Re: [R] Mean using different group for a real r beginner
Hi, Try either: tolerance - read.csv(http://www.ats.ucla.edu/stat/r/examples/alda/data/tolerance1.txt;) aggregate(exposure~male,data=tolerance,mean) # male exposure #1 0 1.246667 #2 1 1.12 #or library(plyr) ddply(tolerance,.(male),summarize,exposure=mean(exposure)) # male exposure #1 0 1.246667 #2 1 1.12 #or with(tolerance,tapply(exposure,list(male),FUN=mean)) # 0 1 #1.246667 1.12 #or with(tolerance,by(exposure,list(male),mean)) #: 0 #[1] 1.246667 # #: 1 #[1] 1.12 #or library(data.table) tolerance1- data.table(tolerance) tolerance1[,list(exposure=mean(exposure)),by=male] # male exposure #1: 0 1.246667 #2: 1 1.12 #or sapply(split(tolerance,tolerance$male),function(x) mean(x$exposure)) # 0 1 #1.246667 1.12 #or library(psych) describeBy(tolerance$exposure,tolerance$male,mat=TRUE)[c(2,5)] # group1 mean #11 0 1.246667 #12 1 1.12 #or library(doBy) summaryBy(exposure~male,data=tolerance,FUN=mean) # male exposure.mean #1 0 1.246667 #2 1 1.12 A.K. Hi there, I'm new to R and I think I have hard time getting used to it. Before asking my question, I looked on several websites, and while they have many informations about it, I did not find clear answer to my question. So, first of all, I use a web-based data set : tolerance - read.csv(http://www.ats.ucla.edu/stat/r/examples/alda/data/tolerance1.txt;) RIght now, I'm able to calculate the mean of the variable exposure. However, I'm unable to find a parsimonious way of obtaining the mean of exposure, separately for male == 0 and male == 1. The only solution I have, for the moment, is: tolerance_male = tolerance [male ==0, c (1:8)] exposure_male = tolerance_male$exposure mean (tolerance_male$exposure) tolerance_female = tolerance [male ==1, c (1:8)] exposure_female = tolerance_female$exposure mean (exposure_female) While this method work well for a simple calculation, it'll be quite unhelpful for more complicated analysis. So, I don't know if someone has a simpler way to do it. Thank you! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 use character in C code?
Characters in R are zero terminated (although I couldn't find that in
the R extensions manual). So, you could use:
void dealWithCharacter(char **chaine, int *size){
Rprintf(The string is '%s'\n, chaine[0]);
}
Jan
On 05/10/2013 03:51 PM, cgenolin wrote:
Hi the list,
I include some C code in a R function using .C. The argument is a
character.
I find how to acces to the characters one by one:
--- 8 --- C
void dealWithCharacter(char **chaine, int *size){
int i=0;
for(i=0;i*size;i++){
Rprintf(Le caractere %i est %c\n,i,chaine[0][i]);
};
}
--- 8 -- R -
ch - zerta
.C(dealWithCharacter,as.character(ch),as.integer(nchar(ch)))
--- 8 --
But is it possible to acces to the full word at once?
Christophe
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[R] Find the indices of non-NA elements of a sequence
Hi, I have a sequence whose 1st, 3rd, 4th, 6th are non-NAs. How could I let R return 1,3,4,6, the indices? I know only how to find the non-NA elements. Thanks, Miao test-c(2,NA,6,8,NA,12) test[is.na(test)==FALSE] [1] 2 6 8 12 [[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] convert a data.frame to matrix
On May 17, 2013, at 01:59 , David Winsemius wrote: On May 16, 2013, at 1:46 PM, Hermann Norpois wrote: Hello, I fail to tranfer data from a dataframe to a matrix. jam is from a dataframe (and belongs still to the class dataframe) and should look like m (see below). jam vec1 vec3 d1 d2 1 172 173 223 356 dput (jam) structure(list(vec1 = 172L, vec3 = 173L, d1 = 223L, d2 = 356L), .Names = c(vec1, vec3, d1, d2), row.names = 1L, class = data.frame) m #THIS IS THE AIM [,1] [,2] [1,] 172 223 [2,] 173 356 dput (m) structure(c(172, 173, 223, 356), .Dim = c(2L, 2L)) How can I transform jam to m? jam - structure(list(vec1 = 172L, vec3 = 173L, d1 = 223L, d2 = 356L), .Names = c(vec1, vec3, d1, d2), row.names = 1L, class = data.frame) jm - data.matrix(jam) dim(jm) - c(2,2) # re-dimension a matrix with column-major order jm [,1] [,2] [1,] 172 223 [2,] 173 356 also matrix(unlist(jam),2) [,1] [,2] [1,] 172 223 [2,] 173 356 -- 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] Find the indices of non-NA elements of a sequence
On 17-05-2013, at 08:36, jpm miao miao...@gmail.com wrote: Hi, I have a sequence whose 1st, 3rd, 4th, 6th are non-NAs. How could I let R return 1,3,4,6, the indices? I know only how to find the non-NA elements. Thanks, Miao test-c(2,NA,6,8,NA,12) test[is.na(test)==FALSE] [1] 2 6 8 12 ?which which(!is.na(test)) Berend __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Selecting A List of Columns
Dear R Helpers, I need help with a slightly unusual situation in which I am trying to select some columns from a data frame. I know how to use the subset statement with column names as in: x=as.data.frame(matrix(c(1,2,3, 1,2,3, 1,2,2, 1,2,2, 1,1,1),ncol=3,byrow=T)) all.cols-colnames(x) to.keep-all.cols[1:2] Kept-subset(x,select=to.keep) Kept However, if I want to select some columns based on a selection of the most important variables from a random forest then I find myself stuck. The example below demonstrates the problem. library(randomForest) data(mtcars) mtcars.rf - randomForest(mpg ~ ., data=mtcars,importance=TRUE) Importance-data.frame(mtcars.rf$importance) Importance MSEImportance-head(Importance[order(Importance$X.IncMSE, decreasing=TRUE),],3) MSEVars-row.names(MSEImportance) MSEVars-data.frame(MSEVars,stringsAsFactors = FALSE) colnames(MSEVars)-Vars NodeImportance-head(Importance[order(Importance$IncNodePurity,decreasing=TRUE),], 3) NodeVars-row.names(NodeImportance) NodeVars-data.frame(NodeVars,stringsAsFactors = FALSE) colnames(NodeVars)-Vars ImportantVars-rbind(MSEVars,NodeVars) ImportantVars-unique(ImportantVars) nrow(ImportantVars) ImportantVars-as.character(ImportantVars) ImportantVars CarsVarsKept-subset(mtcars,select=ImportantVars) Error in `[.data.frame`(x, r, vars, drop = drop) : undefined columns selected Any help on how to select these columns from the data frame would be most appreciated. --John J. Sparks, Ph.D. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] peering inside functions in a package?
On Fri, May 17, 2013 at 2:56 AM, Seth Myers sjmyers3...@gmail.com wrote:
Let's say I would like to look inside the function corBrownian in library
(ape). When I type in the function name I get the following, which is not
nearly the detail that goes into this function. I am wondering how to
begin cracking this function open (and others) so I can learn more about it
and perhaps code my own corClass one day. Thanks.
corBrownian
function (value = 1, phy, form = ~1)
{
if (!inherits(phy, phylo))
stop(object \phy\ is not of class \phylo\)
attr(value, formula) - form
attr(value, fixed) - TRUE
attr(value, tree) - phy
class(value) - c(corBrownian, corPhyl, corStruct)
value
}
environment: namespace:ape
Your premise is wrong! This is exactly all that goes into the corBrownian
function. All that it does is create a thing with some values and most
importantly, a set of class attributes.
It's via that class attribute that your corBrownian objects get their
behaviour via _method_ functions. If you download the source code for ape
from CRAN and have a look in PGLS.R you will see a bunch of functions such
as Initialize.corPhyl and corMatrix.corBrownian which are methods for
Initialize and corMatrix for corPhyl classes and corBrownian classes
respectively.
So when you come to use one of these correlation structures, the calling
code doesn't care how the correlation structure computes its correlation
matrix, it just calls the corMatrix function and the object-oriented magic
calls the specific one for that specific class of model.
So in ape, all the phylogenetic correlation classes are initialised with
Initialize.corPhyl and the Brownian one has corMatrix.corBrownian as the
method for computing its correlation matrix. You can see the source code
for these methods by typing their names on the R command line. Sometimes
methods aren't exported and aren't so easily visible on the command line,
leading to frustration. Download the source package for complete
satisfaction.
The help for ?corClasses in the nlme package has a bit more help, but you
might do well to read the source code of how these methods are created in
the ape package, especially if you are going to be creating a variation on
these phylo-classes.
Good luck. I tried and failed to write a new corClass for nlme a few years
back.
corBlimey.
Barry
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Re: [R] Rcmdr Bug?
Hi, Which command? But I would say that the first row is imported as column names. Regards, Pascal 2013/5/17 Knut Krueger r...@knut-krueger.de may be it is a bug: if a excel file is not containing headlines the first data row will not be imported Knut __** R-help@r-project.org mailing list https://stat.ethz.ch/mailman/**listinfo/r-helphttps://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/** posting-guide.html 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] Find the indices of non-NA elements of a sequence
Hi, For example: which(complete.cases(test)) [1] 1 3 4 6 Probably a more elegant way exists. Regards, Pascal 2013/5/17 jpm miao miao...@gmail.com Hi, I have a sequence whose 1st, 3rd, 4th, 6th are non-NAs. How could I let R return 1,3,4,6, the indices? I know only how to find the non-NA elements. Thanks, Miao test-c(2,NA,6,8,NA,12) test[is.na(test)==FALSE] [1] 2 6 8 12 [[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] read excel files problem [was] Rcmdr Bug
Am 17.05.2013 09:09, schrieb Pascal Oettli: Hi, Which command? But I would say that the first row is imported as column names. maybe its a general problem of importing excel files. First of all I am using only sheeets with headlines ;_) but Students do sometimes not. ;-) So gdata - read.xls is importing sheets default with header=TRUE (its possible to set it false) if not the fist row is interpreted as X+VAlue) means f.e X12.5 thats nearly OK anybody can see that there is a value. readWorksheet the same with header but the name of the Column is Col1 newbies will not realize that there is a value missing and R-C commander has no header Option and the first value will changed to F1. Maybe it would be good to give out a warning in all import functions if header is not set explicit to TRUE f.e Warning: First datarow is used as headlines additionally: output of names(dataset) - follow up to r-de...@r-project.org Regards, Pascal 2013/5/17 Knut Krueger r...@knut-krueger.de mailto:r...@knut-krueger.de may be it is a bug: if a excel file is not containing headlines the first data row will not be imported Knut __ R-help@r-project.org mailto:R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Achtung --- # # Rückantwortadresse ist RH-Help Group # Achtung --- [[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] Selecting A List of Columns
Hello, It works for me if I replace ImportantVars - as.character(ImportantVars) by ImportantVars - ImportantVars$Vars Hope this helps, Pascal 2013/5/17 Sparks, John James jspa...@uic.edu Dear R Helpers, I need help with a slightly unusual situation in which I am trying to select some columns from a data frame. I know how to use the subset statement with column names as in: x=as.data.frame(matrix(c(1,2,3, 1,2,3, 1,2,2, 1,2,2, 1,1,1),ncol=3,byrow=T)) all.cols-colnames(x) to.keep-all.cols[1:2] Kept-subset(x,select=to.keep) Kept However, if I want to select some columns based on a selection of the most important variables from a random forest then I find myself stuck. The example below demonstrates the problem. library(randomForest) data(mtcars) mtcars.rf - randomForest(mpg ~ ., data=mtcars,importance=TRUE) Importance-data.frame(mtcars.rf$importance) Importance MSEImportance-head(Importance[order(Importance$X.IncMSE, decreasing=TRUE),],3) MSEVars-row.names(MSEImportance) MSEVars-data.frame(MSEVars,stringsAsFactors = FALSE) colnames(MSEVars)-Vars NodeImportance-head(Importance[order(Importance$IncNodePurity,decreasing=TRUE),], 3) NodeVars-row.names(NodeImportance) NodeVars-data.frame(NodeVars,stringsAsFactors = FALSE) colnames(NodeVars)-Vars ImportantVars-rbind(MSEVars,NodeVars) ImportantVars-unique(ImportantVars) nrow(ImportantVars) ImportantVars-as.character(ImportantVars) ImportantVars CarsVarsKept-subset(mtcars,select=ImportantVars) Error in `[.data.frame`(x, r, vars, drop = drop) : undefined columns selected Any help on how to select these columns from the data frame would be most appreciated. --John J. Sparks, Ph.D. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] pearson correlation significant level
Hello I am using package Hmisc to calculate the pearson correlation and the significant level for the matrix of t_i and t_r. (temperature minimum and temperature range) However, I have difficulty interpreting the result, even after checking the manual. Please kindly help to indicate if the p-value is zero. Thank you in advance. Elaine The code library(Hmisc) rcorr(as.matrix(datat), type=pearson) # type can be pearson or spearman The result is t_i t_r t_i 1.00 -0.89 t_r -0.89 1.00 n= 4873 P t_i t_r t_i 0 t_r 0 [[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] Selecting A List of Columns
On May 17, 2013, at 12:02 , peter dalgaard wrote: On May 17, 2013, at 08:51 , Sparks, John James wrote: Dear R Helpers, I need help with a slightly unusual situation in which I am trying to select some columns from a data frame. I know how to use the subset statement with column names as in: Notice that subset() is a convenience function for command line use. The non-standard evaluation tricks in it tend to become inconveniences if you try to use subset() in a function (I can say that, I wrote the blasted thing...). Just use normal subseting functions instead and everything behaves much more predictably. If ImportantVars is a vector of column names, use mtcars[ImportantVars] (or mtcars[,ImportantVars], which also works for matrices). Oups, Pascal has the right end of the stick. The above is correct, but the If is important: You need a character vector of names, and that's not what is in ImportantVars. -- 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] Selecting A List of Columns
On May 17, 2013, at 08:51 , Sparks, John James wrote: Dear R Helpers, I need help with a slightly unusual situation in which I am trying to select some columns from a data frame. I know how to use the subset statement with column names as in: Notice that subset() is a convenience function for command line use. The non-standard evaluation tricks in it tend to become inconveniences if you try to use subset() in a function (I can say that, I wrote the blasted thing...). Just use normal subseting functions instead and everything behaves much more predictably. If ImportantVars is a vector of column names, use mtcars[ImportantVars] (or mtcars[,ImportantVars], which also works for matrices). x=as.data.frame(matrix(c(1,2,3, 1,2,3, 1,2,2, 1,2,2, 1,1,1),ncol=3,byrow=T)) all.cols-colnames(x) to.keep-all.cols[1:2] Kept-subset(x,select=to.keep) Kept However, if I want to select some columns based on a selection of the most important variables from a random forest then I find myself stuck. The example below demonstrates the problem. library(randomForest) data(mtcars) mtcars.rf - randomForest(mpg ~ ., data=mtcars,importance=TRUE) Importance-data.frame(mtcars.rf$importance) Importance MSEImportance-head(Importance[order(Importance$X.IncMSE, decreasing=TRUE),],3) MSEVars-row.names(MSEImportance) MSEVars-data.frame(MSEVars,stringsAsFactors = FALSE) colnames(MSEVars)-Vars NodeImportance-head(Importance[order(Importance$IncNodePurity,decreasing=TRUE),], 3) NodeVars-row.names(NodeImportance) NodeVars-data.frame(NodeVars,stringsAsFactors = FALSE) colnames(NodeVars)-Vars ImportantVars-rbind(MSEVars,NodeVars) ImportantVars-unique(ImportantVars) nrow(ImportantVars) ImportantVars-as.character(ImportantVars) ImportantVars CarsVarsKept-subset(mtcars,select=ImportantVars) Error in `[.data.frame`(x, r, vars, drop = drop) : undefined columns selected Any help on how to select these columns from the data frame would be most appreciated. --John J. Sparks, Ph.D. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 https://stat.ethz.ch/mailman/listinfo/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] Repeating sequence elements
I want to create a sequence, repeating each element according to a vector. I have this: v - c(4, 4, 4, 3, 3, 2) And want to create this: 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 5 5 5 6 6 TIA // s R version 3.0.0 (2013-04-03) Platform: x86_64-pc-linux-gnu (64-bit) locale: [1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C [3] LC_TIME=en_US.UTF-8 LC_COLLATE=en_US.UTF-8 [5] LC_MONETARY=en_US.UTF-8 LC_MESSAGES=en_US.UTF-8 [7] LC_PAPER=C LC_NAME=C [9] LC_ADDRESS=C LC_TELEPHONE=C [11] LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C attached base packages: [1] stats graphics grDevices utils datasets methods base [[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] peering inside functions in a package?
On 13-05-16 9:56 PM, Seth Myers wrote:
Let's say I would like to look inside the function corBrownian in library
(ape). When I type in the function name I get the following, which is not
nearly the detail that goes into this function. I am wondering how to
begin cracking this function open (and others) so I can learn more about it
and perhaps code my own corClass one day. Thanks.
corBrownian
function (value = 1, phy, form = ~1)
{
if (!inherits(phy, phylo))
stop(object \phy\ is not of class \phylo\)
attr(value, formula) - form
attr(value, fixed) - TRUE
attr(value, tree) - phy
class(value) - c(corBrownian, corPhyl, corStruct)
value
}
environment: namespace:ape
That's it. Why do you think something is missing?
Duncan Murdoch
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Re: [R] How could I see the source code of functions in an R package?
