If i run the code below with
data - iris[ ,1:4]
The code runs fine, but if i run on a large dataset i get the following
error (showing data structures as matrix is large)
str(cluster.data)
num [1:9985, 1:811] 0 0 0 0 0 0 0 0 0 0 ...
- attr(*, dimnames)=List of 2
..$ : NULL
..$ : chr [1:811] 1073949105 1073930585 1073843224 1073792624 ...
#(This is intended to be chr)
str(iris)
'data.frame': 150 obs. of 5 variables:
$ Sepal.Length: num 5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ...
$ Sepal.Width : num 3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ...
$ Petal.Length: num 1.4 1.4 1.3 1.5 1.4 1.7 1.4 1.5 1.4 1.5 ...
$ Petal.Width : num 0.2 0.2 0.2 0.2 0.2 0.4 0.3 0.2 0.2 0.1 ...
$ Species : Factor w/ 3 levels setosa,versicolor,..: 1 1 1 1 1 1 1
1 1 1 ...
str(as.matrix(iris[,1:4]))
num [1:150, 1:4] 5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ...
- attr(*, dimnames)=List of 2
..$ : NULL
..$ : chr [1:4] Sepal.Length Sepal.Width Petal.Length Petal.Width
n.cols - ncol(data)
n.rows - nrow(data)
X - as.matrix(data)
stepsize - 0.05
c1 - (2 * pi) ** (n.cols / 2)
c2 - n.rows * (smoothing ** (n.cols + 2))
c3 - n.rows * (smoothing ** n.cols)
Kexp - function(sqs){
return (exp((-1 * sqs) / (2 * smoothing ** 2)))
}
FindGradient - function(x){
XmY - t(x - t(X))
sqsum - rowSums(XmY * XmY)
K - sapply(sqsum, Kexp)
dens - ((c1 * c3) ** -1) * sum(K)
grad - -1 * ((c1 * c2) ** -1) * colSums(K * XmY)
return (list(gradient = grad,
density = dens))
}
attractors - matrix(0, n.rows, n.cols)
densities - matrix(0, n.rows)
density.attractors -
sapply(rep(1:n.rows), function(i) {
notconverged - TRUE
# For each row loop through and find the attractor and density value.
x - (X[i, ])
iters - as.integer(1)
# Run gradient ascent for each point to obtain x*
while(notconverged == TRUE) {
find.gradient - FindGradient(x)
next.x - x + stepsize * find.gradient$gradient
change - sqrt(sum((next.x - x) * (next.x - x)))
notconverged - ifelse(change tol, TRUE, FALSE)
x - next.x
iters - iters + 1
}
# store the attractor and density value
return(c(densities[i, ] - find.gradient$density,
attractors[i, ] - x))
})
Error in while (notconverged == TRUE) { :
missing value where TRUE/FALSE needed
Any help would be great
Mike
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__
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and provide commented, minimal, self-contained, reproducible code.
--
Jim Holtman
Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
__
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Re: [R] Help transforming a dist
On Fri, Dec 23, 2011 at 1:06 AM, Taral tar...@gmail.com wrote: Perfect! combn was the trick I needed. Although I'll probably rbind the stuff instead of building a new frame. :) You should try rbind() with the small example I gave before you use it on your data. You'll immediately see why I didn't. :) On Thu, Dec 22, 2011 at 6:48 PM, Sarah Goslee sarah.gos...@gmail.com wrote: I'm not at all sure what you mean with your matrix: is that supposed to be three columns? What about: fakedata - data.frame(X=runif(3), Y = runif(3)) rownames(fakedata) - c(A, B, C) dist(fakedata) A B B 0.8617733 C 0.3813032 0.5124284 data.frame(t(combn(rownames(fakedata), 2)), dist=as.vector(dist(fakedata))) X1 X2 dist 1 A B 0.8617733 2 A C 0.3813032 3 B C 0.5124284 Sarah On Thu, Dec 22, 2011 at 7:03 PM, Taral tar...@gmail.com wrote: I'd like to convert a dist into a table/matrix/thingy of the form: A B value A C value B C value (assuming the dist was over 3 names). Is there a way to do this without using a for loop? __ 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] ggplot2: behaviour with empty datasets
See https://github.com/hadley/ggplot2/issues/31 - I totally agree that it's annoying. Hadley PS. You are more likely to get helpful responses about ggplot2 on the ggplot mailing list. On Fri, Dec 23, 2011 at 7:08 AM, Casper Ti. Vector caspervec...@gmail.com wrote: For example, prepare like this df.0 - data.frame(x = 0, y = 0, note = 1) df.1 - subset(df.0, note == 1) df.2 - subset(df.0, note == 2) Then a call to ggplot() + aes(x = x, y = y) + geom_point(data = df.1) + geom_point(data = df.2) produces the error Error in eval(expr, envir, enclos) : object 'x' not found And a call to ggplot(df.1, aes(x = x, y = y, shape = 4)) + geom_point() + geom_point(aes(shape = 1), df.2) plots both a cross (shape = 4) and circle (shape = 1) on position (0, 0). However, as expected, only a cross should be plotted because `dt.2' is empty. So with the current behaviour, if someone writes a script to plot datasets automatically and distinguish between subsets, he will need to write different code path for empty and non-empty data subsets. I think ggplot2 can just choose to plot nothing for empty data subsets, and only produce error when the dataset (union of a subsets) to plot is completely empty. This can make scripting easier based on ggplot2. -- Using GPG/PGP? Please get my current public key (ID: 0xAEF6A134, valid from 2010 to 2013) from a key server. __ 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. -- Assistant Professor / Dobelman Family Junior Chair Department of Statistics / Rice University 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.
[R] error from character NAs in indexing array
The help page for ?Extract says: When extracting, a numerical, logical or character NA index picks an unknown element and so returns NA in the corresponding element of a logical, integer, numeric, complex or character result, and NULL for a list. (It returns 00 for a raw result.] However, I seem to be getting different behavior when converting NA's to character indices for an array with character dimnames: str(frailmed) num [1:73, 1:2, 1:2] -0.315 -0.31 -0.252 -0.311 -0.29 ... - attr(*, dimnames)=List of 3 ..$ : chr [1:73] 18 19 20 21 ... ..$ : chr [1:2] Female Male ..$ : chr [1:2] FALSE TRUE frailmed[as.character(c(18,18,30)), Female, TRUE] 18 18 30 -0.3477757 -0.3477757 -0.3697707 frailmed[as.character(c(18,18,30,NA)), Female, TRUE] Error: subscript out of bounds I started thinking I might need to use grep or match to convert the numeric value for ageLB to a numeric index for [, but after looking at the manual is seems I shouldn't be forced to do that. as.character(NA) [1] NA frailmed[as.character(NA), Female, TRUE] Error: subscript out of bounds I'm guessing that NA_character_ is not being recognized by the extraction engine: frailmed[NA_character_, Female, TRUE] Error: subscript out of bounds Test subset: test - structure(c(-0.314905119091182, -0.309721294756184, -0.252154582680236, + 0.218589733943967, 0.227724425528112, 0.219274402692938, -0.347775659359612, + -0.351245759870099, -0.361973673972107, 0.15524082878, 0.149765287705234, + 0.150206947093079), .Dim = c(3L, 2L, 2L), .Dimnames = list(c(18, + 19, 20), c(Female, Male), c(FALSE, TRUE))) test , , FALSE Female Male 18 -0.3149051 0.2185897 19 -0.3097213 0.2277244 20 -0.2521546 0.2192744 , , TRUE Female Male 18 -0.3477757 0.1552556 19 -0.3512458 0.1497653 20 -0.3619737 0.1502069 test[NA, 1,1] NA NA NA NA NA NA test[as.character(NA), 1,1] Error: subscript out of bounds Is this intended behavior? I do see from that last successful execution that I may need to attack my problem by just processing complete cases, since the return of a vector of NA's with length 1 will also cause my logic to fail, but this behavior seems different than described. sessionInfo() R version 2.14.0 Patched (2011-11-13 r57650) Platform: x86_64-apple-darwin9.8.0/x86_64 (64-bit) locale: [1] en_US.UTF-8/en_US.UTF-8/en_US.UTF-8/C/en_US.UTF-8/en_US.UTF-8 attached base packages: [1] splines stats graphics grDevices utils datasets [7] methods base other attached packages: [1] rms_3.3-2Hmisc_3.9-0 survival_2.36-10 sos_1.3-1 [5] brew_1.0-6 lattice_0.20-0 loaded via a namespace (and not attached): [1] cluster_1.14.1 grid_2.14.0tools_2.14.0 -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problem with RCOM package
On 23.12.2011 11:02, KUMAR wrote: Hi, I am using R version 2.14.0 on windows 7. Trying to use RCOM package and followed example provided at (http://cran.r-project.org/web/packages/rcom/rcom.pdf) . Please ask on the mailing list that was set up for that package. I wonder: Do you have statconnDCOM installed. Uwe Ligges I am able to load the com object as seen in screenshot 2.png. however I am unable to invoke/get/set property to the object. Please guide me on the same, Regards; Kumar Anand 1.png 2.png __ 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] ggplot2: behaviour with empty datasets
Thanks, I'll post to that list if the encountered problem is just about ggplot2. And from the date of the issue on GitHub it seems that I might need to manually work around the problem some more times :) On Fri, Dec 23, 2011 at 08:10:05AM -0600, Hadley Wickham wrote: See https://github.com/hadley/ggplot2/issues/31 - I totally agree that it's annoying. Hadley PS. You are more likely to get helpful responses about ggplot2 on the ggplot mailing list. -- Using GPG/PGP? Please get my current public key (ID: 0xAEF6A134, valid from 2010 to 2013) from a key server. signature.asc Description: Digital signature __ 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] nlrob problem
Pascal A Niklaus pascal.nikl...@ieu.uzh.ch
on Mon, 19 Dec 2011 20:46:57 +0100 writes:
Dear all, I am not sure if this mail is for R-help or
should be sent to R-devel instead, and therefore post to
both.
