Re: [R] outer() or some other function for regression prediction with 2 IVs
On Jul 10, 2012, at 05:35 , Joseph Clark wrote:
Thanks. I was able to get what I wanted by doing this:
predxn - function(s,d) { coef(m3)[1] + coef(m3)[2]*s + coef(m3)[3]*s^2 +
coef(m3)[4]*d + coef(m3)[5]*d^2 }
But it's not very elegant...
You didn't take Michael's hint:
coef(m3) %*% cbind(1, s, s^2, d, d^2)
or even
predict(m3, newdata=data.frame(x1=s, x2=d))
(in which x1, x2 needs replacement to match the names used in m3).
Also, a quick (but not fast) solution to the generic non-vectorized-function
problem is to Vectorize() it.
// joseph w. clark , phd candidate
\\ usc marshall school of business
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Peter Dalgaard, Professor,
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Email: pd@cbs.dk Priv: pda...@gmail.com
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[R] [R-pkgs] package JM -- version 1.0-0
Dear R-users, I'd like to announce the release of version 1.0-0 of package JM (already available from CRAN) for the joint modeling of longitudinal and time-to-event data using shared parameter models. These models are applicable in mainly two settings. First, when focus is in the survival outcome and we wish to account for the effect of an endogenous (aka internal) time-dependent covariate measured with error. Second, when focus is in the longitudinal outcome and we wish to correct for nonrandom dropout. Some basic features of JM: * it fits joint models for continuous longitudinal responses and allows for several options for the survival submodel, including PH models with Weibull, piecewise-constant, spline-approximated and unspecified baseline hazard functions. The most complete option is the PH model with the spline-approximated baseline hazard, which also allows for the inclusion of stratification factors, competing risks and (exogenous) time-varying covariates; * it allows for several formulations of the association structure between the longitudinal and survival outcomes; * it computes dynamic individualized predictions for the survival and longitudinal outcomes, which are updated as extra longitudinal information is recorded; * it computes time-dependent sensitivity and specificity, and the corresponding ROCs and AUCs with several options for the prediction rule; * several types of residuals are supported for both outcomes; * fast fitting of these models is facilitated with a pseudo-adaptive Gaussian-Hermite rule. The theory and application of this type of models along with a comprehensive overview of the capabilities of the package can be found in the recently published book Joint Models for Longitudinal and Time-to-Event Data, with Applications in R by Chapman and Hall/CRC (http://www.crcpress.com/product/isbn/9781439872864). The code used in the book and additional material are available in the R-forge web site: http://jmr.r-forge.r-project.org/ Additional information can be found in the corresponding help files, and examples at the R wiki web page devoted to JM: http://rwiki.sciviews.org/doku.php?id=packages:cran:jm As always, any kind of feedback (e.g., questions, suggestions, bug-reports, etc.) is more than welcome. Best, Dimitris -- Dimitris Rizopoulos Assistant Professor Department of Biostatistics Erasmus University Medical Center Address: PO Box 2040, 3000 CA Rotterdam, the Netherlands Tel: +31/(0)10/7043478 Fax: +31/(0)10/7043014 Web: http://www.erasmusmc.nl/biostatistiek/ ___ R-packages mailing list r-packa...@r-project.org https://stat.ethz.ch/mailman/listinfo/r-packages __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Read vector as multi-dimensional data in R by row
Hi, Try this: b1-aperm(array(a,dim=c(5,2,2)),perm=c(2,1,3)) b1 , , 1 [,1] [,2] [,3] [,4] [,5] [1,] 1 2 3 4 5 [2,] 6 7 8 9 10 , , 2 [,1] [,2] [,3] [,4] [,5] [1,] 11 12 13 14 15 [2,] 16 17 18 19 20 A.K. - Original Message - From: HJ YAN yhj...@googlemail.com To: r-help@r-project.org Cc: Sent: Monday, July 9, 2012 7:25 PM Subject: [R] Read vector as multi-dimensional data in R by row Dear R users Say I wanted to read a vector into R as multi-dimensional array by row, e.g. a-c(1:20) b-array(a,dim=c(2,5,2)) b , , 1 [,1] [,2] [,3] [,4] [,5] [1,] 1 3 5 7 9 [2,] 2 4 6 8 10 , , 2 [,1] [,2] [,3] [,4] [,5] [1,] 11 13 15 17 19 [2,] 12 14 16 18 20 But actually I wanted... [,1] [,2] [,3] [,4] [,5] [1,] 1 2 3 4 5 [2,] 6 7 8 9 10 , , 2 [,1] [,2] [,3] [,4] [,5] [1,] 11 12 13 14 15 [2,] 16 17 18 19 20 I checked '?array' but there is not an argument or something like 'byrow=T' as the one in 'matrix'. Could anyone help please? Many thanks in advance! HJ [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Skipping lines and incomplete rows
Thanks a lot Rui and Arun. The methods work fine with the data I gave but when I tried the two methods with the following semi-colon separated data using sep = ;. Only the first 3 columnns are read properly rest of the columns are either empty or NAs. ** Remove this line Remove this line Remove this line Time;Actual Speed;Actual Direction;Temp;Press;Value1;Value2 ;[m/s];[°];°C;[hPa];[MWh];[MWh] 1/1/2012;0.0;0;#N/A;#N/A;0.;0. 1/2/2012;0.0;0;#N/A;#N/A;0.;0. 1/3/2012;0.0;0;#N/A;#N/A;1.5651;2.2112 1/4/2012;0.0;0;#N/A;#N/A;1.;2. 1/5/2012;0.0;0;#N/A;#N/A;3.2578;7.5455 *** I used the following code: dat1-read.table(testInput.txt,sep=;,skip=3,fill=TRUE,header=TRUE) dat1-dat1[-1,] row.names(dat1)-1:nrow(dat1) Could you please let me know what is wrong with this approach? Thank you. Ravi -- View this message in context: http://r.789695.n4.nabble.com/Skipping-lines-and-incomplete-rows-tp4635830p4635952.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] How to add marker in Stacked bar plot?
Thanks it works fine. But can i control its width? Regards -- View this message in context: http://r.789695.n4.nabble.com/How-to-add-marker-in-Stacked-bar-plot-tp4635946p4635954.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] How to add marker in Stacked bar plot?
I found arrow is too thin. I have one arrow image. I want to put it there. How can i import external image and merge with stacked bar plot. Regards -- View this message in context: http://r.789695.n4.nabble.com/How-to-add-marker-in-Stacked-bar-plot-tp4635946p4635955.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] How to use external image with R plot?
Hi, I am wokring on stacked bar plot and i need to add one arrow dynamically. http://r.789695.n4.nabble.com/file/n4635959/arrow_glossy_right_red.jpg http://r.789695.n4.nabble.com/file/n4635959/Screenshot.10.png Final image: http://r.789695.n4.nabble.com/file/n4635959/Screenshot.1.png Is there any packae which can be used for merging images? -- View this message in context: http://r.789695.n4.nabble.com/How-to-use-external-image-with-R-plot-tp4635959.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] Package 'MASS' (polr): Error in svd(X) : infinite or missing values in 'x'
Hi Jeremy, newData-data.frame(JVeg5=factor(Jdata[,JVeg5]),scale(Jdata[,c(Elevation,Lat_Y_pos,Coast_dist,Stream_dist)])) Global - polr(JVeg5 ~ Elevation + Lat_Y_pos + Coast_dist + Stream_dist, data=newData, na.action = na.omit, Hess = TRUE) summary(Global) Does this still do what you want? At least it doesn't produce the error like this. greetings Jessi On 09.07.2012, at 11:55, Jeremy Little wrote: Hello, I am trying to run an ordinal logistic regression (polr) using the package 'MASS'. I have successfully run other regression classes (glm, multinom) without much problem, but with the 'polr' class I get the following error: Error in svd(X) : infinite or missing values in 'x' which appears when I run the summary command. The data file is large (585000 rows) and has no NA, - or blank values. My script (in brief) is as follows, with results: library(MASS) ## ADD DATA Jdata- read.delim(/Analysis/20120709 JLittle data file.txt, header=T) attach(Jdata) names(Jdata) [1] POINTID Lat_Y_pos JVeg5 Subregion Rock_U_Nam Rock_Name Elevation Slope Aspect Hillshade Stream_dist Coast_dist Coast_SE [14] Coast_E Wind_310TPI Landform Global - polr(JVeg5 ~ Elevation + Lat_Y_pos + Coast_dist + Stream_dist, data=Jdata) summary(Global) Error in svd(X) : infinite or missing values in 'x' ##Try with omit NA command Global - polr(JVeg5 ~ Elevation + Lat_Y_pos + Coast_dist + Stream_dist, data=Jdata, na.action = na.omit, Hess = TRUE) summary(Global) Error in svd(X) : infinite or missing values in 'x' Does this imply an 'infinite value' and what would this mean? If anyone has any idea how to address this error, I would very much appreciate your response. Thank you in advance. Jeremy Date File Attachment (200 rows): http://r.789695.n4.nabble.com/file/n4635829/20120709_JLittle_data_file.txt 20120709_JLittle_data_file.txt -- View this message in context: http://r.789695.n4.nabble.com/Package-MASS-polr-Error-in-svd-X-infinite-or-missing-values-in-x-tp4635829.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] outer() or some other function for regression prediction with 2 IVs
I saw the hint but didn't know how to implement it, I learn more every day.
Thanks for spelling it out!
I knew there had to be a function like predict!
// joseph w. clark , phd candidate
\\ usc marshall school of business
Subject: Re: [R] outer() or some other function for regression prediction
with 2 IVs
From: pda...@gmail.com
Date: Tue, 10 Jul 2012 08:46:42 +0200
CC: michael.weyla...@gmail.com; r-help@r-project.org
To: joeclar...@hotmail.com
On Jul 10, 2012, at 05:35 , Joseph Clark wrote:
Thanks. I was able to get what I wanted by doing this:
predxn - function(s,d) { coef(m3)[1] + coef(m3)[2]*s + coef(m3)[3]*s^2 +
coef(m3)[4]*d + coef(m3)[5]*d^2 }
But it's not very elegant...
You didn't take Michael's hint:
coef(m3) %*% cbind(1, s, s^2, d, d^2)
or even
predict(m3, newdata=data.frame(x1=s, x2=d))
(in which x1, x2 needs replacement to match the names used in m3).
Also, a quick (but not fast) solution to the generic non-vectorized-function
problem is to Vectorize() it.
// joseph w. clark , phd candidate
\\ usc marshall school of business
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Peter Dalgaard, Professor,
Center for Statistics, Copenhagen Business School
Solbjerg Plads 3, 2000 Frederiksberg, Denmark
Phone: (+45)38153501
Email: pd@cbs.dk Priv: pda...@gmail.com
__
R-help@r-project.org mailing list
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Skipping lines and incomplete rows
Hello, My approach was slightly different, to use readLines to take care of the header and read.table for the data. This works with the new dataset you've posted, but we must use the option comment.char = . Try the following. head - readLines(test.txt, n=4)[4] dat - read.table(test.txt, skip=5, sep=;, stringsAsFactors=FALSE, comment.char=c) names(dat) - unlist(strsplit(head, ;)) dat$Time - as.Date(dat$Time, format=%m/%d/%Y) dat$Temp[dat$Temp == '#N/A'] - NA dat$Press[dat$Press == '#N/A'] - NA dat It works with me, good luck. Rui Barradas Em 10-07-2012 06:41, vioravis escreveu: Thanks a lot Rui and Arun. The methods work fine with the data I gave but when I tried the two methods with the following semi-colon separated data using sep = ;. Only the first 3 columnns are read properly rest of the columns are either empty or NAs. ** Remove this line Remove this line Remove this line Time;Actual Speed;Actual Direction;Temp;Press;Value1;Value2 ;[m/s];[°];°C;[hPa];[MWh];[MWh] 1/1/2012;0.0;0;#N/A;#N/A;0.;0. 1/2/2012;0.0;0;#N/A;#N/A;0.;0. 1/3/2012;0.0;0;#N/A;#N/A;1.5651;2.2112 1/4/2012;0.0;0;#N/A;#N/A;1.;2. 1/5/2012;0.0;0;#N/A;#N/A;3.2578;7.5455 *** I used the following code: dat1-read.table(testInput.txt,sep=;,skip=3,fill=TRUE,header=TRUE) dat1-dat1[-1,] row.names(dat1)-1:nrow(dat1) Could you please let me know what is wrong with this approach? Thank you. Ravi -- View this message in context: http://r.789695.n4.nabble.com/Skipping-lines-and-incomplete-rows-tp4635830p4635952.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] Extracting arithmetic mean for specific values from multiple .txt-files
Dear Mr. Holtman,
but I cannot leave out the value and cannot change the values to 1200.995
manually (for each test subject with a reaction time 1000 ms), because the
first your lead to incomplete data and the latter would be too
time-consuming.
Dear Rui,
here I have three files, which have exactly the same content as
XYZ_34.txt, EXCEPT that the file XYZ_50.txt doesn't have a period in the
first value 1200.9952 IF YOU OPEN IT WITH THE EDITOR (!), maybe because I
didn't change the structure with MS Excel. The other two files should be
identical.
http://r.789695.n4.nabble.com/file/n4635962/XYZ_2.txt XYZ_2.txt
http://r.789695.n4.nabble.com/file/n4635962/XYZ_50.txt XYZ_50.txt
http://r.789695.n4.nabble.com/file/n4635962/XYZ_1112.txt XYZ_1112.txt
R gives me the following output:
fun - function(x){
+ dat - read.table(x, skip=14)
+ dat[ , 8] - as.numeric(gsub(\\., , dat[, 8]))
+ mean(dat[, 8])
+ }
sapply(list.files(pattern=XYZ.*\\.txt), fun)
XYZ_1112.txtXYZ_2.txt XYZ_50.txt
345210.4 345210.4 310112.0
Your second suggestion leads to the same output:
fun - function(x, skip = 14){
+ dat - read.table(x, skip=skip)
+ dat[ , 8] - as.numeric(gsub(\\., , dat[, 8]))
+ mean(dat[, 8])
+ }
sapply(list.files(pattern=XYZ.*\\.txt), fun)
XYZ_1112.txtXYZ_2.txt XYZ_50.txt
345210.4 345210.4 310112.0
Thank you for your replies!
Kind regards
--
View this message in context:
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Re: [R] Skipping lines and incomplete rows
Or maybe it's better to coerce Temp and Press to numeric, if they are variables temperature and presssure. dat$Time - as.Date(dat$Time, format=%m/%d/%Y) dat$Temp - as.numeric(dat$Temp) dat$Press - as.numeric(dat$Press) This makes those '#N/A' values NA. Rui Barradas Em 10-07-2012 09:34, Rui Barradas escreveu: Hello, My approach was slightly different, to use readLines to take care of the header and read.table for the data. This works with the new dataset you've posted, but we must use the option comment.char = . Try the following. head - readLines(test.txt, n=4)[4] dat - read.table(test.txt, skip=5, sep=;, stringsAsFactors=FALSE, comment.char=c) names(dat) - unlist(strsplit(head, ;)) dat$Time - as.Date(dat$Time, format=%m/%d/%Y) dat$Temp[dat$Temp == '#N/A'] - NA dat$Press[dat$Press == '#N/A'] - NA dat It works with me, good luck. Rui Barradas Em 10-07-2012 06:41, vioravis escreveu: Thanks a lot Rui and Arun. The methods work fine with the data I gave but when I tried the two methods with the following semi-colon separated data using sep = ;. Only the first 3 columnns are read properly rest of the columns are either empty or NAs. ** Remove this line Remove this line Remove this line Time;Actual Speed;Actual Direction;Temp;Press;Value1;Value2 ;[m/s];[°];°C;[hPa];[MWh];[MWh] 1/1/2012;0.0;0;#N/A;#N/A;0.;0. 1/2/2012;0.0;0;#N/A;#N/A;0.;0. 1/3/2012;0.0;0;#N/A;#N/A;1.5651;2.2112 1/4/2012;0.0;0;#N/A;#N/A;1.;2. 1/5/2012;0.0;0;#N/A;#N/A;3.2578;7.5455 *** I used the following code: dat1-read.table(testInput.txt,sep=;,skip=3,fill=TRUE,header=TRUE) dat1-dat1[-1,] row.names(dat1)-1:nrow(dat1) Could you please let me know what is wrong with this approach? Thank you. Ravi -- View this message in context: http://r.789695.n4.nabble.com/Skipping-lines-and-incomplete-rows-tp4635830p4635952.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. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] boxplot with cut
Hello,
Maybe this iss what you're looking for. GD is your data.frame.
multi.boxplot - function(x, by, ...){
x - as.data.frame(x)
sp - split(x, by)
len - length(sp) - 1
n - ncol(x)
n1 - n + 1
boxplot(x[[ 1 ]] ~ by, at = 0:len*n1 + 1,
xlim = c(0, (len + 1)*n1), ylim = range(unlist(x)), xaxt = n,
...)
for(i in seq_len(n)[-1])
boxplot(x[[i]] ~ by, at = 0:len*n1 + i, xaxt = n, add = TRUE,
...)
axis(1, at = 0:len*n1 + n1/2, labels = names(sp), tick = TRUE)
}
cols - grep(ReadCount, names(GD))
multi.boxplot(GD[, cols], cut(GD$GeneDensity, breaks=10))
If this is it and you don't like those x-axis tick lables, use
as.integer(cut(...etc...)).
Hope this helps,
Rui Barradas
Em 09-07-2012 20:51, Vining, Kelly escreveu:
Dear UseRs,
I'm making box plots from a data set that looks like this:
Chr Start End GeneDensity ReadCount_Explant ReadCount_Callus
ReadCount_Regen
1 1 1 1 107.82 1.2431.047 1.496
2 1 10001 2 202.50 0.8350.869 0.456
3 1 20001 3 158.80 1.8131.529 1.131
4 1 30001 4 100.53 1.7311.752 1.610
5 1 40001 5 100.53 3.0562.931 3.631
6 1 50001 6 100.53 1.9602.013 2.459
I'm breaking the GeneDensity column into deciles, then making a box plot of the
relationship between the GeneDensity parameter and each of the three ReadCount columns.
Here's an example of one of my boxplot commands:
boxplot(GeneDensity$ReadCount_Explant ~ cut(GeneDensitySorted$GeneDensity, breaks=10), ylim=c(0,40),
ylab=RPKM, xlab=GENE DENSITY (LOW - HIGH), main=INTERNODE EXPLANT)
Right now, I'm making three separate graphs: one for each of the three
ReadCount columns. I'd like to put all three sets on one graph, so that each
decile is represented by three boxes, one for each ReadCount category, but don't know how
to make that work. I tried this:
boxplot(GeneDensitySorted$ReadCount_Explant ~ cut(GeneDensitySorted$GeneDensity, breaks=10),
GeneDensitySorted$ReadCount_Callus ~ cut(GeneDensitySorted$GeneDensity, breaks=10),
GeneDensitySorted$ReadCount_Regen ~ cut(GeneDensitySorted$GeneDensity, breaks=10), ylim=c(0,40),
ylab=RPKM, xlab=GENE DENSITY (LOW - HIGH))
Not surprisingly, I got this error:
Error in as.data.frame.default(data) :
cannot coerce class 'formula' into a data.frame
Does anyone know how to accomplish this box plot?