On 13-05-17 12:01 AM, David Winsemius wrote: Do this search with your favorite search tool Accessing the sources ligges (Uwe Ligges is the author of a comprehensive article on the topic in R News.) That article is also linked from the R help system. After help.start(), look for technical papers. It's one of those. 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] Repeating sequence elements
Hello, At an R prompt, type ?rep Then use v - c(4, 4, 4, 3, 3, 2) rep(1:6, v) Hope this helps, Rui Barradas Em 17-05-2013 11:53, Stefan Petersson escreveu: I want to create a sequence, repeating each element according to a vector. I have this: v - c(4, 4, 4, 3, 3, 2) And want to create this: 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 5 5 5 6 6 TIA // s R version 3.0.0 (2013-04-03) Platform: x86_64-pc-linux-gnu (64-bit) locale: [1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C [3] LC_TIME=en_US.UTF-8 LC_COLLATE=en_US.UTF-8 [5] LC_MONETARY=en_US.UTF-8 LC_MESSAGES=en_US.UTF-8 [7] LC_PAPER=C LC_NAME=C [9] LC_ADDRESS=C LC_TELEPHONE=C [11] LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C attached base packages: [1] stats graphics grDevices utils datasets methods base [[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] Repeating sequence elements
Try rep(1:length(v), v) HTH, Jorge.- On Fri, May 17, 2013 at 8:53 PM, Stefan Petersson ste...@inizio.se wrote: I want to create a sequence, repeating each element according to a vector. I have this: v - c(4, 4, 4, 3, 3, 2) And want to create this: 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 5 5 5 6 6 TIA // s R version 3.0.0 (2013-04-03) Platform: x86_64-pc-linux-gnu (64-bit) locale: [1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C [3] LC_TIME=en_US.UTF-8 LC_COLLATE=en_US.UTF-8 [5] LC_MONETARY=en_US.UTF-8 LC_MESSAGES=en_US.UTF-8 [7] LC_PAPER=C LC_NAME=C [9] LC_ADDRESS=C LC_TELEPHONE=C [11] LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C attached base packages: [1] stats graphics grDevices utils datasets methods base [[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] Rcmdr Bug?
Dear Knut, This is really more a limitation than a bug. Like most of the dialogs in the Rcmdr, the dialog to read Excel files uses existing facilities in R and R packages. I'll see whether it's possible to add an option to read a data set without variable names. The obvious workaround is to add a row of variable names to the spreadsheet. Best, John John Fox Sen. William McMaster Prof. of Social Statistics Department of Sociology McMaster University Hamilton, Ontario, Canada http://socserv.mcmaster.ca/jfox/ On Fri, 17 May 2013 07:40:51 +0200 Knut Krueger r...@knut-krueger.de wrote: may be it is a bug: if a excel file is not containing headlines the first data row will not be imported Knut __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Repeating sequence elements
Hi, rep(seq_along(v),v) #[1] 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 5 5 5 6 6 A.K. - Original Message - From: Stefan Petersson ste...@inizio.se To: r-help@r-project.org Cc: Sent: Friday, May 17, 2013 6:53 AM Subject: [R] Repeating sequence elements I want to create a sequence, repeating each element according to a vector. I have this: v - c(4, 4, 4, 3, 3, 2) And want to create this: 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 5 5 5 6 6 TIA // s R version 3.0.0 (2013-04-03) Platform: x86_64-pc-linux-gnu (64-bit) locale: [1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C [3] LC_TIME=en_US.UTF-8 LC_COLLATE=en_US.UTF-8 [5] LC_MONETARY=en_US.UTF-8 LC_MESSAGES=en_US.UTF-8 [7] LC_PAPER=C LC_NAME=C [9] LC_ADDRESS=C LC_TELEPHONE=C [11] LC_MEASUREMENT=en_US.UTF-8 LC_IDENTIFICATION=C attached base packages: [1] stats graphics grDevices utils datasets methods base [[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] A function that can modify an object? Or at least shows principles how to modify an object?
To make it easier for R-help readers to help you, please you provide the before UUU[2] data using the dput() function. dput(UUU[2]) Jean On Thu, May 16, 2013 at 11:12 AM, Aldi a...@dsgmail.wustl.edu wrote: Hi, If I have an R object UUU, where the second element is U2, based on g column of my.table my.table of UUU is: mmm ggg gindex map Info aaa123 U1 1 1 1 aaa124 U1 1 2 1 bbb1378U2 2 1 1 bbbU2 2 2 0 bbb1389U2 2 3 1 ccc Z3 3 1 1 ccc3Z3 3 2 0 ccc Z3 3 3 1 ccc5Z3 3 4 0 Before (see Before) the analysis I had dropped those mmm rows from my.table of UUU that had Info==0, thus not included in the analysis. After (see After) analysis found that I need to add bbb (in the second position, as defined by map) for U2, ccc and ccc for Z3 and s.o. The bbb, does not have any info in general, but a collaborator is saying that he needs the bbb for documentation in the UUU object. A solution is to rerun the analysis, but the list is very long, the analysis will take 1 day. Is there any ready function that can modify an object? Or at least shows principles how to modify an object? I have labeled elements of U2 as (e) which needs an NA in the second position; (d) needs a name bbb as a second column and as a second row and correspondingly in its rows and columns use value 1.0; NO change for (c); adding 0.0 in the second position for (b); and NO change for (a). (see After) Thank you in advance, Aldi Before: === UUU[2] $U2 $U1$scores (e) [1] -1.946707 -57.970488 $U2$cov (d) bbb1378 bbb1379 bbb1378 1.10362564 -0.01222695 bbb1379 -0.01222695 26.88805020 $U2$n (c) [1] 1802 $U1$maf (b) [1] 0.0002774695 0.0077691454 $U2$sey (a) [1] 13.3867 After: = UUU[2] $U2 $U2$scores (e) [1] -1.946707 NA -57.970488 $U2$cov (d) bbb1378 bbb bbb1379 bbb1378 1.10362564 1.0 -0.01222695 bbb 1.0 1.0 1.0 bbb1379 -0.01222695 1.0 26.88805020 $U2$n (c) [1] 1802 $U2$maf (b) [1] 0.0002774695 0.0 0.0077691454 $U2$sey (a) [1] 13.3867 -- __** R-help@r-project.org mailing list https://stat.ethz.ch/mailman/**listinfo/r-helphttps://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/** posting-guide.html http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] n in pglm() and relogit()
Dear all, How do I obtain the number of countries that my panel models estimated by pglm() and relogit() (the latter one is in the Zelig library) use? I couldn't find any information on that in the help files. Thank you very much in advance. Best wishes __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Homals: Nonlinear PCA
Hello! I'm using the NLPCA to reduce the dimensionality of nine variables (4 nominal / 3 ordinal /2 numeric) to obtain the object-scores to be used as dependent variable in a regression model. I'm using the package homals (http://www.jstatsoft.org/v31/i04/paper). The output is: Call: homals (date = date, Ndim = 1, rank = 1, level = c (numerical, rep (ordinal, 3), numerical, rep (nominal, 4), active = TRUE) Loss: 0.0002050824 Eigenvaluesââ: D1 0.0212 I'm having the following questions: 1) Is it best to consider Ndim = rank = 1 or Ndim = rank = max (rank) to reduce the dimensionality of data? 2) Is there a command to automatically calculate the proportion of variance explained by the first component? Otherwise, how can I calculate it by hand?3) Is it necessary to standardize numeric variables before perfoming homals? If anyone has any thoughts for this, responses would be greatly appreciated. Thanks. Lucia [[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] Using grubbs test for residuals to find outliers
Hi, I am a new user of R. This is a conceptual doubt regarding screeing out outliers from the dataset in regression. I read up that Cook's distance can be used and if we want to remove influential observations, we can use the metric (4/n) (n=no of observations) to remove any outliers. I also came across Grubb's test to identify outliers in univariate distns. (assumed normal) but i was not able to find contexts in Regression where Grubb's test is used (may be I didn't search enough) Is it a good idea to find out Cook's distance and identify outliers. Perform the Grubb's test for each of these outliers and then delete them? Right now, I am only using Cook's distance in my problem but I am uncertain as repeating the procedure with the new datasets (after removing influential observations) subsequently still keeps showing outliers in the plots. One reason maybe, i have only 50 data tuples and around 10 input variables in the Multiple regression equation. Am I going wrong in my fundamentals while using this approach. Thanks and regards, Karthik Srinivasan M.Mgt - Business Analytics Indian Institute of Science, Bangalore [[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] Contour lines in a persp plot
Thanks a lot, that is all i want. If someone is interessed, see the code
below
panel.3d.contour -
function(x, y, z, rot.mat, distance,
nlevels = 20, zlim.scaled, ...) # les3 points de suspension
pour dire les autres paramètres sont ceux données par défaut
{
add.line - trellis.par.get(add.line)
panel.3dwire(x, y, z, rot.mat, distance,
zlim.scaled = zlim.scaled, ...)
clines -
contourLines(x, y, matrix(z, nrow = length(x), byrow = TRUE),
nlevels = nlevels)
for (ll in clines) {
m - ltransform3dto3d(rbind(ll$x, ll$y, zlim.scaled[2]),
rot.mat, distance)
panel.lines(m[1,], m[2,], col = add.line$col,
lty = add.line$lty, lwd = add.line$lwd)
}
}
fn-function(x,y){sin(x)+2*y} #this looks like a corrugated tin roof
x-seq(from=1,to=100,by=2) #generates a list of x values to sample
y-seq(from=1,to=100,by=2) #generates a list of y values to sample
z-outer(x,y,FUN=fn) #applies the funct. across the combos of x and y
wireframe(z,zlim = c(1, 300), nlevels = 10,
aspect = c(1, 0.5), panel.aspect = 0.6,
panel.3d.wireframe = panel.3d.contour,
shade = FALSE ,
screen = list(z = 20, x = -60))
--
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
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[R] Log scales
Dear All,
I have a query about using log scales for filled.contour plots. I have
used log values on the x and y axes but labelled the axes with their
linear (non-log) values. I's like to do the same with the contour
values. I want the range to go from -100% to +100%. To use log I just
added 1000 to the % values before taking the log. I'd like the original
% values to appear on the contour lines and in the contour colourbar
labelling.
I'd appreciate any suggestions you have.
Many thanks,
Emily Gleeson
*
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[R] External access to fft routines
Dear colleagues (especially those with an understanding of the inner workings of R), For some years, I have been using R's fft (Fast Fourier Transform) routines from within C code as part of an R package. With R-3.0.1, fft_factor and fft_work are no longer available as entry points declared in R_ext/Applic.h and this is documented in the Changelog but I lack the skill/knowledge to track down the reason for the change. Does anyone know if (i) there is an alternative route to the fft routines; (ii) there is a reason not to use them; (iii) one should use some other routines instead? Thanks for any help given, Peter Craig -- ## | E-mail: p.s.cr...@durham.ac.uk Telephone: +44-191-3343076 (Work) | | Fax: +44-191-3343051 +44-191-3737415 (Home) | || | WWW: | || | http://www.dur.ac.uk/research/directory/staff/?mode=staffid=455 | || | Snail:Peter Craig, Dept. of Math. Sciences, Univ. of Durham, | | South Road, Durham DH1 3LE, England | ## __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] zigzag confidence interval in a plot
Dear All, When I plot the values and linear regression line for one data set, it is fine. But for another one I see zigzags, when I plot the confidence interval cd Depth CHAOsep12RNA 9,94804 25,06 1476,83 40,04 1540,561404 50,11 1575,17 52,46 349,22 54,92 1941,5 57,29 1053,507042 60,11 1535,1 70,04 2244,963303 79,97 1954,507042 100,31 2679,140625 plot(cd$CHAOsep12RNA,cd$Depth, ylim = rev(range(0:100)), xlab=CHAO, ylab=Depth, pch=15, las=2, main=Sep12-RNA, cex.main=1) lmR - lm(cd$Depth~cd$CHAOsep12RNA) abline(lmR) pconfR - predict(lmR,interval=confidence) matlines(cd$CHAOsep12RNA,pconfR[,c(lwr,upr)], col=1, lty=2) I also tried newx - seq(min(cd$CHAOsep12RNA), max(cd$CHAOsep12RNA), length.out=11) a - predict(lmR, newdata=data.frame(CHAO=newx), interval=c(confidence)) plot(cd$CHAOsep12RNA,cd$Depth, ylim = rev(range(0:100)), xlab=CHAO, ylab=Depth, pch=15, las=2, main=Sep12-RNA, cex.main=1) abline(lmR) lines(cd$CHAOsep12RNA, a[,2], lty=2) But I see both cases kind of zigzags. What can it be the reason? thank you! zigzaging.pdf Description: Adobe PDF document __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R looping help
It's difficult to see where your error is when you provide no accompanying
data for us to test your code out on (X? Y? Stand?).
However, it looks like you are making your code more complex than it needs
to be. Are you simply trying to a fit a separate linear regression to each
subset of your data identified by a unique standID? If so, you can do so
like this:
# a fake version of your data
mydata - data.frame(standID=rep(1:5, rep(20, 5)), X=rnorm(100),
Y=rnorm(100))
# split the data up by stand ID
standlist - split(mydata, mydata$standID)
# fit a regression to each stand and save some results
standEst1 - lapply(standlist, function(df) {
fit - summary(lm(Y ~ X, data=df))
co - coef(fit)
intercept - co[1, 1]
slope - co[2, 1]
pValue - co[2, 4]
mse - fit$sigma^2
rSquare - fit$r.squared
c(standID=df$standID[1], intercept=intercept, slope=slope, mse=mse,
rSquare=rSquare, pValue=pValue)
})
# convert the results to a data frame
standEstimates - data.frame(do.call(rbind, standEst1))
Jean
On Thu, May 16, 2013 at 5:17 AM, rishard or...@uclive.ac.nz wrote:
Hey I'm not really sure what I should put on here, but I am having trouble
with my R code. I am trying to get the p-values, R^2s etc for a number of
different groups of variables that are all in one dataset.
This is the code:
#Stand counter
st-1
#Collections
stands-numeric(67)
slopes-numeric(67)
intercepts-numeric(67)
mses-numeric(67)
rsquares-numeric(67)
pValues-numeric(67)
#Start lists for X and Y values within each stand
xi-numeric(0)
yi-numeric(0)
#Set the first element to the starting X and Y values
xi[1]=X[1]
yi[1]=Y[1]
#Start looping working through your data, record by record
for (i in 2:length(X)) {
#If you are in the same stand as on the last record, continue to
#collect X and Y values
if(Stand[i]==Stand[i-1]) {
xi=cbind(xi,X[i])
yi=cbind(yi,Y[i])
} else {
#If a new stand is encountered make your linear model and
#collect statistics
model-lm(yi~xi)
stands[st]-Stand[i-1]
intercepts[st]-model$coefficients[1]
slopes[st]-model$coefficients[2]
mses[st]-sum(resid(model)^2)/(length(xi)-2)
ssr-var(yi)*(length(xi)-1)-sum(resid(model)^2)
rsquares[st]-ssr/(var(yi)*(length(xi)-1))
fRatio-ssr/mses[st]
pValues[st]-1-pf(fRatio,1,length(xi)-2)
#Increment the stand number, zero the within stand collections,
#and start again
st-st+1
xi-numeric(0)
yi-numeric(0)
xi[1]=X[i]
yi[1]=Y[i]
}
}
#Make your data set
standEstimates-data.frame(standID=stands,intercept=intercepts,slop=slopes,mse=mses,rSquare=rsquares,pValue=pValues)
The standEstimate outputs look like this:
standID intercept slopmse rSquare pValue
1 6833319.2470NA 0 NA NA
2 756 708.7470NA 0 NA NA
3 795 508.2290NA 0 NA NA
4 1249503.1460NA 0 NA NA
5 1331703.0620NA 0 NA NA
6 1417747.7620NA 0 NA NA
7 4715549.3400NA 0 NA NA
8 4850603.9940NA 0 NA NA
9 2105573.3270NA 0 NA NA
Etc etc
and I get these warnings:
1: In rsquares[st] - ssr/(var(yi) * (length(xi) - 1)) :
number of items to replace is not a multiple of replacement length
2: In pValues[st] - 1 - pf(fRatio, 1, length(xi) - 2) :
number of items to replace is not a multiple of replacement length
3: In rsquares[st] - ssr/(var(yi) * (length(xi) - 1)) :
number of items to replace is not a multiple of replacement length
4: In pValues[st] - 1 - pf(fRatio, 1, length(xi) - 2) :
number of items to replace is not a multiple of replacement length
5: In rsquares[st] - ssr/(var(yi) * (length(xi) - 1)) :
number of items to replace is not a multiple of replacement length
6: In pValues[st] - 1 - pf(fRatio, 1, length(xi) - 2) :
number of items to replace is not a multiple of replacement length
7: In rsquares[st] - ssr/(var(yi) * (length(xi) - 1)) :
number of items to replace is not a multiple of replacement length
8: In pValues[st] - 1 - pf(fRatio, 1, length(xi) - 2) :
number of items to replace is not a multiple of replacement length
9: In rsquares[st] - ssr/(var(yi) * (length(xi) - 1)) :
number of items to replace is not a multiple of replacement length
10: In pValues[st] - 1 - pf(fRatio, 1, length(xi) - 2) :
number of items to replace is not a multiple of replacement length
11: In rsquares[st] - ssr/(var(yi) * (length(xi) - 1)) :
number of items to replace is not a multiple of replacement length
12: In pValues[st] - 1 - pf(fRatio, 1, length(xi) - 2) :
number of items to replace is not a multiple of replacement length
13: In rsquares[st] - ssr/(var(yi) * (length(xi) - 1)) :
number of items to replace is not a multiple of replacement length
Re: [R] Rcmdr Bug?
Am 17.05.2013 13:36, schrieb John Fox: Dear Knut, This is really more a limitation than a bug. Like most of the dialogs in the Rcmdr, the dialog to read Excel files uses existing facilities in R and R packages. I'll see whether it's possible to add an option to read a data set without variable names. The obvious workaround is to add a row of variable names to the spreadsheet. dear John, indeed I would never use an excel sheet without variable names ... but students do and then the statistics are wrong As I mentioned in the other answer it would be fine to add a hint that the first row is used as variable name ... for the students ;-) Knut __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] out of the mailing list
Hello, I want to stop receiveng mails from R help forum, thanks. This is my data : Mario Garrido Escudero Dpto. de Biología Animal, Ecología, Parasitología, Edafología y Qca. Agrícola Fac. de Farmacia Campus Unamuno Universidad de Salamanca gaiarr...@usal.es [[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] out of the mailing list
Hello, It is not the correct procedure. Go to the bottom of the following webpage to unsubscribe. https://stat.ethz.ch/mailman/listinfo/r-help Regards, Pascal 2013/5/17 Mario Garrido gaiarr...@usal.es Hello, I want to stop receiveng mails from R help forum, thanks. This is my data : Mario Garrido Escudero Dpto. de Biología Animal, Ecología, Parasitología, Edafología y Qca. Agrícola Fac. de Farmacia Campus Unamuno Universidad de Salamanca gaiarr...@usal.es [[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.
[R] update an array of plots in 'real-time' without drawing lags
Hi, I know R is not made for this, but I still wanted to ask if there are possibilities to do this; I repeatedly collect data from a database for a given time interval. Now I would like to monitor the change of this data with some nice plots. I actually have to draw 15 plots to get the whole picture. Now I can write a loop that, for a given time interval, updates the data and plots them using one quartz window that is split into subparts with the layout function. However, since I redraw every plot every time, the drawing takes some time and it is not an instant update that I see. The plots are slowly redrawn from the upper left corner of the quartz window to the bottom right. That makes me wonder; Is there a way to buffer the graphical device before updating it? Or are there any other solutions for R that enables to smoothly plot / visualize data in 'real-time'? I am grateful for any suggestions! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] mirt package error in ESTIMATION...