While using nlrob from package 'robustbase', I ran into
the following problem:
For psi-functions that can become zero
(e.g. psi.bisquare), weights in the internal call to nls
can become zero. Example:
d - data.frame(x=1:5,y=c(2,3,5,10,9))
d.nlrob - nlrob(y ~ k*x,start=list(k=1),data=d,psi=psi.bisquare)
I think the problem is as follows: After the call to nls, a weighted
residual vector is calculated by dividing by sqrt(w). This generates
NaN's for weights of zero, which leads to problems in the subsequent
call to nls during the next iteration:
for (iiter in 1:maxit) {
...
w - psi(resid/Scale, ...)
...
data$w - sqrt(w)
...
out - nls( ., data=data ... )
...
resid - -residuals(out)/sqrt(w) # NaN for w=0
...
}
I wonder whether this problem shouldn't be dealt with by setting 'w' to
0 for the NaN cases.
I can get a fit by calling nlrob with na.action=na.exclude, but I'd have
intuitively assumed that na.action applies to the NAs in the data set
passed to nlrob, not to the one internally generated by nlrob and passed
to nls.
You are right.
The next version of robustbase (0.8-1) will have this fixed.
Note however, that for me, in your example and in other more
interesting ones, as soon as I use a redescending psi() function,
the IRLS iterations do *not* converge...
but rather strangely ``cycle'' around the true solution.
As a redescending psi() has a non-convex rho(), and non-linear
problems depend quite a bit on feasible/good starting
values, I currently think I would discourage from using such psi().
As others, some possibly more expert in robust non-linear fitting,
may be interested, and have interesting feedback,
I'm crossposting this reply to the R-SIG-robust mailing
list.
(Further replies should typically only go there!)
The second 'issue' is that the weights are passed as 'w', whereas the
documentation of 'nls' says weights should be given as 'weights' :
data: an optional data frame in which to evaluate the variables in
‘formula’ and ‘weights’. Can also be a list or an
environment, but not a matrix.
I think it would be good to mention in the documentation of 'nls' that
weights can be passed as 'w' as well.
You have overlooked that nlrob uses a different *formula* for nls()
--- nowadays unnecessarily ---
and that formula has a 'w'.
So this part of your report is not correct.
Martin Maechler, ETH Zurich
Pascal Niklaus
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[R] data vector to corresonding percentile ranks
I have a problem where I need to calculate the corresponding cohort percentile ranks for each of several variables. Essentially, what I need is a function that will calculate the distribution-free percentiles from each variable's data vector, returning a corresponding vector of percentiles: e.g.: percentile.my.data-/function/(my.data) I tried to make ecdf() perform this task but was unsuccessful. I'd be grateful for any help or advice... -- View this message in context: http://r.789695.n4.nabble.com/data-vector-to-corresonding-percentile-ranks-tp4228971p4228971.html Sent from the R help mailing list archive at Nabble.com. [[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] Long jobs completing without output
I've been running a glmer logit on a very large data set (600k obs). Running on a 10% subset works correctly, but for the complete data set, R completes apparently without error, but does not display the results. Given these jobs take about 200 hours, it's very hard to make progress by trial and error. I append the code and the sample and complete output. As is apparent, I upgraded R during the complete run, but I recall testing on the subsample with the earlier version too. I am also assuming that upgrading R will not affect the running process -- is this true? I'd be grateful for any leads. In the meantime I'll be running with larger subsamples! Regards, Brendan Halpin - code --- library(arm) library(foreign) mlm - read.dta(../workingdata.dta) attach(mlm) gender - as.factor(stu_gend) yr - year - 1998 failure - (lmer(fail ~ 1 + cao + subj1 + subj2 + subj3 + gender + yr + ageentry + as.factor(yrs5) + modsize + meancao + depfemr + (1|deptno) + (1|modinst) + (1|ulid) , na.action = na.exclude, family = binomial (link=logit))) display(failure, digits=5, detail=TRUE) -- - output with 10% sample data R version 2.14.0 (2011-10-31) Copyright (C) 2011 The R Foundation for Statistical Computing ISBN 3-900051-07-0 Platform: i486-pc-linux-gnu (32-bit) R is free software and comes with ABSOLUTELY NO WARRANTY. You are welcome to redistribute it under certain conditions. Type 'license()' or 'licence()' for distribution details. Natural language support but running in an English locale R is a collaborative project with many contributors. Type 'contributors()' for more information and 'citation()' on how to cite R or R packages in publications. Type 'demo()' for some demos, 'help()' for on-line help, or 'help.start()' for an HTML browser interface to help. Type 'q()' to quit R. library(arm) arm (Version 1.4-13, built: 2011-6-19) Working directory is /home/brendan/work/mlmmarks/genderECSR library(foreign) mlm - read.dta(../worksample-random1.dta) attach(mlm) gender - as.factor(stu_gend) yr - year - 1998 failure - (lmer(fail ~ + 1 + cao + subj1 + subj2 + subj3 + gender + yr + ageentry + as.factor(yrs5) + + modsize + meancao + depfemr + (1|deptno) + (1|modinst) + (1|ulid) , na.action = na.exclude, family = binomial (link=logit))) display(failure, digits=5, detail=TRUE) glmer(formula = fail ~ 1 + cao + subj1 + subj2 + subj3 + gender + yr + ageentry + as.factor(yrs5) + modsize + meancao + depfemr + (1 | deptno) + (1 | modinst) + (1 | ulid), family = binomial(link = logit), na.action = na.exclude) coef.est coef.se z value Pr(|z|) (Intercept)2.63826 0.97870 2.69568 0.00702 cao -2.08963 0.11987 -17.43314 0.0 subj1 0.02608 0.23573 0.11064 0.91190 subj2 -0.55668 0.32759 -1.69932 0.08926 subj3 -1.57120 0.30664 -5.12400 0.0 genderM0.36368 0.09188 3.95845 0.8 yr 0.06067 0.01658 3.65996 0.00025 ageentry -0.00720 0.04338 -0.16598 0.86817 as.factor(yrs5)1 -0.25181 0.05712 -4.40806 0.1 as.factor(yrs5)2 -0.54725 0.07601 -7.20005 0.0 as.factor(yrs5)3 -1.07483 0.08660 -12.41184 0.0 as.factor(yrs5)4 -1.22447 0.14373 -8.51932 0.0 as.factor(yrs5)5 -1.55032 0.31342 -4.94653 0.0 modsize0.03387 0.02533 1.33733 0.18112 meancao1.08747 0.10748 10.11780 0.0 depfemr -1.49097 0.49350 -3.02122 0.00252 Error terms: Groups NameStd.Dev. modinst (Intercept) 1.14308 ulid (Intercept) 1.54030 deptno (Intercept) 0.52497 Residual 1.0 --- number of obs: 63254, groups: modinst, 9076; ulid, 2275; deptno, 26 AIC = 30275.2, DIC = 30237.2 deviance = 30237.2 Loading required package: MASS Loading required package: Matrix Loading required package: lattice Attaching package: ‘Matrix’ The following object(s) are masked from ‘package:base’: det Loading required package: lme4 Attaching package: ‘lme4’ The following object(s) are masked from ‘package:stats’: AIC, BIC Loading required package: R2WinBUGS Loading required package: coda Attaching package: ‘coda’ The following object(s) are masked from ‘package:lme4’: HPDinterval Loading required package: abind Loading required package: foreign Attaching package: ‘arm’ The following object(s) are masked from ‘package:coda’: traceplot -- - output with complete data -- R version 2.13.1 (2011-07-08) Copyright (C) 2011 The R Foundation for Statistical Computing ISBN 3-900051-07-0
Re: [R] Help with code
this is how the ouput from the code should be structure(list(HTN = 1:10, HTN_FDR = structure(c(4L, 2L, 1L, 2L, 3L, 1L, 1L, 2L, 3L, 2L), .Label = c(Ctrl_noc, T1D_noc, T1D_oc, T1d_w), class = factor), Dyslipidemia = structure(c(3L, 2L, 1L, 2L, 4L, 1L, 1L, 2L, 4L, 2L), .Label = c(Ctrl_noc, T1D_noc, T1D_oc, T1D_w), class = factor), CAD = structure(c(3L, 2L, 1L, 2L, 3L, 1L, 1L, 2L, 3L, 2L), .Label = c(Ctrl_noc, T1D_noc, T1D_oc), class = factor), CAD_FDR = structure(c(3L, 2L, 1L, 2L, 3L, 1L, 1L, 2L, 3L, 2L), .Label = c(Ctrl_noc, T1D_noc, T1D_oc), class = factor), Prior_MI = structure(c(3L, 2L, 1L, 2L, 3L, 1L, 1L, 2L, 3L, 2L), .Label = c(Ctrl_noc, T1D_noc, T1D_oc), class = factor), t1d_ptype = structure(c(3L, 3L, 2L, 3L, 1L, 1L, 2L, 3L, 3L, 3L), .Label = c(Ctrl, Ctrl_FDR, T1D), class = factor)), .Names = c(HTN, HTN_FDR, Dyslipidemia, CAD, CAD_FDR, Prior_MI, t1d_ptype), class = data.frame, row.names = c(NA, -10L)) -- View this message in context: http://r.789695.n4.nabble.com/Help-with-code-tp4218989p4228759.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] missing value where TRUE/FALSE needed
Apologies, I was using top = 0.0001
I had looked at browser and did show notconverged = NA. But I couldn't
understand why it worked for one and not the other?