Any help is much appreciated.
--Kelly V.
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and provide commented, minimal, self-contained, reproducible code.
[R] how can I show the xlab and ylab information while using layout
hi R-users: I want to draw three plot into one figure by layout and the script has been shown below. But I find R does not show the xlab and ylab information completely as shown the figure attached. How can I midify the script.? thank you . xxlab-paste(cpmd, (,ro,%),sep= ) yylab-paste(rfmd, (,co,%),sep= ) par(mar=c(3,3,1,1)) #layout(matrix(c(2,0,1,3),2,2,byrow=TRUE),widths=lcm(30), heights=lcm(25),TRUE) layout(matrix(c(2,0,1,3),2,2,byrow=TRUE),c(5,1),c(1,5),TRUE) layout.show(3) plot(data_cpmd,data_rfmd,xlab=xxlab,ylab=yylab,xlim=XX,ylim=YY,asp=1) abline(1,1) #rug(side=1,jitter(data_cpmd,5)) #rug(side=2,jitter(data_rfmd,5)) #plot(homo_ana$dism16cpmd,homo_ana$dism16rfmd,main=mtitle,xlab=xxlab,ylab=yylab,xlim=XX,ylim=YY,asp=1) par(mar=c(0,3,1,1)) barplot(data_cpmd, axes=FALSE, ylim=YY, space=0) par(mar=c(3,0,1,1)) barplot(data_rfmd, axes=FALSE,main=mtitle, xlim=XX, space=0, horiz=TRUE) #boxplot(data_cpmd,horizontal = TRUE,xlim=XX,ylim=YY,outline=ifout, xaxt = n) #par(mar=c(3,0,1,1)) #boxplot(data_rfmd,xlim=XX,ylim=YY,outline=ifout,yaxt = n) attachment: homo_jawt_pgtwhr24.png__ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how can I show the xlab and ylab information while using layout
The margins you specified aren't large enough to hold the information you're trying to put in them, so you need to make them larger. Sarah On Tuesday, July 10, 2012, Jie Tang wrote: hi R-users: I want to draw three plot into one figure by layout and the script has been shown below. But I find R does not show the xlab and ylab information completely as shown the figure attached. How can I midify the script.? thank you . xxlab-paste(cpmd, (,ro,%),sep= ) yylab-paste(rfmd, (,co,%),sep= ) par(mar=c(3,3,1,1)) #layout(matrix(c(2,0,1,3),2,2,byrow=TRUE),widths=lcm(30), heights=lcm(25),TRUE) layout(matrix(c(2,0,1,3),2,2,byrow=TRUE),c(5,1),c(1,5),TRUE) layout.show(3) plot(data_cpmd,data_rfmd,xlab=xxlab,ylab=yylab,xlim=XX,ylim=YY,asp=1) abline(1,1) #rug(side=1,jitter(data_cpmd,5)) #rug(side=2,jitter(data_rfmd,5)) #plot(homo_ana$dism16cpmd,homo_ana$dism16rfmd,main=mtitle,xlab=xxlab,ylab=yylab,xlim=XX,ylim=YY,asp=1) par(mar=c(0,3,1,1)) barplot(data_cpmd, axes=FALSE, ylim=YY, space=0) par(mar=c(3,0,1,1)) barplot(data_rfmd, axes=FALSE,main=mtitle, xlim=XX, space=0, horiz=TRUE) #boxplot(data_cpmd,horizontal = TRUE,xlim=XX,ylim=YY,outline=ifout, xaxt = n) #par(mar=c(3,0,1,1)) #boxplot(data_rfmd,xlim=XX,ylim=YY,outline=ifout,yaxt = n) -- Sarah Goslee http://www.stringpage.com http://www.sarahgoslee.com http://www.functionaldiversity.org [[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] Extracting arithmetic mean for specific values from multiple .txt-files
Hello,
Ok, I think that there were two problems.
One, gsub substitutes all (g - global) occurrences of the search
pattern, so both periods were removed.
The other, it would allways consider column 8 as character, but when
there are no values with two periods it's read in with class numeric.
Both are now corrected.
fun - function(x, skip = 14){
dat - read.table(x, skip=skip, stringsAsFactors = FALSE)
if(is.character(dat[, 8])){
len - sapply(strsplit(dat[, 8], \\.), length)
dat[len == 3 , 8] - sub(\\., , dat[len == 3 , 8])
dat[, 8] - as.numeric(dat[, 8])
}
mean(dat[, 8])
}
sapply(list.files(pattern=XYZ.*\\.txt), fun)
Rui Barradas
Em 10-07-2012 09:35, vimmster escreveu:
Dear Mr. Holtman,
but I cannot leave out the value and cannot change the values to 1200.995
manually (for each test subject with a reaction time 1000 ms), because the
first your lead to incomplete data and the latter would be too
time-consuming.
Dear Rui,
here I have three files, which have exactly the same content as
XYZ_34.txt, EXCEPT that the file XYZ_50.txt doesn't have a period in the
first value 1200.9952 IF YOU OPEN IT WITH THE EDITOR (!), maybe because I
didn't change the structure with MS Excel. The other two files should be
identical.
http://r.789695.n4.nabble.com/file/n4635962/XYZ_2.txt XYZ_2.txt
http://r.789695.n4.nabble.com/file/n4635962/XYZ_50.txt XYZ_50.txt
http://r.789695.n4.nabble.com/file/n4635962/XYZ_1112.txt XYZ_1112.txt
R gives me the following output:
fun - function(x){
+ dat - read.table(x, skip=14)
+ dat[ , 8] - as.numeric(gsub(\\., , dat[, 8]))
+ mean(dat[, 8])
+ }
sapply(list.files(pattern=XYZ.*\\.txt), fun)
XYZ_1112.txtXYZ_2.txt XYZ_50.txt
345210.4 345210.4 310112.0
Your second suggestion leads to the same output:
fun - function(x, skip = 14){
+ dat - read.table(x, skip=skip)
+ dat[ , 8] - as.numeric(gsub(\\., , dat[, 8]))
+ mean(dat[, 8])
+ }
sapply(list.files(pattern=XYZ.*\\.txt), fun)
XYZ_1112.txtXYZ_2.txt XYZ_50.txt
345210.4 345210.4 310112.0
Thank you for your replies!
Kind regards
--
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Re: [R] How to use external image with R plot?
There are many ways to do this, here is one using an example from ?barplot
library(jpeg)
x - readJPEG(arrow_glossy_right_red.jpg)
barplot(VADeaths, border = dark blue)
rasterImage(x, 0, 60, 1, 80)
You'll have to change those numbers to suit your plot, par('usr)
gives a quick idea of the extents. You should be able to see that JPEG
is not a good format for drawings like this (and using images like
this is not a good method anyway since you never know what scale you
might want the shape to be drawn at). You might be better just using
?arrows or ?symbols but if you are bound to a particular image source
try to save it as a pixel format that doesn't crush the data (such as
PNG).
Also, I'm not really merging images here, and I wouldn't say that's
the way to go about it. This is using R's plot support for vector and
raster graphics.
Cheers, Mike.
On Tue, Jul 10, 2012 at 5:19 PM, Manish Gupta mandecent.gu...@gmail.com wrote:
Hi,
I am wokring on stacked bar plot and i need to add one arrow dynamically.
http://r.789695.n4.nabble.com/file/n4635959/arrow_glossy_right_red.jpg
http://r.789695.n4.nabble.com/file/n4635959/Screenshot.10.png
Final image:
http://r.789695.n4.nabble.com/file/n4635959/Screenshot.1.png
Is there any packae which can be used for merging images?
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e-mail: mdsum...@gmail.com
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Re: [R] is it possible to insert a figure into into another new figure by r script
Jie, I think the R contributed package, grImport, by Paul Murrell does what you want. See this: http://www.jstatsoft.org/v30/i04/paper/ Tom On Mon, Jul 9, 2012 at 11:14 PM, Mikhail Titov m...@gmx.us wrote: Jie Tang totang...@gmail.com writes: hi R-users Now I have a figure in emf or png or tiff format that have been drawn by other tool and I want to insert this figure into my new figure by R script. I wonder if is possible ? This [1] might be relevant. [1] http://addictedtor.free.fr/graphiques/RGraphGallery.php?graph=168 -- Mikhail __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Thomas E Adams National Weather Service Ohio River Forecast Center 1901 South State Route 134 Wilmington, OH 45177 EMAIL: thomas.ad...@noaa.gov VOICE: 937-383-0528 FAX:937-383-0033 [[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] image.plot transparent?
Hi R users, I have a maybe strange problem. Normaly I do image.plot() with x,y coordinates and add=T and if I have some NA values in my data matrix z, the color will be transparent of these pixels. But now I have a disorted coordinate system and x,y are a matrix. It works also fine, but now NA values are white colored and not transparent anymore. It is problematic if I have a secondary information underlying. Is there any solution for this stuff? Thanks! best regards -- View this message in context: http://r.789695.n4.nabble.com/image-plot-transparent-tp4635976.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] How to add marker in Stacked bar plot?
Yes, you can. See the lwd argument under ?arrows. --JIV On Tue, Jul 10, 2012 at 2:16 AM, Manish Gupta mandecent.gu...@gmail.comwrote: Thanks it works fine. But can i control its width? Regards -- View this message in context: http://r.789695.n4.nabble.com/How-to-add-marker-in-Stacked-bar-plot-tp4635946p4635954.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Predicted values for zero-inflated Poisson
Alain- Thanks again for the response. I guess my question is more related to R, which I'm learning as I go along. Could you provide guidance as to how I would code this in R? Thanks, Laura From: Alain Zuur [via R] [mailto:ml-node+s789695n4635920...@n4.nabble.com] Sent: Monday, July 09, 2012 5:20 PM To: Lee, Laura Subject: Re: Predicted values for zero-inflated Poisson Laura Lee laura.lee at ncdenr.gov Mon Jul 9 22:51:40 CEST 2012 Previous message: [R] Predicted values for zero-inflated Poisson Next message: [R] Lavaan Package - How to Extract Residuals in Data Values Messages sorted by: [ date ] [ thread ] [ subject ] [ author ] Thanks for your reply. I do have a copy of Zero Inflated Models and Generalized Linear Mixed Models with R and have been using that as a guide. I applied the predict function (type=count) to the dataset for which I built the model to compare the predicted bycatch numbers to the observed to ensure I was doing things correctly. When I summed over the new predicted variable, the value is over 500 whereas there were less than 10 observed. I'd appreciate any advice regarding what I'm doing wrong. -- View this message in context: http://r.789695.n4.nabble.com/Predicted-values-for-zero-inflated-Poisson-tp4635861p4635916.html Sent from the R help mailing list archive at Nabble.com. AFZ: Laurano...you don't want to compare the count part of the model with raw datayou need to compare the Exp(Y) = (1-pi) * mu with the raw data. Have a look at the Epilogue chapterit shows the difference between mu and (1-pi) * mu. Alain -- Dr. Alain F. Zuur First author of: 1. Analysing Ecological Data (2007). Zuur, AF, Ieno, EN and Smith, GM. Springer. 680 p. URL: www.springer.com/0-387-45967-7http://www.springer.com/0-387-45967-7 2. Mixed effects models and extensions in ecology with R. (2009). Zuur, AF, Ieno, EN, Walker, N, Saveliev, AA, and Smith, GM. Springer. http://www.springer.com/life+sci/ecology/book/978-0-387-87457-9 3. A Beginner's Guide to R (2009). Zuur, AF, Ieno, EN, Meesters, EHWG. Springer http://www.springer.com/statistics/computational/book/978-0-387-93836-3 4. Zero Inflated Models and Generalized Linear Mixed Models with R. (2012) Zuur, Saveliev, Ieno. http://www.highstat.com/book4.htm Other books: http://www.highstat.com/books.htm Statistical consultancy, courses, data analysis and software Highland Statistics Ltd. 6 Laverock road UK - AB41 6FN Newburgh Tel: 0044 1358 788177 Email: [hidden email]/user/SendEmail.jtp?type=nodenode=4635920i=0 URL: www.highstat.comhttp://www.highstat.com URL: www.brodgar.comhttp://www.brodgar.com [[alternative HTML version deleted]] __ [hidden email]/user/SendEmail.jtp?type=nodenode=4635920i=1 mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Dr. Alain F. Zuur First author of: 1. Analysing Ecological Data (2007). Zuur, AF, Ieno, EN and Smith, GM. Springer. 680 p. URL: www.springer.com/0-387-45967-7http://www.springer.com/0-387-45967-7 2. Mixed effects models and extensions in ecology with R. (2009). Zuur, AF, Ieno, EN, Walker, N, Saveliev, AA, and Smith, GM. Springer. http://www.springer.com/life+sci/ecology/book/978-0-387-87457-9 3. A Beginner's Guide to R (2009). Zuur, AF, Ieno, EN, Meesters, EHWG. Springer http://www.springer.com/statistics/computational/book/978-0-387-93836-3 Other books: http://www.highstat.com/books.htm Statistical consultancy, courses, data analysis and software Highland Statistics Ltd. 6 Laverock road UK - AB41 6FN Newburgh Tel: 0044 1358 788177 Email: highs...@highstat.commailto:highs...@highstat.com URL: www.highstat.comhttp://www.highstat.com URL: www.brodgar.comhttp://www.brodgar.com If you reply to this email, your message will be added to the discussion below: http://r.789695.n4.nabble.com/Predicted-values-for-zero-inflated-Poisson-tp4635861p4635920.html To unsubscribe from Predicted values for zero-inflated Poisson, click herehttp://r.789695.n4.nabble.com/template/NamlServlet.jtp?macro=unsubscribe_by_codenode=4635861code=bGF1cmEubGVlQG5jZGVuci5nb3Z8NDYzNTg2MXw0NTg5ODY5NTk=. NAMLhttp://r.789695.n4.nabble.com/template/NamlServlet.jtp?macro=macro_viewerid=instant_html%21nabble%3Aemail.namlbase=nabble.naml.namespaces.BasicNamespace-nabble.view.web.template.NabbleNamespace-nabble.view.web.template.NodeNamespacebreadcrumbs=notify_subscribers%21nabble%3Aemail.naml-instant_emails%21nabble%3Aemail.naml-send_instant_email%21nabble%3Aemail.naml - Laura M. Lee Senior Stock Assessment Scientist North Carolina Division of Marine Fisheries E-Mail: laura@ncdenr.gov -- View this message in context: http://r.789695.n4.nabble.com/Predicted-values-for-zero-inflated-Poisson-tp4635861p4635978.html Sent from the R help mailing
Re: [R] how can I show the xlab and ylab information while using layout
The mar argument to par(). Please do read ?par and perhaps the posting
guide.
Sarah
On Tuesday, July 10, 2012, Jie Tang wrote:
which parameter ?
2012/7/10 Sarah Goslee sarah.gos...@gmail.com javascript:_e({}, 'cvml',
'sarah.gos...@gmail.com');
The margins you specified aren't large enough to hold the information
you're trying to put in them, so you need to make them larger.
Sarah
On Tuesday, July 10, 2012, Jie Tang wrote:
hi R-users:
I want to draw three plot into one figure by layout and the script has
been shown below.
But I find R does not show the xlab and ylab information completely
as
shown the figure attached.
How can I midify the script.? thank you .
xxlab-paste(cpmd, (,ro,%),sep= )
yylab-paste(rfmd, (,co,%),sep= )
par(mar=c(3,3,1,1))
#layout(matrix(c(2,0,1,3),2,2,byrow=TRUE),widths=lcm(30),
heights=lcm(25),TRUE)
layout(matrix(c(2,0,1,3),2,2,byrow=TRUE),c(5,1),c(1,5),TRUE)
layout.show(3)
plot(data_cpmd,data_rfmd,xlab=xxlab,ylab=yylab,xlim=XX,ylim=YY,asp=1)
abline(1,1)
#rug(side=1,jitter(data_cpmd,5))
#rug(side=2,jitter(data_rfmd,5))
#plot(homo_ana$dism16cpmd,homo_ana$dism16rfmd,main=mtitle,xlab=xxlab,ylab=yylab,xlim=XX,ylim=YY,asp=1)
par(mar=c(0,3,1,1))
barplot(data_cpmd, axes=FALSE, ylim=YY, space=0)
par(mar=c(3,0,1,1))
barplot(data_rfmd, axes=FALSE,main=mtitle, xlim=XX, space=0, horiz=TRUE)
#boxplot(data_cpmd,horizontal = TRUE,xlim=XX,ylim=YY,outline=ifout, xaxt
=
n)
#par(mar=c(3,0,1,1))
#boxplot(data_rfmd,xlim=XX,ylim=YY,outline=ifout,yaxt =
--
Sarah Goslee
http://www.stringpage.com
http://www.sarahgoslee.com
http://www.functionaldiversity.org
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Re: [R] image.plot transparent?
This may be device and OS dependent, so please provide the information requested in the posting guide, at a minimum the output of sessionInfo(). A small reproducible example is also necessary. Sarah On Tuesday, July 10, 2012, Chris82 wrote: Hi R users, I have a maybe strange problem. Normaly I do image.plot() with x,y coordinates and add=T and if I have some NA values in my data matrix z, the color will be transparent of these pixels. But now I have a disorted coordinate system and x,y are a matrix. It works also fine, but now NA values are white colored and not transparent anymore. It is problematic if I have a secondary information underlying. Is there any solution for this stuff? Thanks! best regards -- View this message in context: http://r.789695.n4.nabble.com/image-plot-transparent-tp4635976.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org javascript:; mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Sarah Goslee http://www.stringpage.com http://www.sarahgoslee.com http://www.functionaldiversity.org [[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] RGB components of plot() colours
A quick question: Is there anywhere a listing of the RGB components of the named colours listed by colors()? For example, where would I find the RGB for orange1 or salmon? When I look at an EPS file from R where I have used these colours, it seems that for: salmon: 0.9804 0.5020 0.4471 rgb orange1: 1 0.6471 0 rgb However, this is a tedious way of finding out! With thanks, Ted. - E-Mail: (Ted Harding) ted.hard...@wlandres.net Date: 10-Jul-2012 Time: 14:05:15 This message was sent by XFMail __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] number of decimal places in a number?
Dear R users,
Thank You very much for Your responsiveness. I think the suggestion of arun's
modification of Josh's code works best and it is what I am going to implement.
Cite Ellison:
Surely the issue is not the particular numeric resolution of the numbers but
the idea that the bounding box limits should be integer multiples of the
resolution?
Is that not accomplished more straightforwardly by things like
min - resol * floor( min(lat)/resol )
max - resol * ceil( max(lat)/resol )
?
Dear Ellison,
not the bounding box limits, but the bounding box RANGE needs to be an integer
multiple of the resolution, that is max-min=ineger*resolution. The bbox limits
must just be rounded to the same number of digits as the resolution and include
all available values, of course.
Best regards.
Оригинално писмо
От: arun
Относно: Re: [R] number of decimal places in a number?