Hello everyone, I am trying to undertake an item bifactor analysis of graded response data from a questionnaire. I am using the mirt package, especially the bfactor function.My dataset is called data.items, it contains about 2000 observations and 31 variables (variables represent the items in the questionnaire). The items follow a Likert scale format and represent a level of satisfaction, missing values are coded as NA). I am having trouble at the beginning of my analysis, during the exploratory model fitting. The syntax and the error message are as follows: bfactor(data.items,model=9) Error in ESTIMATION(data = data, model = model, group = rep(all, nrow(data)), : index out of bounds I get the same error when i try with: bfactor(data.items,9,itemtype=graded) or bfactor(data.items,9,prev.cor=cor) where cor is the correlationmatrix that i compute without the missing values. If someone have an idea to suggest, do not hesitate. Thank you. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] how to calculate the mean in a period of time?
Hi,
Try this:
dat1$idx-with(dat1,ifelse(is.na(delais)|delais45
delais20, 1,ifelse(delais60
delais=45,2,ifelse(delais=90 delais=60,3,NA
dat1$idx1-c(dat1$idx[-head(dat1$idx,1)],1)
library(zoo)
res1-do.call(rbind,lapply(split(dat1,dat1$patient_id),function(x)
{x$idx[as.logical(cumsum(is.na(x$idx)))]-NA; x1-x[!is.na(x$idx),];
x1[,6:8]-na.locf(x1[,6:8]);x1$idx1[is.na(x1$idx1)]-1;
x2-x1[rep(seq_len(nrow(x1)),x1$idx1),];
x2$delais[duplicated(x2$delais,fromLast=FALSE)]-0;
x2$t-seq(0,nrow(x2)-1,1);x2[,-c(8:9)]}))
row.names(res1)- 1:nrow(res1)
res2- res1[,c(1:3,8,4:7)]
res2
# patient_id number responsed_at t delais scores1 scores2 scores3
#1 1 1 2010-05-26 0 NA 2.6 0.5 0.7
#2 1 2 2010-07-07 1 42 2.5 0.5 0.7
#3 1 3 2010-08-14 2 38 2.3 0.5 0.7
#4 1 3 2010-08-14 3 0 2.3 0.5 0.7
#5 1 4 2010-10-01 4 48 2.5 0.7 0.6
#6 1 4 2010-10-01 5 0 2.5 0.7 0.6
#7 1 4 2010-10-01 6 0 2.5 0.7 0.6
#8 1 5 2010-12-01 7 61 2.5 0.7 0.6
#9 2 1 2011-07-19 0 NA 2.5 0.8 0.5
#10 2 1 2011-07-19 1 0 2.5 0.8 0.5
#11 2 1 2011-07-19 2 0 2.5 0.8 0.5
#12 2 2 2011-09-22 3 65 2.6 0.8 0.5
#13 2 3 2011-10-26 4 34 2.7 0.8 0.5
#14 3 1 2011-07-17 0 NA 2.8 0.5 0.6
A.K.
From: GUANGUAN LUO guanguan...@gmail.com
To: arun smartpink...@yahoo.com
Sent: Friday, May 17, 2013 9:33 AM
Subject: Re: how to calculate the mean in a period of time?
Hello，
Thank you for your help
the lines added to the tables are the precedent lines but not the followed
lines, if i just change x2-x1[rep(seq_len(nrow(x1)-1), is that ok? and so the
delais should be changed too, isn't it?
GG
2013/5/17 arun smartpink...@yahoo.com
Hi,
No problem.
Arun
From: GUANGUAN LUO guanguan...@gmail.com
To: arun smartpink...@yahoo.com
Sent: Friday, May 17, 2013 4:19 AM
Subject: Re: how to calculate the mean in a period of time?
Ah, yes, that is the wrong thing i have written. Thank you so much. the output
which you have got is right.
Thanks a lot.
GG
2013/5/16 arun smartpink...@yahoo.com
Hi,
The output you showed is not clear especially the for the scores3,
2 2 2011-09-22 3 65 2.6
0.8 0.8
2 3 2011-10-26 4 34 2.7
0.8 0.8
3 1 2011-07-17 0 NA 2.8
0.5 0.6
In the input data, the scores3 column didn't had 0.8.
This is what I got:
dat1- read.table(text=
patient_id number responsed_at delais scores1 scores2
scores3
1 1 2010-05-26 NA 2.6
0.5 0.7
1 2 2010-07-07 42 2.5
NA NA
1 3 2010-08-14 38 2.3
NA NA
1 4 2010-10-01 48 2.5
0.7 0.6
1 5 2010-12-01 61 2.5
NA NA
2 1 2011-07-19 NA 2.5
0.8 0.5
2 2 2011-09-22 65 2.6
NA NA
2 3 2011-10-26 34 2.7
NA NA
3 1 2011-07-17 NA 2.8
0.5 0.6
3 2 2011-10-30 103 2.6
NA NA
3 3 2011-12-23 54 2.5
NA NA
,sep=,header=TRUE,stringsAsFactors=FALSE)
dat1$idx-with(dat1,ifelse(is.na(delais)|delais45 delais20,
1,ifelse(delais60 delais=45,2,ifelse(delais=90 delais=60,3,NA
library(zoo)
res-do.call(rbind,lapply(split(dat1,dat1$patient_id),function(x)
{x$idx[as.logical(cumsum(is.na(x$idx)))]-NA; x1-x[!is.na(x$idx),];
x1[,6:8]-na.locf(x1[,6:8]);x2-x1[rep(seq_len(nrow(x1)),x1$idx),];
x2$delais[duplicated(x2$delais,fromLast=TRUE)]-0;
x2$t-seq(0,nrow(x2)-1,1);x2}))
row.names(res)- 1:nrow(res)
res1- res[,c(1:3,9,4:7)]
res1
# patient_id number responsed_at t delais scores1 scores2 scores3
#1 1 1 2010-05-26 0 NA 2.6 0.5 0.7
#2 1 2 2010-07-07 1 42 2.5 0.5 0.7
#3 1 3 2010-08-14 2 38 2.3 0.5 0.7
#4 1 4
[R] Error with adehabitatHR and kernelbb
Dear all,
I'm trying to get a Brownian bridge kernel (kernelbb) for each combination of
two consecutive animal locations (see commands below) and put them, with a
loop, inside a list. It works well at the beginning but after 42 runs, it
appears the following warning :
Error in seq.default(yli[1], yli[2], by = diff(xg[1:2])) :
invalid (to - from)/by in seq(.)
I looked at the coordinates, at the id, at the time of the run 43 and it's all
good...
I looked on the net and it happened to only one person and there was no answer
to his post.
Someone could help me?
## commands
BBtraj - list()
for (i in 1:(nrow(loc@data)-1)) {
BBtraj[[i]] - kernelbb(as.ltraj(loc@coords[i:(i+1),],
date=loc@data$time[i:(i+1)], id = as.character(loc@data$id[i:(i+1)]),
typeII = TRUE), sig1=as.numeric(as.character(loc@data$sig1[i])), sig2= 5, grid
= 1000)
}
Rémi Lesmerises, biol. M.Sc.,
Candidat Ph.D. en Biologie
Université du Québec à Rimouski
remilesmeri...@yahoo.ca
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[R] R and libre office base
Dear community . I would like to connect r to libre office base. Does anyone know if and how this can be done? I think of the pendant to rodbc for libre office. I am using windows 7. Thanks in advance and best regards johan [[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 and libre office base
On May 17, 2013, at 9:47 AM, Johan Lassen johanlas...@gmail.com wrote: Dear community . I would like to connect r to libre office base. Does anyone know if and how this can be done? I think of the pendant to rodbc for libre office. I am using windows 7. Thanks in advance and best regards johan As an FYI, there is an R list just for database access related issues: https://stat.ethz.ch/mailman/listinfo/r-sig-db Both OpenOffice and LibreOffice use HSQLDB, which is a java based database. There are no ODBC connections supported by HSQLDB. However, I believe that some have had success with JDBC and there is the ODB package on CRAN (http://cran.r-project.org/web/packages/ODB/). Regards, Marc Schwartz __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] points overlay axis
Apologies to John - I should have thought to give an example. However, xpd is what I was looking for. Thanks for the help! On 14 May 2013 14:55, David Carlson dcarl...@tamu.edu wrote: Let's try again after restraining Outlook's desire to use html. set.seed(42) dat - matrix(c(runif(48), 0, 0), 25, 2, byrow=TRUE) # Complete plot symbol on axes, but axis on top plot(dat, xaxs=i, yaxs=i, pch=16, col=red, xpd=TRUE) # Complete plot symbol on axes with symbol on top plot(dat, xaxs=i, yaxs=i, type=n) points(dat, xaxs=i, yaxs=i, pch=16, col=red, xpd=TRUE) David L Carlson Associate Professor of Anthropology Texas AM University College Station, TX 77840-4352 -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of John Kane Sent: Tuesday, May 14, 2013 7:47 AM To: Jonathan Phillips; r-help@r-project.org Subject: Re: [R] points overlay axis Probably but since we don't know what you are doing, it is very hard to give any advice. Please read this for a start https://github.com/hadley/devtools/wiki/Reproducibility and give us a clear statement of the problem Thanks John Kane Kingston ON Canada -Original Message- From: 994p...@gmail.com Sent: Tue, 14 May 2013 13:34:35 +0100 To: r-help@r-project.org Subject: [R] points overlay axis Hi, I'm trying to do quite a simple task, but I'm stuck. I've set xaxs = 'i' as I want the origin to be (0,0), but unfortunately I have points that are sat on the axis. R draws the axis over the points, which hides the points somewhat and looks unsightly. Is there any way of getting a point to be drawn over the axis? Thanks, Jon Phillips [[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. FREE ONLINE PHOTOSHARING - Share your photos online with your friends and family! Visit http://www.inbox.com/photosharing to find out more! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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. [[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] max length of a factor variable
Technically this is correct for raw R functionality. in practice various modules impose their own limits on variables so you have to check. For example the coxreg package truncates all variables to 16 characters. for example the test below res Call: coxreg(formula = Surv(vtime, vstatus) ~ abcdefghijklmnopqrstuvwxyzabcdefghijklmnopqrstuvwxyz1_fx1 + alt_const_fx1 + alt_const_fx2 + strata(vstrata), data = aggregate) Covariate Mean Coef Rel.Risk S.E.Wald p abcdefghijklmnop -0.000NANA 0.000NA alt_const_fx1 -0.045 0.132 1.141 0.044 0.002 alt_const_fx2 -0.024 0.011 1.011 0.045 0.801 -- View this message in context: http://r.789695.n4.nabble.com/max-length-of-a-factor-variable-tp2715509p4667332.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] zigzag confidence interval in a plot
On 2013-05-17 06:03, Ozgul Inceoglu wrote: Dear All, When I plot the values and linear regression line for one data set, it is fine. But for another one I see zigzags, when I plot the confidence interval cd Depth CHAOsep12RNA 9,94804 25,06 1476,83 40,04 1540,561404 50,11 1575,17 52,46 349,22 54,92 1941,5 57,29 1053,507042 60,11 1535,1 70,04 2244,963303 79,97 1954,507042 100,31 2679,140625 plot(cd$CHAOsep12RNA,cd$Depth, ylim = rev(range(0:100)), xlab=CHAO, ylab=Depth, pch=15, las=2, main=Sep12-RNA, cex.main=1) lmR - lm(cd$Depth~cd$CHAOsep12RNA) abline(lmR) pconfR - predict(lmR,interval=confidence) matlines(cd$CHAOsep12RNA,pconfR[,c(lwr,upr)], col=1, lty=2) I also tried newx - seq(min(cd$CHAOsep12RNA), max(cd$CHAOsep12RNA), length.out=11) a - predict(lmR, newdata=data.frame(CHAO=newx), interval=c(confidence)) plot(cd$CHAOsep12RNA,cd$Depth, ylim = rev(range(0:100)), xlab=CHAO, ylab=Depth, pch=15, las=2, main=Sep12-RNA, cex.main=1) abline(lmR) lines(cd$CHAOsep12RNA, a[,2], lty=2) But I see both cases kind of zigzags. What can it be the reason? thank you! Sort your dataframe by x-values (CHAOsep12RNA). Peter Ehlers __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] zigzag confidence interval in a plot
Hello, Try the following. lmR - lm(cd$Depth ~ cd$CHAOsep12RNA) pconfR - predict(lmR,interval=confidence) pcR - pconfR[order(pconfR[, 1]), ] # Add this line plot(cd$CHAOsep12RNA,cd$Depth, ylim = rev(range(0:100)), xlab=CHAO, ylab=Depth, pch=15, las=2, main=Sep12-RNA, cex.main=1) abline(lmR) matlines(sort(cd$CHAOsep12RNA), pcR[,c(lwr, upr)], col=1, lty=2) # changed Hope this helps, Rui Barradas Em 17-05-2013 14:03, Ozgul Inceoglu escreveu: Dear All, When I plot the values and linear regression line for one data set, it is fine. But for another one I see zigzags, when I plot the confidence interval cd Depth CHAOsep12RNA 9,94804 25,06 1476,83 40,04 1540,561404 50,11 1575,17 52,46 349,22 54,92 1941,5 57,29 1053,507042 60,11 1535,1 70,04 2244,963303 79,97 1954,507042 100,31 2679,140625 plot(cd$CHAOsep12RNA,cd$Depth, ylim = rev(range(0:100)), xlab=CHAO, ylab=Depth, pch=15, las=2, main=Sep12-RNA, cex.main=1) lmR - lm(cd$Depth~cd$CHAOsep12RNA) abline(lmR) pconfR - predict(lmR,interval=confidence) matlines(cd$CHAOsep12RNA,pconfR[,c(lwr,upr)], col=1, lty=2) I also tried newx - seq(min(cd$CHAOsep12RNA), max(cd$CHAOsep12RNA), length.out=11) a - predict(lmR, newdata=data.frame(CHAO=newx), interval=c(confidence)) plot(cd$CHAOsep12RNA,cd$Depth, ylim = rev(range(0:100)), xlab=CHAO, ylab=Depth, pch=15, las=2, main=Sep12-RNA, cex.main=1) abline(lmR) lines(cd$CHAOsep12RNA, a[,2], lty=2) But I see both cases kind of zigzags. What can it be the reason? thank you! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problem with ordiellipse coloured factor in Vegan
No data. The list tends to strip out many kinds of attachements including csv
files. I'd suggest trying .txt or parking the data at someplace like
[url=http://www.mediafire.com/][b]MediaFire[/b][/url] or
Dropbox [url=https://www.dropbox.com/][b]Dropbox[/b][/url]
John Kane
Kingston ON Canada
-Original Message-
From: suparna.mitra...@gmail.com
Sent: Fri, 17 May 2013 11:56:52 +0800
To: r-help@r-project.org
Subject: [R] Problem with ordiellipse coloured factor in Vegan
Hello R experts,
I am trying to plot ordiellipse for my data but the col according to
factors.
Metabolites_raw= read.csv(file.choose(), head = TRUE) #file
21Metabolites.csv
Metabolites_t=t(Metabolites_raw[,2:82])
ord - metaMDS(Metabolites_t, distance=bray)
symbol=as.numeric(Metab_metadata$LandType)
col.list - c(red,slategray,seagreen,cyan,pink,brown,black,
blue,yellow,magenta)
palette(col.list)
plot(ord$points, col = Metab_metadata$Day+2,pch=symbol,
xlim=c(-0.3,0.35))
legend(.28,.25, c(0, 8, 16),fill = c(2, 10,18))
# draw dispersion ellipses around data points
groupz -c(2,10,18)
for(i in seq(groupz)) {
ordiellipse(ord, Metab_metadata$Day, kind = sd, label =
T,col=groupz[i],
show.groups=groupz[i])
}
Now here I get the error
Error in text.default(...) : no coordinates were supplied (But sometimes
same code works. I am not sure what am I doing wrong.)
Data files are attached.
Any help will be really great.
