Re: [R] how to analysisi spectrum of a dataset with NA value
This behavior is typical for "fast" Fourier Transform algorithms. You need to decide what theory you want to use in dealing with your missing data, and apply that yourself. There are some packages that can help you (e.g. Lomb-Scargle or discrete slow Fourier transform). Search with package sos or use the RSiteSearch function. Please read the Posting Guide regarding not using HTML formatted email... your code fragment was unreadable as is common with non-plain text. --- Jeff NewmillerThe . . Go Live... DCN:Basics: ##.#. ##.#. 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. Jie Tang wrote: >hi R users > >I have a large 1D dataset and some of them is NA value . > >I found I cound get the spectrum by such a command. > >ua=c��10��30 ��40��50��NA�� >spectrum(ua) > and I could not use na.rm just like mean or sd function > >How could I get the spectrum of ua ? > > 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] multicore and mclapply problem in calculation server
Hello
I am using package multicore for parallel computing in a Altix UltraViolet
1000 server with 64 CPUs and 960 GB of RAM memory. Access is managed by
means of a SGE queue system. This is the first time I am using parallel
computing and my experience with supercomputers is quite limited. So any
help will be much, much appreciated.
My experiment consists of a number of runs (N.runs) each involving a number
of permutations (N. perms). (An excerpt of the code is included below.) The
permutations are very time consuming and I am using mclapply to distribute
the job among a given number of cpus (usually 12 to 24). The problem is that
the system administrators notice that threads keep increasing as the program
is executed to the point that they compromise the functioning of the whole
system and have to abort the job.
I have tried to specify in the bash file sent to the queue RAM limits (using
ulimit) and the number of cpus to be used but it doesn't help.
An example of the code I am using may be
#LOAD LIBRARIES NEEDED:
library(ape)
library(phytools)
library(phangorn)
library(multicore)
#
SOME 100 LINES HERE DEVOTED TO DEFINE FUNCTIONS -- OMITTED FOR BREVITY
###
body <- function (N.perm) { #MAIN BODY = 1 RUN -- IT PUTS
FUNCTIONS TOGETHER
HP <- HPrandomizer(NH,NP,N.assoc)
linH = readLines(conH, n= N.perm)
linP = readLines(conP, n= N.perm)
stat.matrix <- matrix((rep(NA, 6*N.perm)), ,6)
#
wrapper <- function (x) { # THIS FUCNTION IS SUPPOSED TO BE
PARALLELIZED (SEE BELOW)
treeH <- read.tree(text=linH[x])
treeP <- read.tree(text=linP[x])
mrcaL <- MRCALink.simul (treeH, treeP, HP)
stat.matrix[x,] <- three.stat(mrcaL)
}
x <- c(1: length(linH))#NOTE THAT linH IS SUPPOSED TO BE =
N.perm
stat.matrix <- do.call(rbind, mclapply(x, wrapper, mc.cores= 6)) # USE OF
MCLAPPLY
Pstat <- apply(stat.matrix, 2, rank)[1,]/length(linH)
write(c(stat.matrix[1,], Pstat), file =
"/scratch/ba/balbuena/30H30P40.txt", sep ="\t", append =TRUE,ncolumns=12)
}
ptm <- proc.time() # THE PROGRAM STARTS HERE
NH= 30
NP= 30
N.assoc= 40
N.runs = 1000
N.perm = 999
conH = file("/scratch/ba/balbuena/1MH_30.tre", open="rt") # READS TEXT DATA
FROM EXTERNAL FILE
conP = file("/scratch/ba/balbuena/1MP_30.tre", open="rt") #""
"" " "
replicate (N.runs, body(N.perm)) # LOOPING body NUMBER OF RUNS
close(conH)
close(conP)
proc.time() - ptm
Than you very much for your attention
Juan A. Balbuena
--
Dr. Juan A. Balbuena
Marine Zoology Unit
Cavanilles Institute of Biodiversity and Evolutionary Biology
University of
Valencia
[1]http://www.uv.es/~balbuena
P.O. Box 22085
[2]http://www.uv.es/cavanilles/zoomarin/index.htm
46071 Valencia, Spain
[3]http://cetus.uv.es/mullpardb/index.html
e-mail: [4]j.a.balbu...@uv.estel. +34 963 543 658fax +34 963 543 733
NOTE! For shipments by EXPRESS COURIER use the following street address:
C/ Catedrático José Beltrán 2, 46980 Paterna (Valencia), Spain.
References
1. http://www.uv.es/%7Ebalbuena
2. http://www.uv.es/cavanilles/zoomarin/index.htm
3. http://cetus.uv.es/mullpardb/index.html
4. mailto:j.a.balbu...@uv.es
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[R] Tinn-R user list on Google groups
Dears Tinn-R users, Below a link to the new user list on Google groups: https://groups.google.com/forum/#!forum/tinn-r All the best, -- ///\\\///\\\///\\\///\\\///\\\///\\\///\\\///\\\ Jose Claudio Faria Estatistica UESC/DCET/Brasil joseclaudio.faria at gmail.com Telefones: 55(73)3680.5545 - UESC 55(73)9100.7351 - TIM 55(73)8817.6159 - OI ///\\\///\\\///\\\///\\\///\\\///\\\///\\\///\\\ __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] ifelse statement with two vectors of different length
Hi,
Please show a reproducible example.
countrydiff <- c("Albania", "Algeria", "Belarus", "Canada", "Germany")
long_df <- data.frame(country_name = c("Algeria", "Guyana", "Hungary",
"Algeria", "Canada", "Iran", "Iran", "Norway","Uruguay", "Zimbabwe") )
ifelse(long_df$country_name %in% countrydiff,1,0)
# [1] 1 0 0 1 1 0 0 0 0 0
#or
1*(long_df$country_name %in% countrydiff)
# [1] 1 0 0 1 1 0 0 0 0 0
A.K.
Dear list-members,
I have the following problem: I have a vector (countrydiff) with
length 72 and another vector (long_df$country_name) which is about
12000 long. Basically what I want to do is to if the factor level (or
string name) in long_df$country_name appears on the countrydiff, then
long_df$povdat should be equal to 1, if it does not appear on the
countrydiff vector then long_df$povdat should be equal to zero. I have
tried different combinations and read some. The following code should in
my mind do it, but it doesn’t:
long_df$povdat<-ifelse(long_df$country_name == countrydiff, 1, 0)
long_df$povdat<-ifelse(long_df$country_name %in% countrydiff, 1, 0)
Additional information: the factor vector countrydiff contains
unique country names (Albania, Zimbabwe etc.), whereas
long_df$country_name also contains country names albeit not unique since
it is in longform. The unique names that appear in long_df$country_name
is around 200.