До: ted.hard...@wlandres.net
Изпратено на: Събота, 2012, Юли 7 22:50:03 EEST
Hi,
I checked the count for the cases (A) and (F) with a variant of Josh's
function:
decimalnumcount-function(x){stopifnot(class(x)==quot;characterquot;)
x-gsub(quot;(.*)(#92;#92;.)|([0]*#36;)quot;,quot;quot;,x)
nchar(x)
}
A)
x-quot;123456789.123456789quot;
decimalnumcount(x)
#[1] 9
x-quot;923456789.123456789quot;
decimalnumcount(x)
#[1] 9
B)
x-quot;0.012345quot;
decimalnumcount(x)
#[1] 10
c)
x - c(quot;3.14quot;, quot;3.142quot;, quot;3.1400quot;,
quot;123456.123456789quot;, quot;123456789.123456789quot;,pi,sqrt(2))
decimalnumcount(x)
#[1] 2 3 2 9 9 14 13
x
[1] quot;3.14quot; quot;3.142quot;
quot;3.1400quot;
[4] quot;123456.123456789quot; quot;123456789.123456789quot;
quot;3.14159265358979quot;
[7] quot;1.4142135623731quot;
D)
x - c(3.14, 3.142, 3.1400, 123456.123456789, 123456789.123456789,pi,sqrt(2))
decimalnumcount(as.character(x) )
[1] 2 3 2 9 6 14 13
#as.character(x)
#[1] quot;3.14quot; quot;3.142quot;
quot;3.14quot; quot;123456.123456789quot;
#[5] quot;123456789.123457quot; quot;3.14159265358979quot;
quot;1.4142135623731quot;
E)
print(pi,22)
#[1] 3.141592653589793115998
print(sqrt(2),22)
#[1] 1.414213562373095145475
x - c(quot;3.14quot;, quot;3.142quot;, quot;3.1400quot;,
quot;123456.123456789quot;,
quot;123456789.123456789quot;,quot;3.141592653589793115998quot;,quot;1.414213562373095145475quot;)
decimalnumcount(x)
#[1] 2 3 2 9 9 21 21
F)
formatC(pi,format=quot;fquot;,digits=22)
#[1] quot;3.1415926535897931159980quot;
formatC(sqrt(2),format=quot;fquot;,digits=22)
#[1] quot;1.4142135623730951454746quot;
x - c(quot;3.14quot;, quot;3.142quot;, quot;3.1400quot;,
quot;123456.123456789quot;,
quot;123456789.123456789quot;,formatC(pi,format=quot;fquot;,digits=22),formatC(sqrt(2),format=quot;fquot;,digits=22))
decimalnumcount(x)
#[1] 2 3 2 9 9 21 22
G)
# formatC() didn't show the limitations of print()
print(sqrt(2),22)
#[1] 1.414213562373095145475
print(sqrt(2),35)
#Error in print.default(sqrt(2), 35) : invalid 'digits' argument
formatC(sqrt(2),35)
#[1] quot;1.4142135623730951454746218587388285quot;
or,
formatC(sqrt(2),format=quot;fquot;,digits=35)
#[1] quot;1.41421356237309514547462185873882845quot;
#using 22
x - c(quot;3.14quot;, quot;3.142quot;, quot;3.1400quot;,
quot;123456.123456789quot;,
quot;123456789.123456789quot;,formatC(pi,format=quot;fquot;,digits=35),formatC(sqrt(2),format=quot;fquot;,digits=50))
decimalnumcount(x)
#[1] 2 3 2 9 9 35 50
So, I guess it will be better to deal with character strings rather than
using as.character.
A.K.
- Original Message -
From: quot;ted.hard...@wlandres.netquot;
To: r-help@r-project.org
Cc: Martin Ivanov
Sent: Saturday, July 7, 2012 8:12 AM
Subject: Re: [R] number of decimal places in a number?
I had thought of also (as well as my numerical routing) suggesting
a quot;gsub()quot; type solution like Joshua's below, but held back because
the result could depend on how the number arose (keyboard input,
file input, or from computation within R).
However, I now also realise that (again because of binary rounding
errors), the quot;gsub()quot; method has interesting differences from my
numerical method. Example:
[A] (as from my original method):
f(123456789.123456789)
# [1] 7
[B] (the quot;gsub()quot; method)
nchar(gsub(quot;(.*#92;#92;.)|([0]*#36;)quot;, quot;quot;,
as.character(123456789.123456789)))
# [1] 6
Now look at:
[C] (what as.character() does to 123456789.123456789)
as.character(123456789.123456789)
# [1] quot;123456789.123457quot;
[D] (quot;22quot; is the maximum number of decimal digits for print())
print(123456789.123456789,22)
# [1] 123456789.1234568
So as.character() has rounded it to 6 decimal places (agreeing
with [B]), while
[R] define stuff to be only usable in the same file
Hello R-Help! I've looked around and have not found: A simple(short) way to hide functions and variables from the global environment. What i want is for a few of them to only be accessable from the scriptfile they're in. I probably could do fun things with environments , but that seems quite a hassle. As example: I have a file that gets me stuff from the database and creates an R object from the results, plus some functions on that object. Now i want the objectrelated stuff to be global, while the functions and variables i use for accessing the database shall be hidden. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] fill 0-row data.frame with 1 line of NAs
Dear all Is there a simpler method to achieve the following: When I obtain an empty data.frame after subsetting, I need for it to contain one line of NAs. Here's a dummy example: (.xb - iris[ iris$Species=='zz', ]) [1] Sepal.Length Sepal.Width Petal.Length Petal.Width Species 0 rows (or 0-length row.names) dim(.xb) [1] 0 5 (.xa - data.frame(matrix(rep(NA, ncol(.xb)), 1))) X1 X2 X3 X4 X5 1 NA NA NA NA NA names(.xa) - names(.xb) (.xb - .xa) Sepal.Length Sepal.Width Petal.Length Petal.Width Species 1 NA NA NA NA NA The solution I came up with is way too convoluted. Anything simpler? Regards Liviu -- 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] RGB components of plot() colours
On 12-07-10 9:05 AM, (Ted Harding) wrote: A quick question: Is there anywhere a listing of the RGB components of the named colours listed by colors()? For example, where would I find the RGB for orange1 or salmon? When I look at an EPS file from R where I have used these colours, it seems that for: salmon: 0.9804 0.5020 0.4471 rgb orange1: 1 0.6471 0 rgb However, this is a tedious way of finding out! col2rgb(salmon) [,1] red250 green 128 blue 114 Those need to be divided by 255 to get the colours on the 0-1 scale. Duncan Murdoch __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] define stuff to be only usable in the same file
On 12-07-10 9:13 AM, Jessica Streicher wrote: Hello R-Help! I've looked around and have not found: A simple(short) way to hide functions and variables from the global environment. What i want is for a few of them to only be accessable from the scriptfile they're in. I probably could do fun things with environments , but that seems quite a hassle. The simplest and best way to do this is to write a package. You can also use local() around the code in a script, but it gets messy when you want to export more than one thing. Duncan Murdoch As example: I have a file that gets me stuff from the database and creates an R object from the results, plus some functions on that object. Now i want the objectrelated stuff to be global, while the functions and variables i use for accessing the database shall be hidden. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] RGB components of plot() colours
R uses the sandard X11 colors, I believe, and if you're using linux there's a rgb.txt file on your computer that contains them. It's also available here http://cvsweb.xfree86.org/cvsweb/*checkout*/xc/programs/rgb/rgb.txt?rev=1.1 and a less-authoritative but prettier version: http://en.wikipedia.org/wiki/X11_color_names Within R, colors() is what you need. See here for an exhaustive discussion: http://research.stowers-institute.org/efg/R/Color/Chart/ Sarah On Tuesday, July 10, 2012, Ted Harding wrote: A quick question: Is there anywhere a listing of the RGB components of the named colours listed by colors()? For example, where would I find the RGB for orange1 or salmon? When I look at an EPS file from R where I have used these colours, it seems that for: salmon: 0.9804 0.5020 0.4471 rgb orange1: 1 0.6471 0 rgb However, this is a tedious way of finding out! With thanks, Ted. - E-Mail: (Ted Harding) ted.hard...@wlandres.net javascript:; Date: 10-Jul-2012 Time: 14:05:15 This message was sent by XFMail __ R-help@r-project.org javascript:; mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Sarah Goslee http://www.stringpage.com http://www.sarahgoslee.com http://www.functionaldiversity.org [[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] Predicted values for zero-inflated Poisson
On Tue, 10 Jul 2012, Laura Lee wrote: Alain- Thanks again for the response. I guess my question is more related to R, which I'm learning as I go along. Could you provide guidance as to how I would code this in R? That depends on what exactly you want to predict. As Alain said: The type=count predictions are probably not interesting to you. In the JSS paper accompanying zeroinfl, these correspond to exp(x'b) in Equation 8. More interesting is probably the mean (1 - pi) * exp(x'b) which is the expected mean mu_i in Equation 8. This can be obtained as type=response. Additionally, you might be interested in the individual probabilities for counts of 0, 1, 2, ... etc. This corresponds to evaluating Equation 7 for y = 0, 1, 2, ... which can be obtained via type=prob. And from this you could also get the mode or median rather than the mean of the distribution. hth, Z Thanks, Laura From: Alain Zuur [via R] [mailto:ml-node+s789695n4635920...@n4.nabble.com] Sent: Monday, July 09, 2012 5:20 PM To: Lee, Laura Subject: Re: Predicted values for zero-inflated Poisson Laura Lee laura.lee at ncdenr.gov Mon Jul 9 22:51:40 CEST 2012 Previous message: [R] Predicted values for zero-inflated Poisson Next message: [R] Lavaan Package - How to Extract Residuals in Data Values Messages sorted by: [ date ] [ thread ] [ subject ] [ author ] Thanks for your reply. I do have a copy of Zero Inflated Models and Generalized Linear Mixed Models with R and have been using that as a guide. I applied the predict function (type=count) to the dataset for which I built the model to compare the predicted bycatch numbers to the observed to ensure I was doing things correctly. When I summed over the new predicted variable, the value is over 500 whereas there were less than 10 observed. I'd appreciate any advice regarding what I'm doing wrong. -- View this message in context: http://r.789695.n4.nabble.com/Predicted-values-for-zero-inflated-Poisson-tp4635861p4635916.html Sent from the R help mailing list archive at Nabble.com. AFZ: Laurano...you don't want to compare the count part of the model with raw datayou need to compare the Exp(Y) = (1-pi) * mu with the raw data. Have a look at the Epilogue chapterit shows the difference between mu and (1-pi) * mu. Alain -- Dr. Alain F. Zuur First author of: 1. Analysing Ecological Data (2007). Zuur, AF, Ieno, EN and Smith, GM. Springer. 680 p. URL: www.springer.com/0-387-45967-7http://www.springer.com/0-387-45967-7 2. Mixed effects models and extensions in ecology with R. (2009). Zuur, AF, Ieno, EN, Walker, N, Saveliev, AA, and Smith, GM. Springer. http://www.springer.com/life+sci/ecology/book/978-0-387-87457-9 3. A Beginner's Guide to R (2009). Zuur, AF, Ieno, EN, Meesters, EHWG. Springer http://www.springer.com/statistics/computational/book/978-0-387-93836-3 4. Zero Inflated Models and Generalized Linear Mixed Models with R. (2012) Zuur, Saveliev, Ieno. http://www.highstat.com/book4.htm Other books: http://www.highstat.com/books.htm Statistical consultancy, courses, data analysis and software Highland Statistics Ltd. 6 Laverock road UK - AB41 6FN Newburgh Tel: 0044 1358 788177 Email: [hidden email]/user/SendEmail.jtp?type=nodenode=4635920i=0 URL: www.highstat.comhttp://www.highstat.com URL: www.brodgar.comhttp://www.brodgar.com [[alternative HTML version deleted]] __ [hidden email]/user/SendEmail.jtp?type=nodenode=4635920i=1 mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Dr. Alain F. Zuur First author of: 1. Analysing Ecological Data (2007). Zuur, AF, Ieno, EN and Smith, GM. Springer. 680 p. URL: www.springer.com/0-387-45967-7http://www.springer.com/0-387-45967-7 2. Mixed effects models and extensions in ecology with R. (2009). Zuur, AF, Ieno, EN, Walker, N, Saveliev, AA, and Smith, GM. Springer. http://www.springer.com/life+sci/ecology/book/978-0-387-87457-9 3. A Beginner's Guide to R (2009). Zuur, AF, Ieno, EN, Meesters, EHWG. Springer http://www.springer.com/statistics/computational/book/978-0-387-93836-3 Other books: http://www.highstat.com/books.htm Statistical consultancy, courses, data analysis and software Highland Statistics Ltd. 6 Laverock road UK - AB41 6FN Newburgh Tel: 0044 1358 788177 Email: highs...@highstat.commailto:highs...@highstat.com URL: www.highstat.comhttp://www.highstat.com URL: www.brodgar.comhttp://www.brodgar.com If you reply to this email, your message will be added to the discussion below: http://r.789695.n4.nabble.com/Predicted-values-for-zero-inflated-Poisson-tp4635861p4635920.html To unsubscribe from Predicted values for zero-inflated Poisson, click
Re: [R] fill 0-row data.frame with 1 line of NAs
Hello,
If you write a function, it becomes less convoluted...
empty - function(x){
if(NROW(x) == 0){
y - rep(NA, NCOL(x))
names(y) - names(x)
y
}else x
}
(.xb - iris[ iris$Species=='zz', ])
empty(.xb)
Hope this helps,
Rui Barradas
Em 10-07-2012 14:15, Liviu Andronic escreveu:
Dear all
Is there a simpler method to achieve the following: When I obtain an
empty data.frame after subsetting, I need for it to contain one line
of NAs. Here's a dummy example:
(.xb - iris[ iris$Species=='zz', ])
[1] Sepal.Length Sepal.Width Petal.Length Petal.Width Species
0 rows (or 0-length row.names)
dim(.xb)
[1] 0 5
(.xa - data.frame(matrix(rep(NA, ncol(.xb)), 1)))
X1 X2 X3 X4 X5
1 NA NA NA NA NA
names(.xa) - names(.xb)
(.xb - .xa)
Sepal.Length Sepal.Width Petal.Length Petal.Width Species
1 NA NA NA NA NA
The solution I came up with is way too convoluted. Anything simpler? Regards
Liviu
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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Re: [R] Plotting rpart trees with long list of class members
Thanks. Very helpful.
You can use the information from the splits in the first tree, to define a
new grouping variable, which will simplify the plot:
suvar - sort(unique(test_set$list_var))
test_set$var_grp - as.factor(testtree$csplit[match(test_set$list_var,
suvar)])
testtree2 - rpart ( list_val ~ var_grp, data = test_set )
rpart.plot(testtree2, type=3)
Not to other readers, you will need to load these packages, before running
the code:
library(rpart)
library(rpart.plot)
Jean
MarkBeauchene markbeauch...@hotmail.com wrote on 07/09/2012 03:42:32 PM:
Here is some sample code. It generates a class (list_var) that is used
in
rpart. list_val is the dependant variable.
The plot shows all the values of the class, which is a mess and makes
the
plot unuseable. I'd like to either suppress the list entirely or
replace it
with something like Group 1, Group 2, etc.
list_var - rep(NA,2000)
list_val - rep(NA,2000)
for (i in 1:1000) {
list_var[i] - paste(A,i%/%25,sep='')
list_val[i] - runif(1,0,1) }
test_set - data.frame(list_var, list_val )
testtree - rpart ( list_val ~ list_var, data = test_set )
rpart.plot(testtree, type=3)
[[alternative HTML version deleted]]
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Re: [R] image.plot transparent?
On 10/07/2012 13:41, Sarah Goslee wrote:
This may be device and OS dependent, so please provide the information
requested in the posting guide, at a minimum the output of sessionInfo(). A
small reproducible example is also necessary.
I think not. My guess is it is part of a package which we were not told
('fields'), and this is the programmed behaviour in its two subcases.
But we'd need reproducible code (including which package) to be sure.
If this is fields, read ?poly.image.
Sarah
On Tuesday, July 10, 2012, Chris82 wrote:
Hi R users,
I have a maybe strange problem.
Normaly I do image.plot() with x,y coordinates and add=T and if I have some
NA values in my data matrix z, the color will be transparent of these
pixels.
But now I have a disorted coordinate system and x,y are a matrix. It works
also fine, but now NA values are white colored and not transparent anymore.
It is problematic if I have a secondary information underlying.
Is there any solution for this stuff?
Thanks!