Thanks,
Mitra
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Re: [R] pearson correlation significant level
Dear Elaine, One of the elements you obtain the P matrix, which is the matrix of asymptotic p-values. In your case, you get that the asymptotic p-value of the association between t_i and t_r is 0. That is, there would exist a perfect association (look also at the first result, the matrix of correlations: the correlation coefficient between these variables is 1). Hope this helps, José Prof. José Iparraguirre Chief Economist Age UK -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Elaine Kuo Sent: 17 May 2013 10:40 To: r-help@r-project.org Subject: [R] pearson correlation significant level Hello I am using package Hmisc to calculate the pearson correlation and the significant level for the matrix of t_i and t_r. (temperature minimum and temperature range) However, I have difficulty interpreting the result, even after checking the manual. Please kindly help to indicate if the p-value is zero. Thank you in advance. Elaine The code library(Hmisc) rcorr(as.matrix(datat), type=pearson) # type can be pearson or spearman The result is t_i t_r t_i 1.00 -0.89 t_r -0.89 1.00 n= 4873 P t_i t_r t_i 0 t_r 0 [[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. Please donate to the Syria Crisis Appeal by text or online: To donate £5 by mobile, text SYRIA to 70800. To donate online, please visit http://www.ageinternational.org.uk/syria Over one million refugees are desperately in need of water, food, healthcare, warm clothing, blankets and shelter; Age International urgently needs your support to help affected older refugees. Age International is a subsidiary charity of Age UK and a member of the Disasters Emergency Committee (DEC). The DEC launches and co-ordinates national fundraising appeals for public donations on behalf of its member agencies. Texts cost £5 plus one standard rate message. Age International will receive a minimum of £4.96. More info at ageinternational.org.uk/SyriaTerms --- Age UK is a registered charity and company limited by guarantee, (registered charity number 1128267, registered company number 6825798). Registered office: Tavis House, 1-6 Tavistock Square, London WC1H 9NA. For the purposes of promoting Age UK Insurance, Age UK is an Appointed Representative of Age UK Enterprises Limited, Age UK is an Introducer Appointed Representative of JLT Benefit Solutions Limited and Simplyhealth Access for the purposes of introducing potential annuity and health cash plans customers respectively. Age UK Enterprises Limited, JLT Benefit Solutions Limited and Simplyhealth Access are all authorised and regulated by the Financial Services Authority. -- This email and any files transmitted with it are confidential and intended solely for the use of the individual or entity to whom they are addressed. If you receive a message in error, please advise the sender and delete immediately. Except where this email is sent in the usual course of our business, any opinions expressed in this email are those of the author and do not necessarily reflect the opinions of Age UK or its subsidiaries and associated companies. Age UK monitors all e-mail transmissions passing through its network and may block or modify mails which are deemed to be unsuitable. Age Concern England (charity number 261794) and Help the Aged (charity number 272786) and their trading and other associated companies merged on 1st April 2009. Together they have formed the Age UK Group, dedicated to improving the lives of people in later life. The three national Age Concerns in Scotland, Northern Ireland and Wales have also merged with Help the Aged in these nations to form three registered charities: Age Scotland, Age NI, Age Cymru. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Comma separated vector
Hi all, I have a vector of numbers, and to be able to pass it to RMySQL and use the IN clause I need to have this vector to be a single list numeric and comma separated. I saw the post below but it is about strings, which I do not need (I cannot pass strings in this SQL query, I need something like ' where ASSETT in (1,2,3,4,5)' http://stackoverflow.com/questions/6347356/creating-a-comma-separated-vector Any clue? -- View this message in context: http://r.789695.n4.nabble.com/Comma-separated-vector-tp4667340.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] Comma separated vector
is this what you want x - c(1, 2, 3, 4, 5) paste0(where ASSETT in (, paste(x, collapse = ','), )) [1] where ASSETT in (1,2,3,4,5) On Fri, May 17, 2013 at 11:46 AM, Manta mantin...@libero.it wrote: Hi all, I have a vector of numbers, and to be able to pass it to RMySQL and use the IN clause I need to have this vector to be a single list numeric and comma separated. I saw the post below but it is about strings, which I do not need (I cannot pass strings in this SQL query, I need something like ' where ASSETT in (1,2,3,4,5)' http://stackoverflow.com/questions/6347356/creating-a-comma-separated-vector Any clue? -- View this message in context: http://r.789695.n4.nabble.com/Comma-separated-vector-tp4667340.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Jim Holtman Data Munger Guru What is the problem that you are trying to solve? Tell me what you want to do, not how you want to do it. [[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] Comma separated vector
You are wrong... since the SQL query you wish to create is itself a string. Of course, you cannot send a SQL fragment such as you used as an example, so be sure to form a complete, syntactically correct SQL statement before giving it to your database query function. Oh, and if you need more assistance with this topic, then you should probably post in the R-sig-DB mailing list. Please also keep in mind that the actual SQL syntax used for your specific database is basically off-topic in any R forum, so you may need other help resources as well. --- Jeff NewmillerThe . . Go Live... DCN:jdnew...@dcn.davis.ca.usBasics: ##.#. ##.#. Live Go... Live: OO#.. Dead: OO#.. Playing Research Engineer (Solar/BatteriesO.O#. #.O#. with /Software/Embedded Controllers) .OO#. .OO#. rocks...1k --- Sent from my phone. Please excuse my brevity. Manta mantin...@libero.it wrote: Hi all, I have a vector of numbers, and to be able to pass it to RMySQL and use the IN clause I need to have this vector to be a single list numeric and comma separated. I saw the post below but it is about strings, which I do not need (I cannot pass strings in this SQL query, I need something like ' where ASSETT in (1,2,3,4,5)' http://stackoverflow.com/questions/6347356/creating-a-comma-separated-vector Any clue? -- View this message in context: http://r.789695.n4.nabble.com/Comma-separated-vector-tp4667340.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] Problems using lmer {lme4}
Dear R list,
I'm attaching a sample of my data which consists on the presence/absence
(punto6, binomial n=5 occasions)
of different species (sp), on different sites (site) within routes
('route).
First, I want to be able to find if there is autocorrelation of the
response variable between
the sites within each route. For this I start testing 2 models, but when I
try to run the second model,
to test for the random effects of sites within routes R stops working!
I'm not being able to find out why r is crashing... and whay am I doing
wrong.
Thanks!
Andrea
###
#dput(d)
d-structure(list(site = structure(c(55L, 56L, 57L, 58L, 59L, 60L,
55L, 56L, 57L, 58L, 59L, 60L, 55L, 56L, 57L, 58L, 59L, 60L, 55L,
56L, 57L, 58L, 59L, 60L, 55L, 56L, 57L, 58L, 59L, 60L, 55L, 56L,
57L, 58L, 59L, 60L, 55L, 56L, 57L, 58L, 59L, 60L, 229L, 230L,
231L, 232L, 233L, 234L, 229L, 230L, 231L, 232L, 233L, 234L, 229L,
230L, 231L, 232L, 233L, 234L, 229L, 230L, 231L, 232L, 233L, 234L,
229L, 230L, 231L, 232L, 233L, 234L, 229L, 230L, 231L, 232L, 233L,
234L, 229L, 230L, 231L, 232L, 233L, 234L, 331L, 332L, 333L, 334L,
335L, 336L, 331L, 332L, 333L, 334L, 335L, 336L, 331L, 332L, 333L,
334L, 335L, 336L, 331L, 332L, 333L, 334L, 335L, 336L, 331L, 332L,
333L, 334L, 335L, 336L, 331L, 332L, 333L, 334L, 335L, 336L, 331L,
332L, 333L, 334L, 335L, 336L, 389L, 390L, 391L, 392L, 393L, 394L,
389L, 390L, 391L, 392L, 393L, 394L, 389L, 390L, 391L, 392L, 393L,
394L, 389L, 390L, 391L, 392L, 393L, 394L, 389L, 390L, 391L, 392L,
393L, 394L, 389L, 390L, 391L, 392L, 393L, 394L, 389L, 390L, 391L,
392L, 393L, 394L, 205L, 206L, 207L, 208L, 209L, 210L, 205L, 206L,
207L, 208L, 209L, 210L, 205L, 206L, 207L, 208L, 209L, 210L, 205L,
206L, 207L, 208L, 209L, 210L, 205L, 206L, 207L, 208L, 209L, 210L,
205L, 206L, 207L, 208L, 209L, 210L, 205L, 206L, 207L, 208L, 209L,
210L, 163L, 164L, 165L, 166L, 167L, 168L, 163L, 164L, 165L, 166L,
167L, 168L, 163L, 164L, 165L, 166L, 167L, 168L, 163L, 164L, 165L,
166L, 167L, 168L, 163L, 164L, 165L, 166L, 167L, 168L, 163L, 164L,
165L, 166L, 167L, 168L, 163L, 164L, 165L, 166L, 167L, 168L, 247L,
248L, 249L, 250L, 251L, 252L, 247L, 248L, 249L, 250L, 251L, 252L,
247L, 248L, 249L, 250L, 251L, 252L, 247L, 248L, 249L, 250L, 251L,
252L, 247L, 248L, 249L, 250L, 251L, 252L, 247L, 248L, 249L, 250L,
251L, 252L, 247L, 248L, 249L, 250L, 251L, 252L), .Label = c(102-1,
102-2, 102-3, 102-4, 102-5, 102-6, 1023-1, 1023-2,
1023-3, 1023-4, 1023-5, 1023-6, 1027-1, 1027-2, 1027-3,
1027-4, 1027-5, 1027-6, 1028-1, 1028-2, 1028-3, 1028-4,
1028-5, 1028-6, 1032-1, 1032-2, 1032-3, 1032-4, 1032-5,
1032-6, 1034-1, 1034-2, 1034-3, 1034-4, 1034-5, 1034-6,
1036-1, 1036-2, 1036-3, 1036-4, 1036-5, 1036-6, 1041-1,
1041-2, 1041-3, 1041-4, 1041-5, 1041-6, 1046-1, 1046-2,
1046-3, 1046-4, 1046-5, 1046-6, 105-1, 105-2, 105-3,
105-4, 105-5, 105-6, 107-1, 107-2, 107-3, 107-4,
107-5, 107-6, 108-1, 108-2, 108-3, 108-4, 108-5,
108-6, 1101-1, 1101-2, 1101-3, 1101-4, 1101-5, 1101-6,
1104-1, 1104-2, 1104-3, 1104-4, 1104-5, 1104-6, 1108-1,
1108-2, 1108-3, 1108-4, 1108-5, 1108-6, 111-1, 111-2,
111-3, 111-4, 111-5, 111-6, 1113-1, 1113-2, 1113-3,
1113-4, 1113-5, 1113-6, 1116-1, 1116-2, 1116-3, 1116-4,
1116-5, 1116-6, 1121-1, 1121-2, 1121-3, 1121-4, 1121-5,
1121-6, 1204-1, 1204-2, 1204-3, 1204-4, 1204-5, 1204-6,
1205-1, 1205-2, 1205-3, 1205-4, 1205-5, 1205-6, 1207-1,
1207-2, 1207-3, 1207-4, 1207-5, 1207-6, 1212-1, 1212-2,
1212-3, 1212-4, 1212-5, 1212-6, 202-1, 202-2, 202-3,
202-4, 202-5, 202-6, 205-1, 205-2, 205-3, 205-4,
205-5, 205-6, 207-1, 207-2, 207-3, 207-4, 207-5,
207-6, 208-1, 208-2, 208-3, 208-4, 208-5, 208-6,
211-1, 211-2, 211-3, 211-4, 211-5, 211-6, 213-1,
213-2, 213-3, 213-4, 213-5, 213-6, 214-1, 214-2,
214-3, 214-4, 214-5, 214-6, 217-1, 217-2, 217-3,
217-4, 217-5, 217-6, 218-1, 218-2, 218-3, 218-4,
218-5, 218-6, 219-1, 219-2, 219-3, 219-4, 219-5,
219-6, 223-1, 223-2, 223-3, 223-4, 223-5, 223-6,
302-1, 302-2, 302-3, 302-4, 302-5, 302-6, 305-1,
305-2, 305-3, 305-4, 305-5, 305-6, 308-1, 308-2,
308-3, 308-4, 308-5, 308-6, 311-1, 311-2, 311-3,
311-4, 311-5, 311-6, 401-1, 401-2, 401-3, 401-4,
401-5, 401-6, 402-1, 402-2, 402-3, 402-4, 402-5,
402-6, 405-1, 405-2, 405-3, 405-4, 405-5, 405-6,
407-1, 407-2, 407-3, 407-4, 407-5, 407-6, 408-1,
408-2, 408-3, 408-4, 408-5, 408-6, 410-1, 410-2,
410-3, 410-4, 410-5, 410-6, 411-1, 411-2, 411-3,
411-4, 411-5, 411-6, 414-1, 414-2, 414-3, 414-4,
414-5, 414-6, 416-1, 416-2, 416-3, 416-4, 416-5,
416-6, 417-1, 417-2, 417-3, 417-4, 417-5, 417-6,
420-1, 420-2, 420-3, 420-4, 420-5, 420-6, 423-1,
423-2, 423-3, 423-4, 423-5, 423-6, 501-1, 501-2,
501-3, 501-4, 501-5, 501-6, 502-1, 502-2, 502-3,
502-4, 502-5, 502-6, 504-1, 504-2, 504-3, 504-4,
504-5, 504-6, 505-1, 505-2, 505-3, 505-4, 505-5,
505-6, 507-1, 507-2, 507-3, 507-4, 507-5, 507-6,
508-1, 508-2, 508-3, 508-4, 508-5, 508-6, 511-1,
511-2, 511-3, 511-4, 511-5, 511-6, 602-1, 602-2,
602-3, 602-4, 602-5, 602-6, 604-1, 604-2, 604-3,
604-4, 604-5, 604-6, 605-1,
[R] inverse for formula transformations on LHS
This is an R formula handling question. It arose in class. We were working on the Animals data in the MASS package. In order to see a relationship, you need to log brain and body weight. It's a fun one for teaching regression, if you did not try it yet. There are outliers too! Students wanted to make a predicted value plot in the non-logged values of y, for comparison, and I wondered if I couldn't automate this somehow for them. It made me wonder how R manages formulae and if a transformation like log(y) can be be mechanically inverted. So we have something concrete to talk about, suppose x and y are variables in dat, a person fits m1 - lm(log(y) ~ log(x), data = dat) termplot shows log(y) on the vertical. What if I want y on the vertical? Similarly, predict gives values on the log(y) scale, there's no argument like type = untransformed. I want my solution to be a bit general, so that it would give back predicted y for formulae like sqrt(y) or exp(y) or log(y + d) or whatever other math people might throw in there. Here's what I can tell so far about R's insides. The formula handler makes a list out of the formula, I can get that from the terms object that the model generates. The formula list has ~ as element 1, and log(x) becomes element [[2]]. Where in the R source code can I see how R looks at the symbol log(y) and discerns that there is a variable y that needs to be logged? If I could understand that, and if R has a table of inverse functions, then maybe I could see what to do. If you have ideas, I'm very grateful if you share them. pj -- Paul E. Johnson Professor, Political Science Assoc. Director 1541 Lilac Lane, Room 504 Center for Research Methods University of Kansas University of Kansas http://pj.freefaculty.org http://quant.ku.edu [[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] Problems using lmer {lme4}
It would be better to post this on the r-sig-mixed-models list, I think.
-- Bert
On Fri, May 17, 2013 at 10:02 AM, Andrea Goijman
agoij...@cnia.inta.gov.ar wrote:
Dear R list,
I'm attaching a sample of my data which consists on the presence/absence
(punto6, binomial n=5 occasions)
of different species (sp), on different sites (site) within routes
('route).
First, I want to be able to find if there is autocorrelation of the
response variable between
the sites within each route. For this I start testing 2 models, but when I
try to run the second model,
to test for the random effects of sites within routes R stops working!
I'm not being able to find out why r is crashing... and whay am I doing
wrong.
Thanks!
Andrea
###
#dput(d)
d-structure(list(site = structure(c(55L, 56L, 57L, 58L, 59L, 60L,
55L, 56L, 57L, 58L, 59L, 60L, 55L, 56L, 57L, 58L, 59L, 60L, 55L,
56L, 57L, 58L, 59L, 60L, 55L, 56L, 57L, 58L, 59L, 60L, 55L, 56L,
57L, 58L, 59L, 60L, 55L, 56L, 57L, 58L, 59L, 60L, 229L, 230L,
231L, 232L, 233L, 234L, 229L, 230L, 231L, 232L, 233L, 234L, 229L,
230L, 231L, 232L, 233L, 234L, 229L, 230L, 231L, 232L, 233L, 234L,
229L, 230L, 231L, 232L, 233L, 234L, 229L, 230L, 231L, 232L, 233L,
234L, 229L, 230L, 231L, 232L, 233L, 234L, 331L, 332L, 333L, 334L,
335L, 336L, 331L, 332L, 333L, 334L, 335L, 336L, 331L, 332L, 333L,
334L, 335L, 336L, 331L, 332L, 333L, 334L, 335L, 336L, 331L, 332L,
333L, 334L, 335L, 336L, 331L, 332L, 333L, 334L, 335L, 336L, 331L,
332L, 333L, 334L, 335L, 336L, 389L, 390L, 391L, 392L, 393L, 394L,
389L, 390L, 391L, 392L, 393L, 394L, 389L, 390L, 391L, 392L, 393L,
394L, 389L, 390L, 391L, 392L, 393L, 394L, 389L, 390L, 391L, 392L,
393L, 394L, 389L, 390L, 391L, 392L, 393L, 394L, 389L, 390L, 391L,
392L, 393L, 394L, 205L, 206L, 207L, 208L, 209L, 210L, 205L, 206L,
207L, 208L, 209L, 210L, 205L, 206L, 207L, 208L, 209L, 210L, 205L,
206L, 207L, 208L, 209L, 210L, 205L, 206L, 207L, 208L, 209L, 210L,
205L, 206L, 207L, 208L, 209L, 210L, 205L, 206L, 207L, 208L, 209L,
210L, 163L, 164L, 165L, 166L, 167L, 168L, 163L, 164L, 165L, 166L,
167L, 168L, 163L, 164L, 165L, 166L, 167L, 168L, 163L, 164L, 165L,
166L, 167L, 168L, 163L, 164L, 165L, 166L, 167L, 168L, 163L, 164L,
165L, 166L, 167L, 168L, 163L, 164L, 165L, 166L, 167L, 168L, 247L,
248L, 249L, 250L, 251L, 252L, 247L, 248L, 249L, 250L, 251L, 252L,
247L, 248L, 249L, 250L, 251L, 252L, 247L, 248L, 249L, 250L, 251L,
252L, 247L, 248L, 249L, 250L, 251L, 252L, 247L, 248L, 249L, 250L,
251L, 252L, 247L, 248L, 249L, 250L, 251L, 252L), .Label = c(102-1,
102-2, 102-3, 102-4, 102-5, 102-6, 1023-1, 1023-2,
1023-3, 1023-4, 1023-5, 1023-6, 1027-1, 1027-2, 1027-3,
1027-4, 1027-5, 1027-6, 1028-1, 1028-2, 1028-3, 1028-4,
1028-5, 1028-6, 1032-1, 1032-2, 1032-3, 1032-4, 1032-5,
1032-6, 1034-1, 1034-2, 1034-3, 1034-4, 1034-5, 1034-6,
1036-1, 1036-2, 1036-3, 1036-4, 1036-5, 1036-6, 1041-1,
1041-2, 1041-3, 1041-4, 1041-5, 1041-6, 1046-1, 1046-2,
1046-3, 1046-4, 1046-5, 1046-6, 105-1, 105-2, 105-3,
105-4, 105-5, 105-6, 107-1, 107-2, 107-3, 107-4,
107-5, 107-6, 108-1, 108-2, 108-3, 108-4, 108-5,
108-6, 1101-1, 1101-2, 1101-3, 1101-4, 1101-5, 1101-6,
1104-1, 1104-2, 1104-3, 1104-4, 1104-5, 1104-6, 1108-1,
1108-2, 1108-3, 1108-4, 1108-5, 1108-6, 111-1, 111-2,
111-3, 111-4, 111-5, 111-6, 1113-1, 1113-2, 1113-3,
1113-4, 1113-5, 1113-6, 1116-1, 1116-2, 1116-3, 1116-4,
1116-5, 1116-6, 1121-1, 1121-2, 1121-3, 1121-4, 1121-5,
1121-6, 1204-1, 1204-2, 1204-3, 1204-4, 1204-5, 1204-6,
1205-1, 1205-2, 1205-3, 1205-4, 1205-5, 1205-6, 1207-1,
1207-2, 1207-3, 1207-4, 1207-5, 1207-6, 1212-1, 1212-2,
1212-3, 1212-4, 1212-5, 1212-6, 202-1, 202-2, 202-3,
202-4, 202-5, 202-6, 205-1, 205-2, 205-3, 205-4,
205-5, 205-6, 207-1, 207-2, 207-3, 207-4, 207-5,
207-6, 208-1, 208-2, 208-3, 208-4, 208-5, 208-6,
211-1, 211-2, 211-3, 211-4, 211-5, 211-6, 213-1,
213-2, 213-3, 213-4, 213-5, 213-6, 214-1, 214-2,
214-3, 214-4, 214-5, 214-6, 217-1, 217-2, 217-3,
217-4, 217-5, 217-6, 218-1, 218-2, 218-3, 218-4,
218-5, 218-6, 219-1, 219-2, 219-3, 219-4, 219-5,
219-6, 223-1, 223-2, 223-3, 223-4, 223-5, 223-6,
302-1, 302-2, 302-3, 302-4, 302-5, 302-6, 305-1,
305-2, 305-3, 305-4, 305-5, 305-6, 308-1, 308-2,
308-3, 308-4, 308-5, 308-6, 311-1, 311-2, 311-3,
311-4, 311-5, 311-6, 401-1, 401-2, 401-3, 401-4,
401-5, 401-6, 402-1, 402-2, 402-3, 402-4, 402-5,
402-6, 405-1, 405-2, 405-3, 405-4, 405-5, 405-6,
407-1, 407-2, 407-3, 407-4, 407-5, 407-6, 408-1,
408-2, 408-3, 408-4, 408-5, 408-6, 410-1, 410-2,
410-3, 410-4, 410-5, 410-6, 411-1, 411-2, 411-3,
411-4, 411-5, 411-6, 414-1, 414-2, 414-3, 414-4,
414-5, 414-6, 416-1, 416-2, 416-3, 416-4, 416-5,
416-6, 417-1, 417-2, 417-3, 417-4, 417-5, 417-6,
420-1, 420-2, 420-3, 420-4, 420-5, 420-6, 423-1,
423-2, 423-3, 423-4, 423-5, 423-6, 501-1, 501-2,
501-3, 501-4, 501-5, 501-6, 502-1, 502-2, 502-3,
502-4, 502-5, 502-6, 504-1, 504-2, 504-3, 504-4,
504-5, 504-6, 505-1,
Re: [R] Problems using lmer {lme4}
Thanks! I didn't know about that list. I forwarded my question there
Andrea
Bert Gunter wrote
It would be better to post this on the r-sig-mixed-models list, I think.