On Friday, 23 December 2011, jim holtman jholt...@gmail.com wrote:
Does this look similar to the error you are getting:
while(NA == TRUE) 1
Error in while (NA == TRUE) 1 : missing value where TRUE/FALSE needed
SO 'notconverged' is probably equal to NA. BTW, what is the value of
'tol'; I do not see it defined. So when computing 'notconverged' you
have generated an NA. You can test it to see if this is true.
You can use the following command:
options(error=utils::recover)
and then learn how to use the 'browser' to examine variables when the
error occurs.
On Fri, Dec 23, 2011 at 5:44 AM, Michael Pearmain
michael.pearm...@gmail.com wrote:
Merry Xmas to all,
I am writing a function and curiously this runs sometimes on one data set
and fails on another and i cannot figure out why.
Any help much appreciated.
If i run the code below with
data - iris[ ,1:4]
The code runs fine, but if i run on a large dataset i get the following
error (showing data structures as matrix is large)
str(cluster.data)
num [1:9985, 1:811] 0 0 0 0 0 0 0 0 0 0 ...
- attr(*, dimnames)=List of 2
..$ : NULL
..$ : chr [1:811] 1073949105 1073930585 1073843224 1073792624
...
#(This is intended to be chr)
str(iris)
'data.frame': 150 obs. of 5 variables:
$ Sepal.Length: num 5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ...
$ Sepal.Width : num 3.5 3 3.2 3.1 3.6 3.9 3.4 3.4 2.9 3.1 ...
$ Petal.Length: num 1.4 1.4 1.3 1.5 1.4 1.7 1.4 1.5 1.4 1.5 ...
$ Petal.Width : num 0.2 0.2 0.2 0.2 0.2 0.4 0.3 0.2 0.2 0.1 ...
$ Species : Factor w/ 3 levels setosa,versicolor,..: 1 1 1 1 1
1 1
1 1 1 ...
str(as.matrix(iris[,1:4]))
num [1:150, 1:4] 5.1 4.9 4.7 4.6 5 5.4 4.6 5 4.4 4.9 ...
- attr(*, dimnames)=List of 2
..$ : NULL
..$ : chr [1:4] Sepal.Length Sepal.Width Petal.Length
Petal.Width
n.cols - ncol(data)
n.rows - nrow(data)
X - as.matrix(data)
stepsize - 0.05
c1 - (2 * pi) ** (n.cols / 2)
c2 - n.rows * (smoothing ** (n.cols + 2))
c3 - n.rows * (smoothing ** n.cols)
Kexp - function(sqs){
return (exp((-1 * sqs) / (2 * smoothing ** 2)))
}
FindGradient - function(x){
XmY - t(x - t(X))
sqsum - rowSums(XmY * XmY)
K - sapply(sqsum, Kexp)
dens - ((c1 * c3) ** -1) * sum(K)
grad - -1 * ((c1 * c2) ** -1) * colSums(K * XmY)
return (list(gradient = grad,
density = dens))
}
attractors - matrix(0, n.rows, n.cols)
densities - matrix(0, n.rows)
density.attractors -
sapply(rep(1:n.rows), function(i) {
notconverged - TRUE
# For each row loop through and find the attractor and density
value.
x - (X[i, ])
iters - as.integer(1)
# Run gradient ascent for each point to obtain x*
while(notconverged == TRUE) {
find.gradient - FindGradient(x)
next.x - x + stepsize * find.gradient$gradient
change - sqrt(sum((next.x - x) * (next.x - x)))
notconverged - ifelse(change tol, TRUE, FALSE)
x - next.x
iters - iters + 1
}
# store the attractor and density value
return(c(densities[i, ] - find.gradient$density,
attractors[i, ] - x))
})
Error in while (notconverged == TRUE) { :
missing value where TRUE/FALSE needed
Any help would be great
Mike
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Data Munger Guru
What is the problem that you are trying to solve?
Tell me what you want to do, not how you want to do it.
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[R] Applying a function
Hi, I need help writing a function I capture seal pups mutliple times during the lactation season in order to monitor their growth rate. When I release them, the recovery (mother-pup) time is not the same for all individuals. I want to know if individuals that recover their mother the fastest are the ones with the highest growth rates. So, I noted at every release if the pup reovered his mother before we leave (yes or no). My dataframe looks like this Capture nb individual individual capture motherrecovery growth rate 1 1 1 n 0.5 2 1 2 y 0.5 3 1 3 y 0.5 4 1 4 y 0.5 5 1 5 n 0.5 6 2 1 y 0.3 7 2 2 y 0.3 8 3 1 y 0.4 9 3 2 n 0.4 10 3 3 y 0.4 ... I want to calculate a rate of mother recovery by individual, i.e. nb of recoveries (y)/nb of captures. So for indivial 1 it would be 3/5 = 0.6. I want to write a function that does this for all the individuals in my dataframe, i.e. around 400 individuals (this is why I want to write a function, it would be too long by hand) Thank you, Joanie Van de Walle Ãtudiante à la Maîtrise en Biologie Département de Biologie 1045, avenue de la Médecine Pavillon Vachon, bureau 2044 Université Laval Québec, Canada, G1V 0A6 __ 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] nlrob problem
Please ignore my previous posting, in the meanwhile I have realised that I did not look carefully enough how nlrob works (irls), although this is pretty obvious from the code. The only issue that remains is the case of zero weight for redescending M estimators, in which case NaN's are produced. I think it would be better if one could avoid the NaN's produced by dividing by zero when psi becomes zero. Something along the line of replacing resid - -residuals(out)/sqrt(w) by resid - ifelse(w0,-residuals(out)/sqrt(w),0); There probably is a more elegant way to do this. This would avoid that one has to use na.action=na.exclude for cases where there are no NAs in the data set originally passed to nlrob (and the NAs only occur 'internally' in the code of nlrob). Hope I haven't missed something else this time... Pascal Niklaus -- View this message in context: http://r.789695.n4.nabble.com/nlrob-problem-tp4215473p4229016.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] missing value where TRUE/FALSE needed
Given that the maximum floating point value is:
$double.xmax
[1] 1.797693e+308
and the number you are trying to calculate is 5E323 you are exceeding
the size of numbers you can process.
Have a happy holiday and glad I could help.
On Fri, Dec 23, 2011 at 11:24 AM, Michael Pearmain
michael.pearm...@gmail.com wrote:
Thanks for al your help with Jim,
I realise this is really a case of debugging (and so falls to the code
writer) using your help i have traced the NaN to be from the c1 calculation
in the FindGradient function.
there is a place where we use the calculation of c1...
which at the start was calculated as
c1 - (2*pi) ** (n.cols/2)
when i use
c1 - (2*pi) ** (811/2)
[1] Inf
So i think i have found a ceiling on the number of cols i can use to make a
calcuation.
Many thanks again for your help, especially on 23rd Dec!
Mike
On 23 December 2011 16:01, jim holtman jholt...@gmail.com wrote:
The next step is to now evaluate the objects used in the calculation;
e.g., look at the objects in the statement:
change - sqrt(sum((next.x - x) * (next.x - x)))
What do each of them show? Most likely you have something in the data
that causing this statement to evaluate to NaN (not a number) and this
is caused by the values of one of these as not being what you think it
is. From the value of 'i', it looks like the first iteration (or at
least when it has the value of 1). Notice that X is defined as a
matrix with NAs as the initial values and you are picking up the first
row (X[i,]) which on the initial pass is an NA. So start by examining
the values of all the objects used up to that point and what the
values of each expression is.
On Fri, Dec 23, 2011 at 10:38 AM, Michael Pearmain
michael.pearm...@gmail.com wrote:
Thanks for guidance;
the return from the dump is below:
So i can see that the NaN is causing NA, as its evaluating to NA in the
ifelse.
So i guess my next question is why this happens for my large data set
but
not he iris dataset?
Thank you again for your help on this, i really appreciate you taking
the
time to help
1: DensitiesAndAttractors(data = cluster.data, smoothing = 0.34, tol =
1e-05,
2: denclue_par.R#136: sfSapply(rep(1:n.rows), function(i) {
3: sapply(x, fun, ..., simplify = simplify, USE.NAMES = USE.NAMES)
4: lapply(X, FUN, ...)
5: FUN(1:9985[[1]], ...)
Selection: 5
Called from: function ()
{
if (.isMethodsDispatchOn()) {
tState - tracingState(FALSE)
on.exit(tracingState(tState))
}
calls - sys.calls()
from - 0L
n - length(calls)
if (identical(sys.function(n), recover))
n - n - 1L
for (i in rev(seq_len(n))) {
calli - calls[[i]]
fname - calli[[1L]]
if (!is.na(match(deparse(fname)[1L], c(methods::.doTrace,
.doTrace {
from - i - 1L
break
}
}
if (from == 0L)
for (i in rev(seq_len(n))) {
calli - calls[[i]]
fname - calli[[1L]]
if (!is.name(fname) || is.na(match(as.character(fname),
c(recover, stop, Stop {
from - i
break
}
}
if (from 0L) {
if (!interactive()) {
try(dump.frames())
cat(gettext(recover called non-interactively; frames
dumped,
use debugger() to view\n))
return(NULL)
}
else if (identical(getOption(show.error.messages),
FALSE))
return(NULL)
calls - limitedLabels(calls[1L:from])
repeat {
which - menu(calls, title = \nEnter a frame number, or 0
to
exit )
if (which)
eval(substitute(browser(skipCalls = skip), list(skip = 7
-
which)), envir = sys.frame(which))
else break
}
}
else cat(gettext(No suitable frames for recover()\n))
}()
Browse[1] ls()
[1] change find.gradient i iters
[5] next.x notconverged x
Browse[1] change
[1] NaN
Browse[1] i
[1] 1
Browse[1]
On 23 December 2011 15:22, jim holtman jholt...@gmail.com wrote:
According to your dump, if you look in section 5 (this is the function
being call by 'sapply') you should do an 'ls()' and see the object
'notconverged' and if you type its name, I am guessing that is that
the value NA. Also look at value of 'change' and 'tol' to determine
how the 'ifelse' was evaluated.