Any suggestions?
Thanks in advance.
Best
Adel
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[R] ifelse statement with two vectors of different length
Dear list-members, I have the following problem: I have a vector (countrydiff) with length 72 and another vector (long_df$country_name) which is about 12000 long. Basically what I want to do is to if the factor level (or string name) in long_df$country_name appears on the countrydiff, then long_df$povdat should be equal to 1, if it does not appear on the countrydiff vector then long_df$povdat should be equal to zero. I have tried different combinations and read some. The following code should in my mind do it, but it doesn’t: long_df$povdat<-ifelse(long_df$country_name == countrydiff, 1, 0) long_df$povdat<-ifelse(long_df$country_name %in% countrydiff, 1, 0) Additional information: the factor vector countrydiff contains unique country names (Albania, Zimbabwe etc.), whereas long_df$country_name also contains country names albeit not unique since it is in longform. The unique names that appear in long_df$country_name is around 200. Any suggestions? Thanks in advance. Best Adel -- View this message in context: http://r.789695.n4.nabble.com/ifelse-statement-with-two-vectors-of-different-length-tp4682401.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] calculating power function
Hello, i know there are several functions on r to calculate the power of a test. f.e the functions of pwr packet. In this packet the use the power function (1-\beta) to calculate. My question is how we get this function for anova (pwr.anova.test)? I know that 1-\beta=P(H1|H1) = P(F>F_[1-\alpha,k-1,N-k]) = 1-P(Fhttp://r.789695.n4.nabble.com/calculating-power-function-tp4682396.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] 3D Surface plot
Hi everyone, I am a very new user of r (doing most of my previous stuff in vba). I am now mandated to draw a 3-d surface of a mine pit hole. I have all the location points (around 5000 points) of the pit in a CSV file under 3 column X, Y & Z. However, going from page to page on the web, I could not figure out how to code the figure. Something I noted is that most of the time X and Y have to be in ascending order but the data I have are not (since they are finite points of the pit, if X gets in ascending order, Y and Z are not). Also with the data, sometimes 2 rows fallowing each ohter have the same X and Y value with a different Z. That would be nice if things could be modeled with rgl function in order to have a rotating pit, but I would be glad to only have a 3d surface to start! Hope you can help. Sincerly, Simon Delay-Fortier __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] ifelse statement with two vectors of different length
Hi, Suggestion 1: read http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example and bookmark it for future reference. Suggestion 2: set.seed(123) countrydiff <- letters[1:5] long_df <- data.frame(country_name = sample(letters[1:8], 20, replace=TRUE)) long_df$povdat <- as.numeric(long_df$country_name %in% countrydiff) Sarah On Wed, Dec 18, 2013 at 8:57 AM, Adel wrote: > > Dear list-members, > > I have the following problem: I have a vector (countrydiff) with length 72 > and another vector (long_df$country_name) which is about 12000 long. > Basically what I want to do is to if the factor level (or string name) in > long_df$country_name appears on the countrydiff, then long_df$povdat should > be equal to 1, if it does not appear on the countrydiff vector then > long_df$povdat should be equal to zero. I have tried different combinations > and read some. The following code should in my mind do it, but it doesn’t: > > long_df$povdat<-ifelse(long_df$country_name == countrydiff, 1, 0) > > long_df$povdat<-ifelse(long_df$country_name %in% countrydiff, 1, 0) > > Additional information: the factor vector countrydiff contains unique > country names (Albania, Zimbabwe etc.), whereas long_df$country_name also > contains country names albeit not unique since it is in longform. The unique > names that appear in long_df$country_name is around 200. > > > Any suggestions? > Thanks in advance. > > Best > Adel > > -- Sarah Goslee http://www.functionaldiversity.org __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] ANOVA repeated mesures
Hi to everyone, I am tring to make a Anova with repeated measures,my data set
looks like:
participantes <- c("1", "2", "3", "4", "5", "6", "7", "8", "1", "2", "3", "4",
"5", "6", "7", "8", "1", "2", "3", "4", "5", "6", "7", "8")
grupo <- factor(c(rep("A", 8), rep("B", 8), rep("C", 8)))valor <- c(1, 2, 4, 1,
1, 2, 2, 3, 3, 4, 4, 2, 3, 4, 4, 3, 4, 5, 3, 5, 5, 3, 4, 6)df2 <-
data.frame(participantes, grupo, valor)
I want to find if there is differences in the "grupo" in each participant. I
don't know how to find this p-value for each participant, could you help me,
please??Thanks a lot
[[alternative HTML version deleted]]
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Re: [R] read ".slk" file
Hi Extension does not specify file format. You can rename any file with any extension without changing its nature. However slk stays for symbolic link and therefore it just brings actual file to Excel. Maybe you could start to play with http://stat.ethz.ch/R-manual/R-devel/library/base/html/files.html Regards Petr > -Original Message- > From: r-help-boun...@r-project.org [mailto:r-help-bounces@r- > project.org] On Behalf Of Santosh > Sent: Tuesday, December 17, 2013 12:11 AM > To: r-help > Subject: [R] read ".slk" file > > Dear Rxperts.. > > I recently received a data file with the extension ".slk". If I save > the file as MS Excel file, I am able to read in R without issues. Is > it possible to read this ".slk" file without converting into another R- > readable data format? > > Regards, > Santosh > > [[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] ifelse statement with two vectors of different length
Dear Arun
Thanks for your reply, it made me realize that the problem was not in the
code but in the levels() of the factors. Some countries had some extra
spacing which made the ifelse() function not work. So if I modify your code
(added space to countrydiff), it will then look something like this:
countrydiff <- c("Albania", "Algeria", "Belarus", "Canada ",
"Germany ")
long_df <- data.frame(country_name = c("Algeria", "Guyana", "Hungary",
"Algeria", "Canada", "Iran", "Iran", "Norway","Uruguay", "Zimbabwe") )
I had to use the gsub to fix this first.