-- Bert
On Fri, May 17, 2013 at 10:02 AM, Andrea Goijman
lt;
agoijman@.gov
gt; wrote:
Dear R list,
I'm attaching a sample of my data which consists on the presence/absence
(punto6, binomial n=5 occasions)
of different species (sp), on different sites (site) within routes
('route).
First, I want to be able to find if there is autocorrelation of the
response variable between
the sites within each route. For this I start testing 2 models, but when
I
try to run the second model,
to test for the random effects of sites within routes R stops working!
I'm not being able to find out why r is crashing... and whay am I doing
wrong.
Thanks!
Andrea
###
#dput(d)
d-structure(list(site = structure(c(55L, 56L, 57L, 58L, 59L, 60L,
55L, 56L, 57L, 58L, 59L, 60L, 55L, 56L, 57L, 58L, 59L, 60L, 55L,
56L, 57L, 58L, 59L, 60L, 55L, 56L, 57L, 58L, 59L, 60L, 55L, 56L,
57L, 58L, 59L, 60L, 55L, 56L, 57L, 58L, 59L, 60L, 229L, 230L,
231L, 232L, 233L, 234L, 229L, 230L, 231L, 232L, 233L, 234L, 229L,
230L, 231L, 232L, 233L, 234L, 229L, 230L, 231L, 232L, 233L, 234L,
229L, 230L, 231L, 232L, 233L, 234L, 229L, 230L, 231L, 232L, 233L,
234L, 229L, 230L, 231L, 232L, 233L, 234L, 331L, 332L, 333L, 334L,
335L, 336L, 331L, 332L, 333L, 334L, 335L, 336L, 331L, 332L, 333L,
334L, 335L, 336L, 331L, 332L, 333L, 334L, 335L, 336L, 331L, 332L,
333L, 334L, 335L, 336L, 331L, 332L, 333L, 334L, 335L, 336L, 331L,
332L, 333L, 334L, 335L, 336L, 389L, 390L, 391L, 392L, 393L, 394L,
389L, 390L, 391L, 392L, 393L, 394L, 389L, 390L, 391L, 392L, 393L,
394L, 389L, 390L, 391L, 392L, 393L, 394L, 389L, 390L, 391L, 392L,
393L, 394L, 389L, 390L, 391L, 392L, 393L, 394L, 389L, 390L, 391L,
392L, 393L, 394L, 205L, 206L, 207L, 208L, 209L, 210L, 205L, 206L,
207L, 208L, 209L, 210L, 205L, 206L, 207L, 208L, 209L, 210L, 205L,
206L, 207L, 208L, 209L, 210L, 205L, 206L, 207L, 208L, 209L, 210L,
205L, 206L, 207L, 208L, 209L, 210L, 205L, 206L, 207L, 208L, 209L,
210L, 163L, 164L, 165L, 166L, 167L, 168L, 163L, 164L, 165L, 166L,
167L, 168L, 163L, 164L, 165L, 166L, 167L, 168L, 163L, 164L, 165L,
166L, 167L, 168L, 163L, 164L, 165L, 166L, 167L, 168L, 163L, 164L,
165L, 166L, 167L, 168L, 163L, 164L, 165L, 166L, 167L, 168L, 247L,
248L, 249L, 250L, 251L, 252L, 247L, 248L, 249L, 250L, 251L, 252L,
247L, 248L, 249L, 250L, 251L, 252L, 247L, 248L, 249L, 250L, 251L,
252L, 247L, 248L, 249L, 250L, 251L, 252L, 247L, 248L, 249L, 250L,
251L, 252L, 247L, 248L, 249L, 250L, 251L, 252L), .Label = c(102-1,
102-2, 102-3, 102-4, 102-5, 102-6, 1023-1, 1023-2,
1023-3, 1023-4, 1023-5, 1023-6, 1027-1, 1027-2, 1027-3,
1027-4, 1027-5, 1027-6, 1028-1, 1028-2, 1028-3, 1028-4,
1028-5, 1028-6, 1032-1, 1032-2, 1032-3, 1032-4, 1032-5,
1032-6, 1034-1, 1034-2, 1034-3, 1034-4, 1034-5, 1034-6,
1036-1, 1036-2, 1036-3, 1036-4, 1036-5, 1036-6, 1041-1,
1041-2, 1041-3, 1041-4, 1041-5, 1041-6, 1046-1, 1046-2,
1046-3, 1046-4, 1046-5, 1046-6, 105-1, 105-2, 105-3,
105-4, 105-5, 105-6, 107-1, 107-2, 107-3, 107-4,
107-5, 107-6, 108-1, 108-2, 108-3, 108-4, 108-5,
108-6, 1101-1, 1101-2, 1101-3, 1101-4, 1101-5, 1101-6,
1104-1, 1104-2, 1104-3, 1104-4, 1104-5, 1104-6, 1108-1,
1108-2, 1108-3, 1108-4, 1108-5, 1108-6, 111-1, 111-2,
111-3, 111-4, 111-5, 111-6, 1113-1, 1113-2, 1113-3,
1113-4, 1113-5, 1113-6, 1116-1, 1116-2, 1116-3, 1116-4,
1116-5, 1116-6, 1121-1, 1121-2, 1121-3, 1121-4, 1121-5,
1121-6, 1204-1, 1204-2, 1204-3, 1204-4, 1204-5, 1204-6,
1205-1, 1205-2, 1205-3, 1205-4, 1205-5, 1205-6, 1207-1,
1207-2, 1207-3, 1207-4, 1207-5, 1207-6, 1212-1, 1212-2,
1212-3, 1212-4, 1212-5, 1212-6, 202-1, 202-2, 202-3,
202-4, 202-5, 202-6, 205-1, 205-2, 205-3, 205-4,
205-5, 205-6, 207-1, 207-2, 207-3, 207-4, 207-5,
207-6, 208-1, 208-2, 208-3, 208-4, 208-5, 208-6,
211-1, 211-2, 211-3, 211-4, 211-5, 211-6, 213-1,
213-2, 213-3, 213-4, 213-5, 213-6, 214-1, 214-2,
214-3, 214-4, 214-5, 214-6, 217-1, 217-2, 217-3,
217-4, 217-5, 217-6, 218-1, 218-2, 218-3, 218-4,
218-5, 218-6, 219-1, 219-2, 219-3, 219-4, 219-5,
219-6, 223-1, 223-2, 223-3, 223-4, 223-5, 223-6,
302-1, 302-2, 302-3, 302-4, 302-5, 302-6, 305-1,
305-2, 305-3, 305-4, 305-5, 305-6, 308-1, 308-2,
308-3, 308-4, 308-5, 308-6, 311-1, 311-2, 311-3,
311-4, 311-5, 311-6, 401-1, 401-2, 401-3, 401-4,
401-5, 401-6, 402-1, 402-2, 402-3, 402-4, 402-5,
402-6, 405-1, 405-2, 405-3, 405-4, 405-5, 405-6,
407-1, 407-2, 407-3, 407-4, 407-5, 407-6, 408-1,
408-2, 408-3, 408-4, 408-5, 408-6, 410-1, 410-2,
410-3, 410-4, 410-5, 410-6, 411-1, 411-2, 411-3,
411-4, 411-5, 411-6, 414-1, 414-2, 414-3, 414-4,
414-5, 414-6, 416-1, 416-2, 416-3, 416-4, 416-5,
416-6, 417-1, 417-2, 417-3, 417-4, 417-5, 417-6,
420-1, 420-2, 420-3, 420-4, 420-5, 420-6, 423-1,
423-2, 423-3, 423-4, 423-5, 423-6, 501-1, 501-2,
501-3, 501-4, 501-5,
Re: [R] update an array of plots in 'real-time' without drawing lags
On May 17, 2013, at 7:04 AM, Martin Batholdy wrote: Hi, I know R is not made for this, but I still wanted to ask if there are possibilities to do this; I repeatedly collect data from a database for a given time interval. Now I would like to monitor the change of this data with some nice plots. I actually have to draw 15 plots to get the whole picture. Now I can write a loop that, for a given time interval, updates the data and plots them using one quartz window that is split into subparts with the layout function. However, since I redraw every plot every time, the drawing takes some time and it is not an instant update that I see. It is not clear whether this is an asynchronous process that uses new data to draw some of the plots but not others. The plots are slowly redrawn from the upper left corner of the quartz window to the bottom right. That makes me wonder; Is there a way to buffer the graphical device before updating it? Or are there any other solutions for R that enables to smoothly plot / visualize data in 'real-time'? You can split 'screens' on a graphics device to which you can send plotting output. You already know that you can construct 'layout's. You can have multiple interactive output devices. The split-screen method is incompatible with layout. ?Devices ?dev.cur ?screen -- David Winsemius Alameda, CA, USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Error with adehabitatHR and kernelbb
On May 17, 2013, at 7:44 AM, Rémi Lesmerises wrote:
Dear all,
I'm trying to get a Brownian bridge kernel (kernelbb) for each combination of
two consecutive animal locations (see commands below) and put them, with a
loop, inside a list. It works well at the beginning but after 42 runs, it
appears the following warning :
Error in seq.default(yli[1], yli[2], by = diff(xg[1:2])) :
invalid (to - from)/by in seq(.)
I looked at the coordinates, at the id, at the time of the run 43 and it's
all good...
I looked on the net and it happened to only one person and there was no
answer to his post.
I wonder if that posting (like yours) had no data on which to display the
problem or to test potential solutions? I would think you would want to post
any setup objects and then data for items 40-45.
Someone could help me?
## commands
BBtraj - list()
for (i in 1:(nrow(loc@data)-1)) {
BBtraj[[i]] - kernelbb(as.ltraj(loc@coords[i:(i+1),],
date=loc@data$time[i:(i+1)], id = as.character(loc@data$id[i:(i+1)]),
typeII = TRUE), sig1=as.numeric(as.character(loc@data$sig1[i])), sig2= 5,
grid = 1000)
}
--
David Winsemius
Alameda, CA, USA
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Re: [R] pearson correlation significant level
On 2013-05-17 08:37, Jose Iparraguirre wrote: Dear Elaine, One of the elements you obtain the P matrix, which is the matrix of asymptotic p-values. In your case, you get that the asymptotic p-value of the association between t_i and t_r is 0. That is, there would exist a perfect association (look also at the first result, the matrix of correlations: the correlation coefficient between these variables is 1). Hope this helps, José Actually, the coefficient is -0.89. Not surprisingly, with n = 4873, this indicates a significant correlation. For details of the t-test involved in the calculation, see any intro stats text or look at the code of cor.test.default(). Peter Ehlers Prof. José Iparraguirre Chief Economist Age UK -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Elaine Kuo Sent: 17 May 2013 10:40 To: r-help@r-project.org Subject: [R] pearson correlation significant level Hello I am using package Hmisc to calculate the pearson correlation and the significant level for the matrix of t_i and t_r. (temperature minimum and temperature range) However, I have difficulty interpreting the result, even after checking the manual. Please kindly help to indicate if the p-value is zero. Thank you in advance. Elaine The code library(Hmisc) rcorr(as.matrix(datat), type=pearson) # type can be pearson or spearman The result is t_i t_r t_i 1.00 -0.89 t_r -0.89 1.00 n= 4873 P t_i t_r t_i 0 t_r 0 [[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. Please donate to the Syria Crisis Appeal by text or online: To donate £5 by mobile, text SYRIA to 70800. To donate online, please visit http://www.ageinternational.org.uk/syria Over one million refugees are desperately in need of water, food, healthcare, warm clothing, blankets and shelter; Age International urgently needs your support to help affected older refugees. Age International is a subsidiary charity of Age UK and a member of the Disasters Emergency Committee (DEC). The DEC launches and co-ordinates national fundraising appeals for public donations on behalf of its member agencies. Texts cost £5 plus one standard rate message. Age International will receive a minimum of £4.96. More info at ageinternational.org.uk/SyriaTerms --- Age UK is a registered charity and company limited by guarantee, (registered charity number 1128267, registered company number 6825798). Registered office: Tavis House, 1-6 Tavistock Square, London WC1H 9NA. For the purposes of promoting Age UK Insurance, Age UK is an Appointed Representative of Age UK Enterprises Limited, Age UK is an Introducer Appointed Representative of JLT Benefit Solutions Limited and Simplyhealth Access for the purposes of introducing potential annuity and health cash plans customers respectively. Age UK Enterprises Limited, JLT Benefit Solutions Limited and Simplyhealth Access are all authorised and regulated by the Financial Services Authority. -- This email and any files transmitted with it are confidential and intended solely for the use of the individual or entity to whom they are addressed. If you receive a message in error, please advise the sender and delete immediately. Except where this email is sent in the usual course of our business, any opinions expressed in this email are those of the author and do not necessarily reflect the opinions of Age UK or its subsidiaries and associated companies. Age UK monitors all e-mail transmissions passing through its network and may block or modify mails which are deemed to be unsuitable. Age Concern England (charity number 261794) and Help the Aged (charity number 272786) and their trading and other associated companies merged on 1st April 2009. Together they have formed the Age UK Group, dedicated to improving the lives of people in later life. The three national Age Concerns in Scotland, Northern Ireland and Wales have also merged with Help the Aged in these nations to form three registered charities: Age Scotland, Age NI, Age Cymru. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Error with adehabitatHR and kernelbb
This is the problematic data, especially the line 43, but when I removed that
line, it bugs at line 62 and so on.
sig1 time id UTMnorthin
UTMeasting
40 4.5766 2012.05.30 08:00:00 Ade-55419576 390052
41 4.5766 2012.05.30 10:00:00 Ade-55419581 390058
42 4.5766 2012.05.30 12:00:00 Ade-55419560 390045
43 4.5766 2012.05.30 14:00:00 Ade-55419574 390051
44 4.5766 2012.05.30 16:00:00 Ade-55419490 390051
45 4.5766 2012.05.30 18:00:00 Ade-55419435 390293
46 4.5766 2012.05.30 20:00:00 Ade-55419661 390876
47 4.5766 2012.05.30 22:00:00 Ade-55419934 390673
48 4.5766 2012.05.31 02:00:00 Ade-55420636 389777
49 4.5766 2012.06.05 02:00:00 Ade-75419275 391206
50 4.5766 2012.06.05 04:00:00 Ade-75419276 391202
This is a data frame so before the loop (see commands in the previous mail
below), I transform with the following commands:
coordinates(fece)-~UTMeasting+UTMnorthin
fece@data$time - as.POSIXct(strptime(as.character(fece@data$time),%Y.%m.%d
%H:%M:%S))
Rémi Lesmerises, biol. M.Sc.,
PH.D candidate
Université du Québec à Rimouski
De : David Winsemius dwinsem...@comcast.net
À : Rémi Lesmerises remilesmeri...@yahoo.ca
Cc : r-help@r-project.org r-help@r-project.org
Envoyé le : vendredi 17 mai 2013 13h53
Objet : Re: [R] Error with adehabitatHR and kernelbb
On May 17, 2013, at 7:44 AM, Rémi Lesmerises wrote:
Dear all,
I'm trying to get a Brownian bridge kernel (kernelbb) for each combination of
two consecutive animal locations (see commands below) and put them, with a
loop, inside a list. It works well at the beginning but after 42 runs, it
appears the following warning :
Error in seq.default(yli[1], yli[2], by = diff(xg[1:2])) :
invalid (to - from)/by in seq(.)
I looked at the coordinates, at the id, at the time of the run 43 and it's
all good...
I looked on the net and it happened to only one person and there was no
answer to his post.
I wonder if that posting (like yours) had no data on which to display the
problem or to test potential solutions? I would think you would want to post
any setup objects and then data for items 40-45.
Someone could help me?
## commands
BBtraj - list()
for (i in 1:(nrow(loc@data)-1)) {
BBtraj[[i]] - kernelbb(as.ltraj(loc@coords[i:(i+1),],
date=loc@data$time[i:(i+1)], id = as.character(loc@data$id[i:(i+1)]),
typeII = TRUE), sig1=as.numeric(as.character(loc@data$sig1[i])), sig2= 5,
grid = 1000)
}
--
David Winsemius
Alameda, CA, USA
[[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] Problems using lmer {lme4}
Hi Andrea,
I'm not exactly sure what you're trying to do, but you've included a
random effect for a site coefficient that's not even in your list of
fixed effects... you're basically allowing the coefficient for site to
vary across routes, but you're never including the coefficient in the
first place. Given the appropriate nested structure (I haven't looked
at your data), a model that should run would be:
m2 - lmer(cbind(punto6,5) ~ sp + site + (site|route) ,family=binomial,data = d)
The problem here is I'm not sure this is what you mean to do... If you
could way simplify your example (for example, by including a small
number of mock observations with only the relevant variables, in a
table instead of in R syntax), that would be helpful.