Let me know how it goes.
On Fri, Dec 23, 2011 at 9:35 AM, Michael Pearmain
michael.pearm...@gmail.com wrote:
Hi Jim,
Sorry to email directly, i wasn't sure what i was looking for, can
you
give
me a little guidance then i can post back to the list? Many thanks
density.data - DensitiesAndAttractors(data = cluster.data,
Re: [R] data vector to corresonding percentile ranks
I'm not sure I understand the question, but does quantile() do what you want? On Fri, Dec 23, 2011 at 10:28 AM, Steve Jones sjone...@jhmi.edu wrote: I have a problem where I need to calculate the corresponding cohort percentile ranks for each of several variables. Essentially, what I need is a function that will calculate the distribution-free percentiles from each variable's data vector, returning a corresponding vector of percentiles: e.g.: percentile.my.data-/function/(my.data) I tried to make ecdf() perform this task but was unsuccessful. I'd be grateful for any help or advice... -- Sarah Goslee http://www.functionaldiversity.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help with code
On Fri, Dec 23, 2011 at 9:25 AM, 1Rnwb sbpuro...@gmail.com wrote: this is how the ouput from the code should be What code might that be? Readers of the R-help email list have no idea whatsoever what you're asking. Please include context, as requested in the posting guide. Sarah structure(list(HTN = 1:10, HTN_FDR = structure(c(4L, 2L, 1L, 2L, 3L, 1L, 1L, 2L, 3L, 2L), .Label = c(Ctrl_noc, T1D_noc, T1D_oc, T1d_w), class = factor), Dyslipidemia = structure(c(3L, 2L, 1L, 2L, 4L, 1L, 1L, 2L, 4L, 2L), .Label = c(Ctrl_noc, T1D_noc, T1D_oc, T1D_w), class = factor), CAD = structure(c(3L, 2L, 1L, 2L, 3L, 1L, 1L, 2L, 3L, 2L), .Label = c(Ctrl_noc, T1D_noc, T1D_oc), class = factor), CAD_FDR = structure(c(3L, 2L, 1L, 2L, 3L, 1L, 1L, 2L, 3L, 2L), .Label = c(Ctrl_noc, T1D_noc, T1D_oc), class = factor), Prior_MI = structure(c(3L, 2L, 1L, 2L, 3L, 1L, 1L, 2L, 3L, 2L), .Label = c(Ctrl_noc, T1D_noc, T1D_oc), class = factor), t1d_ptype = structure(c(3L, 3L, 2L, 3L, 1L, 1L, 2L, 3L, 3L, 3L), .Label = c(Ctrl, Ctrl_FDR, T1D), class = factor)), .Names = c(HTN, HTN_FDR, Dyslipidemia, CAD, CAD_FDR, Prior_MI, t1d_ptype), class = data.frame, row.names = c(NA, -10L)) -- Sarah Goslee http://www.functionaldiversity.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] data vector to corresonding percentile ranks
On Dec 23, 2011, at 10:28 AM, Steve Jones wrote: I have a problem where I need to calculate the corresponding cohort percentile ranks for each of several variables. Essentially, what I need is a function that will calculate the distribution-free percentiles from each variable's data vector, returning a corresponding vector of percentiles: e.g.: percentile.my.data-/function/(my.data) I tried to make ecdf() perform this task but was unsuccessful. Unsuccessful? How? Seems like a reasonable strategy: set.seed(123) x - rnorm(1000) xCdist - ecdf(x) Seems to give sensible results. x[1] [1] -0.7104066 100*xCdist(x[1]) [1] 23.4 x[2] [1] 0.2568837 100*xCdist(x[2]) [1] 60 I'd be grateful for any help or advice... My advice would be to post what code you were trying so that you can get help understand what difficulties you need to overcome. -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] data vector to corresonding percentile ranks
It's far more useful to send your explanation to the list than it is to send it just to me. I've taken the liberty of doing so. But this does sound like a job for ecdf() - what did you do, and what went wrong? On Fri, Dec 23, 2011 at 12:23 PM, Steven Jones sjone...@jhmi.edu wrote: Actually, what I am trying to do is very simple. I want to take a large vector of data, about 1.5 million observations and return a corresponding vector of percentiles, each corresponding to the elements of the original data vector in a 1:1 correspondence. Original data vector: x--(x1,x2,x3,...xn) ^ ^ ^ | | | Derived percentile vector: p--(p1,p2,p3,...pn) Where pn--is the percentile value of xn determined from the rank in the data vector x. Ideally, there is a function which could return a vector p from my original data vector x. My understanding, which may be incorrect, is that quantile() function returns the value of x for a specified percentile rank in a vector. Steve -Original Message- From: Sarah Goslee [mailto:sarah.gos...@gmail.com] Sent: Friday, December 23, 2011 12:11 PM To: Steven Jones Cc: r-help@r-project.org Subject: Re: [R] data vector to corresonding percentile ranks I'm not sure I understand the question, but does quantile() do what you want? On Fri, Dec 23, 2011 at 10:28 AM, Steve Jones sjone...@jhmi.edu wrote: I have a problem where I need to calculate the corresponding cohort percentile ranks for each of several variables. Essentially, what I need is a function that will calculate the distribution-free percentiles from each variable's data vector, returning a corresponding vector of percentiles: e.g.: percentile.my.data-/function/(my.data) I tried to make ecdf() perform this task but was unsuccessful. I'd be grateful for any help or advice... __ 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] black and white in qplot? layout 4 graphs in one screen
You might find the ggplot mailing list a friendlier place to ask questions about ggplot2. Hadley On Wed, Dec 21, 2011 at 2:16 PM, rachaelohde cox.rach...@gmail.com wrote: Hello, I am trying to plot means and standard errors conditioned by a factor, using qplot. I am successful at getting the bar graph I want with a error bar, however I have tried many things and cannot get the bars to change colors. Currently showing as red and blue, but need it to be black and white for publication. Any suggestions please? Using a data set June, which is str: 'data.frame': 21 obs. of 6 variables: $ BLCK : Factor w/ 4 levels ,B,I,W: 3 4 2 3 2 2 2 4 3 4 ... $ PLOT : int 3 3 6 1 2 5 1 1 2 2 ... $ TRT : Factor w/ 5 levels ,crop,ten,..: 2 2 2 3 3 3 4 4 4 5 ... $ Date : Factor w/ 8 levels , 6/16/11,..: 2 2 2 2 2 2 2 2 2 2 ... $ habitat : Factor w/ 3 levels ,C,P: 2 2 2 2 2 2 2 2 2 2 ... $ Abundance: num 0.333 0 1.333 0 1.667 ... Current code is: Ab.avg-ddply(June, c(TRT, habitat), function(df) return(c(Ab.avg=mean(df$Abundance), Ab.sd=sd(df$Abundance avg.plot-qplot(TRT, Ab.avg, fill=factor(habitat), data=Ab.avg, geom=bar, position=dodge) dodge - position_dodge(width=0.9) avg.plot++geom_linerange(aes(ymax=Ab.avg+Ab.sd, ymin=Ab.avg-Ab.sd), position=dodge)+theme_bw() http://r.789695.n4.nabble.com/file/n4223035/june_bar_graph.png Also, would like to plot 4 of these bar graphs (for four dates) on the same screen, I cannot get the par() or layout() function to work with qplot. Is there another way? Thank you! -- View this message in context: http://r.789695.n4.nabble.com/black-and-white-in-qplot-layout-4-graphs-in-one-screen-tp4223035p4223035.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. -- Assistant Professor / Dobelman Family Junior Chair Department of Statistics / Rice University 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] Stacked area plot for time series
You are more likely to get a helpful response if you provide a reproducible example - without that I can only guess that you need to use approx so you get y values at same x values. Hadley On Wed, Dec 21, 2011 at 8:13 AM, UncleFish bpn...@ucsd.edu wrote: I wish to make a stacked area chart of a time series with three variables. The observations are very irregular, covering 500 years, with a few historical observations in the range 1500-1850, and then more regular observations since 1880 or so. If I plot this in Excell it truncates the time series, but it can make a nice stacked area plot. My question is how can I have a correct x axis with a stacked plot in R? I thave tried the data in a zoo as well as a ts, and have tried ggplot2 and plotrix, but cannot get beyond a lattice plot. -- View this message in context: http://r.789695.n4.nabble.com/Stacked-area-plot-for-time-series-tp4221849p4221849.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. -- Assistant Professor / Dobelman Family Junior Chair Department of Statistics / Rice University 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] vif function using lm object
On Wed, Dec 21, 2011 at 9:28 AM, arunkumar akpbond...@gmail.com wrote: Hi, can anyone please explain why the vif should have more than 2 terms. *vif.lm(lmobj) : model contains fewer than 2 terms* why it is throwng error if it is one variable. The _VIF_ is a measure of _multicollinearity_, which occurs between two or more predictors in a regression model. In other words, you cannot have collinearity in a single predictor. Also, to compute the VIF you need to regress one regressor on all the remaining regressors. If you have a single regressor, then you cannot estimate a VIF. See the related articles on Wikipedia. Regards Liviu - Thanks in Advance Arun -- View this message in context: http://r.789695.n4.nabble.com/vif-function-using-lm-object-tp4220904p4220904.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. -- Do you know how to read? http://www.alienetworks.com/srtest.cfm http://goodies.xfce.org/projects/applications/xfce4-dict#speed-reader Do you know how to write? http://garbl.home.comcast.net/~garbl/stylemanual/e.htm#e-mail __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] data vector to corresonding percentile ranks