Interestingly, the setdiff() function did not react on spacing difference
which I used before coming to the ifelse statement and therefore I did not
react on this in the first place
#no reaction from R on spacing diff.
setdiff(countrydiff, long_df$country_name)
Nevertheless, thanks again for being helpful!
Adel
--
View this message in context:
http://r.789695.n4.nabble.com/ifelse-statement-with-two-vectors-of-different-length-tp4682401p4682403.html
Sent from the R help mailing list archive at Nabble.com.
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] ifelse statement with two vectors of different length
Hi Adel,
If the problem is the spacing, then
library(stringr)
1*(long_df$country_name %in% str_trim(countrydiff))
# [1] 1 0 0 1 1 0 0 0 0 0
A.K.
Dear Arun
Thanks for your reply, it made me realize that the problem was
not in the code but in the levels() of the factors. Some countries had
some extra spacing which made the ifelse() function not work. So if I
modify your code (added space to countrydiff), it will then look
something like this:
countrydiff <- c("Albania ", "Algeria ", "Belarus ", "Canada ",
"Germany ")
long_df <- data.frame(country_name = c("Algeria", "Guyana",
"Hungary", "Algeria", "Canada", "Iran", "Iran", "Norway","Uruguay",
"Zimbabwe") )
I had to use the gsub to fix this first.
Interestingly, the setdiff() function did not react on
spacing difference which I used before coming to the ifelse statement
and therefore I did not react on this in the first place
#no reaction from R on spacing diff.
setdiff(countrydiff, long_df$country_name)
Nevertheless, thanks again for being helpful!
Adel
On Wednesday, December 18, 2013 9:58 AM, Adel
wrote:
Dear list-members,
I have the following problem: I have a vector (countrydiff) with length 72
and another vector (long_df$country_name) which is about 12000 long.
Basically what I want to do is to if the factor level (or string name) in
long_df$country_name appears on the countrydiff, then long_df$povdat should
be equal to 1, if it does not appear on the countrydiff vector then
long_df$povdat should be equal to zero. I have tried different combinations
and read some. The following code should in my mind do it, but it doesn’t:
long_df$povdat<-ifelse(long_df$country_name == countrydiff, 1, 0)
long_df$povdat<-ifelse(long_df$country_name %in% countrydiff, 1, 0)
Additional information: the factor vector countrydiff contains unique
country names (Albania, Zimbabwe etc.), whereas long_df$country_name also
contains country names albeit not unique since it is in longform. The unique
names that appear in long_df$country_name is around 200.
Any suggestions?
Thanks in advance.
Best
Adel
--
View this message in context:
http://r.789695.n4.nabble.com/ifelse-statement-with-two-vectors-of-different-length-tp4682401.html
Sent from the R help mailing list archive at Nabble.com.
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[R] 3D Surface Plot
Hi everyone,I am a very new user of r. I am now mandated to draw a 3-d surface (and possibly rotating) of a mine pit hole. I have all the location points (around 5000 points) of the pit in a CSV file under 3 column X, Y & Z. The fllowing gives a 3d scatter but i would like to have a surface. attach(EASTPITCREST) par(bg="grey6", col.lab="white", col.axis="white", col.main="white", col.sub="white") scatterplot3d(X, Y, Z, color = "red", pch=19, main="Mine and Drilling Map") I noted is that most of the time X and Y have to be in ascending order but the data I have are not (since they are finite points of the pit, if X gets in ascending order, Y and Z are not). Also with the data, sometimes 2 rows fallowing each ohter have the same X and Y value with a different Z. Hope you can help. Sincerly, Simon Delay-Fortier __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] 3D Surface plot
On Wed, Dec 18, 2013 at 2:52 PM, Simon Delay-Fortier wrote: > Hi everyone, > > I am a very new user of r (doing most of my previous stuff in vba). I am now > mandated to draw a 3-d surface of a mine pit hole. I have all the location > points (around 5000 points) of the pit in a CSV file under 3 column X, Y & Z. > However, going from page to page on the web, I could not figure out how to > code the figure. Something I noted is that most of the time X and Y have to > be in ascending order but the data I have are not (since they are finite > points of the pit, if X gets in ascending order, Y and Z are not). Also with > the data, sometimes 2 rows fallowing each ohter have the same X and Y value > with a different Z. That would be nice if things could be modeled with rgl > function in order to have a rotating pit, but I would be glad to only have a > 3d surface to start! How "3d" is your surface here? Because there's "3d" and there's what's known as "2.5d" A 3d surface could be something like the surface of a sphere, or a cave, or an overhanging cliff, whereas a 2.5d surface is a single-valued function of x and y. If you have a true 3d surface then that's beyond me and involves working out surface normals and all sorts of other clever stuff which I don't know about. If its really 2.5d then yes, you do need to compute the values of your surface on a grid in order to display it using rgl's surface3d function. You can do this with some 2d interpolation code, such as kriging or inverse distance weighting, using assorted packages such as automap (which makes it easy but beware) or gstat. Since this seems to be a real-world geography problem, you might haul over to the r-sig-geo mailing list and as there where the spatial guys hang out. Barry __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Help using mapply to run multiple models
Thanks! that works, more or less. Although the wonky behaviour of mapply that
David pointed out is irritating. I tried deleting the $call item from the
models produced and passing them to stargazer for reporting the results, but
stargazer won't recognize the results even though the class is explicitly "glm
lm".
Does anyone know why mapply produces such weird results?
On 2013-12-18, at 3:29 AM, Dennis Murphy wrote:
> Hi:
>
> Here's a way to generate a list of model objects. Once you have the
> list, you can write or call functions to extract useful pieces of
> information from each model object and use lapply() to call each list
> component recursively.
>
> sample.df<-data.frame(var1=rbinom(50, size=1, prob=0.5),
> var2=rbinom(50, size=2, prob=0.5),
> var3=rbinom(50, size=3, prob=0.5),
> var4=rbinom(50, size=2, prob=0.5),
> var5=rbinom(50, size=2, prob=0.5))
>
> # vector of x-variable names
> xvars <- names(sample.df)[-1]
>
> # function to paste a variable x into a formula object and
> # then pass it to glm()
> f <- function(x)
> {
>form <- as.formula(paste("var1", x, sep = " ~ "))
>glm(form, data = sample.df)
> }
>
> # Apply the function f to each variable in xvars
> modlist <- lapply(xvars, f)
>
> To give you an idea of some of the things you can do with the list:
>
> sapply(modlist, class)# return class of each component
> lapply(modlist, summary) # return the summary of each model
>
> # combine the model coefficients into a two-column matrix
> do.call(rbind, lapply(modlist, coef))
>
>
> You'd probably want to rename the second column since the slopes are
> associated with different x variables.