--
Brian D. Ripley, rip...@stats.ox.ac.uk
Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/
University of Oxford, Tel: +44 1865 272861 (self)
1 South Parks Road, +44 1865 272866 (PA)
Oxford OX1 3TG, UKFax: +44 1865 272595
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[R] customize packages' help index ( 00index.html file )
Hi all, I'm writing my packages helps files and I'm not really satisfied by the visual results. I'm would like to make subsections in a package function help index file. I would like for example to put all S4 object documentation link together, then all the getters function.. and so on.. Does someone know if it's possible to do it? Is it possible to define by myself the html/00index.html file that will be use in my package? If it's not possible, how could I add the alphabetic subsections that exist in most of packages index help files? Best, Damien __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] identify.hclust() doesn't cut tree at the vertical position of the mouse pointer
Dear All According to the identify.hclust documentation the function cuts the tree at the vertical position of the pointer and highlights the cluster containing the horizontal position of the pointer. When I carry out this, the tree isn't cut where I click - in fact, there seems to be a limit below which I cannot go. Consider the following code: mat - matrix(rnorm(5000), ncol=5) hc - hclust(dist(mat)) plot(hc) identify(hc) No matter where I click on the tree, I cannot cut below around about 5. I can cut above that value, but not below. Any help is much appreciated. sessionInfo() R version 2.15.1 (2012-06-22) Platform: x86_64-pc-mingw32/x64 (64-bit) locale: [1] LC_COLLATE=English_United Kingdom.1252 LC_CTYPE=English_United Kingdom.1252LC_MONETARY=English_United Kingdom.1252 LC_NUMERIC=C [5] LC_TIME=English_United Kingdom.1252 attached base packages: [1] grid stats graphics grDevices utils datasets methods base other attached packages: [1] geneplotter_1.34.0 lattice_0.20-6 annotate_1.34.1 AnnotationDbi_1.18.1 Biobase_2.16.0 BiocGenerics_0.2.0 BiocInstaller_1.4.7 gplots_2.11.0MASS_7.3-18 [10] KernSmooth_2.23-7caTools_1.13 bitops_1.0-4.1 gdata_2.11.0 gtools_2.7.0 loaded via a namespace (and not attached): [1] DBI_0.2-5 IRanges_1.14.4 RColorBrewer_1.0-5 RSQLite_0.11.1 stats4_2.15.1 tools_2.15.1 XML_3.9-4.1xtable_1.7-0 Thanks Mick -- The University of Edinburgh is a charitable body, registered in Scotland, with registration number SC005336. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Use of Sappy and Tappy for Mathematical Calculation
Hi, i have a matrix like this, ABCXYZ... . - -- 1220 ... . 2435 ... . 3040 ... . Here, i need to get Sum of each columns, Mean of each columns, median of each columns, mode of each columns, Standard deviation of each columns, variance of each columns, range of each columns, count of each columns, max of each columns, min of each columns Can i get output using sappy or tappy functions ? because there have alots of records. Could you please help me fast its kind of urgent ! - Thanks Antony -- View this message in context: http://r.789695.n4.nabble.com/Use-of-Sappy-and-Tappy-for-Mathematical-Calculation-tp4635969.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] fitting power growth
Dear all I am using the x and y vectors as defined below and want do to a power law regression: y = a x^b using lm(log(y)~log(x)) gives reasonable values (b=1.23) but is not very popular due to biases of back-transformation from log to non-log values. Using nls(y~a*x^b,start=list(a=100,b=1.23)) is statistically more correct but gives a too large a value and a too small b value. Doe anybody have a better way to solve the above power-law regression (using for instance maximum likely hood or anything else). Kind regards for your help Thomas x [1] 744.90 806.40 838.00 910.70 1818.60 2870.10 4070.00 4476.80 4857.60 4858.10 [11] 5916.40 13970.80 27306.60 28226.60 2532.10 2658.40 18863.10 758.0054.0079.00 [21] 139.0046.70 1003.0024.00 106.00 186.00 1503.00 228.0010.24 162.00 [31] 381.70 312.60 209.00 246.00 221.20 1151.55 y [1] 1.500e+08 2.850e+08 1.800e+08 1.800e+08 6.300e+08 7.200e+08 1.170e+09 1.095e+09 1.620e+09 [10] 4.650e+09 1.575e+09 4.200e+09 7.755e+09 8.745e+09 9.900e+08 6.600e+08 1.077e+10 3.450e+08 [19] 1.350e+07 2.550e+07 6.600e+07 6.000e+06 3.300e+07 1.500e+06 4.500e+06 7.500e+06 2.415e+08 [28] 6.900e+07 9.000e+05 9.450e+06 3.510e+07 4.880e+07 3.100e+06 1.930e+07 2.270e+07 5.270e+07 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Mac OS X R uninstallation question
Hi, I've been using R for a number of years and have always installed the newest version when released. However I've just noticed that old versions of R are taking up quite a lot of disk space. lap-alastair:/ alastair$ du -h -d 1 /Library/Frameworks/R.framework/Versions/ 266M/Library/Frameworks/R.framework/Versions/2.10 204M/Library/Frameworks/R.framework/Versions/2.11 459M/Library/Frameworks/R.framework/Versions/2.12 511M/Library/Frameworks/R.framework/Versions/2.13 478M/Library/Frameworks/R.framework/Versions/2.14 217M/Library/Frameworks/R.framework/Versions/2.15 32K /Library/Frameworks/R.framework/Versions/2.5 5.5M/Library/Frameworks/R.framework/Versions/2.6 84M /Library/Frameworks/R.framework/Versions/2.8 2.2G/Library/Frameworks/R.framework/Versions/ Do I need to keep any of the versions before the current (2.15). If I delete all the previous versions will that have any impact on the current installation, or is each version entirely self-contained? Thanks. -- View this message in context: http://r.789695.n4.nabble.com/Mac-OS-X-R-uninstallation-question-tp4635971.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] Questions about doing analysis based on time
Thanks to everyone for their help so far. It's been greatly appreciated. I have a new, but similar problem: I have data that I have broken down by hour (median/mean for each hour). I would like to break it down further, by each half hour (0:00-0:29, 0:30-0:59, 1:00-1:29, 1:30-1:59, etc). I thought cut.dates() from the chron package would be able to do it, but I can't find anything in the chron package documention. I'm fairly certain that if I can figure out how to break down the data by half hour that I can do all the other analysis just fine (median/mean for each half hour, etc) Here is a small sample of the data: dput(head(SundayData, 100)) structure(list(SunDate = structure(c(1273377600, 1307851200, 1213502640, 1217736420, 1311480420, 1211688540, 1337487060, 1255839120, 1268543520, 1293945120, 1280635980, 1309061640, 1322975640, 1297574280, 1221970740, 1253420340, 1218946800, 1329024060, 1290316920, 1224994980, 1218342420, 1269750420, 1257658080, 1322371680, 1214108940, 1312086540, 1260077400, 1228023060, 1315110660, 1281241920, 1272774960, 1224995820, 1275194220, 1246768860, 1302410460, 1234071780, 1305434580, 1232257500, 1243140300, 1284871500, 1247373960, 1265521560, 1273985160, 1310273160, 1226209620, 1270356420, 1330235280, 1222577400, 1310878200, 1324187400, 1242535860, 1336279860, 1283057520, 1291528320, 1324187580, 1330840380, 1298786100, 1307854500, 1236491880, 1298786280, 1233468180, 1280034240, 1230444300, 1213506360, 1251608760, 1215320820, 1304226420, 1320556080, 1299391740, 1286687580, 1296972780, 1296972780, 1321164780, 1260684960, 1315113420, 1287292680, 1292134800, 1303017600, 1307251200, 1278825720, 1238304180, 1212902640, 1231655100, 1254029100, 1311485100, 1295159160, 1220160420, 1297578540, 1300599000, 1241933640, 1225604100, 1269149880, 1283665140, 1244958120, 1245562980, 1289716980, 1235890020, 1282456080, 1279432140, 1279432140), class = c(POSIXct, POSIXt), tzone = ), SunTime = structure(c(1L, 1L, 2L, 3L, 3L, 4L, 5L, 6L, 6L, 6L, 7L, 8L, 8L, 9L, 10L, 10L, 11L, 12L, 13L, 14L, 15L, 15L, 16L, 16L, 17L, 17L, 18L, 19L, 19L, 20L, 21L, 22L, 22L, 23L, 23L, 24L, 24L, 25L, 25L, 25L, 26L, 26L, 26L, 26L, 27L, 27L, 28L, 29L, 29L, 29L, 30L, 30L, 31L, 31L, 32L, 32L, 33L, 33L, 34L, 34L, 35L, 36L, 37L, 38L, 38L, 39L, 39L, 40L, 41L, 42L, 42L, 42L, 42L, 43L, 44L, 45L, 46L, 46L, 46L, 47L, 48L, 49L, 50L, 50L, 50L, 51L, 52L, 53L, 54L, 55L, 56L, 57L, 58L, 59L, 60L, 60L, 61L, 62L, 63L, 63L), .Label = c(0:00, 0:04, 0:07, 0:09, 0:11, 0:12, 0:13, 0:14, 0:18, 0:19, 0:20, 0:21, 0:22, 0:23, 0:27, 0:28, 0:29, 0:30, 0:31, 0:32, 0:36, 0:37, 0:41, 0:43, 0:45, 0:46, 0:47, 0:48, 0:50, 0:51, 0:52, 0:53, 0:55, 0:58, 1:03, 1:04, 1:05, 1:06, 1:07, 1:08, 1:09, 1:13, 1:16, 1:17, 1:18, 1:20, 1:22, 1:23, 1:24, 1:25, 1:26, 1:27, 1:29, 1:30, 1:34, 1:35, 1:38, 1:39, 1:42, 1:43, 1:47, 1:48, 1:49, 1:52, 1:54, 1:55, 1:56, 1:57, 1:59, 10:00, 10:04, 10:07, 10:08, 10:09, 10:10, 10:11, 10:12, 10:14, 10:15, 10:16, 10:18, 10:20, 10:22, 10:23, 10:24, 10:25, 10:26, 10:27, 10:28, 10:30, 10:31, 10:32, 10:33, 10:34, 10:35, 10:36, 10:37, 10:38, 10:39, 10:40, 10:41, 10:43, 10:44, 10:45, 10:47, 10:48, 10:49, 10:50, 10:51, 10:53, 10:54, 10:55, 10:56, 10:58, 10:59, 11:01, 11:02, 11:05, 11:06, 11:07, 11:09, 11:10, 11:12, 11:14, 11:15, 11:16, 11:20, 11:21, 11:22, 11:23, 11:24, 11:26, 11:27, 11:29, 11:30, 11:31, 11:33, 11:34, 11:35, 11:36, 11:37, 11:38, 11:39, 11:40, 11:43, 11:44, 11:46, 11:47, 11:49, 11:52, 11:56, 11:58, 11:59, 12:00, 12:01, 12:02, 12:03, 12:04, 12:05, 12:06, 12:07, 12:08, 12:09, 12:10, 12:11, 12:13, 12:14, 12:15, 12:17, 12:19, 12:21, 12:22, 12:24, 12:25, 12:26, 12:27, 12:28, 12:30, 12:31, 12:32, 12:34, 12:36, 12:37, 12:38, 12:39, 12:41, 12:45, 12:46, 12:47, 12:48, 12:49, 12:50, 12:51, 12:53, 12:54, 12:55, 12:56, 12:57, 12:58, 12:59, 13:00, 13:01, 13:02, 13:03, 13:04, 13:05, 13:06, 13:07, 13:08, 13:09, 13:10, 13:12, 13:13, 13:14, 13:15, 13:17, 13:18, 13:19, 13:20, 13:21, 13:23, 13:25, 13:26, 13:27, 13:30, 13:31, 13:32, 13:34, 13:35, 13:36, 13:38, 13:39, 13:40, 13:44, 13:45, 13:46, 13:47, 13:48, 13:49, 13:50, 13:51, 13:52, 13:53, 13:54, 13:55, 13:57, 13:58, 13:59, 14:00, 14:01, 14:02, 14:04, 14:05, 14:06, 14:07, 14:08, 14:11, 14:12, 14:13, 14:14, 14:15, 14:16, 14:17, 14:18, 14:20, 14:21, 14:22, 14:23, 14:25, 14:26, 14:28, 14:29, 14:30, 14:31, 14:32, 14:34, 14:35, 14:36, 14:37, 14:38, 14:40, 14:41, 14:42, 14:43, 14:45, 14:46, 14:47, 14:49, 14:50, 14:51, 14:52, 14:53, 14:54, 14:56, 14:57, 14:58, 14:59, 15:01, 15:02, 15:03, 15:04, 15:05, 15:07, 15:11, 15:12, 15:13, 15:14, 15:15, 15:17, 15:18, 15:19, 15:20, 15:22, 15:23, 15:24, 15:25, 15:26, 15:28, 15:29, 15:30, 15:31, 15:33, 15:34, 15:35, 15:36, 15:37, 15:38, 15:40, 15:41, 15:42, 15:43,
[R] Count of elements in coulmns of a matrix
Could you please tell me what is the function or method to get count of elements in all the columns in a matrix ? for eg :- ABC XYZPQR -- - -- 234 4 5 54 3 2 Result will be like ABC XYZPQR -- - -- 2 43 Could you please help me ? -- View this message in context: http://r.789695.n4.nabble.com/Count-of-elements-in-coulmns-of-a-matrix-tp4635979.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] gdata: Problem reading excel document containing non-US characters
I am using the gdata package to read in an Excel document. read.xls chokes on a “foreign” character. Here's the original code: require(gdata) dendro - read.xls(/tmp/avitot.xlsv3WiXg,fileEncoding=Latin1) Now, the fileEncoding=Latin1 ought to work, because if i copy the code for read.xls, and edit retval - read.csv(con, na.strings = na.strings, ...) appropriately, to include fileEncoding=Latin1 and include the definition to override read.xls (to those who wish to test this: you need to include findPerl as well), then it works. but from the code, seems plain that read.xls should pass the ‘fileEncoding=Latin1’ all the way to when read.csv is called. So I don't know if this is a bug, or if I have misuderstood something. Is there a better way to achieve this than to edit and override read.xls? Any and all suggestions are welcome. Thank you, -- Rolf Marvin Bøe Lindgren r...@grendel.no http:/www.grendel.no/ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Count of elements in coulmns of a matrix
It depends: what's in those empty space? Some combination of apply() and something else, depending on what your matrix *actually* looks like, and here dput() would be vastly preferable to copy and paste of something that didn't even come from an R session. The something else might involve length() or is.na() or, well, other possibilities but my telepathy is on the fritz today. Sarah On Tue, Jul 10, 2012 at 8:31 AM, Rantony antony.akk...@ge.com wrote: Could you please tell me what is the function or method to get count of elements in all the columns in a matrix ? for eg :- ABC XYZPQR -- - -- 234 4 5 54 3 2 Result will be like ABC XYZPQR -- - -- 2 43 Could you please help me ? -- 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] define stuff to be only usable in the same file
On 10.07.2012, at 15:24, Duncan Murdoch wrote: On 12-07-10 9:13 AM, Jessica Streicher wrote: Hello R-Help! I've looked around and have not found: A simple(short) way to hide functions and variables from the global environment. What i want is for a few of them to only be accessable from the scriptfile they're in. I probably could do fun things with environments , but that seems quite a hassle. The simplest and best way to do this is to write a package. After an hour i can say it is neither simple nor short nor will it work at all at the moment ERROR cannot change to directory 'testpack' - tried to use the build command from pretty much everywhere You can also use local() around the code in a script, but it gets messy when you want to export more than one thing. I think local isn't quite what i imagined. Duncan Murdoch As example: I have a file that gets me stuff from the database and creates an R object from the results, plus some functions on that object. Now i want the objectrelated stuff to be global, while the functions and variables i use for accessing the database shall be hidden. [[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] Mac OS X R uninstallation question
Please ask Mac-specific questions on R-sig-mac. In particular, it is a little odd that these are not getting deleted when you install a new version. But *if all your packages are up to date* you do not need earlier versions of R.framework, so run update.packages(checkBuilt=TRUE) first. On 10/07/2012 11:55, Alastair wrote: Hi, I've been using R for a number of years and have always installed the newest version when released. However I've just noticed that old versions of R are taking up quite a lot of disk space. lap-alastair:/ alastair$ du -h -d 1 /Library/Frameworks/R.framework/Versions/ 266M/Library/Frameworks/R.framework/Versions/2.10 204M/Library/Frameworks/R.framework/Versions/2.11 459M/Library/Frameworks/R.framework/Versions/2.12 511M/Library/Frameworks/R.framework/Versions/2.13 478M/Library/Frameworks/R.framework/Versions/2.14 217M/Library/Frameworks/R.framework/Versions/2.15 32K /Library/Frameworks/R.framework/Versions/2.5 5.5M/Library/Frameworks/R.framework/Versions/2.6 84M /Library/Frameworks/R.framework/Versions/2.8 2.2G/Library/Frameworks/R.framework/Versions/ Do I need to keep any of the versions before the current (2.15). If I delete all the previous versions will that have any impact on the current installation, or is each version entirely self-contained? Thanks. -- Brian D. Ripley, rip...@stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] fill 0-row data.frame with 1 line of NAs
On 2012-07-10 06:57, Rui Barradas wrote:
Hello,
If you write a function, it becomes less convoluted...
empty - function(x){
if(NROW(x) == 0){
y - rep(NA, NCOL(x))
names(y) - names(x)
y
}else x
}
(.xb - iris[ iris$Species=='zz', ])
empty(.xb)
Both this and Liviu's original solution destroy the
factor nature of 'Species' (which may not matter, of
course). How about
(.xb - iris[ iris$Species=='zz', ])
.xb - .xb[1, ] # this probably shouldn't work, but it does.
?
Peter Ehlers
Hope this helps,
Rui Barradas
Em 10-07-2012 14:15, Liviu Andronic escreveu:
Dear all
Is there a simpler method to achieve the following: When I obtain an
empty data.frame after subsetting, I need for it to contain one line
of NAs. Here's a dummy example:
(.xb - iris[ iris$Species=='zz', ])
[1] Sepal.Length Sepal.Width Petal.Length Petal.Width Species
0 rows (or 0-length row.names)
dim(.xb)
[1] 0 5
(.xa - data.frame(matrix(rep(NA, ncol(.xb)), 1)))
X1 X2 X3 X4 X5
1 NA NA NA NA NA
names(.xa) - names(.xb)
(.xb - .xa)
Sepal.Length Sepal.Width Petal.Length Petal.Width Species
1 NA NA NA NA NA
The solution I came up with is way too convoluted. Anything simpler? Regards
Liviu
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Re: [R] define stuff to be only usable in the same file
On 10.07.2012, at 16:45, Jessica Streicher wrote: On 10.07.2012, at 15:24, Duncan Murdoch wrote: On 12-07-10 9:13 AM, Jessica Streicher wrote: Hello R-Help! I've looked around and have not found: A simple(short) way to hide functions and variables from the global environment. What i want is for a few of them to only be accessable from the scriptfile they're in. I probably could do fun things with environments , but that seems quite a hassle. The simplest and best way to do this is to write a package. After an hour i can say it is neither simple nor short nor will it work at all at the moment ERROR cannot change to directory 'testpack' - tried to use the build command from pretty much everywhere Forget about that, i'm stupid and can't use the tools available... You can also use local() around the code in a script, but it gets messy when you want to export more than one thing. I think local isn't quite what i imagined. Duncan Murdoch As example: I have a file that gets me stuff from the database and creates an R object from the results, plus some functions on that object. Now i want the objectrelated stuff to be global, while the functions and variables i use for accessing the database shall be hidden. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] fill 0-row data.frame with 1 line of NAs
On 7/10/2012 7:53 AM, Peter Ehlers wrote:
On 2012-07-10 06:57, Rui Barradas wrote:
Hello,
If you write a function, it becomes less convoluted...
empty - function(x){
if(NROW(x) == 0){
y - rep(NA, NCOL(x))
names(y) - names(x)
y
}else x
}
(.xb - iris[ iris$Species=='zz', ])
empty(.xb)
Both this and Liviu's original solution destroy the
factor nature of 'Species' (which may not matter, of
course). How about
(.xb - iris[ iris$Species=='zz', ])
.xb - .xb[1, ] # this probably shouldn't work, but it does.
Using NA subscripting seems even better
empty - function(x) {
if(NROW(x) == 0) {
x[NA,]
} else {
x
}
}
It even preserves the factor nature of things:
empty(iris[iris$Specis=='zz',])
Sepal.Length Sepal.Width Petal.Length Petal.Width Species
NA NA NA NA NANA
str(empty(iris[iris$Specis=='zz',]))
'data.frame': 1 obs. of 5 variables:
$ Sepal.Length: num NA
$ Sepal.Width : num NA
$ Petal.Length: num NA
$ Petal.Width : num NA
$ Species : Factor w/ 3 levels setosa,versicolor,..: NA
?
Peter Ehlers
Hope this helps,
Rui Barradas
Em 10-07-2012 14:15, Liviu Andronic escreveu:
Dear all
Is there a simpler method to achieve the following: When I obtain an
empty data.frame after subsetting, I need for it to contain one line
of NAs. Here's a dummy example:
(.xb - iris[ iris$Species=='zz', ])
[1] Sepal.Length Sepal.Width Petal.Length Petal.Width Species
0 rows (or 0-length row.names)
dim(.xb)
[1] 0 5
(.xa - data.frame(matrix(rep(NA, ncol(.xb)), 1)))
X1 X2 X3 X4 X5
1 NA NA NA NA NA
names(.xa) - names(.xb)
(.xb - .xa)
Sepal.Length Sepal.Width Petal.Length Petal.Width Species
1 NA NA NA NA NA
The solution I came up with is way too convoluted. Anything simpler?
Regards
Liviu
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
--
Brian S. Diggs, PhD
Senior Research Associate, Department of Surgery
Oregon Health Science University
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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[R] Help with vectors and rollapply
Hello
I have a vector a =(-2,0,0,0,1,0,0,3,0,0,-4)
I want to replace all zeros into previous non-zero state. So for instance the
above vector should be converted into:
a= (-2,-2,-2,-2,1,1,1,3,3,3,-4)
I tried many things and finally concluded that probably(?) rollapply may be the
best way?