Patrick
2013/5/17 Andrea Goijman agoij...@cnia.inta.gov.ar:
Dear R list,
I'm attaching a sample of my data which consists on the presence/absence
(punto6, binomial n=5 occasions)
of different species (sp), on different sites (site) within routes
('route).
First, I want to be able to find if there is autocorrelation of the
response variable between
the sites within each route. For this I start testing 2 models, but when I
try to run the second model,
to test for the random effects of sites within routes R stops working!
I'm not being able to find out why r is crashing... and whay am I doing
wrong.
Thanks!
Andrea
###
#dput(d)
d-structure(list(site = structure(c(55L, 56L, 57L, 58L, 59L, 60L,
55L, 56L, 57L, 58L, 59L, 60L, 55L, 56L, 57L, 58L, 59L, 60L, 55L,
56L, 57L, 58L, 59L, 60L, 55L, 56L, 57L, 58L, 59L, 60L, 55L, 56L,
57L, 58L, 59L, 60L, 55L, 56L, 57L, 58L, 59L, 60L, 229L, 230L,
231L, 232L, 233L, 234L, 229L, 230L, 231L, 232L, 233L, 234L, 229L,
230L, 231L, 232L, 233L, 234L, 229L, 230L, 231L, 232L, 233L, 234L,
229L, 230L, 231L, 232L, 233L, 234L, 229L, 230L, 231L, 232L, 233L,
234L, 229L, 230L, 231L, 232L, 233L, 234L, 331L, 332L, 333L, 334L,
335L, 336L, 331L, 332L, 333L, 334L, 335L, 336L, 331L, 332L, 333L,
334L, 335L, 336L, 331L, 332L, 333L, 334L, 335L, 336L, 331L, 332L,
333L, 334L, 335L, 336L, 331L, 332L, 333L, 334L, 335L, 336L, 331L,
332L, 333L, 334L, 335L, 336L, 389L, 390L, 391L, 392L, 393L, 394L,
389L, 390L, 391L, 392L, 393L, 394L, 389L, 390L, 391L, 392L, 393L,
394L, 389L, 390L, 391L, 392L, 393L, 394L, 389L, 390L, 391L, 392L,
393L, 394L, 389L, 390L, 391L, 392L, 393L, 394L, 389L, 390L, 391L,
392L, 393L, 394L, 205L, 206L, 207L, 208L, 209L, 210L, 205L, 206L,
207L, 208L, 209L, 210L, 205L, 206L, 207L, 208L, 209L, 210L, 205L,
206L, 207L, 208L, 209L, 210L, 205L, 206L, 207L, 208L, 209L, 210L,
205L, 206L, 207L, 208L, 209L, 210L, 205L, 206L, 207L, 208L, 209L,
210L, 163L, 164L, 165L, 166L, 167L, 168L, 163L, 164L, 165L, 166L,
167L, 168L, 163L, 164L, 165L, 166L, 167L, 168L, 163L, 164L, 165L,
166L, 167L, 168L, 163L, 164L, 165L, 166L, 167L, 168L, 163L, 164L,
165L, 166L, 167L, 168L, 163L, 164L, 165L, 166L, 167L, 168L, 247L,
248L, 249L, 250L, 251L, 252L, 247L, 248L, 249L, 250L, 251L, 252L,
247L, 248L, 249L, 250L, 251L, 252L, 247L, 248L, 249L, 250L, 251L,
252L, 247L, 248L, 249L, 250L, 251L, 252L, 247L, 248L, 249L, 250L,
251L, 252L, 247L, 248L, 249L, 250L, 251L, 252L), .Label = c(102-1,
102-2, 102-3, 102-4, 102-5, 102-6, 1023-1, 1023-2,
1023-3, 1023-4, 1023-5, 1023-6, 1027-1, 1027-2, 1027-3,
1027-4, 1027-5, 1027-6, 1028-1, 1028-2, 1028-3, 1028-4,
1028-5, 1028-6, 1032-1, 1032-2, 1032-3, 1032-4, 1032-5,
1032-6, 1034-1, 1034-2, 1034-3, 1034-4, 1034-5, 1034-6,
1036-1, 1036-2, 1036-3, 1036-4, 1036-5, 1036-6, 1041-1,
1041-2, 1041-3, 1041-4, 1041-5, 1041-6, 1046-1, 1046-2,
1046-3, 1046-4, 1046-5, 1046-6, 105-1, 105-2, 105-3,
105-4, 105-5, 105-6, 107-1, 107-2, 107-3, 107-4,
107-5, 107-6, 108-1, 108-2, 108-3, 108-4, 108-5,
108-6, 1101-1, 1101-2, 1101-3, 1101-4, 1101-5, 1101-6,
1104-1, 1104-2, 1104-3, 1104-4, 1104-5, 1104-6, 1108-1,
1108-2, 1108-3, 1108-4, 1108-5, 1108-6, 111-1, 111-2,
111-3, 111-4, 111-5, 111-6, 1113-1, 1113-2, 1113-3,
1113-4, 1113-5, 1113-6, 1116-1, 1116-2, 1116-3, 1116-4,
1116-5, 1116-6, 1121-1, 1121-2, 1121-3, 1121-4, 1121-5,
1121-6, 1204-1, 1204-2, 1204-3, 1204-4, 1204-5, 1204-6,
1205-1, 1205-2, 1205-3, 1205-4, 1205-5, 1205-6, 1207-1,
1207-2, 1207-3, 1207-4, 1207-5, 1207-6, 1212-1, 1212-2,
1212-3, 1212-4, 1212-5, 1212-6, 202-1, 202-2, 202-3,
202-4, 202-5, 202-6, 205-1, 205-2, 205-3, 205-4,
205-5, 205-6, 207-1, 207-2, 207-3, 207-4, 207-5,
207-6, 208-1, 208-2, 208-3, 208-4, 208-5, 208-6,
211-1, 211-2, 211-3, 211-4, 211-5, 211-6, 213-1,
213-2, 213-3, 213-4, 213-5, 213-6, 214-1, 214-2,
214-3, 214-4, 214-5, 214-6, 217-1, 217-2, 217-3,
217-4, 217-5, 217-6, 218-1, 218-2, 218-3, 218-4,
218-5, 218-6, 219-1, 219-2, 219-3, 219-4, 219-5,
219-6, 223-1, 223-2, 223-3, 223-4, 223-5, 223-6,
302-1, 302-2, 302-3, 302-4, 302-5, 302-6, 305-1,
305-2, 305-3, 305-4, 305-5, 305-6, 308-1, 308-2,
308-3, 308-4, 308-5, 308-6, 311-1, 311-2, 311-3,
311-4, 311-5, 311-6,
Re: [R] pearson correlation significant level
Just stating the obvious that Peter left unsaid: The OP calculated a matrix, whose diagonal is the correlations between each variable and itself, with the off-diagonal entries being the ones of interest. On 17-May-13, at 11:06 AM, Peter Ehlers wrote: On 2013-05-17 08:37, Jose Iparraguirre wrote: Dear Elaine, One of the elements you obtain the P matrix, which is the matrix of asymptotic p-values. In your case, you get that the asymptotic p-value of the association between t_i and t_r is 0. That is, there would exist a perfect association (look also at the first result, the matrix of correlations: the correlation coefficient between these variables is 1). Hope this helps, José Actually, the coefficient is -0.89. Not surprisingly, with n = 4873, this indicates a significant correlation. For details of the t-test involved in the calculation, see any intro stats text or look at the code of cor.test.default(). Peter Ehlers Prof. José Iparraguirre Chief Economist Age UK -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- project.org] On Behalf Of Elaine Kuo Sent: 17 May 2013 10:40 To: r-help@r-project.org Subject: [R] pearson correlation significant level Hello I am using package Hmisc to calculate the pearson correlation and the significant level for the matrix of t_i and t_r. (temperature minimum and temperature range) However, I have difficulty interpreting the result, even after checking the manual. Please kindly help to indicate if the p-value is zero. Thank you in advance. Elaine The code library(Hmisc) rcorr(as.matrix(datat), type=pearson) # type can be pearson or spearman The result is t_i t_r t_i 1.00 -0.89 t_r -0.89 1.00 n= 4873 P t_i t_r t_i 0 t_r 0 [[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. Please donate to the Syria Crisis Appeal by text or online: To donate £5 by mobile, text SYRIA to 70800. To donate online, please visit http://www.ageinternational.org.uk/syria Over one million refugees are desperately in need of water, food, healthcare, warm clothing, blankets and shelter; Age International urgently needs your support to help affected older refugees. Age International is a subsidiary charity of Age UK and a member of the Disasters Emergency Committee (DEC). The DEC launches and co-ordinates national fundraising appeals for public donations on behalf of its member agencies. Texts cost £5 plus one standard rate message. Age International will receive a minimum of £4.96. More info at ageinternational.org.uk/SyriaTerms --- Age UK is a registered charity and company limited by guarantee, (registered charity number 1128267, registered company number 6825798). Registered office: Tavis House, 1-6 Tavistock Square, London WC1H 9NA. For the purposes of promoting Age UK Insurance, Age UK is an Appointed Representative of Age UK Enterprises Limited, Age UK is an Introducer Appointed Representative of JLT Benefit Solutions Limited and Simplyhealth Access for the purposes of introducing potential annuity and health cash plans customers respectively. Age UK Enterprises Limited, JLT Benefit Solutions Limited and Simplyhealth Access are all authorised and regulated by the Financial Services Authority. -- This email and any files transmitted with it are confidential and intended solely for the use of the individual or entity to whom they are addressed. If you receive a message in error, please advise the sender and delete immediately. Except where this email is sent in the usual course of our business, any opinions expressed in this email are those of the author and do not necessarily reflect the opinions of Age UK or its subsidiaries and associated companies. Age UK monitors all e-mail transmissions passing through its network and may block or modify mails which are deemed to be unsuitable. Age Concern England (charity number 261794) and Help the Aged (charity number 272786) and their trading and other associated companies merged on 1st April 2009. Together they have formed the Age UK Group, dedicated to improving the lives of people in later life. The three national Age Concerns in Scotland, Northern Ireland and Wales have also merged with Help the Aged in these nations to form three registered charities: Age Scotland, Age NI, Age Cymru. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting- guide.html and provide commented, minimal,
Re: [R] Error with adehabitatHR and kernelbb
There was some mistakes in my previous sending. The following are correct.
This is the problematic data, especially the line 43, but when I removed that
line, it bugs at line 62 and so on.
sig1 time id
UTMnorthin UTMeasting
40 4.5766 2012.05.30 08:00:00 Ade-5 5419576 390052
41 4.5766 2012.05.30 10:00:00 Ade-5 5419581 390058
42 4.5766 2012.05.30 12:00:00 Ade-5 5419560 390045
43 4.5766 2012.05.30 14:00:00 Ade-5 5419574 390051
44 4.5766 2012.05.30 16:00:00 Ade-5 5419490 390051
45 4.5766 2012.05.30 18:00:00 Ade-5 5419435 390293
46 4.5766 2012.05.30 20:00:00 Ade-5 5419661 390876
47 4.5766 2012.05.30 22:00:00 Ade-5 5419934 390673
48 4.5766 2012.05.31 02:00:00 Ade-5 5420636 389777
49 4.5766 2012.06.05 02:00:00 Ade-7 5419275 391206
50 4.5766 2012.06.05 04:00:00 Ade-7 5419276 391202
This is a data frame so before the loop (see commands in the previous mail
below), I transform with the following commands:
coordinates(fece)-~UTMeasting+UTMnorthin
loc@data$time - as.POSIXct(strptime(as.character(loc@data$time),%Y.%m.%d
%H:%M:%S))
Rémi Lesmerises, biol. M.Sc.,
PH.D candidate
Université du Québec à Rimouski
De : David Winsemius dwinsem...@comcast.net
À : Rémi Lesmerises remilesmeri...@yahoo.ca
Cc : r-help@r-project.org r-help@r-project.org
Envoyé le : vendredi 17 mai 2013 13h53
Objet : Re: [R] Error with adehabitatHR and kernelbb
On May 17, 2013, at 7:44 AM, Rémi Lesmerises wrote:
Dear all,
I'm trying to get a Brownian bridge kernel (kernelbb) for each combination of
two consecutive animal locations (see commands below) and put them, with a
loop, inside a list. It works well at the beginning but after 42 runs, it
appears the following warning :
Error in seq.default(yli[1], yli[2], by = diff(xg[1:2])) :
invalid (to - from)/by in seq(.)
I looked at the coordinates, at the id, at the time of the run 43 and it's
all good...
I looked on the net and it happened to only one person and there was no
answer to his post.
I wonder if that posting (like yours) had no data on which to display the
problem or to test potential solutions? I would think you would want to post
any setup objects and then data for items 40-45.
Someone could help me?