>
> Dennis
>
> On Tue, Dec 17, 2013 at 5:53 PM, Simon Kiss wrote:
>> I think I'm missing something. I have a data frame that looks below.
>> sample.df<-data.frame(var1=rbinom(50, size=1, prob=0.5), var2=rbinom(50,
>> size=2, prob=0.5), var3=rbinom(50, size=3, prob=0.5), var4=rbinom(50,
>> size=2, prob=0.5), var5=rbinom(50, size=2, prob=0.5))
>>
>> I'd like to run a series of univariate general linear models where var1 is
>> always the dependent variable and each of the other variables is the
>> independent. Then I'd like to summarize each in a table.
>> I've tried :
>>
>> sample.formula=list(var1~var2, var1 ~var3, var1 ~var4, var1~var5)
>> mapply(glm, formula=sample.formula, data=list(sample.df), family='binomial')
>>
>> And that works pretty well, except, I'm left with a matrix that contains all
>> the information I need. I can't figure out how to use summary() properly on
>> this information to usefully report that information.
>>
>> Thank you for any suggestions.
>>
>> *
>> Simon J. Kiss, PhD
>> Assistant Professor, Wilfrid Laurier University
>> 73 George Street
>> Brantford, Ontario, Canada
>> N3T 2C9
>>
>> __
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
*
Simon J. Kiss, PhD
Assistant Professor, Wilfrid Laurier University
73 George Street
Brantford, Ontario, Canada
N3T 2C9
Cell: +1 905 746 7606
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Re: [R] Help using mapply to run multiple models
Dennis, how would your function be modified to allow it to be more flexible in
future.
I'm thinking like:
> f <- function(x='Dependent variable', y='List of Independent Variables',
> z='Data Frame')
> {
>form <- as.formula(paste(x, y, sep = " ~ "))
>glm(form, data =z)
> }
I tried that then using
modlist <- lapply(xvars, f), but it didn't work.
On 2013-12-18, at 3:29 AM, Dennis Murphy wrote:
> Hi:
>
> Here's a way to generate a list of model objects. Once you have the
> list, you can write or call functions to extract useful pieces of
> information from each model object and use lapply() to call each list
> component recursively.
>
> sample.df<-data.frame(var1=rbinom(50, size=1, prob=0.5),
> var2=rbinom(50, size=2, prob=0.5),
> var3=rbinom(50, size=3, prob=0.5),
> var4=rbinom(50, size=2, prob=0.5),
> var5=rbinom(50, size=2, prob=0.5))
>
> # vector of x-variable names
> xvars <- names(sample.df)[-1]
>
> # function to paste a variable x into a formula object and
> # then pass it to glm()
> f <- function(x)
> {
>form <- as.formula(paste("var1", x, sep = " ~ "))
>glm(form, data = sample.df)
> }
>
> # Apply the function f to each variable in xvars
> modlist <- lapply(xvars, f)
>
> To give you an idea of some of the things you can do with the list:
>
> sapply(modlist, class)# return class of each component
> lapply(modlist, summary) # return the summary of each model
>
> # combine the model coefficients into a two-column matrix
> do.call(rbind, lapply(modlist, coef))
>
>
> You'd probably want to rename the second column since the slopes are
> associated with different x variables.
>
> Dennis
>
> On Tue, Dec 17, 2013 at 5:53 PM, Simon Kiss wrote:
>> I think I'm missing something. I have a data frame that looks below.
>> sample.df<-data.frame(var1=rbinom(50, size=1, prob=0.5), var2=rbinom(50,
>> size=2, prob=0.5), var3=rbinom(50, size=3, prob=0.5), var4=rbinom(50,
>> size=2, prob=0.5), var5=rbinom(50, size=2, prob=0.5))
>>
>> I'd like to run a series of univariate general linear models where var1 is
>> always the dependent variable and each of the other variables is the
>> independent. Then I'd like to summarize each in a table.
>> I've tried :
>>
>> sample.formula=list(var1~var2, var1 ~var3, var1 ~var4, var1~var5)
>> mapply(glm, formula=sample.formula, data=list(sample.df), family='binomial')
>>
>> And that works pretty well, except, I'm left with a matrix that contains all
>> the information I need. I can't figure out how to use summary() properly on
>> this information to usefully report that information.
>>
>> Thank you for any suggestions.
>>
>> *
>> Simon J. Kiss, PhD
>> Assistant Professor, Wilfrid Laurier University
>> 73 George Street
>> Brantford, Ontario, Canada
>> N3T 2C9
>>
>> __
>> R-help@r-project.org mailing list
>> https://stat.ethz.ch/mailman/listinfo/r-help
>> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
>> and provide commented, minimal, self-contained, reproducible code.
*
Simon J. Kiss, PhD
Assistant Professor, Wilfrid Laurier University
73 George Street
Brantford, Ontario, Canada
N3T 2C9
Cell: +1 905 746 7606
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Re: [R] Help using mapply to run multiple models
Folks:
1. Haven't closely followed the thread. I'm responding only to Simon's post.
2. ?formula ## Especially note the use of "."
So just make an appropriately constructed data frame for the data
argument of glm:
## example
> df <- data.frame(y=rnorm(9),x1=runif(9), x2=1:9)
> glm(y~.,data=df)
## y does not need to be in the data frame.
Another way to handle the OP is via substitute() or bquote(). I'll
leave that to thers to explain.
-- Bert
On Wed, Dec 18, 2013 at 9:11 AM, Simon Kiss wrote:
> Dennis, how would your function be modified to allow it to be more flexible
> in future.
> I'm thinking like:
>> f <- function(x='Dependent variable', y='List of Independent Variables',
>> z='Data Frame')
>> {
>>form <- as.formula(paste(x, y, sep = " ~ "))
>>glm(form, data =z)
>> }
>
> I tried that then using
> modlist <- lapply(xvars, f), but it didn't work.