I tried
f= function(x){
ifelse(x==0,Lag(x),x)
}
And then, rollappy(a,1,f) and that didn't work. Can someone help please?
Thx
R
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Re: [R] multiple comparisons with generalised least squares
racmar wrote I have also been searching various forums and books to see if there are any methods I could use and have only found people, such as yourself, asking the same question. I was looking into this recently, as well, and found that the problem has to do with building the model.matrix/terms/model.frame for gls objects when using the glht function. I ended up creating three gls-specific functions and was able to get estimates/confidence intervals for the toy example below. You may find these functions useful (under the caveat that I only checked that I got estimates and not that the estimates were correct ;) ). Ariel Example: library(nlme) Orthodont$fage - factor(Orthodont$age) #toy example with Orthodont data using age as a factor fitgls - gls( distance ~ fage, data=Orthodont, weights = varIdent(form =~1|fage)) library(multcomp) #notice the error about the model.matrix for gls objects when using glht confint( glht (fitgls, mcp(fage=Tukey) )) #create model.frame, terms, and model.matrix functions specifically for gls objects model.matrix.gls - function(object, ...) model.matrix(terms(object), data = getData(object), ...) model.frame.gls - function(object, ...) model.frame(formula(object), data = getData(object), ...) terms.gls - function(object, ...) terms(model.frame(object),...) #now run glht again confint( glht( fitgls, mcp(fage=Tukey) )) -- View this message in context: http://r.789695.n4.nabble.com/multiple-comparisons-with-generalised-least-squares-tp3441513p4636009.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] Count of elements in coulmns of a matrix
Hi, Try this: list1-list(ABC=c(2,5),XYZ=c(3,4,4,2),PQR=c(4,5,3)) lapply(list1,function(x) length(x)) $ABC [1] 2 $XYZ [1] 4 $PQR [1] 3 list2-lapply(list1,function(x) length(x)) dat2-data.frame(list2) dat2 ABC XYZ PQR 1 2 4 3 A.K. - Original Message - From: Rantony antony.akk...@ge.com To: r-help@r-project.org Cc: Sent: Tuesday, July 10, 2012 8:31 AM Subject: [R] Count of elements in coulmns of a matrix Could you please tell me what is the function or method to get count of elements in all the columns in a matrix ? for eg :- ABC XYZ PQR -- - -- 2 3 4 4 5 5 4 3 2 Result will be like ABC XYZ PQR -- - -- 2 4 3 Could you please help me ? -- View this message in context: http://r.789695.n4.nabble.com/Count-of-elements-in-coulmns-of-a-matrix-tp4635979.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.
[R] RGL 3D curvilinear shapes
Dear useRs, I'm trying to simply fill in the area under a curve using RGL. Here' the set up: x - c(0.75,75.75,150.75,225.75,300.75,375.75,450.75,525.75,600.75,675.75, 0.5,50.5,100.5,150.5,200.5,250.5,300.5,350.5,400.5,450.5, 0.25,25.25,50.25,75.25,100.25,125.25,150.25,175.25,200.25,225.25) y - c(0.05,4.91,9.78,14.64,19.51,24.38,29.24,34.11,38.97,43.84, 0.1,9.83,19.56,29.29,39.02,48.75,58.48,68.21,77.94,87.67, 0.15,14.74,29.34,43.93,58.53,73.13,87.72,102.32,116.91,131.51) z - c(0.05,0.55,0.7,0.78,0.83,0.87,0.9,0.92,0.93,0.94, 0,0.32,0.59,0.77,0.87,0.93,0.96,0.98,0.99,1, 0,0.39,0.66,0.82,0.9,0.95,0.97,0.99,0.99,1) dat - data.frame(x = x, y = y, z = z, ID = c(rep(c(1,2,3),each=10))) plot3d(dat, type = n, ylab = , xlab = , zlab = , axes = F, ylim = c(0,200)) lines3d(dat[1:10,]) lines3d(dat[11:20,]) lines3d(dat[21:30,]) axes3d(edge = c(x--, y-+, z--), nticks = 5, ylim = c(0,200)) bbox3d(color = c(black, white), lit = F, back = line) Any ideas/tips on how to do this? thanks in advance, Patrick -- View this message in context: http://r.789695.n4.nabble.com/RGL-3D-curvilinear-shapes-tp4636011.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] Help with vectors and rollapply
It looks like you already have the zoo package loaded so you can use its
na.locf(),
which replaces NA's with the last non-NA value. Convert the 0s to NAs with
replace() and feed the result into na.locf():
a - c(-2,0,0,0,1,0,0,3,0,0,-4)
aOut - c(-2,-2,-2,-2,1,1,1,3,3,3,-4)
na.locf(replace(a, a==0, NA) )
# [1] -2 -2 -2 -2 1 1 1 3 3 3 -4
all.equal(aOut, .Last.value)
# [1] TRUE
If you need to treat NA and 0 differently you will need to do more work.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Raghuraman Ramachandran
Sent: Tuesday, July 10, 2012 8:23 AM
To: r-help@r-project.org
Subject: [R] Help with vectors and rollapply
Hello
I have a vector a =(-2,0,0,0,1,0,0,3,0,0,-4)
I want to replace all zeros into previous non-zero state. So for instance the
above
vector should be converted into:
a= (-2,-2,-2,-2,1,1,1,3,3,3,-4)
I tried many things and finally concluded that probably(?) rollapply may be
the best
way?
I tried
f= function(x){
ifelse(x==0,Lag(x),x)
}
And then, rollappy(a,1,f) and that didn't work. Can someone help please?
Thx
R
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Re: [R] Predicted values for zero-inflated Poisson
I want to predict the number of turtles for different levels of effort and combinations of covariates. So, for my dataset from which I built the model, would I compare sum(predict(ZIP,type=response)) to the observed bycatch to compare numbers? In order to predict for the new data (called effort), would I use sum(predict(ZIP,newdata=effort,type=response))? I want to be certain I am understanding the coding--this is my first time using the predict function. Thanks, Laura - Laura M. Lee Senior Stock Assessment Scientist North Carolina Division of Marine Fisheries E-Mail: laura@ncdenr.gov -- View this message in context: http://r.789695.n4.nabble.com/Predicted-values-for-zero-inflated-Poisson-tp4635861p4636016.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] Use of Sappy and Tappy for Mathematical Calculation
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Rantony
Sent: Tuesday, July 10, 2012 3:17 AM
To: r-help@r-project.org
Subject: [R] Use of Sappy and Tappy for Mathematical Calculation
Hi,
i have a matrix like this,
ABCXYZ... .
- --
1220 ... .
2435 ... .
3040 ... .
Here, i need to get
Sum of each columns,
Mean of each columns,
median of each columns,
mode of each columns,
Standard deviation of each columns,
variance of each columns,
range of each columns,
count of each columns,
max of each columns,
min of each columns
Can i get output using sappy or tappy functions ? because there have
alots
of records.
Could you please help me fast its kind of urgent !
- Thanks
Antony
Here is some code to get you started. You can add in the other functions that
you want. You will need to figure out what you want to do if there are missing
values. There are built-in functions for most everything you want. You get
the range from the min and the max, and you need to decide what to do if a
variable has 2 or more modes (you will also need to determine how you are
going to get the mode).
# here is sample matrix
mat - matrix(1:100,nrow=10)
colnames(mat) - LETTERS[1:10]
# define summarize function
summarize - function(m) {
sums - apply(m, 2, sum)
counts - apply(m, 2, length)
means - apply(m, 2, mean)
return(rbind(sums, counts, means))
}
# summarize your matrix
summarize(mat)
Hope this is helpful,
Dan
Daniel J. Nordlund
Washington State Department of Social and Health Services
Planning, Performance, and Accountability
Research and Data Analysis Division
Olympia, WA 98504-5204
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Re: [R] Questions about doing analysis based on time
Hello, You can use cut.POSIXt from package base. cut(dat$SunDate, breaks=30 mins) Hope this helps, Rui Barradas Em 10-07-2012 13:43, APOCooter escreveu: Thanks to everyone for their help so far. It's been greatly appreciated. I have a new, but similar problem: I have data that I have broken down by hour (median/mean for each hour). I would like to break it down further, by each half hour (0:00-0:29, 0:30-0:59, 1:00-1:29, 1:30-1:59, etc). I thought cut.dates() from the chron package would be able to do it, but I can't find anything in the chron package documention. I'm fairly certain that if I can figure out how to break down the data by half hour that I can do all the other analysis just fine (median/mean for each half hour, etc) Here is a small sample of the data: dput(head(SundayData, 100)) structure(list(SunDate = structure(c(1273377600, 1307851200, 1213502640, 1217736420, 1311480420, 1211688540, 1337487060, 1255839120, 1268543520, 1293945120, 1280635980, 1309061640, 1322975640, 1297574280, 1221970740, 1253420340, 1218946800, 1329024060, 1290316920, 1224994980, 1218342420, 1269750420, 1257658080, 1322371680, 1214108940, 1312086540, 1260077400, 1228023060, 1315110660, 1281241920, 1272774960, 1224995820, 1275194220, 1246768860, 1302410460, 1234071780, 1305434580, 1232257500, 1243140300, 1284871500, 1247373960, 1265521560, 1273985160, 1310273160, 1226209620, 1270356420, 1330235280, 1222577400, 1310878200, 1324187400, 1242535860, 1336279860, 1283057520, 1291528320, 1324187580, 1330840380, 1298786100, 1307854500, 1236491880, 1298786280, 1233468180, 1280034240, 1230444300, 1213506360, 1251608760, 1215320820, 1304226420, 1320556080, 1299391740, 1286687580, 1296972780, 1296972780, 1321164780, 1260684960, 1315113420, 1287292680, 1292134800, 1303017600, 1307251200, 1278825720, 1238304180, 1212902640, 1231655100, 1254029100, 1311485100, 1295159160, 1220160420, 1297578540, 1300599000, 1241933640, 1225604100, 1269149880, 1283665140, 1244958120, 1245562980, 1289716980, 1235890020, 1282456080, 1279432140, 1279432140), class = c(POSIXct, POSIXt), tzone = ), SunTime = structure(c(1L, 1L, 2L, 3L, 3L, 4L, 5L, 6L, 6L, 6L, 7L, 8L, 8L, 9L, 10L, 10L, 11L, 12L, 13L, 14L, 15L, 15L, 16L, 16L, 17L, 17L, 18L, 19L, 19L, 20L, 21L, 22L, 22L, 23L, 23L, 24L, 24L, 25L, 25L, 25L, 26L, 26L, 26L, 26L, 27L, 27L, 28L, 29L, 29L, 29L, 30L, 30L, 31L, 31L, 32L, 32L, 33L, 33L, 34L, 34L, 35L, 36L, 37L, 38L, 38L, 39L, 39L, 40L, 41L, 42L, 42L, 42L, 42L, 43L, 44L, 45L, 46L, 46L, 46L, 47L, 48L, 49L, 50L, 50L, 50L, 51L, 52L, 53L, 54L, 55L, 56L, 57L, 58L, 59L, 60L, 60L, 61L, 62L, 63L, 63L), .Label = c(0:00, 0:04, 0:07, 0:09, 0:11, 0:12, 0:13, 0:14, 0:18, 0:19, 0:20, 0:21, 0:22, 0:23, 0:27, 0:28, 0:29, 0:30, 0:31, 0:32, 0:36, 0:37, 0:41, 0:43, 0:45, 0:46, 0:47, 0:48, 0:50, 0:51, 0:52, 0:53, 0:55, 0:58, 1:03, 1:04, 1:05, 1:06, 1:07, 1:08, 1:09, 1:13, 1:16, 1:17, 1:18, 1:20, 1:22, 1:23, 1:24, 1:25, 1:26, 1:27, 1:29, 1:30, 1:34, 1:35, 1:38, 1:39, 1:42, 1:43, 1:47, 1:48, 1:49, 1:52, 1:54, 1:55, 1:56, 1:57, 1:59, 10:00, 10:04, 10:07, 10:08, 10:09, 10:10, 10:11, 10:12, 10:14, 10:15, 10:16, 10:18, 10:20, 10:22, 10:23, 10:24, 10:25, 10:26, 10:27, 10:28, 10:30, 10:31, 10:32, 10:33, 10:34, 10:35, 10:36, 10:37, 10:38, 10:39, 10:40, 10:41, 10:43, 10:44, 10:45, 10:47, 10:48, 10:49, 10:50, 10:51, 10:53, 10:54, 10:55, 10:56, 10:58, 10:59, 11:01, 11:02, 11:05, 11:06, 11:07, 11:09, 11:10, 11:12, 11:14, 11:15, 11:16, 11:20, 11:21, 11:22, 11:23, 11:24, 11:26, 11:27, 11:29, 11:30, 11:31, 11:33, 11:34, 11:35, 11:36, 11:37, 11:38, 11:39, 11:40, 11:43, 11:44, 11:46, 11:47, 11:49, 11:52, 11:56, 11:58, 11:59, 12:00, 12:01, 12:02, 12:03, 12:04, 12:05, 12:06, 12:07, 12:08, 12:09, 12:10, 12:11, 12:13, 12:14, 12:15, 12:17, 12:19, 12:21, 12:22, 12:24, 12:25, 12:26, 12:27, 12:28, 12:30, 12:31, 12:32, 12:34, 12:36, 12:37, 12:38, 12:39, 12:41, 12:45, 12:46, 12:47, 12:48, 12:49, 12:50, 12:51, 12:53, 12:54, 12:55, 12:56, 12:57, 12:58, 12:59, 13:00, 13:01, 13:02, 13:03, 13:04, 13:05, 13:06, 13:07, 13:08, 13:09, 13:10, 13:12, 13:13, 13:14, 13:15, 13:17, 13:18, 13:19, 13:20, 13:21, 13:23, 13:25, 13:26, 13:27, 13:30, 13:31, 13:32, 13:34, 13:35, 13:36, 13:38, 13:39, 13:40, 13:44, 13:45, 13:46, 13:47, 13:48, 13:49, 13:50, 13:51, 13:52, 13:53, 13:54, 13:55, 13:57, 13:58, 13:59, 14:00, 14:01, 14:02, 14:04, 14:05, 14:06, 14:07, 14:08, 14:11, 14:12, 14:13, 14:14, 14:15, 14:16, 14:17, 14:18, 14:20, 14:21, 14:22, 14:23, 14:25, 14:26, 14:28, 14:29, 14:30, 14:31, 14:32, 14:34, 14:35, 14:36, 14:37, 14:38, 14:40, 14:41, 14:42, 14:43, 14:45, 14:46, 14:47, 14:49, 14:50, 14:51, 14:52, 14:53, 14:54, 14:56, 14:57, 14:58, 14:59, 15:01, 15:02, 15:03, 15:04, 15:05, 15:07, 15:11, 15:12, 15:13, 15:14, 15:15, 15:17, 15:18, 15:19,
Re: [R] Extracting arithmetic mean for specific values from multiple .txt-files
Dear Rui, thank you very much. Your solution works perfectly. One last question: I need to write a function, with ONE value (here: a ratio) for the correct reactions divided per trials or trialCount, respectively, FOR EACH test subject. / means divided by in the following. I need the ratio correct (reactions)/trial or correct (reactions)/trialCount, respectively (because trial and trialCount are the same WITHIN test SUBJECTS; BUT they differ in length between BETWEEN test SUBJECTS!). It would be very helpful, if I had a data frame in the end in R, with one column for trialCount/trial, one column for correct reactions(= 1) AND (more importantly) one column for correct (= 1) answers / trialCount. legend (just as additional information) for the variable correct: 1 = correct reaction 2 = false reaction 3 = reaction too slow 4 = reaction too fast 5 = more than one button pressed 6 = no reaction within RT window I would be very thankful for an answer! Sorry for the questions, but I am doing this for the first time! Kind regards -- View this message in context: http://r.789695.n4.nabble.com/Extracting-arithmetic-mean-for-specific-values-from-multiple-txt-files-tp4635809p4636020.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] calculating the difference between days?
Hi List, I have one column of beginning dates and one column of ending dates, I want to find their difference. And I want to ignore the trailing zeros, basically everything after the first colon mark. Begin_date End_date 01JAN2000:00:00:00:000 02FEB2002:00:00:00:000 24MAR2012:00:00:00:000 18MAY2012:00:00:00:000 01OCT2003:00:00:00:000 02FEB2004:00:00:00:000 01JAN2000:00:00:00:000 02FEB2000:00:00:00:000 01JAN2000:00:00:00:000 02FEB2000:00:00:00:000 Thanks, Mike [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R to winbugs interface
On 09.07.2012 19:27, PRAGYA SUR wrote:
Yes that was the problem. Thank you very much. Can anyone tell me the
meaning of
The following object(s) are masked _by_ '.GlobalEnv':
beta
It means you have two instances of beta, one in your workspace
('.GlobalEnv') and one that is probably in some data.frame that was
attached to the search path.
Best,
Uwe Ligges
i was shown this notification after the results were printed.
On Mon, Jul 9, 2012 at 1:15 PM, Uwe Ligges
lig...@statistik.tu-dortmund.de
mailto:lig...@statistik.tu-dortmund.de wrote:
On 09.07.2012 18:19, PRAGYA SUR wrote:
I ran the same program in a different computer where it had run
proper a
week ago. This time around in the log file in WinBUGS the
program stopped
at a line which said :
save(C:/DOCUME~1/ADMINI~1/__LOCALS~1/Temp/RtmpU3u46p/log.__odc)
save(C:/DOCUME~1/ADMINI~1/__LOCALS~1/Temp/RtmpU3u46p/log.__txt)
and did not proceed any further.
Can anyone tell me what migh tbe the possible error here?
Although unstated, I guess it was finished and did not close because
you used bugs(., debug=TRUE)?
Uwe Ligges
On Mon, Jul 9, 2012 at 7:04 AM, S Ellison
s.elli...@lgcgroup.com mailto:s.elli...@lgcgroup.com wrote:
-Original Message-
Error in file(con, wb) : cannot open the connection In
addition: Warning messages:
1: In file.create(to[okay]) :
cannot create file 'c:/Program
Files/WinBUGS14//System/Rsrc/__Registry_Rsave.odc', reason
'Permission denied'
This tells you that you do not have operating system
permission to create
a file in the program files area referred to.
2: In file(con, wb) :
cannot open file 'c:/Program
Files/WinBUGS14//System/Rsrc/__Registry.odc':
Permission denied
.. and this tells you you don't have permission to open a
file for writing
(mode 'w' in the same location
Conclusion; you're trying to write to an area you don;t have
permission to
write to.
Either change the permissions for that area (insecure) or
use a different
file location for temporary files.