This seems like what you are interested in: x - rnorm(50) ecdf(x)(x) Michael On Fri, Dec 23, 2011 at 11:26 AM, Sarah Goslee sarah.gos...@gmail.com wrote: It's far more useful to send your explanation to the list than it is to send it just to me. I've taken the liberty of doing so. But this does sound like a job for ecdf() - what did you do, and what went wrong? On Fri, Dec 23, 2011 at 12:23 PM, Steven Jones sjone...@jhmi.edu wrote: Actually, what I am trying to do is very simple. I want to take a large vector of data, about 1.5 million observations and return a corresponding vector of percentiles, each corresponding to the elements of the original data vector in a 1:1 correspondence. Original data vector: x--(x1,x2,x3,...xn) ^ ^ ^ | | | Derived percentile vector: p--(p1,p2,p3,...pn) Where pn--is the percentile value of xn determined from the rank in the data vector x. Ideally, there is a function which could return a vector p from my original data vector x. My understanding, which may be incorrect, is that quantile() function returns the value of x for a specified percentile rank in a vector. Steve -Original Message- From: Sarah Goslee [mailto:sarah.gos...@gmail.com] Sent: Friday, December 23, 2011 12:11 PM To: Steven Jones Cc: r-help@r-project.org Subject: Re: [R] data vector to corresonding percentile ranks I'm not sure I understand the question, but does quantile() do what you want? On Fri, Dec 23, 2011 at 10:28 AM, Steve Jones sjone...@jhmi.edu wrote: I have a problem where I need to calculate the corresponding cohort percentile ranks for each of several variables. Essentially, what I need is a function that will calculate the distribution-free percentiles from each variable's data vector, returning a corresponding vector of percentiles: e.g.: percentile.my.data-/function/(my.data) I tried to make ecdf() perform this task but was unsuccessful. I'd be grateful for any help or advice... __ 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] Problem with RCOM package
All questions related to rcom, statconnDCOM, and RExcel should be posted on the mailing list of the statconnDCOM project. You can subscribe at http://rcom.univie.ac.at On 12/23/2011 11:02 AM, KUMAR wrote: Hi, I am using R version 2.14.0 on windows 7. Trying to use RCOM package and followed example provided at (http://cran.r-project.org/web/packages/rcom/rcom.pdf) . I am able to load the com object as seen in screenshot 2.png. however I am unable to invoke/get/set property to the object. Please guide me on the same, Regards; Kumar Anand __ 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] cast in reshape and reshape2
library(reshape2)
x = melt(airquality, id=c('month', 'day'))
With reshape I can cast with multiple functions:
library(reshape)
cast(x, month+variable~., c(mean,sd))
month variable mean sd
1 5ozone 23.615385 22.224449
2 5 solar.r 181.296296 115.075499
3 5 wind 11.622581 3.531450
4 5 temp 65.548387 6.854870
5 6ozone 29.44 18.207904
6 6 solar.r 190.17 92.882975
7 6 wind 10.27 3.769234
8 6 temp 79.10 6.598589
9 7ozone 59.115385 31.635837
10 7 solar.r 216.483871 80.568344
11 7 wind 8.941935 3.035981
12 7 temp 83.903226 4.315513
13 8ozone 59.961538 39.681210
14 8 solar.r 171.857143 76.834943
15 8 wind 8.793548 3.225930
16 8 temp 83.967742 6.585256
17 9ozone 31.448276 24.141822
18 9 solar.r 167.43 79.118280
19 9 wind 10.18 3.461254
20 9 temp 76.90 8.355671
Is there a way to do the same job with reshape2?
--
Kaiyin Zhong
--
Neuroscience Research Institute, Peking University, Beijing, P.R.China 100038
__
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] Applying a function
First, it may be a good idea to use 1 for âyesâ and 0 for ânoâ in the motherrecovery column. Then if you name your table e.g tab1, you may try something like this.. sum(tab1[,2]==1*tab1[,4])/sum(tab1[,2]==1) 2011/12/23 Joanie Van De Walle joanie.van-de-wall...@ulaval.ca Hi, I need help writing a function I capture seal pups mutliple times during the lactation season in order to monitor their growth rate. When I release them, the recovery (mother-pup) time is not the same for all individuals. I want to know if individuals that recover their mother the fastest are the ones with the highest growth rates. So, I noted at every release if the pup reovered his mother before we leave (yes or no). My dataframe looks like this Capture nb individual individual capture motherrecovery growth rate 1 1 1 n 0.5 2 1 2 y 0.5 3 1 3 y 0.5 4 1 4 y 0.5 5 1 5 n 0.5 6 2 1 y 0.3 7 2 2 y 0.3 8 3 1 y 0.4 9 3 2 n 0.4 10 3 3 y 0.4 ... I want to calculate a rate of mother recovery by individual, i.e. nb of recoveries (y)/nb of captures. So for indivial 1 it would be 3/5 = 0.6. I want to write a function that does this for all the individuals in my dataframe, i.e. around 400 individuals (this is why I want to write a function, it would be too long by hand) Thank you, Joanie Van de Walle Ãâ°tudiante à la Maîtrise en Biologie Département de Biologie 1045, avenue de la Médecine Pavillon Vachon, bureau 2044 Université Laval Québec, Canada, G1V 0A6 __ 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] Custom XML Readers
I need to construct a custom XML reader, the files I'm working with are in funky XML format: str name=authorPaul H/str str name=countryUSA/str date name=created_date2010-02-16/date I want to read the file so it looks like: author = Paul H country = USA created_date=2010-02-16 Does any one know how to go about this problem, or know of good references i could access? Thanks, Andy -- View this message in context: http://r.789695.n4.nabble.com/Custom-XML-Readers-tp4229614p4229614.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] Stacked area plot for time series
Yes, thanks for the direction. I am new to the list and new to R, and I can see that was not a very helpful description. I have since resolved my issue. My main issue was that stackpoly does not (I think) interact with XOO objects or ts object, and my secondary issue was that my time series was highly irregular (with inferred indigenous historical consumption rates of fish, separated by large gaps). I resolved this by taking my data, making it a zoo to order it, transforming to a ts to interpolate missing values, and transforming to a matrix to do the stackpoly plot. I then resolved the issue of too many tick marks on the x axis by making a new axis. I did seem to have difficulty with some plot functions apparently do not to function in plotrix (specifically xaxt='n'). I am aware that stacked area charts might not an acceptable graphical representation of the data in some opinions, and I tend to agree. However, this output had to match the format of other graphs in the work, and was thus desired in this specific format. Thanks for providing this stackpoly tool - very handy! Code follows, and a final graphic is attached: library(zoo) library(plotrix) setwd(~/Documents/Loren/RawData) CaribData = read.csv(FLA_Loren.csv, header = TRUE) attach(CaribData) ## Create ZOO to order by time z1=zoo(Commercial,Year) z2=zoo(Recreational,Year) z3=zoo(Subsistence,Year) z4-merge(z1,z2,z3) ##Convert to ts to interpolate missing NA values ts1-as.ts(z1) ts1Ex-na.approx(ts1) ts2-as.ts(z2) ts2Ex-na.approx(ts2) ts3-as.ts(z3) ts3Ex-na.approx(ts3) ## Merge into one object ts4Ex-cbind(ts1Ex, ts2Ex, ts3Ex, deparse.level = 1) ##Test plots to examine series plot(ts1Ex) plot(ts2Ex) plot(ts3Ex) plot(ts4Ex) ## Convert for stacked area plotting and plot CaribMatrix-as.matrix(ts4Ex) stackpoly(CaribMatrix, stack=TRUE, xlab=Year,ylab=Catch in tons, main=Historical Florida reef fisheries catch by sector, axis4=FALSE) stackpoly(CaribMatrix, stack=TRUE, xlab=Year,ylab=Catch in tons, col.main=red, font.main=4, axis4=FALSE, xat=seq(50,760,by=100), xaxlab=c(1300, 1400, 1500, 1600, 1700, 1800, 1900, 2000)) -Original Message- From: h.wick...@gmail.com on behalf of Hadley Wickham Sent: Fri 12/23/2011 9:41 AM To: UncleFish Cc: r-help@r-project.org Subject: Re: [R] Stacked area plot for time series You are more likely to get a helpful response if you provide a reproducible example - without that I can only guess that you need to use approx so you get y values at same x values. Hadley On Wed, Dec 21, 2011 at 8:13 AM, UncleFish bpn...@ucsd.edu wrote: I wish to make a stacked area chart of a time series with three variables. The observations are very irregular, covering 500 years, with a few historical observations in the range 1500-1850, and then more regular observations since 1880 or so. If I plot this in Excell it truncates the time series, but it can make a nice stacked area plot. My question is how can I have a correct x axis with a stacked plot in R? I thave tried the data in a zoo as well as a ts, and have tried ggplot2 and plotrix, but cannot get beyond a lattice plot. -- View this message in context: http://r.789695.n4.nabble.com/Stacked-area-plot-for-time-series-tp4221849p4221849.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. -- Assistant Professor / Dobelman Family Junior Chair Department of Statistics / Rice University 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] Cuzick's test for trend
I just wrote this up for a project, perhaps 5 years is better than never...