>
> On 2013-12-18, at 3:29 AM, Dennis Murphy wrote:
>
>> Hi:
>>
>> Here's a way to generate a list of model objects. Once you have the
>> list, you can write or call functions to extract useful pieces of
>> information from each model object and use lapply() to call each list
>> component recursively.
>>
>> sample.df<-data.frame(var1=rbinom(50, size=1, prob=0.5),
>> var2=rbinom(50, size=2, prob=0.5),
>> var3=rbinom(50, size=3, prob=0.5),
>> var4=rbinom(50, size=2, prob=0.5),
>> var5=rbinom(50, size=2, prob=0.5))
>>
>> # vector of x-variable names
>> xvars <- names(sample.df)[-1]
>>
>> # function to paste a variable x into a formula object and
>> # then pass it to glm()
>> f <- function(x)
>> {
>>form <- as.formula(paste("var1", x, sep = " ~ "))
>>glm(form, data = sample.df)
>> }
>>
>> # Apply the function f to each variable in xvars
>> modlist <- lapply(xvars, f)
>>
>> To give you an idea of some of the things you can do with the list:
>>
>> sapply(modlist, class)# return class of each component
>> lapply(modlist, summary) # return the summary of each model
>>
>> # combine the model coefficients into a two-column matrix
>> do.call(rbind, lapply(modlist, coef))
>>
>>
>> You'd probably want to rename the second column since the slopes are
>> associated with different x variables.
>>
>> Dennis
>>
>> On Tue, Dec 17, 2013 at 5:53 PM, Simon Kiss wrote:
>>> I think I'm missing something. I have a data frame that looks below.
>>> sample.df<-data.frame(var1=rbinom(50, size=1, prob=0.5), var2=rbinom(50,
>>> size=2, prob=0.5), var3=rbinom(50, size=3, prob=0.5), var4=rbinom(50,
>>> size=2, prob=0.5), var5=rbinom(50, size=2, prob=0.5))
>>>
>>> I'd like to run a series of univariate general linear models where var1 is
>>> always the dependent variable and each of the other variables is the
>>> independent. Then I'd like to summarize each in a table.
>>> I've tried :
>>>
>>> sample.formula=list(var1~var2, var1 ~var3, var1 ~var4, var1~var5)
>>> mapply(glm, formula=sample.formula, data=list(sample.df), family='binomial')
>>>
>>> And that works pretty well, except, I'm left with a matrix that contains
>>> all the information I need. I can't figure out how to use summary()
>>> properly on this information to usefully report that information.
>>>
>>> Thank you for any suggestions.
>>>
>>> *
>>> Simon J. Kiss, PhD
>>> Assistant Professor, Wilfrid Laurier University
>>> 73 George Street
>>> Brantford, Ontario, Canada
>>> N3T 2C9
>>>
>>> __
>>> R-help@r-project.org mailing list
>>> https://stat.ethz.ch/mailman/listinfo/r-help
>>> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
>>> and provide commented, minimal, self-contained, reproducible code.
>
> *
> Simon J. Kiss, PhD
> Assistant Professor, Wilfrid Laurier University
> 73 George Street
> Brantford, Ontario, Canada
> N3T 2C9
> Cell: +1 905 746 7606
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
--
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
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[R] How to manipulate this data?
Hi, I want my output to be like this: Value BXR abc DHH abc DHK abc def DSL def ghi DSM abc def ghi DSS def ghi DST ghi DIW abc DIL abc ghi My input dataset is this with colnames name and Value: name Value abc BXR abc DHH abc DHK def DHK def DSL ghi DSL abc DSM def DSM ghi DSM def DSS ghi DSS ghi DST abc DIW abc DIL ghi DIL [[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] estimating survival function from Cox model
The standard error of the curve cannot be extracted from the summary information you have. The variance is made up of two terms, one of which is a sum over all the death times, of a quadratic term per death time. That term involves the variance matrix of the Cox model coefficients, the target value for x (the curve you want to calculate) and the average value of x at that time in the data set from which the Cox model was created. Just like linear regression, the se are higher when you predict "far from the center" of the original data set. Terry Therneau On 12/18/2013 05:00 AM, r-help-requ...@r-project.org wrote: Hi, I have some questions on how to estimate the survival function from a Cox model. I know how to do this in R using survfit(). But let's say the model was done is another software, and I was only given the estimate of baseline cumulative hazard "A0(t=10)" at the specified time "t=10" (baseline cumulative hazard refers to when covariate X=0)and the beta estimate "b" for the covariate used in Cox model "X". So the survival function at time 10 for a given covariate value x can be calculated as: A(t=10|X=x) = exp(b*x)*A0(t=10) where A is cumulative hazard when X=x S(t=10|X=x) = exp(-A(t=10|X=x)) where S is survival function to be calculated Now I want to calculate confidence interval for S(t=10|X=x). I think I need to calculate the CI for cumulative hazard A(t=10|X=x) first and then exponentiate it to get CI for S, based on the relationship S = exp(-A). To get CI for A, I need to calculate the estimate of standard error of A. I know the other software can give me the standard error of A0, the baseline cumulative hazard. Based on the relationship A = exp(b*x)*A0, I guess I'll need the standard error for b. But how do I calculate the standard error for A based on standard errors for A0 and b? Any insights would be greatly appreciated! John __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to manipulate this data?