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This email and any attachments are confidential. Any
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Predicted values for zero-inflated Poisson
*Laura Lee* laura.lee at ncdenr.gov mailto:r-help%40r-project.org?Subject=Re%3A%20%5BR%5D%20Predicted%20values%20for%20zero-inflated%20PoissonIn-Reply-To=%3C1341937636301-4636016.post%40n4.nabble.com%3E /Tue Jul 10 18:27:16 CEST 2012/ I want to predict the number of turtles for different levels of effort and combinations of covariates. So, for my dataset from which I built the model, would I compare sum(predict(ZIP,type=response)) to the observed bycatch to compare numbers? In order to predict for the new data (called effort), would I use sum(predict(ZIP,newdata=effort,type=response))? I want to be certain I am understanding the coding--this is my first time using the predict function. Thanks, Laura Laura Why do you use the sum? If you use: PredY - predict(ZIP, type = response) then you have predicted values for each of the rows in your effort data frame. Job done. You have an offset in your model, isn't it? You will need to choose values for this in the data frame effort as well. Also double check that the offset is only in the count partat least that is what I would do. Note that using an offset means that you assume that if sampling effort is doubled, your fish (?) numbers double. If you fully want to understand what predict is doing, try to do it manually. Below is R code from Chapter 7 (Zero Inflation and GLMM with R) M3 - zeroinfl(ParrotFish ~ Depth + Slope + SQDistRck + DistSed + Swell + Chla + SST, dist = poisson, link = logit, data = PF2) Betas.logistic - coef(M3, model = zero) X.logistic - model.matrix(M3, model = zero) eta.logistic - X.logistic %*% Betas.logistic p - exp(eta.logistic) / (1 + exp(eta.logistic)) Betas.log - coef(M3, model = count) X.log - model.matrix(M3, model = count) eta.log- X.log %*% Betas.log mu - exp(eta.log) ExpY - mu * (1 - p) VarY - (1 - p) * (mu + p * mu^2) Instead of using model.matrix(M3), you could specify your own data frame with covariates. Your effort. Something like: M4 - zeroinfl(ParrotFish ~ Depth + Slope | SST, dist = poisson, link = logit, data = PF2) betapois - coef(M4, model = count) betaBin - coef(M4, model = zero) MyDataPois - data.frame(Depth = blah blah, Slope = Blah blah) MyDataBin - data.frame(SST = blah) Xpois - model.matrix(~ blah blah, data = MyDataPois) Xbin - model.matrix(~ blah blah, data = MyDataBin) eta.Pois - Xpois %*% betapois eta.Bin - blah blah mu = blah blah pi = blah blah ExpY = ... Doing it like this means you fully understand it..:-) Alain -- Dr. Alain F. Zuur First author of: 1. Analysing Ecological Data (2007). Zuur, AF, Ieno, EN and Smith, GM. Springer. 680 p. URL: www.springer.com/0-387-45967-7 2. Mixed effects models and extensions in ecology with R. (2009). Zuur, AF, Ieno, EN, Walker, N, Saveliev, AA, and Smith, GM. Springer. http://www.springer.com/life+sci/ecology/book/978-0-387-87457-9 3. A Beginner's Guide to R (2009). Zuur, AF, Ieno, EN, Meesters, EHWG. Springer http://www.springer.com/statistics/computational/book/978-0-387-93836-3 4. Zero Inflated Models and Generalized Linear Mixed Models with R. (2012) Zuur, Saveliev, Ieno. http://www.highstat.com/book4.htm Other books: http://www.highstat.com/books.htm Statistical consultancy, courses, data analysis and software Highland Statistics Ltd. 6 Laverock road UK - AB41 6FN Newburgh Tel: 0044 1358 788177 Email: highs...@highstat.com URL: www.highstat.com URL: www.brodgar.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.
Re: [R] Extracting arithmetic mean for specific values from multiple .txt-files
Hello, I'm glad it help. As for this second question, you should explain yourself better. 1. What is a test subject, which column records its id? vpNum? 2. You say divided per trials or trialCount. Does this mean per trial number (example: divide by 1, by 2, by 3, etc, by 149) or per number of trials (149 in the previous example) 3. 'correct' now seems to be categorical. Divide WHAT by trial or trialCount? Hint: post a small data example with three or four subjects and the wanted output. Rui Barradas Em 10-07-2012 18:06, vimmster escreveu: Dear Rui, thank you very much. Your solution works perfectly. One last question: I need to write a function, with ONE value (here: a ratio) for the correct reactions divided per trials or trialCount, respectively, FOR EACH test subject. / means divided by in the following. I need the ratio correct (reactions)/trial or correct (reactions)/trialCount, respectively (because trial and trialCount are the same WITHIN test SUBJECTS; BUT they differ in length between BETWEEN test SUBJECTS!). It would be very helpful, if I had a data frame in the end in R, with one column for trialCount/trial, one column for correct reactions(= 1) AND (more importantly) one column for correct (= 1) answers / trialCount. legend (just as additional information) for the variable correct: 1 = correct reaction 2 = false reaction 3 = reaction too slow 4 = reaction too fast 5 = more than one button pressed 6 = no reaction within RT window I would be very thankful for an answer! Sorry for the questions, but I am doing this for the first time! Kind regards -- View this message in context: http://r.789695.n4.nabble.com/Extracting-arithmetic-mean-for-specific-values-from-multiple-txt-files-tp4635809p4636020.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] fill 0-row data.frame with 1 line of NAs
On 2012-07-10 08:50, Brian Diggs wrote:
On 7/10/2012 7:53 AM, Peter Ehlers wrote:
On 2012-07-10 06:57, Rui Barradas wrote:
Hello,
If you write a function, it becomes less convoluted...
empty - function(x){
if(NROW(x) == 0){
y - rep(NA, NCOL(x))
names(y) - names(x)
y
}else x
}
(.xb - iris[ iris$Species=='zz', ])
empty(.xb)
Both this and Liviu's original solution destroy the
factor nature of 'Species' (which may not matter, of
course). How about
(.xb - iris[ iris$Species=='zz', ])
.xb - .xb[1, ] # this probably shouldn't work, but it does.
Using NA subscripting seems even better
Yes, you can subset with NA or any real number greater than 1.
Peter Ehlers
empty - function(x) {
if(NROW(x) == 0) {
x[NA,]
} else {
x
}
}
It even preserves the factor nature of things:
empty(iris[iris$Specis=='zz',])
Sepal.Length Sepal.Width Petal.Length Petal.Width Species
NA NA NA NA NANA
str(empty(iris[iris$Specis=='zz',]))
'data.frame': 1 obs. of 5 variables:
$ Sepal.Length: num NA
$ Sepal.Width : num NA
$ Petal.Length: num NA
$ Petal.Width : num NA
$ Species : Factor w/ 3 levels setosa,versicolor,..: NA
?
Peter Ehlers
Hope this helps,
Rui Barradas
Em 10-07-2012 14:15, Liviu Andronic escreveu:
Dear all
Is there a simpler method to achieve the following: When I obtain an
empty data.frame after subsetting, I need for it to contain one line
of NAs. Here's a dummy example:
(.xb - iris[ iris$Species=='zz', ])
[1] Sepal.Length Sepal.Width Petal.Length Petal.Width Species
0 rows (or 0-length row.names)
dim(.xb)
[1] 0 5
(.xa - data.frame(matrix(rep(NA, ncol(.xb)), 1)))
X1 X2 X3 X4 X5
1 NA NA NA NA NA
names(.xa) - names(.xb)
(.xb - .xa)
Sepal.Length Sepal.Width Petal.Length Petal.Width Species
1 NA NA NA NA NA
The solution I came up with is way too convoluted. Anything simpler?
Regards
Liviu
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] fill 0-row data.frame with 1 line of NAs
Hello,
Em 10-07-2012 18:59, Peter Ehlers escreveu:
On 2012-07-10 08:50, Brian Diggs wrote:
On 7/10/2012 7:53 AM, Peter Ehlers wrote:
On 2012-07-10 06:57, Rui Barradas wrote:
Hello,
If you write a function, it becomes less convoluted...
empty - function(x){
if(NROW(x) == 0){
y - rep(NA, NCOL(x))
names(y) - names(x)
y
}else x
}
(.xb - iris[ iris$Species=='zz', ])
empty(.xb)
Both this and Liviu's original solution destroy the
factor nature of 'Species' (which may not matter, of
course). How about
(.xb - iris[ iris$Species=='zz', ])
.xb - .xb[1, ] # this probably shouldn't work, but it does.
Using NA subscripting seems even better
Yes, you can subset with NA or any real number greater than 1.
Peter Ehlers
Good to know, was completely unaware of this indexing possibility.
Rui Barradas
empty - function(x) {
if(NROW(x) == 0) {
x[NA,]
} else {
x
}
}
It even preserves the factor nature of things:
empty(iris[iris$Specis=='zz',])
Sepal.Length Sepal.Width Petal.Length Petal.Width Species
NA NA NA NA NANA
str(empty(iris[iris$Specis=='zz',]))
'data.frame': 1 obs. of 5 variables:
$ Sepal.Length: num NA
$ Sepal.Width : num NA
$ Petal.Length: num NA
$ Petal.Width : num NA
$ Species : Factor w/ 3 levels setosa,versicolor,..: NA
?
Peter Ehlers
Hope this helps,
Rui Barradas
Em 10-07-2012 14:15, Liviu Andronic escreveu:
Dear all
Is there a simpler method to achieve the following: When I obtain an
empty data.frame after subsetting, I need for it to contain one line
of NAs. Here's a dummy example:
(.xb - iris[ iris$Species=='zz', ])
[1] Sepal.Length Sepal.Width Petal.Length Petal.Width Species
0 rows (or 0-length row.names)
dim(.xb)
[1] 0 5
(.xa - data.frame(matrix(rep(NA, ncol(.xb)), 1)))
X1 X2 X3 X4 X5
1 NA NA NA NA NA
names(.xa) - names(.xb)
(.xb - .xa)
Sepal.Length Sepal.Width Petal.Length Petal.Width Species
1 NA NA NA NA NA
The solution I came up with is way too convoluted. Anything simpler?
Regards
Liviu
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] Predicted values for zero-inflated Poisson
Alain- Thanks again for your reply. Yes, the offset for effort is only in the count part of the model. Sorry I wasn't clear about why I was using 'sum'...my effort data set contains records of trips with the effort given for each trip. I thought using sum would get me the total number of turtles predicted for the new effort data. I did also log-transform the effort in the new effort data before applying predict (correct?). I will try the manual method as you recommended to help me understand the code. Thanks, Laura From: Alain Zuur [via R] [mailto:ml-node+s789695n4636025...@n4.nabble.com] Sent: Tuesday, July 10, 2012 1:52 PM To: Lee, Laura Subject: Re: Predicted values for zero-inflated Poisson *Laura Lee* laura.lee at ncdenr.gov mailto:r-help%40r-project.org?Subject=Re%3A%20%5BR%5D%20Predicted%20values%20for%20zero-inflated%20PoissonIn-Reply-To=%3C1341937636301-4636016.post%40n4.nabble.com%3E /Tue Jul 10 18:27:16 CEST 2012/ I want to predict the number of turtles for different levels of effort and combinations of covariates. So, for my dataset from which I built the model, would I compare sum(predict(ZIP,type=response)) to the observed bycatch to compare numbers? In order to predict for the new data (called effort), would I use sum(predict(ZIP,newdata=effort,type=response))? I want to be certain I am understanding the coding--this is my first time using the predict function. Thanks, Laura Laura Why do you use the sum? If you use: PredY - predict(ZIP, type = response) then you have predicted values for each of the rows in your effort data frame. Job done. You have an offset in your model, isn't it? You will need to choose values for this in the data frame effort as well. Also double check that the offset is only in the count partat least that is what I would do. Note that using an offset means that you assume that if sampling effort is doubled, your fish (?) numbers double. If you fully want to understand what predict is doing, try to do it manually. Below is R code from Chapter 7 (Zero Inflation and GLMM with R) M3 - zeroinfl(ParrotFish ~ Depth + Slope + SQDistRck + DistSed + Swell + Chla + SST, dist = poisson, link = logit, data = PF2) Betas.logistic - coef(M3, model = zero) X.logistic - model.matrix(M3, model = zero) eta.logistic - X.logistic %*% Betas.logistic p - exp(eta.logistic) / (1 + exp(eta.logistic)) Betas.log - coef(M3, model = count) X.log - model.matrix(M3, model = count) eta.log- X.log %*% Betas.log mu - exp(eta.log) ExpY - mu * (1 - p) VarY - (1 - p) * (mu + p * mu^2) Instead of using model.matrix(M3), you could specify your own data frame with covariates. Your effort. Something like: M4 - zeroinfl(ParrotFish ~ Depth + Slope | SST, dist = poisson, link = logit, data = PF2) betapois - coef(M4, model = count) betaBin - coef(M4, model = zero) MyDataPois - data.frame(Depth = blah blah, Slope = Blah blah) MyDataBin - data.frame(SST = blah) Xpois - model.matrix(~ blah blah, data = MyDataPois) Xbin - model.matrix(~ blah blah, data = MyDataBin) eta.Pois - Xpois %*% betapois eta.Bin - blah blah mu = blah blah pi = blah blah ExpY = ... Doing it like this means you fully understand it..:-) Alain -- Dr. Alain F. Zuur First author of: 1. Analysing Ecological Data (2007). Zuur, AF, Ieno, EN and Smith, GM. Springer. 680 p. URL: www.springer.com/0-387-45967-7http://www.springer.com/0-387-45967-7 2. Mixed effects models and extensions in ecology with R. (2009). Zuur, AF, Ieno, EN, Walker, N, Saveliev, AA, and Smith, GM. Springer. http://www.springer.com/life+sci/ecology/book/978-0-387-87457-9 3. A Beginner's Guide to R (2009). Zuur, AF, Ieno, EN, Meesters, EHWG. Springer http://www.springer.com/statistics/computational/book/978-0-387-93836-3 4. Zero Inflated Models and Generalized Linear Mixed Models with R. (2012) Zuur, Saveliev, Ieno. http://www.highstat.com/book4.htm Other books: http://www.highstat.com/books.htm Statistical consultancy, courses, data analysis and software Highland Statistics Ltd. 6 Laverock road UK - AB41 6FN Newburgh Tel: 0044 1358 788177 Email: [hidden email]/user/SendEmail.jtp?type=nodenode=4636025i=0 URL: www.highstat.comhttp://www.highstat.com URL: www.brodgar.comhttp://www.brodgar.com [[alternative HTML version deleted]] __ [hidden email]/user/SendEmail.jtp?type=nodenode=4636025i=1 mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Dr. Alain F. Zuur First author of: 1. Analysing Ecological Data (2007). Zuur, AF, Ieno, EN and Smith, GM. Springer. 680 p. URL:
Re: [R] Specify model with polynomial interaction terms up to degree n
Yep, that code is verbatim what I typed in, using version 2.14 ... seems weird. -- View this message in context: http://r.789695.n4.nabble.com/Specify-model-with-polynomial-interaction-terms-up-to-degree-n-tp4635130p4636031.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] Use of Sappy and Tappy for Mathematical Calculation
Hi, dat1-data.frame(ABC=c(12,24,30),XYZ=c(20,35,40)) dat2-as.matrix(dat1) #mean dat2mean-apply(dat2,2,mean) dat2mean # ABC XYZ #22.0 31.7 dat2sum-apply(dat2,2,sum) dat2median-apply(dat2,2,median) dat2max-apply(dat2,2,max) dat2min-apply(dat2,2,min) dat2range-apply(dat2,2,range) dat2count-apply(dat2,2,length) dat2sd-apply(dat2,2,sd) dat2var-apply(dat2,2,var) A.K. - Original Message - From: Rantony antony.akk...@ge.com To: r-help@r-project.org Cc: Sent: Tuesday, July 10, 2012 6:16 AM Subject: [R] Use of Sappy and Tappy for Mathematical Calculation Hi, i have a matrix like this, ABC XYZ .. . . - -- 12 20 .. . . 24 35 .. . . 30 40 .. . . Here, i need to get Sum of each columns, Mean of each columns, median of each columns, mode of each columns, Standard deviation of each columns, variance of each columns, range of each columns, count of each columns, max of each columns, min of each columns Can i get output using sappy or tappy functions ? because there have alots of records. Could you please help me fast its kind of urgent ! - Thanks Antony -- View this message in context: http://r.789695.n4.nabble.com/Use-of-Sappy-and-Tappy-for-Mathematical-Calculation-tp4635969.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.
[R] HELP me please with import of csv to R
Hey, I am having problems with importing a csv file to R. I could read the file by typing: read.csv(file=/Users/kama/Desktop/skatter.csv, header=TRUE, sep=;) However, i can not analyze the skatter - for ex, when i type: skatter = read.csv(skatter.csv) i get this message: Error in file(file, rt) : cannot open the connection In addition: Warning message: In file(file, rt) : What i need is to import this file and analyze it using for example histogram. I have Mac(update) and the file is saved in csv file... and I'm quite new user of R. Thank you very much! -- View this message in context: http://r.789695.n4.nabble.com/HELP-me-please-with-import-of-csv-to-R-tp4636019.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] problem for installing rgdal
Hello, I run [R] 2.14 on Mac OS 10.6 and I've been desperately trying to install rpy2 in order to compute spatial statistics on QGIS 1.7.3. rpy2 is dependent upon the rgdal [R]package that I've been unable to install in spite of up to date versions of GDAL and PROJ. More precisely, [R] console writes Error: gdal-config not found. As a matter of fact, typing gdal.config on a Terminal shell doesn't work either... Yet, the file is actually present in /Library/Frameworks/GDAL.framework/Versions/1.9/Programs I then tryed to follow suggestions with: --configure-args='--with-gdal-config=/Library/Frameworks/GDAL.framework/Versions/1.9/Programs/gdal-config' I received no such file or directory I also found this option described here (http://www.r-bloggers.com/installing-rgdal-on-mac-os-x-2/), but couldn't get through it... As anyone an idea? Thanks for your help stan [[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] R code help to change table format
I am trying to input an OTU table into EstimateS, however, the format of the OTU table has to be changed to fit the format EstimateS will accept. In R, I would like to change the format of the OTU table (from excel). Here is what I need to do, take Example 1 and create Example 2. The problem is that I have hundreds of OTUs, so I can't do this by hand (and I'd love to have a code that I could use for different OTU tables). Thanks! Example 1 Example 2 Species Abundance Species 1 3 1 2 2 1 3 2 1 4 2 2 5 4 2 3 3 4 4 5 5 5 5 5 -- View this message in context: http://r.789695.n4.nabble.com/R-code-help-to-change-table-format-tp4636022.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] Skipping lines and incomplete rows
Hello Ravi, I was not aware that your dataset have special character # before NA. If it was just plain NA, it would have worked. So, It's not because of sep= ;. See below: #Without # dat1-read.table(text= Remove this line Remove this line Remove this line Time;Actual Speed;Actual Direction;Temp;Press;Value1;Value2 ;[m/s];[°];°C;[hPa];[MWh];[MWh] 1/1/2012;0.0;0;NA;NA;0.;0. 1/2/2012;0.0;0;NA;NA;0.;0. 1/3/2012;0.0;0;NA;NA;1.5651;2.2112 1/4/2012;0.0;0;NA;NA;1.;2. 1/5/2012;0.0;0;NA;NA;3.2578;7.5455 ,sep=;,header=TRUE,fill=TRUE,skip=4,stringsAsFactors=FALSE) dat1 Time Actual.Speed Actual.Direction Temp Press Value1 Value2 1 [m/s] [°] °C [hPa] [MWh] [MWh] 2 1/1/2012 0.0 0 NA NA 0. 0. 3 1/2/2012 0.0 0 NA NA 0. 0. 4 1/3/2012 0.0 0 NA NA 1.5651 2.2112 5 1/4/2012 0.0 0 NA NA 1. 2. 6 1/5/2012 0.0 0 NA NA 3.2578 7.5455 #With #: Reading data from the .txt file. # In the documentation (http://stat.ethz.ch/R-manual/R-devel/library/utils/html/read.table.html), comment.char=# is an option in the read.table, but unfortunately it shows only blank columns after the first three columns. #I think Rui's method of reading header separately using readLines might be a good option. Or if you know the columnheadings, then you can do this: dat2-read.table(dat2.txt,skip=4,col.names=c(Time,Actual Speed,Actual Direction, Temp,Press,Value1,Value2),fill=TRUE,sep=;,comment.char=c) dat2 Time Actual.Speed Actual.Direction Temp Press Value1 Value2 1 [m/s] [°] °C [hPa] [MWh] [MWh] 2 1/1/2012 0.0 0 #NA #NA 0. 0. 3 1/2/2012 0.0 0 #NA #NA 0. 0. 4 1/3/2012 0.0 0 #NA #NA 1.5651 2.2112 5 1/4/2012 0.0 0 #NA #NA 1. 2. 6 1/5/2012 0.0 0 #NA #NA 3.2578 7.5455 A.K. - Original Message - From: vioravis viora...@gmail.com To: r-help@r-project.org Cc: Sent: Tuesday, July 10, 2012 1:41 AM Subject: Re: [R] Skipping lines and incomplete rows Thanks a lot Rui and Arun. The methods work fine with the data I gave but when I tried the two methods with the following semi-colon separated data using sep = ;. Only the first 3 columnns are read properly rest of the columns are either empty or NAs. ** Remove this line Remove this line Remove this line Time;Actual Speed;Actual Direction;Temp;Press;Value1;Value2 ;[m/s];[°];°C;[hPa];[MWh];[MWh] 1/1/2012;0.0;0;#N/A;#N/A;0.;0. 1/2/2012;0.0;0;#N/A;#N/A;0.;0. 1/3/2012;0.0;0;#N/A;#N/A;1.5651;2.2112 1/4/2012;0.0;0;#N/A;#N/A;1.;2. 1/5/2012;0.0;0;#N/A;#N/A;3.2578;7.5455 *** I used the following code: dat1-read.table(testInput.txt,sep=;,skip=3,fill=TRUE,header=TRUE) dat1-dat1[-1,] row.names(dat1)-1:nrow(dat1) Could you please let me know what is wrong with this approach? Thank you. Ravi -- View this message in context: http://r.789695.n4.nabble.com/Skipping-lines-and-incomplete-rows-tp4635830p4635952.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] calculating the difference between days?