code
cuzick = function(x,z,test.type=c(two.sided, upper, lower)) {
N = length(z)
n = unique(z)
ranks=rank(x)
T = sum(ranks*z)
p = (table(z)/N)
E_Z = sum(unique(z)*p)
E_T = 0.5*N*(N+1)*E_Z
Var_Z = sum(unique(z)^2*p) - E_Z^2
Var_T = N^2*(N+1)/12*Var_Z
Zscore = (T-E_T)/sqrt(Var_T)
if(test.type == two.sided) {
pval = 2*pnorm(-abs(Zscore))
} else if(test.type == upper) {
pval = pnorm(Zscore,lower.tail=F)
} else pval = pnorm(Zscore,lower.tail=T)
out = data.frame(cbind(Zscore,pval,test.type))
colnames(out) = c(Z,p,testType)
return(out)
}
/code
With data from the Cuzick 1985 paper:
code
z = c(rep(1, 8), rep(2,10), rep(3,9), rep(4,9),rep(5,9))
x = c(0, 0, 1, 1, 2, 2, 4, 9,
0, 0, 5, 7, 8, 11, 13, 23, 25, 97,
2, 3, 6, 9, 10, 11, 11, 12, 21,
0, 3, 5, 6, 10, 19, 56, 100, 132,
2, 4, 6, 6, 6, 7, 18, 39, 60)
cuzick(x,z,two.sided)
Z p testType
1 2.11 0.0348 two.sided
/code
arin basu-2 wrote
Hi All:
I was looking for, but could not locate in the packages, or in the R
archive searches if there exists an R implementation of Cuzick's test of
trend. The test is described as follows:
An extension of the Wilcoxon rank-sum test is developed to handle the
situation in which a variable is measured for individuals in three or more
(ordered) groups and a non-parametric test for trend across these groups
is desired.
Reference:
Cuzick J. A Wilcoxon-type test for trend. Stat Med. 1985
Jan-Mar;4(1):87-90
Would greatly appreciate your insights. The R version I use is R-2.3.1
Best,
Arin Basu
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Re: [R] Stacked area plot for time series
Meant to say ZOO objects . . . -Original Message- From: Ben Neal [mailto:bn...@spg.ucsd.edu] Sent: Fri 12/23/2011 10:15 AM To: Hadley Wickham; UncleFish Cc: r-help@r-project.org Subject: RE: [R] Stacked area plot for time series Yes, thanks for the direction. I am new to the list and new to R, and I can see that was not a very helpful description. I have since resolved my issue. My main issue was that stackpoly does not (I think) interact with XOO objects or ts object, and my secondary issue was that my time series was highly irregular (with inferred indigenous historical consumption rates of fish, separated by large gaps). I resolved this by taking my data, making it a zoo to order it, transforming to a ts to interpolate missing values, and transforming to a matrix to do the stackpoly plot. I then resolved the issue of too many tick marks on the x axis by making a new axis. I did seem to have difficulty with some plot functions apparently do not to function in plotrix (specifically xaxt='n'). I am aware that stacked area charts might not an acceptable graphical representation of the data in some opinions, and I tend to agree. However, this output had to match the format of other graphs in the work, and was thus desired in this specific format. Thanks for providing this stackpoly tool - very handy! Code follows, and a final graphic is attached: library(zoo) library(plotrix) setwd(~/Documents/Loren/RawData) CaribData = read.csv(FLA_Loren.csv, header = TRUE) attach(CaribData) ## Create ZOO to order by time z1=zoo(Commercial,Year) z2=zoo(Recreational,Year) z3=zoo(Subsistence,Year) z4-merge(z1,z2,z3) ##Convert to ts to interpolate missing NA values ts1-as.ts(z1) ts1Ex-na.approx(ts1) ts2-as.ts(z2) ts2Ex-na.approx(ts2) ts3-as.ts(z3) ts3Ex-na.approx(ts3) ## Merge into one object ts4Ex-cbind(ts1Ex, ts2Ex, ts3Ex, deparse.level = 1) ##Test plots to examine series plot(ts1Ex) plot(ts2Ex) plot(ts3Ex) plot(ts4Ex) ## Convert for stacked area plotting and plot CaribMatrix-as.matrix(ts4Ex) stackpoly(CaribMatrix, stack=TRUE, xlab=Year,ylab=Catch in tons, main=Historical Florida reef fisheries catch by sector, axis4=FALSE) stackpoly(CaribMatrix, stack=TRUE, xlab=Year,ylab=Catch in tons, col.main=red, font.main=4, axis4=FALSE, xat=seq(50,760,by=100), xaxlab=c(1300, 1400, 1500, 1600, 1700, 1800, 1900, 2000)) -Original Message- From: h.wick...@gmail.com on behalf of Hadley Wickham Sent: Fri 12/23/2011 9:41 AM To: UncleFish Cc: r-help@r-project.org Subject: Re: [R] Stacked area plot for time series You are more likely to get a helpful response if you provide a reproducible example - without that I can only guess that you need to use approx so you get y values at same x values. Hadley On Wed, Dec 21, 2011 at 8:13 AM, UncleFish bpn...@ucsd.edu wrote: I wish to make a stacked area chart of a time series with three variables. The observations are very irregular, covering 500 years, with a few historical observations in the range 1500-1850, and then more regular observations since 1880 or so. If I plot this in Excell it truncates the time series, but it can make a nice stacked area plot. My question is how can I have a correct x axis with a stacked plot in R? I thave tried the data in a zoo as well as a ts, and have tried ggplot2 and plotrix, but cannot get beyond a lattice plot. -- View this message in context: http://r.789695.n4.nabble.com/Stacked-area-plot-for-time-series-tp4221849p4221849.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. -- Assistant Professor / Dobelman Family Junior Chair Department of Statistics / Rice University 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] Axis manipulation in Stackpoly (Plotrix)
Thank you. That helps me understand the issue better. I came up with a similar solution . . . but yours is more elegant (I just wrote out the labels . . ). My solution follows: stackpoly(CaribMatrix, stack=TRUE, xlab=Year,ylab=Catch in tons, col.main=red, font.main=4, axis4=FALSE, xat=seq(50,760,by=100), xaxlab=c(1300, 1400, 1500, 1600, 1700, 1800, 1900, 2000)) -Original Message- From: Jim Lemon [mailto:j...@bitwrit.com.au] Sent: Fri 12/23/2011 2:53 AM To: Ben Neal Cc: r-help@r-project.org Subject: Re: [R] Axis manipulation in Stackpoly (Plotrix) On 12/23/2011 07:52 AM, Ben Neal wrote: Have made a stacked area plot, but now want to manipulate the x axis: make in even increments (50) in order to suppress the tick marks (forming solid bar under plot). However, the plot functions do not seem to work, and I cannot find documentation for use of xat in stackpoly. This is a time series of data covering 579 years, and has been successfully converted from zoo to matrix. Data plotting fine, jsut want to change axis - CaribMatrix-as.matrix(ts4Ex) stackpoly(CaribMatrix, stack=TRUE, xlab=Year,ylab=Catch in tons, xaxlab=(1250:2008), main=Historical Florida reef fisheries catch by sector, axis4=FALSE) Any assistance appreciated! Thanks very much, Benjamin P Neal, Scripps Inst. of Oceanogrpahy Hi Ben, In the stackpoly function, you can specify the positions of the axis ticks (xat=...) and the labels for those positions (xaxlab=...). You don't seem to have supplied the x values, so the values in CaribMatrix will be placed display at 1:dim(CaribMatrix)[1]. I'll have to make up CaribMatrix... CaribMatrix-matrix(order(sample(200:500,579,TRUE),nrow=579,ncol=3) stackpoly(CaribMatrix,stack=TRUE,ylim=c(0,1500),xat=seq(1,600,by=50), xaxlab=seq(1450,2000,by=50),staxx=TRUE) This seems to give a reasonable plot for me. 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] black and white in qplot? layout 4 graphs in one screen
Hi, On 22 December 2011 09:16, rachaelohde cox.rach...@gmail.com wrote: Hello, I am trying to plot means and standard errors conditioned by a factor, using qplot. I am successful at getting the bar graph I want with a error bar, however I have tried many things and cannot get the bars to change colors. Currently showing as red and blue, but need it to be black and white for publication. Any suggestions please? Have a look at scale_fill_grey() and grid.arrange() d= data.frame(x=sample(letters[1:4], 10, replace=TRUE), f=gl(2, 10)) ( p = qplot(x, data=d, position=dodge, fill=f) + scale_fill_grey() ) library(gridExtra) grid.arrange(p, p, p , p, ncol=2) HTH, baptiste Using a data set June, which is str: 'data.frame': 21 obs. of 6 variables: $ BLCK : Factor w/ 4 levels ,B,I,W: 3 4 2 3 2 2 2 4 3 4 ... $ PLOT : int 3 3 6 1 2 5 1 1 2 2 ... $ TRT : Factor w/ 5 levels ,crop,ten,..: 2 2 2 3 3 3 4 4 4 5 ... $ Date : Factor w/ 8 levels , 6/16/11,..: 2 2 2 2 2 2 2 2 2 2 ... $ habitat : Factor w/ 3 levels ,C,P: 2 2 2 2 2 2 2 2 2 2 ... $ Abundance: num 0.333 0 1.333 0 1.667 ... Current code is: Ab.avg-ddply(June, c(TRT, habitat), function(df) return(c(Ab.avg=mean(df$Abundance), Ab.sd=sd(df$Abundance avg.plot-qplot(TRT, Ab.avg, fill=factor(habitat), data=Ab.avg, geom=bar, position=dodge) dodge - position_dodge(width=0.9) avg.plot++geom_linerange(aes(ymax=Ab.avg+Ab.sd, ymin=Ab.avg-Ab.sd), position=dodge)+theme_bw() http://r.789695.n4.nabble.com/file/n4223035/june_bar_graph.png Also, would like to plot 4 of these bar graphs (for four dates) on the same screen, I cannot get the par() or layout() function to work with qplot. Is there another way? Thank you! -- View this message in context: http://r.789695.n4.nabble.com/black-and-white-in-qplot-layout-4-graphs-in-one-screen-tp4223035p4223035.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.