Hi, Try: dat1<- read.table(text="name Value abc BXR abc DHH abc DHK def DHK def DSL ghi DSL abc DSM def DSM ghi DSM def DSS ghi DSS ghi DST abc DIW abc DIL ghi DIL",sep="",header=TRUE,stringsAsFactors=FALSE) aggregate(name~Value,data=dat1,paste,collapse=" ") #or library(plyr) ddply(dat1,.(Value),summarize, name=lapply(list(Value),paste,collapse=" ")) A.K. Hi, I want my output to be like this: Value BXR abc DHH abc DHK abc def DSL def ghi DSM abc def ghi DSS def ghi DST ghi DIW abc DIL abc ghi My input dataset is this with colnames name and Value: name Value abc BXR abc DHH abc DHK def DHK def DSL ghi DSL abc DSM def DSM ghi DSM def DSS ghi DSS ghi DST abc DIW abc DIL ghi DIL __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Predicting response from fitted linear model with incomplete new sample data
I would like to predict a new response from a fitted linear model where the new data is a single case with a missing value. My reading of the help on predict() is inconclusive on whether this is possible. Leaving out the missing value or setting it to NA both fail but differently, see example code below. > y <- runif(50) > x1 <- rnorm(50) > x2 <- rnorm(50) > dat <- data.frame(y, x1, x2) > mod <- lm(y~.,data=dat) > summary(mod) Call: lm(formula = y ~ ., data = dat) Residuals: Min 1Q Median 3Q Max -0.50467 -0.28997 0.01457 0.27970 0.47791 Coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 0.500980.04577 10.945 1.6e-14 *** x1 -0.017620.04172 -0.4220.675 x2 -0.027530.04920 -0.5600.578 --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Residual standard error: 0.3177 on 47 degrees of freedom Multiple R-squared: 0.009301, Adjusted R-squared: -0.03286 F-statistic: 0.2206 on 2 and 47 DF, p-value: 0.8028 > predict(mod, newdata=data.frame(x1=0.1, x2=0.3)) #OK as expected 1 0.4909624 > predict(mod, newdata=data.frame(x1=0.1)) # x2 missing Error in model.frame.default(Terms, newdata, na.action = na.action, xlev = object$xlevels) : variable lengths differ (found for 'x2') In addition: Warning message: 'newdata' had 1 row but variables found have 50 rows > predict(mod, newdata=data.frame(x1=0.1, x2=NA)) #x2=NA Error: variable 'x2' was fitted with type "numeric" but type "logical" was supplied > Thanks Chris __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] plot() function: color transparency
I found all the color transparency was defined with character color, or rgb color. What if I have number code and still try to modify the transparency? For example: >x=c(1:5) > color=c(2,2,3,4,5) > plot(x, col=color) > plot(x, col=color,pch=20) here I defined color by numbers, how can I modify the transparency? Thanks a lot for any input! [[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] Coxph convergence
I'll re-enter the fray. The data set is an example where coxph is incorrect; due to round off error it is treating a 5 column rank 3 matrix as if it were rank 4. This of course results in 0 digits of precision. Immediate fix, for the user, is to add "toler.chol= 1e-10" to their coxph call. It is very likely that they will never ever have to change this value. I will look into changing the default from its current value of .Machine$double.eps^.75. However, I first have to check that this does not break anything else. All "epsilon" constants are a delicate dance between mutiple data sets, and anyone with long experience in numerical anlysis will tell you that it is impossible to find constants that will work for every data set. This is true for linear models, logistic, Cox, ... you name it. In summary: I appreciate the example. I'll add to my list of "nasty" problems. I may be able to fix long term, and maybe not. Changing the constant may break something else. I've given a good short term fix. Terry T. On 12/18/2013 05:00 AM, r-help-requ...@r-project.org wrote: Your comprehension of the issue seem to be entirely wrong. Between r11513 and r11516, some tuning of internal parmeters were done, so the process of finding the rank of a singular matrix no longer converges (within the time/tolerance implicitly specified). There are warnings issued, but then there are misc warnings before and after (and one gets "desensitised" about them). Also the nature of the problem, which is to test for possibility of interactions - or lacking thereof - outcome ~ factor A + factor B + factor A x factor B or just extra terms in "outcome ~ factor A + factor B + ..." as an exploration of auxiliary effects, more often than not extra terms won't make any difference and the matrix involved just isn't the nicest to manipulate; it is in the nature of that kind of exploratory work. Professor Therneau replied that it is possible to get the older convergent behaviour by manual tuning of some of the convergence criteria parameters; I have responded that while that is possible, often one is simultaneously exploring many models with many possible auxiliary effects (and lacking thereof), manual tuning for each is neither feasible nor appropriate; and we sort of left it at that. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] plot() function: color transparency
On 13-12-18 2:08 PM, capricy gao wrote: I found all the color transparency was defined with character color, or rgb color. What if I have number code and still try to modify the transparency? For example: x=c(1:5) color=c(2,2,3,4,5) plot(x, col=color) plot(x, col=color,pch=20) here I defined color by numbers, how can I modify the transparency? See the examples for ?palette. 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] Predicting response from fitted linear model with incomplete new sample data
As far as I can discern, your question makes no sense at all.
Suppose you *know* that y = 2 + 3*x1 + 4*x2.
Now what should you predict when x1 = 6 (with x2 "missing"/unknown)?
See fortune("magic").
On 19/12/13 07:18, Chris Wilkinson wrote:
I would like to predict a new response from a fitted linear model where the
new data is a single case with a missing value. My reading of the help on
predict() is inconclusive on whether this is possible.
Leaving out the missing value or setting it to NA both fail but differently,
see example code below.
y <- runif(50)
x1 <- rnorm(50)
x2 <- rnorm(50)
dat <- data.frame(y, x1, x2)
mod <- lm(y~.,data=dat)
summary(mod)
Call:
lm(formula = y ~ ., data = dat)
Residuals:
Min 1Q Median 3Q Max
-0.50467 -0.28997 0.01457 0.27970 0.47791
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 0.500980.04577 10.945 1.6e-14 ***
x1 -0.017620.04172 -0.4220.675
x2 -0.027530.04920 -0.5600.578
---
Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1
Residual standard error: 0.3177 on 47 degrees of freedom
Multiple R-squared: 0.009301, Adjusted R-squared: -0.03286
F-statistic: 0.2206 on 2 and 47 DF, p-value: 0.8028
predict(mod, newdata=data.frame(x1=0.1, x2=0.3)) #OK as expected
1
0.4909624
predict(mod, newdata=data.frame(x1=0.1)) # x2 missing
Error in model.frame.default(Terms, newdata, na.action = na.action, xlev =
object$xlevels) :
variable lengths differ (found for 'x2')
In addition: Warning message:
'newdata' had 1 row but variables found have 50 rows
predict(mod, newdata=data.frame(x1=0.1, x2=NA)) #x2=NA
Error: variable 'x2' was fitted with type "numeric" but type "logical" was
supplied
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Re: [R] Using assign with mapply
The take home message that you should be learning from your struggles
is to "Not Use The 'assign' Function!" and "Do Not Use Global
Variables Like This".
R has lists (and environments) that make working with objects that are
associated with each other much simpler and fits better with the
functional programming style of R.