Hi, Try this: dat3-read.table(text= Begin_date End_date 01JAN2000:00:00:00:000 02FEB2002:00:00:00:000 24MAR2012:00:00:00:000 18MAY2012:00:00:00:000 01OCT2003:00:00:00:000 02FEB2004:00:00:00:000 01JAN2000:00:00:00:000 02FEB2000:00:00:00:000 01JAN2000:00:00:00:000 02FEB2000:00:00:00:000 ,sep=,header=TRUE) dat3$Begin_date-strptime(dat3[,1],format=%d%b%Y:%H:%M:%S) dat3$End_date-strptime(dat3[,2],format=%d%b%Y:%H:%M:%S) difftime(dat3[,1],dat3[,2]) Time differences in days [1] -763. -55. -124.0417 -32. -32. attr(,tzone) [1] A.K. - Original Message - From: C W tmrs...@gmail.com To: r-help r-help@r-project.org Cc: Sent: Tuesday, July 10, 2012 1:22 PM Subject: [R] calculating the difference between days? Hi List, I have one column of beginning dates and one column of ending dates, I want to find their difference. And I want to ignore the trailing zeros, basically everything after the first colon mark. Begin_date End_date 01JAN2000:00:00:00:000 02FEB2002:00:00:00:000 24MAR2012:00:00:00:000 18MAY2012:00:00:00:000 01OCT2003:00:00:00:000 02FEB2004:00:00:00:000 01JAN2000:00:00:00:000 02FEB2000:00:00:00:000 01JAN2000:00:00:00:000 02FEB2000:00:00:00:000 Thanks, Mike [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Times Series Data using GLS
I am trying to use regression to determine the interaction between a couple of variables while correcting for autocorrelation. Thus far, I have created the code: model - gls(yvar~xvar1*xvar2, correlation = corARMA (p=2), method = ML, data = data) I'm having a difficult time understanding the different correlation structure classes and when to use the correct ones. Also, with regards to method, I am not sure if REML or ML is the correct option. Thanks to anyone who can give me help with this. I really appreciate it. -- View this message in context: http://r.789695.n4.nabble.com/Times-Series-Data-using-GLS-tp4636026.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] Changing x-axis values displayed on histogram
Is it possible to change the x-axis values in a histogram to reflect binned values? Here are my data: histexample-c(6,7,7,8,8,8,9,9,9,9,9,10,10,10,10,10,10,10,11,11,11,11,11,11,12,12,12,12,13,13,13,14,14,14,15,16) hist(histexample) Now, I'll bin pairs of adjacent values together (e.g., 5-6, 7-8, 9-10, 11-12, 13-14, 15-16) using the following bins-c(4.5,6.5,8.5,10.5,12.5,14.5,16.5) hist(histexample,breaks=bins) The displayed x-axis values are 6, 8, 10, 12, 14, and 16. I'd like the x-axis values to reflect the values in each bin (e.g., 5-6, 7-8, 9-10, 11-12, 13-14, 15-16). Any suggestions would be greatly appreciated! Many thanks in advance. John -- View this message in context: http://r.789695.n4.nabble.com/Changing-x-axis-values-displayed-on-histogram-tp4636032.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] Need HELP: how find and use a csv file?
Hey, I am having some problems with importing a csv file into R and then saving it for analyzing. I got a csv file ( skater.csv) which i could read by typing: read.csv(file=/Users/kama/Desktop/skatter.csv, header=TRUE, sep=;) However, when i enter:skatter.csv-read.csv(skatter.csv, header=TRUE) i get this message: Error in file(file, rt) : cannot open the connection In addition: Warning message: In file(file, rt) : I have tried with: skatter.csv-file.choose() and other codes to find the file but it does not work. Please help me fix this problem, i have been sitting with this one in 4 hours.. What i need is to import this file and analyze it using for example histogram. I have Mac(update) and the file is saved in csv file... and I'm quite new user of R. Thank you very much! [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Package 'MASS' (polr): Error in svd(X) : infinite or missing values in 'x'
Hi Jeremy, I think Jessica is right that probably you could make polr converge and produce a Hessian if the data are better scaled, but there might also be other things not allowing you to get the Hessian/vcov. Could be insightful if you showed us the result of str(Jdata) Also, I am thinking that perhaps the another implementation of ordinal regression models might avoid the problem. You could try the ordinal package (of which I am the author) -- the following should reproduce the MASS::polr results: install.packages(ordinal) library(ordinal) Global - clm(JVeg5 ~ Elevation + Lat_Y_pos + Coast_dist + Stream_dist, data=Jdata) summary(Global) Another option would be Frank Harrell's lrm function in the rms package. HTH, Rune On 9 July 2012 11:55, Jeremy Little jeremy.lit...@my.jcu.edu.au wrote: Hello, I am trying to run an ordinal logistic regression (polr) using the package 'MASS'. I have successfully run other regression classes (glm, multinom) without much problem, but with the 'polr' class I get the following error: Error in svd(X) : infinite or missing values in 'x' which appears when I run the summary command. The data file is large (585000 rows) and has no NA, - or blank values. My script (in brief) is as follows, with results: library(MASS) ## ADD DATA Jdata- read.delim(/Analysis/20120709 JLittle data file.txt, header=T) attach(Jdata) names(Jdata) [1] POINTID Lat_Y_pos JVeg5 Subregion Rock_U_Nam Rock_Name Elevation Slope Aspect Hillshade Stream_dist Coast_dist Coast_SE [14] Coast_E Wind_310TPI Landform Global - polr(JVeg5 ~ Elevation + Lat_Y_pos + Coast_dist + Stream_dist, data=Jdata) summary(Global) Error in svd(X) : infinite or missing values in 'x' ##Try with omit NA command Global - polr(JVeg5 ~ Elevation + Lat_Y_pos + Coast_dist + Stream_dist, data=Jdata, na.action = na.omit, Hess = TRUE) summary(Global) Error in svd(X) : infinite or missing values in 'x' Does this imply an 'infinite value' and what would this mean? If anyone has any idea how to address this error, I would very much appreciate your response. Thank you in advance. Jeremy Date File Attachment (200 rows): http://r.789695.n4.nabble.com/file/n4635829/20120709_JLittle_data_file.txt 20120709_JLittle_data_file.txt -- View this message in context: http://r.789695.n4.nabble.com/Package-MASS-polr-Error-in-svd-X-infinite-or-missing-values-in-x-tp4635829.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. -- Rune Haubo Bojesen Christensen Ph.D. Student, M.Sc. Eng. Phone: (+45) 45 25 33 63 Mobile: (+45) 30 26 45 54 DTU Informatics, Section for Statistics Technical University of Denmark, Build. 305, Room 122, DK-2800 Kgs. Lyngby, Denmark __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] HELP me please with import of csv to R
Hi, On Tue, Jul 10, 2012 at 12:48 PM, F86 farad...@gmail.com wrote: Hey, I am having problems with importing a csv file to R. I could read the file by typing: read.csv(file=/Users/kama/Desktop/skatter.csv, header=TRUE, sep=;) So that command does work? However, i can not analyze the skatter - for ex, when i type: skatter = read.csv(skatter.csv) Then you need the above command, not what you type here: skatter - read.csv(file=/Users/kama/Desktop/skatter.csv, header=TRUE, sep=;) But note that if sep=; then you don't have a csv file and should properly use read.table() instead. i get this message: Error in file(file, rt) : cannot open the connection In addition: Warning message: In file(file, rt) : What i need is to import this file and analyze it using for example histogram. I have Mac(update) and the file is saved in csv file... and I'm quite new user of R. That error means that R can't find the file where you told it to look. Specifying the full and complete path as in your first example should work. If you're having problems with paths (which are a Mac issue and not at all an R issue), you could also try read.table(file.choose(), header=TRUE, sep=;) I think that file.choose() should work on Mac. The Intro to R document that came with R might also be of use. Sarah -- 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.
[R] Understanding cenros Error
Before reading water chemistry into a data frame I removed all missing data. Yet when I try to run cenros() to summarize a specific chemical I get an error that I do not understand: with( subset(chem, param=='Ag'), cenros(quant,ceneq1) ) Error in lm.fit(x, y, offset = offset, singular.ok = singular.ok, ...) : NA/NaN/Inf in 'y' I would like to learn what I did incorrectly so I can avoid these errors in the future. The data frame structure is str(chem) 'data.frame': 120309 obs. of 8 variables: $ site: Factor w/ 65 levels ;Influent,D-1,..: 2 2 2 2 2 2 ... $ sampdate: Date, format: 2007-12-12 2007-12-12 ... $ preeq0 : logi TRUE TRUE TRUE TRUE TRUE TRUE ... $ param : Factor w/ 37 levels Ag,Al,Alk_tot,..: 1 2 8 17 3 9 ... $ quant : num 0 0.106 1 231 231 0.011 0 0.002 0 100 ... $ ceneq1 : logi FALSE FALSE TRUE FALSE FALSE FALSE ... $ floor : num 0 0.106 0 231 231 0.011 0 0 0 100 ... $ ceiling : Factor w/ 3909 levels 0.000,0.000),..: 1 116 841 1771 ... I ran dput() on the data frame but cannot make sense of the output (a 5.5M ASCII text file). Pointers appreciated. Rich __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] calculating the difference between days?
When the days and time are identical, difftime() gives difference in secs. I still want difference in days. Say, below my last row is identical dat3-read.table(text= Begin_date End_date 01JAN2000:00:00:00:000 02FEB2002:00:00:00:000 24MAR2012:00:00:00:000 18MAY2012:00:00:00:000 01OCT2003:00:00:00:000 02FEB2004:00:00:00:000 01JAN2000:00:00:00:000 02FEB2000:00:00:00:000 01JAN2000:00:00:00:000 02FEB2000:00:00:00:000 24DEC2012:00:00:00:000 24DEC2012:00:00:00:000 ,sep=,header=TRUE) -M On Tue, Jul 10, 2012 at 1:52 PM, arun smartpink...@yahoo.com wrote: Hi, Try this: dat3-read.table(text= Begin_date End_date 01JAN2000:00:00:00:000 02FEB2002:00:00:00:000 24MAR2012:00:00:00:000 18MAY2012:00:00:00:000 01OCT2003:00:00:00:000 02FEB2004:00:00:00:000 01JAN2000:00:00:00:000 02FEB2000:00:00:00:000 01JAN2000:00:00:00:000 02FEB2000:00:00:00:000 ,sep=,header=TRUE) dat3$Begin_date-strptime(dat3[,1],format=%d%b%Y:%H:%M:%S) dat3$End_date-strptime(dat3[,2],format=%d%b%Y:%H:%M:%S) difftime(dat3[,1],dat3[,2]) Time differences in days [1] -763. -55. -124.0417 -32. -32. attr(,tzone) [1] A.K. - Original Message - From: C W tmrs...@gmail.com To: r-help r-help@r-project.org Cc: Sent: Tuesday, July 10, 2012 1:22 PM Subject: [R] calculating the difference between days? Hi List, I have one column of beginning dates and one column of ending dates, I want to find their difference. And I want to ignore the trailing zeros, basically everything after the first colon mark. Begin_date End_date 01JAN2000:00:00:00:000 02FEB2002:00:00:00:000 24MAR2012:00:00:00:000 18MAY2012:00:00:00:000 01OCT2003:00:00:00:000 02FEB2004:00:00:00:000 01JAN2000:00:00:00:000 02FEB2000:00:00:00:000 01JAN2000:00:00:00:000 02FEB2000:00:00:00:000 Thanks, Mike [[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. [[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] Need HELP: how find and use a csv file?
You don't actually have to post more than once. Really. skatter - read.table(file.choose(), header=TRUE, sep=;) or skatter - read.table(/Users/kama/Desktop/skatter.csv, header=TRUE, sep=;) or whatever the actual path to the file is. As a new user of R, you should read the Introduction to R that came with R, and also the R-help posting guide. Both are full of useful information. Sarah On Tue, Jul 10, 2012 at 2:11 PM, Faradj Koliev farad...@gmail.com wrote: Hey, I am having some problems with importing a csv file into R and then saving it for analyzing. I got a csv file ( skater.csv) which i could read by typing: read.csv(file=/Users/kama/Desktop/skatter.csv, header=TRUE, sep=;) However, when i enter:skatter.csv-read.csv(skatter.csv, header=TRUE) i get this message: Error in file(file, rt) : cannot open the connection In addition: Warning message: In file(file, rt) : I have tried with: skatter.csv-file.choose() and other codes to find the file but it does not work. Please help me fix this problem, i have been sitting with this one in 4 hours.. What i need is to import this file and analyze it using for example histogram. I have Mac(update) and the file is saved in csv file... and I'm quite new user of R. -- 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] Need HELP: how find and use a csv file?
On Tue, 10 Jul 2012, Faradj Koliev wrote:
I got a csv file ( skater.csv) which i could read by typing:
read.csv(file=/Users/kama/Desktop/skatter.csv, header=TRUE, sep=;)
Try:
skatter - read.csv('/Users/kama/Desktop/skatter.csv', header = T, sep =
';')
Rich
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] calculating the difference between days?
Actually, when specifying unit=days inside difftime() will do. -M On Tue, Jul 10, 2012 at 3:55 PM, C W tmrs...@gmail.com wrote: When the days and time are identical, difftime() gives difference in secs. I still want difference in days. Say, below my last row is identical dat3-read.table(text= Begin_date End_date 01JAN2000:00:00:00:000 02FEB2002:00:00:00:000 24MAR2012:00:00:00:000 18MAY2012:00:00:00:000 01OCT2003:00:00:00:000 02FEB2004:00:00:00:000 01JAN2000:00:00:00:000 02FEB2000:00:00:00:000 01JAN2000:00:00:00:000 02FEB2000:00:00:00:000 24DEC2012:00:00:00:000 24DEC2012:00:00:00:000 ,sep=,header=TRUE) -M On Tue, Jul 10, 2012 at 1:52 PM, arun smartpink...@yahoo.com wrote: Hi, Try this: dat3-read.table(text= Begin_date End_date 01JAN2000:00:00:00:000 02FEB2002:00:00:00:000 24MAR2012:00:00:00:000 18MAY2012:00:00:00:000 01OCT2003:00:00:00:000 02FEB2004:00:00:00:000 01JAN2000:00:00:00:000 02FEB2000:00:00:00:000 01JAN2000:00:00:00:000 02FEB2000:00:00:00:000 ,sep=,header=TRUE) dat3$Begin_date-strptime(dat3[,1],format=%d%b%Y:%H:%M:%S) dat3$End_date-strptime(dat3[,2],format=%d%b%Y:%H:%M:%S) difftime(dat3[,1],dat3[,2]) Time differences in days [1] -763. -55. -124.0417 -32. -32. attr(,tzone) [1] A.K. - Original Message - From: C W tmrs...@gmail.com To: r-help r-help@r-project.org Cc: Sent: Tuesday, July 10, 2012 1:22 PM Subject: [R] calculating the difference between days? Hi List, I have one column of beginning dates and one column of ending dates, I want to find their difference. And I want to ignore the trailing zeros, basically everything after the first colon mark. Begin_date End_date 01JAN2000:00:00:00:000 02FEB2002:00:00:00:000 24MAR2012:00:00:00:000 18MAY2012:00:00:00:000 01OCT2003:00:00:00:000 02FEB2004:00:00:00:000 01JAN2000:00:00:00:000 02FEB2000:00:00:00:000 01JAN2000:00:00:00:000 02FEB2000:00:00:00:000 Thanks, Mike [[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. [[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] Changing x-axis values displayed on histogram
Hi, Thanks for providing a small reproducible example. You can disable the default axis and make your own custom version: hist(histexample,breaks=bins, xaxt=n) axis(1, at=seq(5.5, 15.5, by=2), labels=c(5-6, 7-8, 9-10, 11-12, 13-14, 15-16)) Sarah On Tue, Jul 10, 2012 at 3:34 PM, jlwoodard john.wood...@wayne.edu wrote: Is it possible to change the x-axis values in a histogram to reflect binned values? Here are my data: histexample-c(6,7,7,8,8,8,9,9,9,9,9,10,10,10,10,10,10,10,11,11,11,11,11,11,12,12,12,12,13,13,13,14,14,14,15,16) hist(histexample) Now, I'll bin pairs of adjacent values together (e.g., 5-6, 7-8, 9-10, 11-12, 13-14, 15-16) using the following bins-c(4.5,6.5,8.5,10.5,12.5,14.5,16.5) hist(histexample,breaks=bins) The displayed x-axis values are 6, 8, 10, 12, 14, and 16. I'd like the x-axis values to reflect the values in each bin (e.g., 5-6, 7-8, 9-10, 11-12, 13-14, 15-16). Any suggestions would be greatly appreciated! Many thanks in advance. John -- 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] Specify model with polynomial interaction terms up to degree n
a) Please supply some context in your mail message. Not everyone reads R-help
via nabble.
b) poly(raw=TRUE, x, degree=degree) was changed for 2.15.0 to allow it to output
a non-full-rank matrix. See the NEWS file in 2.15.0 or after:
# in R-2.15.1
n - news()
n[grepl(poly, n$Text),]
Changes in version 2.15.0:
NEW FEATURES
o poly(raw = TRUE) no longer requires more unique points than the
degree. (Requested by John Fox.)