Re: [R] cast in reshape and reshape2
On Dec 23, 2011, at 2:58 PM, Kaiyin Zhong wrote:
library(reshape2)
x = melt(airquality, id=c('month', 'day'))
With reshape I can cast with multiple functions:
library(reshape)
cast(x, month+variable~., c(mean,sd))
month variable mean sd
1 5ozone 23.615385 22.224449
2 5 solar.r 181.296296 115.075499
3 5 wind 11.622581 3.531450
4 5 temp 65.548387 6.854870
5 6ozone 29.44 18.207904
6 6 solar.r 190.17 92.882975
7 6 wind 10.27 3.769234
8 6 temp 79.10 6.598589
9 7ozone 59.115385 31.635837
10 7 solar.r 216.483871 80.568344
11 7 wind 8.941935 3.035981
12 7 temp 83.903226 4.315513
13 8ozone 59.961538 39.681210
14 8 solar.r 171.857143 76.834943
15 8 wind 8.793548 3.225930
16 8 temp 83.967742 6.585256
17 9ozone 31.448276 24.141822
18 9 solar.r 167.43 79.118280
19 9 wind 10.18 3.461254
20 9 temp 76.90 8.355671
Is there a way to do the same job with reshape2?
Have you looked at the .summarise argument to dcast? That seems to
deliver the same sort of results one gets with base::aggregate.
--
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West Hartford, CT
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Re: [R] cast in reshape and reshape2
On Dec 23, 2011, at 5:58 PM, David Winsemius wrote:
On Dec 23, 2011, at 2:58 PM, Kaiyin Zhong wrote:
library(reshape2)
x = melt(airquality, id=c('month', 'day'))
With reshape I can cast with multiple functions:
library(reshape)
cast(x, month+variable~., c(mean,sd))
month variable mean sd
1 5ozone 23.615385 22.224449
2 5 solar.r 181.296296 115.075499
3 5 wind 11.622581 3.531450
4 5 temp 65.548387 6.854870
5 6ozone 29.44 18.207904
6 6 solar.r 190.17 92.882975
7 6 wind 10.27 3.769234
8 6 temp 79.10 6.598589
9 7ozone 59.115385 31.635837
10 7 solar.r 216.483871 80.568344
11 7 wind 8.941935 3.035981
12 7 temp 83.903226 4.315513
13 8ozone 59.961538 39.681210
14 8 solar.r 171.857143 76.834943
15 8 wind 8.793548 3.225930
16 8 temp 83.967742 6.585256
17 9ozone 31.448276 24.141822
18 9 solar.r 167.43 79.118280
19 9 wind 10.18 3.461254
20 9 temp 76.90 8.355671
Is there a way to do the same job with reshape2?
Have you looked at the .summarise argument to dcast? That seems to
deliver the same sort of results one gets with base::aggregate.
Actually I see after looking at examples on the plyr-reshape-
googlegroups group that it is not '.summarise' but rather 'summarise'.
Unfortunately there are no links in the help pages that seem to
describe its proper use ... a not uncommon failing for that package in
my experience.
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[R] Latent class multinomial (or conditional) logit using R?
Hi everyone? Does anybody know how can I estimate a Latent class multinomial (or conditional) logit using R? I have tried flexmix, poLCA, and they do not seem to support this model. thanks in advance adan -- View this message in context: http://r.789695.n4.nabble.com/Latent-class-multinomial-or-conditional-logit-using-R-tp4230083p4230083.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] Latent class multinomial (or conditional) logit using R?
You may have a look at lcmixed/flexmixedruns in package fpc. This provides flexmix methods that can be used to fit latent class models with locally independent multinomials (no logits, though, and I'm not sure that for categorical data only this is much different from what poLCA offers). Best regards, Christian On Fri, 23 Dec 2011, adanmartinez wrote: Hi everyone? Does anybody know how can I estimate a Latent class multinomial (or conditional) logit using R? I have tried flexmix, poLCA, and they do not seem to support this model. thanks in advance adan -- View this message in context: http://r.789695.n4.nabble.com/Latent-class-multinomial-or-conditional-logit-using-R-tp4230083p4230083.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. *** --- *** Christian Hennig University College London, Department of Statistical Science Gower St., London WC1E 6BT, phone +44 207 679 1698 chr...@stats.ucl.ac.uk, www.homepages.ucl.ac.uk/~ucakche __ 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] Help with code
Hello,
There are so many people posting answers that I'm curious and decided to try
one.
I don't know if this is it but it doesn't give an error and it reformats
your data
according to the rules in your original code.
#
nr - dim(c1)[1]
nc - dim(c1)[2]
c2 - NULL
c2_row - rep(, nc-1)
for(i in 1:nr){
ptype- as.character(c1$t1d_ptype[i])
stype- substr(ptype,1,4)
num_comp - sum(c1[i,] == Y)
for(j in 1:(nc-1)){
c2_row[j] -
if(num_comp 0){
c1_tmp - as.integer(c1[i ,j])
if(ptype == T1D){
if(c1_tmp == 1) c2_row[j] - T1D_oc
if(c1_tmp == 2) c2_row[j] - T1D_w
}
if(stype == Ctrl){
if(c1_tmp == 1) c2_row[j] - Ctrl_oc
if(c1_tmp == 2) c2_row[j] - Ctrl_w
}
}
else{
if(ptype == T1D) c2_row[j] - T1D_noc
if(stype == Ctrl) c2_row[j] - Ctrl_noc
}
}
c2 - rbind(c2, c(c2_row, ptype))
}
c2 - data.frame(c2)
colnames(c2) - colnames(c1)
c2
I bet there's a way to work on entire objects. This is C-like.
Rui Barradas
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Re: [R] Applying a function
Joanie Van De Walle wrote Hi, I need help writing a function I capture seal pups mutliple times during the lactation season in order to monitor their growth rate. When I release them, the recovery (mother-pup) time is not the same for all individuals. I want to know if individuals that recover their mother the fastest are the ones with the highest growth rates. So, I noted at every release if the pup reovered his mother before we leave (yes or no). My dataframe looks like this Capture nb individual individual capture motherrecovery growth rate 1 1 1 n 0.5 2 1 2 y 0.5 3 1 3 y 0.5 4 1 4 y 0.5 5 1 5 n 0.5 6 2 1 y 0.3 7 2 2 y 0.3 8 3 1 y 0.4 9 3 2 n 0.4 10 3 3 y 0.4 ... I want to calculate a rate of mother recovery by individual, i.e. nb of recoveries (y)/nb of captures. So for indivial 1 it would be 3/5 = 0.6. I want to write a function that does this for all the individuals in my dataframe, i.e. around 400 individuals (this is why I want to write a function, it would be too long by hand) Thank you, __ R-help@ 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. Hello, Here is a one line function recovery.rate - function(x) unlist(lapply(split(tab1, tab1[,2]), function(x) mean(x[[4]]==y))) It works with that table. I hope it helps Merry Christmas Rui Barradas -- View this message in context: http://r.789695.n4.nabble.com/Applying-a-function-tp4229277p4230100.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] if statement problem
Hello,
I want to do fisher test for the rows in data file which has value less than
5 otherwise chi square test .The p values from both test should be stored in
one resulted file. but there is some problem with bold if statement. I don't
know how
implement this line properly.