For example you can create a list from your data frame quickly and
easily with code like:
mydata <- as.list(kkk$vals)
names(mydata) <- kkk$vars
or
mydata <- setNames( as.list(kkk$vals), kkk$vars )
then you will have you variables (with names) inside the list (mydata
in this example, but name it whatever you want)
This list can then be passed out of a function or otherwise used.
To access a specific variable by name you can do:
mydata$var1
or
mydata[['var2']]
or
with(mydata, var3)
but you can also do things like (and this is often a follow-up
question to questions like yours):
varname <- 'var3'
mydata[[ varname ]]
and you can also use lapply and sapply to do the same action on every
variable in your list:
sapply( mydata, function(x) x + 5 )
instead of having to loop through a bunch of global variables.
And if you want to save or delete these, now you just save or delete
the entire list rather than having to loop through the set of global
variables.
If you tell us more about how you want to use these variables we can
give more suggestions, but the main point is that in the long run you
will be happier learning to use lists (and possibly environments) in
place of trying to create and work with global variables like you
asked about.
On Mon, Dec 16, 2013 at 8:55 AM, Julio Sergio Santana
wrote:
> Julio Sergio Santana gmail.com> writes:
>
>>
>> I have a data frame whose first colum contains the names of the variables
>> and whose second colum contains the values to assign to them:
>>
>>: kkk <- data.frame(vars=c("var1", "var2", "var3"),
>> vals=c(10, 20, 30), stringsAsFactors=F)
>>
>
> For those interested in the problem this is how I solved the problem:
>
>
> I want to have something similar to:
> #
> # var1 <- 10
> # var2 <- 20
> # var3 <- 30
>
> my first trial was:
>
>mapply(assign, kkk$vars, kkk$vals)
> ## var1 var2 var3
> ## 10 20 30
> #
>
> This is, however, what I got:
>
>var1
> ## Error: object 'var1' not found
>
> David Winsemius suggested me something similar to
>
>
>mapply(assign, kkk$vars, kkk$vals, MoreArgs = list(pos = 1))
> # or:
>mapply(assign, kkk$vars, kkk$vals, MoreArgs = list(envir = .GlobalEnv))
>
> var1
> ## [1] 10
>
> This almost works, but what if this construction is used inside a function?
>
>example <- function () {
> var1 <- 250
> kkk <- data.frame(vars=c("var1", "var2", "var3"),
> vals=c(10, 20, 30), stringsAsFactors=F)
> mapply(assign, kkk$vars, kkk$vals, MoreArgs = list(pos = 1))
> print (var2)
> print (var1)
>}
>
>example()
> ## [1] 20
> ## [1] 250
>
> var1, which was defined inside the function, isn't modified
>
> To fix this, I defined the function as follows:
>
>example <- function () {
> var1 <- 250
> kkk <- data.frame(vars=c("var1", "var2", "var3"),
>vals=c(10, 20, 30), stringsAsFactors=F)
> mapply(assign, kkk$vars, kkk$vals,
> MoreArgs = list(pos = sys.frame(sys.nframe(
> # sys.nframe() is the number of the frame created inside the function
> # and sys.frame() establishes it as the one assign uses to set values
> print (var2)
> print (var1)
>}
>
>example()
> ## [1] 20
> ## [1] 10
>
> And the purpose is got
>
> Thanks,
>
> -Sergio.
>
> __
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> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
--
Gregory (Greg) L. Snow Ph.D.
538...@gmail.com
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Re: [R] Help using mapply to run multiple models
Try something like the following. Because lm() evaluates many
of its arguments in nonstandard ways, f() manipulates the call
and then evaluates it in the frame from which f() was called.
It also puts that environment on the formula that it creates so
it can refer to variables in that environment.
f <- function (responseName, predictorNames, data, ..., envir =
parent.frame())
{
call <- match.call()
call$formula <- formula(envir = envir, paste(responseName, sep = " ~ ",
paste0("`", predictorNames, "`", collapse = " + ")))
call[[1]] <- quote(glm) # 'f' -> 'glm'
call$responseName <- NULL # omit responseName=
call$predictorNames <- NULL # omit 'predictorNames='
eval(call, envir = envir)
}
as in
z <- lapply(list(c("hp","drat"), c("cyl"), c("am","gear")),
FUN=function(preds)f("carb", preds, data=mtcars, family=poisson))
lapply(z, summary)
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Behalf
> Of Simon Kiss
> Sent: Wednesday, December 18, 2013 9:11 AM
> To: Dennis Murphy
> Cc: r-help@r-project.org
> Subject: Re: [R] Help using mapply to run multiple models
>
> Dennis, how would your function be modified to allow it to be more flexible
> in future.
> I'm thinking like:
> > f <- function(x='Dependent variable', y='List of Independent Variables',
> > z='Data Frame')
> > {
> >form <- as.formula(paste(x, y, sep = " ~ "))
> >glm(form, data =z)
> > }
>
> I tried that then using
> modlist <- lapply(xvars, f), but it didn't work.
>
> On 2013-12-18, at 3:29 AM, Dennis Murphy wrote:
>
> > Hi:
> >
> > Here's a way to generate a list of model objects. Once you have the
> > list, you can write or call functions to extract useful pieces of
> > information from each model object and use lapply() to call each list
> > component recursively.
> >
> > sample.df<-data.frame(var1=rbinom(50, size=1, prob=0.5),
> > var2=rbinom(50, size=2, prob=0.5),
> > var3=rbinom(50, size=3, prob=0.5),
> > var4=rbinom(50, size=2, prob=0.5),
> > var5=rbinom(50, size=2, prob=0.5))
> >
> > # vector of x-variable names
> > xvars <- names(sample.df)[-1]
> >
> > # function to paste a variable x into a formula object and
> > # then pass it to glm()
> > f <- function(x)
> > {
> >form <- as.formula(paste("var1", x, sep = " ~ "))
> >glm(form, data = sample.df)
> > }
> >
> > # Apply the function f to each variable in xvars
> > modlist <- lapply(xvars, f)
> >
> > To give you an idea of some of the things you can do with the list:
> >
> > sapply(modlist, class)# return class of each component
> > lapply(modlist, summary) # return the summary of each model
> >
> > # combine the model coefficients into a two-column matrix
> > do.call(rbind, lapply(modlist, coef))
> >
> >
> > You'd probably want to rename the second column since the slopes are
> > associated with different x variables.