## commands
BBtraj - list()
for (i in 1:(nrow(loc@data)-1)) {
BBtraj[[i]] - kernelbb(as.ltraj(loc@coords[i:(i+1),],
date=loc@data$time[i:(i+1)], id = as.character(loc@data$id[i:(i+1)]),
typeII = TRUE), sig1=as.numeric(as.character(loc@data$sig1[i])), sig2= 5,
grid = 1000)
}
--
David Winsemius
Alameda, CA, USA
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] #Keeping row names when using as.data.frame.matrix
#question I have the following data set: Date-c(9/7/2010,9/7/2010,9/7/2010,9/7/2010,9/7/2010,9/7/2010,9/8/2010) EstimatedQuantity-c(3535,2772,3279,3411,3484,3274,3305) ScowNo-c(4001,3002,4002,BR 8,4002,BR 8,4001) dataset- data.frame(EstimatedQuantity,Date,ScowNo) #I'm trying to convert the data set into a contingency table and then back into a regular data frame: xtabdata-as.data.frame.matrix(xtabs(EstimatedQuantity~Date+ScowNo,data=dataset), row.names=(dataset$Date),optional=F) #I'm trying to keep the row names (in xtabsdata) as the dates. #But the row names keep coming up as integers. #How can I preserve the row names as dates when #the table is converted back to a data frame? -- View this message in context: http://r.789695.n4.nabble.com/Keeping-row-names-when-using-as-data-frame-matrix-tp4667344.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] time-series aggregation of information
I have following data for which I need to calculate the weighted aggregate value of the parameter at each time. Date,Parameter,Weight 2012-01-31,90,200 2012-01-31,80,400 2012-01-31,70,500 2012-01-31,60,800 2012-02-29,120,220 2012-02-29,110,410 2012-02-29,75,520 2012-02-29,65,840 2012-03-31,115,210 2012-03-31,100,405 2012-03-31,70,500 2012-03-31,60,800 So for the above sample the solution looks like: Date,Weighted Parameter 2012-01-31,70 2012-02-29,82.96482412 2012-03-31,77.10182768 Could I potentially use tapply / aggregate for this? Would like to avoid a for loop if possible. Thank you! The information transmitted is intended solely for the individual or entity to which it is addressed and may contain confidential and/or privileged material. Any review, retransmission, dissemination or other use of or taking action in reliance upon this information by persons or entities other than the intended recipient is prohibited. If you have received this email in error please contact the sender and delete the material from any computer. Any information contained herein is neither an offer to sell nor a solicitation to buy any interest in any investment fund. An offer can only be made by the approved offering memorandum, which contains important information concerning risk factors and other material information and must be read carefully before any decision to invest is made. Securities and derivatives trading are speculative and involve a risk of substantial loss. Past results are not necessarily indicative of future performance. All incoming and outgoing e-mails are archived and may be reviewed and/or produced at the request of regulators or in connection with civil litigation. IRON Holdings, LLC. accepts no liability for any errors or omissions arising as a result of transmission. [[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
Hello, I fail to tranfer data from a dataframe to a matrix. var is from a dataframe (and belongs still to the class dataframe) and should look like m (see below). var vec1 vec3 d1 d2 1 172 173 223 356 dput (var) structure(list(vec1 = 172L, vec3 = 173L, d1 = 223L, d2 = 356L), .Names = c(vec1, vec3, d1, d2), row.names = 1L, class = data.frame) m #THIS IS THE AIM [,1] [,2] [1,] 172 223 [2,] 173 356 dput (m) structure(c(172, 173, 223, 356), .Dim = c(2L, 2L)) How can I transform var to m? Thanks akhgar __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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
hi deer all Estimate KM survival probabilities for each categories of RX means ‘treatment’ and ‘placebo’ separately surv.Rx-survfit(Surv(SURVT,STATUS)~strata(RX),data=rem.data) when write that command. it doesnt run. what should i do? thanks __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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
On 05/17/2013 02:34 PM, masumeh akhgar wrote: hi deer all Estimate KM survival probabilities for each categories of RX means ‘treatment’ and ‘placebo’ separately surv.Rx-survfit(Surv(SURVT,STATUS)~strata(RX),data=rem.data) when write that command. it doesnt run. what should i do? thanks What do you mean it doesn't run? Is there an error message? What is it? Also, you do not need to put RX in strata() in the survfit() function. -- Kevin E. Thorpe Head of Biostatistics, Applied Health Research Centre (AHRC) Li Ka Shing Knowledge Institute of St. Michael's Assistant Professor, Dalla Lana School of Public Health University of Toronto email: kevin.tho...@utoronto.ca Tel: 416.864.5776 Fax: 416.864.3016 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] #Keeping row names when using as.data.frame.matrix
On May 17, 2013, at 9:46 AM, Tim wrote: #question I have the following data set: Date-c(9/7/2010,9/7/2010,9/7/2010,9/7/2010,9/7/2010,9/7/2010,9/8/2010) EstimatedQuantity-c(3535,2772,3279,3411,3484,3274,3305) ScowNo-c(4001,3002,4002,BR 8,4002,BR 8,4001) dataset- data.frame(EstimatedQuantity,Date,ScowNo) #I'm trying to convert the data set into a contingency table and then back into a regular data frame: xtabdata-as.data.frame.matrix(xtabs(EstimatedQuantity~Date+ScowNo,data=dataset), row.names=(dataset$Date),optional=F) #I'm trying to keep the row names (in xtabsdata) as the dates. #But the row names keep coming up as integers. #How can I preserve the row names as dates when #the table is converted back to a data frame? It's a factor-problem (see the FAQ …. read all of section 7, i'm too lazy to look it up again.): xtabdata-as.data.frame.matrix(xtabs(EstimatedQuantity~Date+ScowNo,data=dataset), row.names=sort( unique( as.character(dataset$Date))), optional=F) -- View this message in context: http://r.789695.n4.nabble.com/Keeping-row-names-when-using-as-data-frame-matrix-tp4667344.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. David Winsemius Alameda, CA, USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] help
On 17-05-2013, at 20:15, masumeh akhgar akhgar.masu...@gmail.com wrote: Hello, I fail to tranfer data from a dataframe to a matrix. var is from a dataframe (and belongs still to the class dataframe) and should look like m (see below). var vec1 vec3 d1 d2 1 172 173 223 356 dput (var) structure(list(vec1 = 172L, vec3 = 173L, d1 = 223L, d2 = 356L), .Names = c(vec1, vec3, d1, d2), row.names = 1L, class = data.frame) m #THIS IS THE AIM [,1] [,2] [1,] 172 223 [2,] 173 356 dput (m) structure(c(172, 173, 223, 356), .Dim = c(2L, 2L)) How can I transform var to m? Thanks akhgar This question appears to be an almost exact copy of this: http://r.789695.n4.nabble.com/convert-a-data-frame-to-matrix-td4667256.html See the answers there. Spam? Berend __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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
m - as.matrix(var) John Kane Kingston ON Canada -Original Message- From: akhgar.masu...@gmail.com Sent: Fri, 17 May 2013 22:45:22 +0430 To: r-help@r-project.org Subject: [R] help Hello, I fail to tranfer data from a dataframe to a matrix. var is from a dataframe (and belongs still to the class dataframe) and should look like m (see below). var vec1 vec3 d1 d2 1 172 173 223 356 dput (var) structure(list(vec1 = 172L, vec3 = 173L, d1 = 223L, d2 = 356L), .Names = c(vec1, vec3, d1, d2), row.names = 1L, class = data.frame) m #THIS IS THE AIM [,1] [,2] [1,] 172 223 [2,] 173 356 dput (m) structure(c(172, 173, 223, 356), .Dim = c(2L, 2L)) How can I transform var to m? Thanks akhgar __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Share photos screenshots in seconds... TRY FREE IM TOOLPACK at http://www.imtoolpack.com/default.aspx?rc=if1 Works in all emails, instant messengers, blogs, forums and social networks. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] time-series aggregation of information
On May 17, 2013, at 11:48 AM, Chirag Maru wrote: I have following data for which I need to calculate the weighted aggregate value of the parameter at each time. Date,Parameter,Weight 2012-01-31,90,200 2012-01-31,80,400 2012-01-31,70,500 2012-01-31,60,800 2012-02-29,120,220 2012-02-29,110,410 2012-02-29,75,520 2012-02-29,65,840 2012-03-31,115,210 2012-03-31,100,405 2012-03-31,70,500 2012-03-31,60,800 So for the above sample the solution looks like: Date,Weighted Parameter 2012-01-31,70 2012-02-29,82.96482412 2012-03-31,77.10182768 Could I potentially use tapply / aggregate for this? Would like to avoid a for loop if possible. by(dat, dat[1], FUN=function(d) weighted.mean(d[[Parameter]], w=d[[Weight]]) ) Date: 2012-01-31 [1] 70 Date: 2012-02-29 [1] 82.96482 Date: 2012-03-31 [1] 77.10183 It's a bit of a shame that there is no as.data.frame.by function. You can create a dataframe from the by object with as.data.frame.table with the only defect in the naming of the second column as.data.frame.table(by(dat, dat[1], FUN=function(d) weighted.mean(d[[Parameter]], w=d[[Weight]]) )) Date Freq 1 2012-01-31 70.0 2 2012-02-29 82.96482 3 2012-03-31 77.10183 setNames(as.data.frame.table(by(dat, dat[1], FUN=function(d) weighted.mean(d[[Parameter]], w=d[[Weight]]) )), c(Dts, wtdmeans)) Dts wtdmeans 1 2012-01-31 70.0 2 2012-02-29 82.96482 3 2012-03-31 77.10183 -- David Winsemius Alameda, CA, USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] help
hi all this command used tt function for all variables. How can i define a different function for each variable? exCox.all-coxph(Surv(SURVT,STATUS) ~ RX+LOGWBC+SEX+tt(RX)+tt(LOGWBC)+tt(SEX), data=rem.data,tt=function(x,t,...) log(t)*x)) thank you __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] time-series aggregation of information
Hi, May be this helps: dat- read.table(text= Date,Parameter,Weight 2012-01-31,90,200 2012-01-31,80,400 2012-01-31,70,500 2012-01-31,60,800 2012-02-29,120,220 2012-02-29,110,410 2012-02-29,75,520 2012-02-29,65,840 2012-03-31,115,210 2012-03-31,100,405 2012-03-31,70,500 2012-03-31,60,800 ,sep=,,header=TRUE,stringsAsFactors=FALSE) library(plyr) ddply(dat,.(Date), summarize, wtdmeans=weighted.mean(Parameter,Weight)) # Date wtdmeans #1 2012-01-31 70.0 #2 2012-02-29 82.96482 #3 2012-03-31 77.10183 A.K. - Original Message - From: Chirag Maru chirag.m...@ironfinancial.com To: r-help@R-project.org r-help@r-project.org Cc: Sent: Friday, May 17, 2013 2:48 PM Subject: [R] time-series aggregation of information I have following data for which I need to calculate the weighted aggregate value of the parameter at each time. Date,Parameter,Weight 2012-01-31,90,200 2012-01-31,80,400 2012-01-31,70,500 2012-01-31,60,800 2012-02-29,120,220 2012-02-29,110,410 2012-02-29,75,520 2012-02-29,65,840 2012-03-31,115,210 2012-03-31,100,405 2012-03-31,70,500 2012-03-31,60,800 So for the above sample the solution looks like: Date,Weighted Parameter 2012-01-31,70 2012-02-29,82.96482412 2012-03-31,77.10182768 Could I potentially use tapply / aggregate for this? Would like to avoid a for loop if possible. Thank you! The information transmitted is intended solely for the individual or entity to which it is addressed and may contain confidential and/or privileged material. Any review, retransmission, dissemination or other use of or taking action in reliance upon this information by persons or entities other than the intended recipient is prohibited. If you have received this email in error please contact the sender and delete the material from any computer. Any information contained herein is neither an offer to sell nor a solicitation to buy any interest in any investment fund. An offer can only be made by the approved offering memorandum, which contains important information concerning risk factors and other material information and must be read carefully before any decision to invest is made. Securities and derivatives trading are speculative and involve a risk of substantial loss. Past results are not necessarily indicative of future performance. All incoming and outgoing e-mails are archived and may be reviewed and/or produced at the request of regulators or in connection with civil litigation. IRON Holdings, LLC. accepts no liability for any errors or omissions arising as a result of transmission. [[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] help
On May 17, 2013, at 11:55 AM, masumeh akhgar wrote: hi all this command used tt function for all variables. How can i define a different function for each variable? exCox.all-coxph(Surv(SURVT,STATUS) ~ RX+LOGWBC+SEX+tt(RX)+tt(LOGWBC)+tt(SEX), data=rem.data,tt=function(x,t,...) log(t)*x)) After reading the help page it seems pretty clear that the 'tt' argument has a very special purpose and that it is not designed to inserting transformations of anything other than time-related transformations. Applying a tt() function to 'logwbc' or to 'SEX' makes no sense at all. In the help page the `tt` function is applied to the age variable and is intended to advance a subject's age value forward as they …. wait for it…. age (used now as an English verb rather than a numeric value). Other transformations would need to be built in the usual manner in the formula or by constructing transformed variables in the input dataset. -- David Winsemius Alameda, CA, USA __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Bivariate - multivariate linear regression
Hi there,
I want to do several bivariate linear regressions and, than, do a
multivariate linear regression including only variables significantly
associated *(p 0.15)* with y in bivariate analysis, without having to
look manually to those p values.
So, here what I got for the moment.
First, I use this data set:
tolerance - read.csv(
http://www.ats.ucla.edu/stat/r/examples/alda/data/tolerance1.txt;).
Second, I used this command, allowing me to extract p-values later:
lmp - function (modelobject) {
if (class(modelobject) != lm) stop(Not an object of class
'lm' )
f - summary(modelobject)$fstatistic
p - pf(f[1],f[2],f[3],lower.tail=F)
attributes(p) - NULL
return(p)}
Third, I did my bivariate linear regressions:
fit = lm(exposure~tol11, data = tolerance)
fit_2 = lm(exposure~tol12, data= tolerance)
fit_3 = lm(exposure~tol13, data= tolerance)
fit_4 = lm(exposure~tol14, data= tolerance)
fit_5 = lm(exposure~tol15, data= tolerance)
Fourth, I extracted p-values:
lmp(fit)
lmp(fit_2)
lmp(fit_3)
lmp(fit_4)
lmp(fit_5)
Firth, I confirmed that p-values were OK (just to be sure, it's the first
time I used the above procedure) :
summary (fit)
summary (fit_2)
summary (fit_3)
summary (fit_4)
summary (fit_5)
And now, Im, I dont know what to do.
The multivariate linear regression (if all variables were included) is:
fit_multi = lm (exposure ~ tol11 + tol12 + tol13 + tol14 + tol15, data=
tolerance)
I would like to be able to do something like:
fit_multi = lm (exposure ~ tol11 [include only if lmp( fit) 0.15] +
tol12 [include only if lmp(fit_2) 0.15] + tol13 [include only if
lmp(fit_3) 0.15] + tol14 [include only if lmp(fit_4) 0.15] +
tol15 [include
only if lmp(fit_4) 0.15], data= tolerance)
Any idea?
Thank you!
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[R] formatting column names of data frame
Is there any way to format the headers of data frames, for printing? I am using Sweave to generate formatted reports. In Sweave, I read in a data.frame: result - read.table(path.to.table); then display it: print.data.frame(result); This gives me what I expect in the eventual final output: RecordAverage Maximum 1 34 899 2 14 15 3 433 1003 ... ... ... What I am hoping to do is distinguish one or more of the column headers, for example, I want Average or Maximum to be bold, underlined, etc., just some way to make the column name stand out visually. Any idea if this is possible? Any suggestions appreciated. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] image and color gradient
Hello,
I have a nice function that makes an image of an matrix
e.g.:
qt[1:3,1:3]
rs655246 rs943795 rs955612
rs655246 NA NA NA
rs943795 9.610070e-04 NA NA
rs955612 5.555616e-05 7.915982e-07 NA
myimage - function(x, cex.axis = 0.7, ...){
opar - par(mar=c(5,4,4,6),
pty ='s')
on.exit(par(opar))
image(x, axes = FALSE, ...)
ats - 0:(nrow(x)-1)/(nrow(x)-1)
axis(1, at=ats, lab=rownames(x), cex.axis=cex.axis, las=2)
axis(4, at=ats, lab=colnames(x), cex.axis=cex.axis, las=2)
box()
}
The ranges in my matrix are from 1 to 1e-08. But in my image there is no
difference between, for instance 1e-05 and 1e-06 or -07 etc.
How can I ameliorate my function myimage to do so. I guess it has something
to do with breaks but I do not understand how to handle.
Thanks
Hermann
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Re: [R] inverse for formula transformations on LHS
Paul, Inverting log(y) is just the beginning of the problem, after that you need to teach predict.lm() that E(y |x) = exp(x'betahat + .5*sigmahat^2) and then further lessons are required to get it to understand how to adapt its confidence and prediction bands… and then you need to generalize all of this to other transformations. Quite a project! Best, Roger Roger Koenker rkoen...@illinois.edu On May 17, 2013, at 12:21 PM, Paul Johnson wrote: This is an R formula handling question. It arose in class. We were working on the Animals data in the MASS package. In order to see a relationship, you need to log brain and body weight. It's a fun one for teaching regression, if you did not try it yet. There are outliers too! Students wanted to make a predicted value plot in the non-logged values of y, for comparison, and I wondered if I couldn't automate this somehow for them. It made me wonder how R manages formulae and if a transformation like log(y) can be be mechanically inverted. So we have something concrete to talk about, suppose x and y are variables in dat, a person fits m1 - lm(log(y) ~ log(x), data = dat) termplot shows log(y) on the vertical. What if I want y on the vertical? Similarly, predict gives values on the log(y) scale, there's no argument like type = untransformed. I want my solution to be a bit general, so that it would give back predicted y for formulae like sqrt(y) or exp(y) or log(y + d) or whatever other math people might throw in there. Here's what I can tell so far about R's insides. The formula handler makes a list out of the formula, I can get that from the terms object that the model generates. The formula list has ~ as element 1, and log(x) becomes element [[2]]. Where in the R source code can I see how R looks at the symbol log(y) and discerns that there is a variable y that needs to be logged? If I could understand that, and if R has a table of inverse functions, then maybe I could see what to do. If you have ideas, I'm very grateful if you share them. pj -- Paul E. Johnson Professor, Political Science Assoc. Director 1541 Lilac Lane, Room 504 Center for Research Methods University of Kansas University of Kansas http://pj.freefaculty.org http://quant.ku.edu [[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] Problems using lmer {lme4}
Hi Patrick,
Thanks for you reply. I tried adding site fixed effect as you told me, but
the program failed again (R stopped working).
Basically, what I am trying to do is to test for auto-correlation between
sites within routes. My survey takes place in routes, but each route is
divided in segments (or sites), and the presence of the species is
monitored at a site level.
A simple table for one species only would be as follows:
Route site Visit1 Visit2 Visit3 Visit4 Visit5 #detections (punto6) 1 1-1
1 0 0 0 1 2 1 1-2 1 1 1 1 0 4 1 1-3 0 1 1 1 1 4 1 1-4 0 1 1 0 0 2 1 1-5
1 0 0 0 1 2 2 2-1 0 0 1 1 1 3 2 2-2 1 1 1 1 1 5 2 2-3 0 0 0 1 0 1 2 2-4
0 0 1 1 1 3 2 2-5 0 0 1 0 1 2 3 3-1 1 1 0 1 0 3 3 3-2 1 0 1 1 1 4 3 3-3
1 0 1 1 1 4 3 3-4 0 1 1 1 0 3 3 3-5 0 0 0 1 1 2
On Fri, May 17, 2013 at 2:15 PM, Patrick Coulombe
patrick.coulo...@gmail.com wrote:
Hi Andrea,
I'm not exactly sure what you're trying to do, but you've included a
random effect for a site coefficient that's not even in your list of
fixed effects... you're basically allowing the coefficient for site to
vary across routes, but you're never including the coefficient in the
first place. Given the appropriate nested structure (I haven't looked
at your data), a model that should run would be:
m2 - lmer(cbind(punto6,5) ~ sp + site + (site|route)
,family=binomial,data = d)
The problem here is I'm not sure this is what you mean to do... If you
could way simplify your example (for example, by including a small
number of mock observations with only the relevant variables, in a
table instead of in R syntax), that would be helpful.
Patrick
2013/5/17 Andrea Goijman agoij...@cnia.inta.gov.ar:
Dear R list,
I'm attaching a sample of my data which consists on the presence/absence
(punto6, binomial n=5 occasions)
of different species (sp), on different sites (site) within routes
('route).
First, I want to be able to find if there is autocorrelation of the
response variable between
the sites within each route. For this I start testing 2 models, but when
I
try to run the second model,
to test for the random effects of sites within routes R stops working!
I'm not being able to find out why r is crashing... and whay am I doing
wrong.
Thanks!
Andrea
###
#dput(d)
d-structure(list(site = structure(c(55L, 56L, 57L, 58L, 59L, 60L,
55L, 56L, 57L, 58L, 59L, 60L, 55L, 56L, 57L, 58L, 59L, 60L, 55L,
56L, 57L, 58L, 59L, 60L, 55L, 56L, 57L, 58L, 59L, 60L, 55L, 56L,
57L, 58L, 59L, 60L, 55L, 56L, 57L, 58L, 59L, 60L, 229L, 230L,
231L, 232L, 233L, 234L, 229L, 230L, 231L, 232L, 233L, 234L, 229L,
230L, 231L, 232L, 233L, 234L, 229L, 230L, 231L, 232L, 233L, 234L,
229L, 230L, 231L, 232L, 233L, 234L, 229L, 230L, 231L, 232L, 233L,
234L, 229L, 230L, 231L, 232L, 233L, 234L, 331L, 332L, 333L, 334L,
335L, 336L, 331L, 332L, 333L, 334L, 335L, 336L, 331L, 332L, 333L,
334L, 335L, 336L, 331L, 332L, 333L, 334L, 335L, 336L, 331L, 332L,
333L, 334L, 335L, 336L, 331L, 332L, 333L, 334L, 335L, 336L, 331L,
332L, 333L, 334L, 335L, 336L, 389L, 390L, 391L, 392L, 393L, 394L,
389L, 390L, 391L, 392L, 393L, 394L, 389L, 390L, 391L, 392L, 393L,
394L, 389L, 390L, 391L, 392L, 393L, 394L, 389L, 390L, 391L, 392L,
393L, 394L, 389L, 390L, 391L, 392L, 393L, 394L, 389L, 390L, 391L,
392L, 393L, 394L, 205L, 206L, 207L, 208L, 209L, 210L, 205L, 206L,
207L, 208L, 209L, 210L, 205L, 206L, 207L, 208L, 209L, 210L, 205L,
206L, 207L, 208L, 209L, 210L, 205L, 206L, 207L, 208L, 209L, 210L,
205L, 206L, 207L, 208L, 209L, 210L, 205L, 206L, 207L, 208L, 209L,
210L, 163L, 164L, 165L, 166L, 167L, 168L, 163L, 164L, 165L, 166L,
167L, 168L, 163L, 164L, 165L, 166L, 167L, 168L, 163L, 164L, 165L,
166L, 167L, 168L, 163L, 164L, 165L, 166L, 167L, 168L, 163L, 164L,
165L, 166L, 167L, 168L, 163L, 164L, 165L, 166L, 167L, 168L, 247L,
248L, 249L, 250L, 251L, 252L, 247L, 248L, 249L, 250L, 251L, 252L,
247L, 248L, 249L, 250L, 251L, 252L, 247L, 248L, 249L, 250L, 251L,
252L, 247L, 248L, 249L, 250L, 251L, 252L, 247L, 248L, 249L, 250L,
251L, 252L, 247L, 248L, 249L, 250L, 251L, 252L), .Label = c(102-1,
102-2, 102-3, 102-4, 102-5, 102-6, 1023-1, 1023-2,
1023-3, 1023-4, 1023-5, 1023-6, 1027-1, 1027-2, 1027-3,
1027-4, 1027-5, 1027-6, 1028-1, 1028-2, 1028-3, 1028-4,
1028-5, 1028-6, 1032-1, 1032-2, 1032-3, 1032-4, 1032-5,
1032-6, 1034-1, 1034-2, 1034-3, 1034-4, 1034-5, 1034-6,
1036-1, 1036-2, 1036-3, 1036-4, 1036-5, 1036-6, 1041-1,
1041-2, 1041-3, 1041-4, 1041-5, 1041-6, 1046-1, 1046-2,
1046-3, 1046-4, 1046-5, 1046-6, 105-1, 105-2, 105-3,
105-4, 105-5, 105-6, 107-1, 107-2, 107-3, 107-4,
107-5, 107-6, 108-1, 108-2, 108-3, 108-4, 108-5,
108-6, 1101-1, 1101-2, 1101-3, 1101-4, 1101-5, 1101-6,
1104-1, 1104-2, 1104-3, 1104-4, 1104-5, 1104-6, 1108-1,
1108-2, 1108-3, 1108-4, 1108-5, 1108-6, 111-1, 111-2,
111-3, 111-4, 111-5, 111-6, 1113-1, 1113-2, 1113-3,
1113-4,
Re: [R] image and color gradient
On May 17, 2013, at 12:58 PM, Hermann Norpois wrote:
Hello,
I have a nice function that makes an image of an matrix
e.g.:
qt[1:3,1:3]
rs655246 rs943795 rs955612
rs655246 NA NA NA
rs943795 9.610070e-04 NA NA
rs955612 5.555616e-05 7.915982e-07 NA
If you had used dput to offer that test case I would have tested my
suggestions. As it is I will simply suggest:
... , breaks=10^-(0:8), ...