...
c) Error messages include the innermost R function call before the error. To
see more call traceback(),
which will show the call stack from the innermost call back to the call you
made:
# in R-2.14.1
m - matrix(1:6, ncol=2)
poly(m, degree=6, raw=TRUE)
Error in poly(dots[[1L]], degree, raw = raw) :
'degree' must be less than number of unique points
traceback()
6: stop('degree' must be less than number of unique points)
5: poly(dots[[1L]], degree, raw = raw)
4: cbind(1, poly(dots[[1L]], degree, raw = raw))
3: polym(V1 = 1:3, V2 = 4:6, degree = 6, raw = TRUE)
2: do.call(polym, c(m, degree = degree, raw = raw))
1: poly(m, degree = 6, raw = TRUE)
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of YTP
Sent: Tuesday, July 10, 2012 11:23 AM
To: r-help@r-project.org
Subject: Re: [R] Specify model with polynomial interaction terms up to degree
n
Yep, that code is verbatim what I typed in, using version 2.14 ... seems
weird.
--
View this message in context:
http://r.789695.n4.nabble.com/Specify-model-with-
polynomial-interaction-terms-up-to-degree-n-tp4635130p4636031.html
Sent from the R help mailing list archive at Nabble.com.
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Re: [R] Use of Sappy and Tappy for Mathematical Calculation
On Jul 10, 2012, at 11:30 AM, Nordlund, Dan (DSHS/RDA) nord...@dshs.wa.gov
wrote:
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Rantony
Sent: Tuesday, July 10, 2012 3:17 AM
To: r-help@r-project.org
Subject: [R] Use of Sappy and Tappy for Mathematical Calculation
Hi,
i have a matrix like this,
ABCXYZ... .
- --
1220 ... .
2435 ... .
3040 ... .
Here, i need to get
Sum of each columns,
Mean of each columns,
median of each columns,
mode of each columns,
Standard deviation of each columns,
variance of each columns,
range of each columns,
count of each columns,
max of each columns,
min of each columns
Can i get output using sappy or tappy functions ? because there have
alots
of records.
I like Dan's solution a lot and it teaches inportant idioms, but I might
suggest that you use the col***() functions from the MatrixStats package if
speed/data-size is of issue.
Best,
Michael
Could you please help me fast its kind of urgent !
- Thanks
Antony
Here is some code to get you started. You can add in the other functions
that you want. You will need to figure out what you want to do if there are
missing values. There are built-in functions for most everything you want.
You get the range from the min and the max, and you need to decide what to do
if a variable has 2 or more modes (you will also need to determine how you
are going to get the mode).
# here is sample matrix
mat - matrix(1:100,nrow=10)
colnames(mat) - LETTERS[1:10]
# define summarize function
summarize - function(m) {
sums - apply(m, 2, sum)
counts - apply(m, 2, length)
means - apply(m, 2, mean)
return(rbind(sums, counts, means))
}
# summarize your matrix
summarize(mat)
Hope this is helpful,
Dan
Daniel J. Nordlund
Washington State Department of Social and Health Services
Planning, Performance, and Accountability
Research and Data Analysis Division
Olympia, WA 98504-5204
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Re: [R] calculating the difference between days?
What's wrong with manipulating the results arun got? dat3-read.table(text= Begin_date End_date 01JAN2000:00:00:00:000 02FEB2002:00:00:00:000 24MAR2012:00:00:00:000 18MAY2012:00:00:00:000 01OCT2003:00:00:00:000 02FEB2004:00:00:00:000 01JAN2000:00:00:00:000 02FEB2000:00:00:00:000 01JAN2000:00:00:00:000 02FEB2000:00:00:00:000 ,sep=,header=TRUE) dat3$Begin_date-strptime(dat3[,1],format=%d%b%Y:%H:%M:%S) dat3$End_date-strptime(dat3[,2],format=%d%b%Y:%H:%M:%S) # difference in days result - difftime(dat3[,1],dat3[,2]) # difference in seconds ddays - as.numeric(result/(24*3600)) ddays HTH, Jorge.- On Tue, Jul 10, 2012 at 3:55 PM, C W wrote: When the days and time are identical, difftime() gives difference in secs. I still want difference in days. Say, below my last row is identical dat3-read.table(text= Begin_date End_date 01JAN2000:00:00:00:000 02FEB2002:00:00:00:000 24MAR2012:00:00:00:000 18MAY2012:00:00:00:000 01OCT2003:00:00:00:000 02FEB2004:00:00:00:000 01JAN2000:00:00:00:000 02FEB2000:00:00:00:000 01JAN2000:00:00:00:000 02FEB2000:00:00:00:000 24DEC2012:00:00:00:000 24DEC2012:00:00:00:000 ,sep=,header=TRUE) -M On Tue, Jul 10, 2012 at 1:52 PM, arun smartpink...@yahoo.com wrote: Hi, Try this: dat3-read.table(text= Begin_date End_date 01JAN2000:00:00:00:000 02FEB2002:00:00:00:000 24MAR2012:00:00:00:000 18MAY2012:00:00:00:000 01OCT2003:00:00:00:000 02FEB2004:00:00:00:000 01JAN2000:00:00:00:000 02FEB2000:00:00:00:000 01JAN2000:00:00:00:000 02FEB2000:00:00:00:000 ,sep=,header=TRUE) dat3$Begin_date-strptime(dat3[,1],format=%d%b%Y:%H:%M:%S) dat3$End_date-strptime(dat3[,2],format=%d%b%Y:%H:%M:%S) difftime(dat3[,1],dat3[,2]) Time differences in days [1] -763. -55. -124.0417 -32. -32. attr(,tzone) [1] A.K. - Original Message - From: C W tmrs...@gmail.com To: r-help r-help@r-project.org Cc: Sent: Tuesday, July 10, 2012 1:22 PM Subject: [R] calculating the difference between days? Hi List, I have one column of beginning dates and one column of ending dates, I want to find their difference. And I want to ignore the trailing zeros, basically everything after the first colon mark. Begin_date End_date 01JAN2000:00:00:00:000 02FEB2002:00:00:00:000 24MAR2012:00:00:00:000 18MAY2012:00:00:00:000 01OCT2003:00:00:00:000 02FEB2004:00:00:00:000 01JAN2000:00:00:00:000 02FEB2000:00:00:00:000 01JAN2000:00:00:00:000 02FEB2000:00:00:00:000 Thanks, Mike [[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. [[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. [[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] Changing x-axis values displayed on histogram
Perfect! Thanks so much, Sarah! -- View this message in context: http://r.789695.n4.nabble.com/Changing-x-axis-values-displayed-on-histogram-tp4636032p4636051.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R code help to change table format
Hello, You should say what is the package you are using, EstimateS returns hundreds of hits. [ package sos, findFn() ]. As for the question, try sp - 1:5 ab - c(3, 2, 2, 2, 4) rep(sp, ab) Hope this helps, Rui Barradas Em 10-07-2012 18:23, peziza escreveu: I am trying to input an OTU table into EstimateS, however, the format of the OTU table has to be changed to fit the format EstimateS will accept. In R, I would like to change the format of the OTU table (from excel). Here is what I need to do, take Example 1 and create Example 2. The problem is that I have hundreds of OTUs, so I can't do this by hand (and I'd love to have a code that I could use for different OTU tables). Thanks! Example 1 Example 2 Species Abundance Species 1 3 1 2 2 1 3 2 1 4 2 2 5 4 2 3 3 4 4 5 5 5 5 5 -- View this message in context: http://r.789695.n4.nabble.com/R-code-help-to-change-table-format-tp4636022.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] Need HELP: how find and use a csv file?
On Jul 10, 2012, at 2:56 PM, Rich Shepard rshep...@appl-ecosys.com wrote:
On Tue, 10 Jul 2012, Faradj Koliev wrote:
I got a csv file ( skater.csv) which i could read by typing:
read.csv(file=/Users/kama/Desktop/skatter.csv, header=TRUE, sep=;)
Try:
skatter - read.csv('/Users/kama/Desktop/skatter.csv', header = T, sep =
';')
That seems paradoxical...
Michael
Rich
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Re: [R] Need HELP: how find and use a csv file?
On 10.07.2012 20:11, Faradj Koliev wrote: However, when i enter:skatter.csv-read.csv(skatter.csv, header=TRUE) i get this message: Error in file(file, rt) : cannot open the connection In addition: Warning message: In file(file, rt) : I have tried with: skatter.csv-file.choose() and other codes to find the file but it does not work. Could it be a problem with your Working directory? use getwd() to find out whether your working directory is set to your Desktop where your file seems to be and use setwd(/Path/to/Desktop) to go there if you aren't. Then you can use read.csv() without an absolute path for your file as you did above Regards, Moritz -- GnuPG Key: 0x7340821E __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Mann-Whitney by group
This works very well--thanks so much. By way of extension: how would one extract elements from the result object? For example: thing=apply(Dtb[,3:10], 2, function(x) wilcox.test(x~Dtb$Group)) summary(thing)$p.value Does not provide a list of p-values as it would in a regression object. Ideally, I would like to be able to extract the W score and p-value by A,B,C,... Any ideas greatly appreciated! -- View this message in context: http://r.789695.n4.nabble.com/Mann-Whitney-by-group-tp4635618p4636055.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] -1.1 - 0.1 + 1.2 is NOT null! Why?
Hello, I fear its a stupid question,..but here it is: If I do this simple calculation with the R console, I surprisingly do not get a zero. Why? -1.1-0.1+1.2 [1] -2.220446e-16 greetings, Ole -- View this message in context: http://r.789695.n4.nabble.com/1-1-0-1-1-2-is-NOT-null-Why-tp4636053.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] Download large file from https url with progress meter
Dear useRs, I would like to download a large (15Mb) file from a github https url with a progress meter. My first attempt was: zip_url=https://github.com/jefferis/AnalysisSuite/zipball/master; zip_file=tempfile() download.file(zip_url,zip_file) Error in download.file(zip_url, zip_file) : unsupported URL scheme Then I tried install.packages(httr) require(httr) system.time(request-GET(zip_url, config(ssl.verifypeer = FALSE))) stop_for_status(request) writeBin(content(request),zip_file) user system elapsed 1.234 0.236 27.685 I can't seem to find a way for the httr package to show progress. Can anyone suggest an alternative approach? With many thanks, Greg Jefferis. -- Gregory Jefferis, PhD Division of Neurobiology MRC Laboratory of Molecular Biology, Hills Road, Cambridge, CB2 0QH, UK. http://www2.mrc-lmb.cam.ac.uk/group-leaders/h-to-m/g-jefferis http://www.neuroscience.cam.ac.uk/directory/profile.php?gsxej2 http://flybrain.stanford.edu __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] -1.1 - 0.1 + 1.2 is NOT null! Why?
This is the behavior of the floating point number representation. Decimal fractions do not come out even in binary notation. Please see FAQ 7.31 On Tue, Jul 10, 2012 at 4:17 PM, ollestrat stratm...@gmx.de wrote: Hello, I fear its a stupid question,..but here it is: If I do this simple calculation with the R console, I surprisingly do not get a zero. Why? -1.1-0.1+1.2 [1] -2.220446e-16 greetings, Ole -- View this message in context: http://r.789695.n4.nabble.com/1-1-0-1-1-2-is-NOT-null-Why-tp4636053.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. [[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] -1.1 - 0.1 + 1.2 is NOT null! Why?
It is for the same reason that if you must work with numbers stored with 2 significant decimal digits 1-(1/3+1/3+1/3) is 0.01 (== 10 ^ -2). Double precision numbers, supported by your computer hardware and used by R, are stored using 52 significant binary digits and 2^-52 is about -2.220446e-16. (By the way, in R zero and NULL are different things: the former is numeric and the latter is not.) Bill Dunlap Spotfire, TIBCO Software wdunlap tibco.com -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of ollestrat Sent: Tuesday, July 10, 2012 1:17 PM To: r-help@r-project.org Subject: [R] -1.1 - 0.1 + 1.2 is NOT null! Why? Hello, I fear its a stupid question,..but here it is: If I do this simple calculation with the R console, I surprisingly do not get a zero. Why? -1.1-0.1+1.2 [1] -2.220446e-16 greetings, Ole -- View this message in context: http://r.789695.n4.nabble.com/1-1-0-1-1-2-is-NOT- null-Why-tp4636053.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-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] -1.1 - 0.1 + 1.2 is NOT null! Why?
In addition to taking cognisance of Richard Heiberger's reply you should also learn to distinguish between the concept of null and zero. They are not at all the same thing. cheers, Rolf Turner On 11/07/12 08:17, ollestrat wrote: Hello, I fear its a stupid question,..but here it is: If I do this simple calculation with the R console, I surprisingly do not get a zero. Why? -1.1-0.1+1.2 [1] -2.220446e-16 greetings, Ole __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Thanks! RE: boxplot with cut
Thanks for your help, Rui! That works and will save me a lot of trouble.
--Kelly
-Original Message-
From: Rui Barradas [mailto:ruipbarra...@sapo.pt]
Sent: Tuesday, July 10, 2012 2:24 AM
To: Vining, Kelly
Cc: r-help@r-project.org
Subject: Re: [R] boxplot with cut
Hello,
Maybe this iss what you're looking for. GD is your data.frame.
multi.boxplot - function(x, by, ...){
x - as.data.frame(x)
sp - split(x, by)
len - length(sp) - 1
n - ncol(x)
n1 - n + 1
boxplot(x[[ 1 ]] ~ by, at = 0:len*n1 + 1,
xlim = c(0, (len + 1)*n1), ylim = range(unlist(x)), xaxt = n,
...)
for(i in seq_len(n)[-1])
boxplot(x[[i]] ~ by, at = 0:len*n1 + i, xaxt = n, add = TRUE,
...)
axis(1, at = 0:len*n1 + n1/2, labels = names(sp), tick = TRUE) }
cols - grep(ReadCount, names(GD))
multi.boxplot(GD[, cols], cut(GD$GeneDensity, breaks=10))
If this is it and you don't like those x-axis tick lables, use
as.integer(cut(...etc...)).
Hope this helps,
Rui Barradas
Em 09-07-2012 20:51, Vining, Kelly escreveu:
Dear UseRs,
I'm making box plots from a data set that looks like this:
Chr Start End GeneDensity ReadCount_Explant ReadCount_Callus
ReadCount_Regen
1 1 1 1 107.82 1.2431.047
1.496
2 1 10001 2 202.50 0.8350.869
0.456
3 1 20001 3 158.80 1.8131.529
1.131
4 1 30001 4 100.53 1.7311.752
1.610
5 1 40001 5 100.53 3.0562.931
3.631
6 1 50001 6 100.53 1.9602.013
2.459
I'm breaking the GeneDensity column into deciles, then making a box plot of
the relationship between the GeneDensity parameter and each of the three
ReadCount columns. Here's an example of one of my boxplot commands:
boxplot(GeneDensity$ReadCount_Explant ~
cut(GeneDensitySorted$GeneDensity, breaks=10), ylim=c(0,40),
ylab=RPKM, xlab=GENE DENSITY (LOW - HIGH), main=INTERNODE
EXPLANT)
Right now, I'm making three separate graphs: one for each of the three
ReadCount columns. I'd like to put all three sets on one graph, so that
each decile is represented by three boxes, one for each ReadCount category,
but don't know how to make that work. I tried this:
boxplot(GeneDensitySorted$ReadCount_Explant ~
cut(GeneDensitySorted$GeneDensity, breaks=10),
GeneDensitySorted$ReadCount_Callus ~
cut(GeneDensitySorted$GeneDensity, breaks=10),
GeneDensitySorted$ReadCount_Regen ~ cut(GeneDensitySorted$GeneDensity,
breaks=10), ylim=c(0,40), ylab=RPKM, xlab=GENE DENSITY (LOW -
HIGH))
Not surprisingly, I got this error:
Error in as.data.frame.default(data) :
cannot coerce class 'formula' into a data.frame
Does anyone know how to accomplish this box plot?
Any help is much appreciated.
--Kelly V.
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Re: [R] -1.1 - 0.1 + 1.2 is NOT null! Why?
But R-ish NULL is *NOT* equal to R-ish zero, and that's what counts here. cheers, Rolf Turner On 11/07/12 09:19, Erdal Karaca wrote: german Null == english zero :-) 2012/7/10 Rolf Turner rolf.tur...@xtra.co.nz mailto:rolf.tur...@xtra.co.nz In addition to taking cognisance of Richard Heiberger's reply you should also learn to distinguish between the concept of null and zero. They are not at all the same thing. cheers, Rolf Turner On 11/07/12 08:17, ollestrat wrote: Hello, I fear its a stupid question,..but here it is: If I do this simple calculation with the R console, I surprisingly do not get a zero. Why? -1.1-0.1+1.2 [1] -2.220446e-16 greetings, Ole [[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] Mann-Whitney by group
Untested, I think you need to lapply() over thing with some sort of extractor: lapply(thing, function(x) x[['p.value']]) Michael On Jul 10, 2012, at 3:45 PM, Oxenstierna david.chert...@gmail.com wrote: This works very well--thanks so much. By way of extension: how would one extract elements from the result object? For example: thing=apply(Dtb[,3:10], 2, function(x) wilcox.test(x~Dtb$Group)) summary(thing)$p.value Does not provide a list of p-values as it would in a regression object. Ideally, I would like to be able to extract the W score and p-value by A,B,C,... Any ideas greatly appreciated! -- View this message in context: http://r.789695.n4.nabble.com/Mann-Whitney-by-group-tp4635618p4636055.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] HELP me please with import of csv to R
On Jul 10, 2012, at 21:44 , Sarah Goslee wrote: But note that if sep=; then you don't have a csv file and should properly use read.table() instead. That's not actually true. In a substantial part of the world, csv files are semicolon separated. That's what read.csv2() is for. (Yes, it is silly, please don't get me started...) -- Peter Dalgaard, Professor, Center for Statistics, Copenhagen Business School Solbjerg Plads 3, 2000 Frederiksberg, Denmark Phone: (+45)38153501 Email: pd@cbs.dk Priv: pda...@gmail.com __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] -1.1 - 0.1 + 1.2 is NOT null! Why?
Il 7/10/12 4:17 PM, ollestrat ha scritto: Hello, I fear its a stupid question,..but here it is: If I do this simple calculation with the R console, I surprisingly do not get a zero. Why? -1.1-0.1+1.2 [1] -2.220446e-16 greetings, Ole -- View this message in context: http://r.789695.n4.nabble.com/1-1-0-1-1-2-is-NOT-null-Why-tp4636053.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. ... Python : -1.1-0.1+1.2 -2.220446049250313e-16 -1.2-0.2+1.4 0.0 R : -1.1-0.1+1.2 [1] -2.220446e-16 -1.2-0.2+1.4 [1] 0 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