x = cbind(obs1,obs2,exp1,exp2)
a = matrix(c(0,0,0,0), ncol=2, byrow =TRUE)#matrix with initialized
values
for (i in 1: length(x[,1]))
{
*if((x[i,1] || x[i,2] || x[i,3] || x[i,4]) 5)*
{
a[1,1] - x[i,1];
a[1,2] - x[i,2];
a[2,1] - x[i,3];
a[2,2] - x[i,4];
result - fisher.test(a)
write.table(result[[p.value]],file=results.txt,
sep=\n, append=TRUE, col.names=FALSE, row.names=FALSE);
}
else
{
a[1,1] - x[i,1];
a[1,2] - x[i,2];
a[2,1] - x[i,3];
a[2,2] - x[i,4];
result - chisq.test(a)
write.table(result[[p.value]],file=results.txt,
sep=\n, append=TRUE, col.names=FALSE, row.names=FALSE);}
}
Regards
R
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Re: [R] Trouble converting hourly data into daily data
Hi Jean, Thanks for the help. I couldn't quite get the results I needed with the merge command, but I ended up using the following work-around: Weather - read.csv(Weather.csv) Weather$diff.time - abs(.5 - Weather$TimeNumeric) agg - aggregate(diff.time ~ Date, data = Weather, FUN = which.min) n.obs - cumsum(rle(as.double(Weather$Date))$lengths) n.obs - c(0, n.obs[1:(length(n.obs) - 1)]) noon.ind - agg$diff.time + n.obs subset - Weather[noon.ind,] Cheers, Sean On Mon, Dec 19, 2011 at 6:03 AM, Jean V Adams jvad...@usgs.gov wrote: Sean Baumgarten wrote on 12/14/2011 06:38:08 PM: Hello, I have a data frame with hourly or sub-hourly weather records that span several years, and from that data frame I'm trying to select only the records taken closest to noon for each day. Here's what I've done so far: #Add a column to the data frame showing the difference between noon and the observation time (I converted time to a 0-1 scale so 0.5 represents noon): data$Diff_from_noon - abs(0.5-data$Time) #Find the minimum value of Diff_from_noon for each Date: aggregated - aggregate(Diff_from_noon ~ Date, data, FUN=min) The problem is that the aggregated data frame only has two columns: Date and Diff_from_noon. I can't figure out how to get the columns with the actual weather variables to carry over from the original data frame. Any suggestions you have would be much appreciated. Thanks, Sean You don't provide any example data, so I will use data from R datasets, airquality. After using the aggregate() function to find the minimum Day for each Month, merge the resulting data frame with the original data frame to see all the columns corresponding to the selected minimums. aggregated - aggregate(Day ~ Month, airquality, FUN=min) aggregated Month Day 1 5 1 2 6 1 3 7 1 4 8 1 5 9 1 merge(aggregated, airquality) Month Day Ozone Solar.R Wind Temp 1 5 141 190 7.4 67 2 6 1NA 286 8.6 78 3 7 1 135 269 4.1 84 4 8 139 83 6.9 81 5 9 196 167 6.9 91 For your data, the code would look like this: aggregated - aggregate(Diff_from_noon ~ Date, data, FUN=min) merge(aggregated, data) I recommend that you use a name other than data for your data frame, since data() is a built in R function. Jean [[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] Applying a function
Sorry, in the function body, NO 'tab1', use 'x' only: recovery.rate - function(x) unlist(lapply(split(x, x[,2]), function(x) mean(x[[4]]==y))) The error is because 'tab1' existed in the environment and the function would find it. This time, tested after removing 'tab1'. Rui Barradas -- View this message in context: http://r.789695.n4.nabble.com/Applying-a-function-tp4229277p4230106.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] if statement problem
reena wrote
Hello,
I want to do fisher test for the rows in data file which has value less
than 5 otherwise chi square test .The p values from both test should be
stored in one resulted file. but there is some problem with bold if
statement. I don't know how
implement this line properly.
x = cbind(obs1,obs2,exp1,exp2)
a = matrix(c(0,0,0,0), ncol=2, byrow =TRUE)#matrix with
initialized values
for (i in 1: length(x[,1]))
{
*if((x[i,1] || x[i,2] || x[i,3] || x[i,4]) 5)*
Hello,
Try
*if(any(x[i,] 5))*
Merry Christmas
Rui Barradas
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Re: [R] Latent class multinomial (or conditional) logit using R?
Hi Christian, Thanks a lot for the suggestion. As you expected, lcmixed does not provide more functionality than poLCA for the case of only categorical data... Do you know whether there is an explanation (some technical difficulty I may be missing, I am Economist so it is very likely) of why I can fit a Poisson, a logit latent class model but not a conditional logit? In fact, I started using R when I realize that my (very rudimentary) EM script written in Matlab had tons of problems to converge. Now I start wondering if there is some specific situation I should be aware of. Thanks a lot again -- View this message in context: http://r.789695.n4.nabble.com/Latent-class-multinomial-or-conditional-logit-using-R-tp4230083p4230194.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] Custom XML Readers
Hi Andy, On Dec 23, 2011, at 2:51 PM, pl.r...@gmail.com wrote: I need to construct a custom XML reader, the files I'm working with are in funky XML format: str name=authorPaul H/str str name=countryUSA/str date name=created_date2010-02-16/date I want to read the file so it looks like: author = Paul H country = USA created_date=2010-02-16 Does any one know how to go about this problem, or know of good references i could access? Have you tried Duncan Temple Lang's XML package for R? It works very well for parsing and building XML formatted data. http://www.omegahat.org/RSXML/ Cheers, Ben Thanks, Andy -- View this message in context: http://r.789695.n4.nabble.com/Custom-XML-Readers-tp4229614p4229614.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. Ben Tupper Bigelow Laboratory for Ocean Sciences 180 McKown Point Rd. P.O. Box 475 West Boothbay Harbor, Maine 04575-0475 http://www.bigelow.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] cast in reshape and reshape2
On Fri, Dec 23, 2011 at 1:58 PM, Kaiyin Zhong kindlych...@gmail.com wrote:
library(reshape2)
x = melt(airquality, id=c('month', 'day'))
With reshape I can cast with multiple functions:
library(reshape)
cast(x, month+variable~., c(mean,sd))
month variable mean sd
1 5 ozone 23.615385 22.224449
2 5 solar.r 181.296296 115.075499
3 5 wind 11.622581 3.531450
4 5 temp 65.548387 6.854870
5 6 ozone 29.44 18.207904
6 6 solar.r 190.17 92.882975
7 6 wind 10.27 3.769234
8 6 temp 79.10 6.598589
9 7 ozone 59.115385 31.635837
10 7 solar.r 216.483871 80.568344
11 7 wind 8.941935 3.035981
12 7 temp 83.903226 4.315513
13 8 ozone 59.961538 39.681210
14 8 solar.r 171.857143 76.834943
15 8 wind 8.793548 3.225930
16 8 temp 83.967742 6.585256
17 9 ozone 31.448276 24.141822
18 9 solar.r 167.43 79.118280
19 9 wind 10.18 3.461254
20 9 temp 76.90 8.355671
Is there a way to do the same job with reshape2?
No, this is something that reshape2 does not support.
Hadley
--
Assistant Professor / Dobelman Family Junior Chair
Department of Statistics / Rice University
http://had.co.nz/
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Re: [R] cast in reshape and reshape2
Have you looked at the .summarise argument to dcast? That seems to deliver the same sort of results one gets with base::aggregate. Actually I see after looking at examples on the plyr-reshape-googlegroups group that it is not '.summarise' but rather 'summarise'. Unfortunately there are no links in the help pages that seem to describe its proper use ... a not uncommon failing for that package in my experience. ?plyr::summarise seems pretty helpful to me. If you can do better, please submit a patch - they are very much appreciated. Hadley -- Assistant Professor / Dobelman Family Junior Chair Department of Statistics / Rice University 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] Help creating a symmetric matrix?
Matt Considine wrote Hi, I am trying to work with the output of the MINE analysis routine found at http://www.exploredata.net Specifically, I am trying to read the results into a matrix (ideally an n x n x 6 matrix, but I'll settle right now for getting one column into a matrix.) The problem I have is not knowing how to take what amounts to being one half of a symmetric matrix - excluding the diagonal - and getting it into a matrix. I have tried using lower.tri as found here https://stat.ethz.ch/pipermail/r-help/2008-September/174516.html but it appears to only partially fill in the matrix. My code and an example of the output is below. Can anyone point me to an example that shows how to create a matrix with this sort of input? Thank you in advance, Matt #v-newx[,3] #or, for the sake of this example v-c(0.33740, 0.26657, 0.23388, 0.23122, 0.21476, 0.20829, 0.20486, 0.19439, 0.19237, 0.18633, 0.17298, 0.17174, 0.16822, 0.16480, 0.15027) z-diag(6) ind - lower.tri(z) z[ind] - t(v)[ind] z [,1][,2] [,3] [,4] [,5] [,6] [1,] 1.0 0.00000 [2,] 0.26657 1.00000 [3,] 0.23388 0.192371000 [4,] 0.23122 0.18633 NA100 [5,] 0.21476 0.17298 NA NA10 [6,] 0.20829 0.17174 NA NA NA1 [[alternative HTML version deleted]] __ R-help@ 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. Hello, Aren't you complicating? In the last line of your code, why use 'v[ind]' if 'ind' indexes the matrix, not the vector? z-diag(6) ind - lower.tri(z) z[ind] - v#This works z Rui Barradas -- View this message in context: http://r.789695.n4.nabble.com/Help-creating-a-symmetric-matrix-tp4227301p4230335.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] cast in reshape and reshape2
On Dec 23, 2011, at 9:23 PM, Hadley Wickham wrote: Have you looked at the .summarise argument to dcast? That seems to deliver the same sort of results one gets with base::aggregate. Actually I see after looking at examples on the plyr-reshape- googlegroups group that it is not '.summarise' but rather 'summarise'. Unfortunately there are no links in the help pages that seem to describe its proper use ... a not uncommon failing for that package in my experience. ?plyr::summarise seems pretty helpful to me. If you can do better, please submit a patch - they are very much appreciated. My failure to find it stemmed from it not being mentioned in any way in package reshape2's help files, but maybe I was mistaken that it was meant to be used in that context? -- David Winsemius, MD West Hartford, CT __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Optimising timeboxing in xts
I don't know if timeboxing is the correct term to use to accomplish
what I'm attempting, so allow me to explain. I have a set n of tagged
observations in time series t. What I'm interested in is taking i
seconds before and after every n. My code is below:
# observations.xts is an xts time series and arg is the number of
seconds to for the timebox
timeboxes - sapply(c(1:as.numeric(last(index(observations.xts))-arg)),
function(x) { return(observations.xts[index(x) - arg, index(x)+arg])},
simplify=TRUE)
What I'd like to see in timeboxes is observations.xts elements with
timestamps within the range of abs(arg).
--
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Envoyait de mon portable
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