You will need to specify the colors to match the breaks.
myimage - function(x, cex.axis = 0.7, ...){
opar - par(mar=c(5,4,4,6),
pty ='s')
on.exit(par(opar))
image(x, axes = FALSE, ...)
ats - 0:(nrow(x)-1)/(nrow(x)-1)
axis(1, at=ats, lab=rownames(x), cex.axis=cex.axis, las=2)
axis(4, at=ats, lab=colnames(x), cex.axis=cex.axis, las=2)
box()
}
The ranges in my matrix are from 1 to 1e-08. But in my image there is no
difference between, for instance 1e-05 and 1e-06 or -07 etc.
How can I ameliorate my function myimage to do so. I guess it has something
to do with breaks but I do not understand how to handle.
Thanks
Hermann
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Re: [R] formatting column names of data frame
Have you tried xtable?
library( xtable )
x - structure(list(Record = 1:3, Average = c(34L, 14L, 433L), Maximum =
c(899L,
15L, 1003L)), .Names = c(Record, Average, Maximum), class = data.frame,
row.names = c(NA,
-3L))
x - xtable( x )
print( x )
% latex table generated in R 2.15.2 by xtable 1.7-1 package
% Fri May 17 22:22:00 2013
\begin{table}[ht]
\centering
\begin{tabular}{}
\hline
Record Average Maximum \\
\hline
11 34 899 \\
22 14 15 \\
33 433 1003 \\
\hline
\end{tabular}
\end{table}
On Friday 17 May 2013 12:53:12 Patrick Leyshock wrote:
Is there any way to format the headers of data frames, for printing?
I am using Sweave to generate formatted reports. In Sweave, I read in a
data.frame:
result - read.table(path.to.table);
then display it:
print.data.frame(result);
This gives me what I expect in the eventual final output:
RecordAverage Maximum
1 34 899
2 14 15
3 433 1003
... ... ...
What I am hoping to do is distinguish one or more of the column headers,
for example, I want Average or Maximum to be bold, underlined, etc.,
just some way to make the column name stand out visually.
Any idea if this is possible? Any suggestions appreciated.
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Re: [R] #Keeping row names when using as.data.frame.matrix
Hi, library(plyr) res-dcast(dataset,Date~ScowNo,sum,value.var=EstimatedQuantity) rownames(res)- res[,1] res[,-1] # 3002 4001 4002 BR 8 #9/7/2010 2772 3535 6763 6685 #9/8/2010 0 3305 0 0 A.K. - Original Message - From: Tim t...@mde.state.md.us To: r-help@r-project.org Cc: Sent: Friday, May 17, 2013 12:46 PM Subject: [R] #Keeping row names when using as.data.frame.matrix #question I have the following data set: Date-c(9/7/2010,9/7/2010,9/7/2010,9/7/2010,9/7/2010,9/7/2010,9/8/2010) EstimatedQuantity-c(3535,2772,3279,3411,3484,3274,3305) ScowNo-c(4001,3002,4002,BR 8,4002,BR 8,4001) dataset- data.frame(EstimatedQuantity,Date,ScowNo) #I'm trying to convert the data set into a contingency table and then back into a regular data frame: xtabdata-as.data.frame.matrix(xtabs(EstimatedQuantity~Date+ScowNo,data=dataset), row.names=(dataset$Date),optional=F) #I'm trying to keep the row names (in xtabsdata) as the dates. #But the row names keep coming up as integers. #How can I preserve the row names as dates when #the table is converted back to a data frame? -- View this message in context: http://r.789695.n4.nabble.com/Keeping-row-names-when-using-as-data-frame-matrix-tp4667344.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] filter rows by value
Hey All, I want to delete rows based on the last 2 digits on the value in one column but I dont know how to do that. Suppose my data looks like this: Var Time 1 51 2 151 3 251 *4234* *5 331* 6351 I want to delete the rows that the value in column Time, the last 2 digit is not 51, in this case the rows highlighted will be removed. Thanks for your help! [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] filter rows by value
Hello, Try the following. dat - read.table(text = Var Time 1 51 2 151 3 251 4234 5 331 6351 , header = TRUE) dat[dat$Time %% 100 == 51, ] Em 17-05-2013 22:01, Ye Lin escreveu: Hey All, I want to delete rows based on the last 2 digits on the value in one column but I dont know how to do that. Suppose my data looks like this: Var Time 1 51 2 151 3 251 *4234* *5 331* 6351 I want to delete the rows that the value in column Time, the last 2 digit is not 51, in this case the rows highlighted will be removed. Thanks for your help! [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Bivariate - multivariate linear regression
On 2013-05-17 12:45, Jesse Gervais wrote:
Hi there,
I want to do several bivariate linear regressions and, than, do a
multivariate linear regression including only variables significantly
associated *(p 0.15)* with y in bivariate analysis, without having to
look manually to those p values.
So, here what I got for the moment.
First, I use this data set:
tolerance - read.csv(
http://www.ats.ucla.edu/stat/r/examples/alda/data/tolerance1.txt;).
Second, I used this command, allowing me to extract p-values later:
lmp - function (modelobject) {
if (class(modelobject) != lm) stop(Not an object of class
'lm' )
f - summary(modelobject)$fstatistic
p - pf(f[1],f[2],f[3],lower.tail=F)
attributes(p) - NULL
return(p)}
Third, I did my bivariate linear regressions:
fit = lm(exposure~tol11, data = tolerance)
fit_2 = lm(exposure~tol12, data= tolerance)
fit_3 = lm(exposure~tol13, data= tolerance)
fit_4 = lm(exposure~tol14, data= tolerance)
fit_5 = lm(exposure~tol15, data= tolerance)
Fourth, I extracted p-values:
lmp(fit)
lmp(fit_2)
lmp(fit_3)
lmp(fit_4)
lmp(fit_5)
Firth, I confirmed that p-values were OK (just to be sure, it's the first
time I used the above procedure) :
summary (fit)
summary (fit_2)
summary (fit_3)
summary (fit_4)
summary (fit_5)
And now, I’m, I don’t know what to do.
The multivariate linear regression (if all variables were included) is:
fit_multi = lm (exposure ~ tol11 + tol12 + tol13 + tol14 + tol15, data=
tolerance)
I would like to be able to do something like:
fit_multi = lm (exposure ~ tol11 [include only if lmp( fit) 0.15] +
tol12 [include only if lmp(fit_2) 0.15] + tol13 [include only if
lmp(fit_3) 0.15] + tol14 [include only if lmp(fit_4) 0.15] +
tol15 [include
only if lmp(fit_4) 0.15], data= tolerance)
Any idea?
(Thanks for providing reproducible code!)
It seems to me that you're just missing two things:
1. a way to determine the names of the variables to be included
in the multiple (not 'multivariate' to be nitpicky) regression;
2. a way to build the formula for the multiple regression once
you know which predictors to include.
To get the variables:
varnames - names(tolerance)[2:6]
pvec - c(lmp(fit), lmp(fit_2), lmp(fit_3), lmp(fit_4), lmp(fit_5))
use - varnames[pvec 0.15]
use
#[1] tol14 tol15
To construct the formula:
rhs - paste(use, collapse = + )
form - paste(exposure ~, rhs)
And then use it:
fit_multi - lm(formula = form, data = tolerance)
Peter Ehlers
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Re: [R] filter rows by value
Hi,
dat1- read.table(text=
Var Time
1 51
2 151
3 251
4 234
5 331
6 351
,sep=,header=TRUE)
dat1[!is.na(match(gsub(.*(\\d{2})$,\\1,dat1$Time),51)),]
# Var Time
#1 1 51
#2 2 151
#3 3 251
#6 6 351
#or
dat1[substr(dat1$Time,nchar(dat1$Time)-1,nchar(dat1$Time))==51,]
# Var Time
#1 1 51
#2 2 151
#3 3 251
#6 6 351
#or
library(stringr)
dat1[str_detect(dat1$Time,51$),]
# Var Time
#1 1 51
#2 2 151
#3 3 251
#6 6 351
#or
dat1[grepl(51$,dat1$Time),]
# Var Time
#1 1 51
#2 2 151
#3 3 251
#6 6 351
#or
dat1[str_sub(dat1$Time, start=-2)%in%51,]
Var Time
#1 1 51
#2 2 151
#3 3 251
#6 6 351
A.K.
- Original Message -
From: Ye Lin ye...@lbl.gov
To: R help r-help@r-project.org
Cc:
Sent: Friday, May 17, 2013 5:01 PM
Subject: [R] filter rows by value
Hey All,
I want to delete rows based on the last 2 digits on the value in one column
but I dont know how to do that.
Suppose my data looks like this:
Var Time
1 51
2 151
3 251
*4 234*
*5 331*
6 351
I want to delete the rows that the value in column Time, the last 2 digit
is not 51, in this case the rows highlighted will be removed.
Thanks for your help!
[[alternative HTML version deleted]]
__
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
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Re: [R] filter rows by value
it works!Thanks! On Fri, May 17, 2013 at 2:28 PM, Rui Barradas ruipbarra...@sapo.pt wrote: Hello, Try the following. dat - read.table(text = Var Time 1 51 2 151 3 251 4234 5 331 6351 , header = TRUE) dat[dat$Time %% 100 == 51, ] Em 17-05-2013 22:01, Ye Lin escreveu: Hey All, I want to delete rows based on the last 2 digits on the value in one column but I dont know how to do that. Suppose my data looks like this: Var Time 1 51 2 151 3 251 *4234* *5 331* 6351 I want to delete the rows that the value in column Time, the last 2 digit is not 51, in this case the rows highlighted will be removed. Thanks for your help! [[alternative HTML version deleted]] __** R-help@r-project.org mailing list https://stat.ethz.ch/mailman/**listinfo/r-helphttps://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/** posting-guide.html 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] Error with adehabitatHR and kernelbb
I presume it's related to the fact that the X coordinate is duplciated
in the two records, and the grid generated by adehabitatHR is a single
column,
Here's a test:
require(adehabitatHR)
m - matrix(c(5419574 ,390051, 5419490 , 390051), ncol = 2,
byrow = TRUE)
tt - as.POSIXct(c(2012-05-30 14:00:00, 2012-05-30 16:00:00))
id - c(Ade=5, Ade-5)
x - kernelbb(as.ltraj(m, date = tt, id = id, typeII=TRUE), sig1 =
4.5766, sig2 = 5, grid = 1000)
This degenerate Y dimension affects downstream methods (like image),
and I'd say this is worth reporting to the package author as an issue:
dim(x[[1]])
[1] 10001
dim(x[[2]])
[1] 10001
The clue comes from points2grid in sp:
Warning messages:
1: In points2grid(points, tolerance, round) :
cell size from constant coordinate 2 possibly taken from other coordinate
You could override the auto-generated grid by passing in your own,
probably something you will want anyway so that your collection share
the same extent and resolution. See ?kernelbb, the grid argument can
be a Spatial object rather than a pixel size.
I would really wonder about what is the point in generating summaries
from single line-segments, but that is off-topic I guess.
Cheers, Mike.
On Sat, May 18, 2013 at 4:47 AM, Rémi Lesmerises
remilesmeri...@yahoo.ca wrote:
There was some mistakes in my previous sending. The following are correct.
This is the problematic data, especially the line 43, but when I removed that
line, it bugs at line 62 and so on.
sig1timeid
UTMnorthin UTMeasting
40 4.5766 2012.05.30 08:00:00 Ade-5 5419576 390052
41 4.5766 2012.05.30 10:00:00 Ade-5 5419581 390058
42 4.5766 2012.05.30 12:00:00 Ade-5 5419560 390045
43 4.5766 2012.05.30 14:00:00 Ade-5 5419574 390051
44 4.5766 2012.05.30 16:00:00 Ade-5 5419490 390051
45 4.5766 2012.05.30 18:00:00 Ade-5 5419435 390293
46 4.5766 2012.05.30 20:00:00 Ade-5 5419661 390876
47 4.5766 2012.05.30 22:00:00 Ade-5 5419934 390673
48 4.5766 2012.05.31 02:00:00 Ade-5 5420636 389777
49 4.5766 2012.06.05 02:00:00 Ade-7 5419275 391206
50 4.5766 2012.06.05 04:00:00 Ade-7 5419276 391202
This is a data frame so before the loop (see commands in the previous mail
below), I transform with the following commands:
coordinates(fece)-~UTMeasting+UTMnorthin
loc@data$time - as.POSIXct(strptime(as.character(loc@data$time),%Y.%m.%d
%H:%M:%S))
Rémi Lesmerises, biol. M.Sc.,
PH.D candidate
Université du Québec à Rimouski
De : David Winsemius dwinsem...@comcast.net
À : Rémi Lesmerises remilesmeri...@yahoo.ca
Cc : r-help@r-project.org r-help@r-project.org
Envoyé le : vendredi 17 mai 2013 13h53
Objet : Re: [R] Error with adehabitatHR and kernelbb
On May 17, 2013, at 7:44 AM, Rémi Lesmerises wrote:
Dear all,
I'm trying to get a Brownian bridge kernel (kernelbb) for each combination
of two consecutive animal locations (see commands below) and put them, with
a loop, inside a list. It works well at the beginning but after 42 runs, it
appears the following warning :
Error in seq.default(yli[1], yli[2], by = diff(xg[1:2])) :
invalid (to - from)/by in seq(.)
I looked at the coordinates, at the id, at the time of the run 43 and it's
all good...
I looked on the net and it happened to only one person and there was no
answer to his post.
I wonder if that posting (like yours) had no data on which to display the
problem or to test potential solutions? I would think you would want to post
any setup objects and then data for items 40-45.
Someone could help me?
## commands
BBtraj - list()
for (i in 1:(nrow(loc@data)-1)) {
BBtraj[[i]] - kernelbb(as.ltraj(loc@coords[i:(i+1),],
date=loc@data$time[i:(i+1)], id = as.character(loc@data$id[i:(i+1)]),
typeII = TRUE), sig1=as.numeric(as.character(loc@data$sig1[i])), sig2= 5,
grid = 1000)
}
--
David Winsemius
Alameda, CA, USA
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--
Michael Sumner
Hobart, Australia
e-mail: mdsum...@gmail.com
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[R] Heterogeneous negative binomial
I have seen several queries about parameterizing the negative binomial scale parameter. This is called the heterogeneous negative binomial. I have written a function called nbinomial which is in the msme package on CRAN. Type ?nbinomial to see the help file. The default model is a negative binomial for which the dispersion parameter is directly related to mu, which is how Stata, SAS, SPSS, Limdep, and so forth parameterize the negative binomial. The direct parameterization make sense in that the more variation or correlation there is in a Poisson model, the greater is the value of the dispersion parameter which is adjusting for the excessive variation. With this parameterization the dispersion parameter is directly related to both mu, as well as the dispersion statistic, or Pearson Chi2/(residual DOF). A dispersion parameter of 0 is Poisson, which is equidispersed. When the dispersion parameter for other mixture models such as generalized Poisson and Poisson inverse Gaussian is zero, the models reduce to Poisson. I also provide an option so that the output is similar to glm.nb, for which the dispersion parameter is indirectly related to the mean. I have also provided the abililty of nbinomial to parameterize the dispersion parameter, providing Coefficients, SEs, CIs etc for the predictors of the dispersion, as there are coefficients etc for the mean parameter. The output look nearly identical to glm.nb, except that I also display a summary of Pearson residuals, As well as the null and residual Pearson Chi2, and dispersion statistic. The dispersion parameter is listed At the bottom of the table of coefficients, with SE, Z, p-value and confidence intervals. You may select any variable(s) in the data to be a predictor(s) of the dispersion. Predictors of the dispersion parameter, if positive and significant, indicate that they influence the extra variability of diswhich likely have a bearn I also provide a number of saved post-estimation statistics when nbinomial is run, which the analyst may use in additional analysis. The function is one of a number of functions that are included in Hilbe and Robinson, Methods of Statistical Model Estimation, Chapman Hall/CRC, which is due to be published in the next two weeks. The msme Package should be thought of as an adjunct package to the COUNT package, which is on CRAN and provides the data sets, functions and a host of scripts for Hilbe, Negative Binomial Regression, 2nd edition, Cambridge University Press (2011). Best, J. Hilbe Joseph M. Hilbe, PhD Emer Prof, Univ of Hawaii Adj Prof of Statistics, Arizona St Univ; SSA Program, NASA/Jet Propulsion Laboratory, Caltech President, International Astrostatistics Association Coordinating editor, Cambridge Univ Press Series on Predictive Analytics Email: hi...@asu.edu or jhi...@aol.com URL: http://works.bepress.com/joseph_hilbe/ [[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.