> >
> > Dennis
> >
> > On Tue, Dec 17, 2013 at 5:53 PM, Simon Kiss wrote:
> >> I think I'm missing something. I have a data frame that looks below.
> >> sample.df<-data.frame(var1=rbinom(50, size=1, prob=0.5), var2=rbinom(50,
> >> size=2,
> prob=0.5), var3=rbinom(50, size=3, prob=0.5), var4=rbinom(50, size=2,
> prob=0.5),
> var5=rbinom(50, size=2, prob=0.5))
> >>
> >> I'd like to run a series of univariate general linear models where var1 is
> >> always the
> dependent variable and each of the other variables is the independent. Then
> I'd like to
> summarize each in a table.
> >> I've tried :
> >>
> >> sample.formula=list(var1~var2, var1 ~var3, var1 ~var4, var1~var5)
> >> mapply(glm, formula=sample.formula, data=list(sample.df),
> >> family='binomial')
> >>
> >> And that works pretty well, except, I'm left with a matrix that contains
> >> all the
> information I need. I can't figure out how to use summary() properly on this
> information
> to usefully report that information.
> >>
> >> Thank you for any suggestions.
> >>
> >> *
> >> Simon J. Kiss, PhD
> >> Assistant Professor, Wilfrid Laurier University
> >> 73 George Street
> >> Brantford, Ontario, Canada
> >> N3T 2C9
> >>
> >> __
> >> R-help@r-project.org mailing list
> >> https://stat.ethz.ch/mailman/listinfo/r-help
> >> PLEASE do read the posting guide
> >> http://www.R-project.org/posting-guide.html
> >> and provide commented, minimal, self-contained, reproducible code.
>
> *
> Simon J. Kiss, PhD
> Assistant Professor, Wilfrid Laurier University
> 73 George Street
> Brantford, Ontario, Canada
> N3T 2C9
> Cell: +1 905 746 7606
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the pos
Re: [R] Exporting R graphics into Word without losing graph quality
Another option to consider if your goal is to create a word file with 1 or more plots in it (possibly intermingled with text and other output) is to use the knitr or pander packages (or odfWeave or sweave or ...). This way you can create a script (or template file) that sets a couple of options up front (width, height, resolution, file type) and creates the graphs (and possibly other output to be included), run the script (and maybe the external program pandoc) and you have a word document with the plots without needing to copy/paste or import. This becomes a real time saver when the client comes back to you and says there was a typo in the data, the 13 on line 27 needs to be an 18 and can you rerun everything with that change? (if something along those lines has not happened to you yet, it will). On Mon, Dec 16, 2013 at 9:33 PM, david hamer wrote: > Thanks to everyone for the helpful suggestions. -- David. > > > On Mon, Dec 16, 2013 at 7:23 PM, Steve Taylor wrote: > >> > From: Duncan Murdoch... >> >> > Don't use a bitmap format (png). >> I disagree. Each vector format comes with its own problems. >> >> > Don't produce your graph in one format (screen display), then convert to >> > another (png). Open the device in the format you want for the final >> file. >> Agreed. >> >> > Use a vector format for output. >> Why? Sure, that's good advice in the ideal (pdflatex) world, but not >> necessarily the best of advice for Word users. >> >> > I don't know what kinds Word supports, but >> > EPS or PDF would likely be best; if it can't read those, then Windows >> metafile >> > (via windows() to open the device) would be best. >> None of these works well, if at all, in my experience with Word. >> >> > Don't use Word. >> Some of us don't really have a choice. >> >> >> > > [[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. -- Gregory (Greg) L. Snow Ph.D. 538...@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] Problems with Installing R ('failure to expand shell folder constant "userdocs")
 Hello,  Iâm trying to install R 3.02 for windows on a Windows 7 machine (we were just upgraded from Windows XP). Itâs a networked machine in a work environment. I have administrator rights. Program files go on C; data is stored on D (it's a partitioned drive).  I get the message 'failure to expand shell folder constant "userdocs"  I searched, and found some stuff on the Microsoft website, which made me believe that I needed to unlock a folder with my user name. There are two, one in âC:\Usersâ, which has a padlock symbol on it, and another in âD:\âm which doesnât have a padlock icon. Both had been set to read only; I removed that.  Iâve not been able to find anything on the CRAN website about this.  Has anybody run into this problem before?   Thank you very much,  Barry DeCicco [[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] plot() function: color transparency
I checked as you suggested. However, I found that the number in those functions are the number of colors. In contrast, my number here means a specific color, for example, 2 in my code means "red", 3 in my code means "green" On Wednesday, December 18, 2013 1:18 PM, Duncan Murdoch wrote: On 13-12-18 2:08 PM, capricy gao wrote: > I found all the color transparency was defined with character color, or rgb > color. What if I have number code and still try to modify the transparency? > > For example: > >> x=c(1:5) >> color=c(2,2,3,4,5) >> plot(x, col=color) >> plot(x, col=color,pch=20) > > here I defined color by numbers, how can I modify the transparency? See the examples for ?palette. Duncan Murdoch [[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] plot() function: color transparency
you will need to specify colours as RGB values and then set transparency via the "alpha" argument. e.g.: color=rgb(0,0,0,alpha=0.3) # will give black (0,0,0) and a transparency of 30%. Best wishes Christoph On 18/12/2013 23:23, capricy gao wrote: I checked as you suggested. However, I found that the number in those functions are the number of colors. In contrast, my number here means a specific color, for example, 2 in my code means "red", 3 in my code means "green" On Wednesday, December 18, 2013 1:18 PM, Duncan Murdoch wrote: On 13-12-18 2:08 PM, capricy gao wrote: I found all the color transparency was defined with character color, or rgb color. What if I have number code and still try to modify the transparency? For example: x=c(1:5) color=c(2,2,3,4,5) plot(x, col=color) plot(x, col=color,pch=20) here I defined color by numbers, how can I modify the transparency? See the examples for ?palette. Duncan Murdoch [[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] Exporting R graphics into Word without losing graph quality
Another nice thing about your solution is that circles look like circles,
and not like diamonds (when viewed on screen).
Thanks.
Kevin Wright
On Mon, Dec 16, 2013 at 8:02 PM, Steve Taylor wrote:
> Unfortunately the win.metafile() device does not support semi-transparent
> colours, which I like using.
>
> In my experience, the best way to get R graphics into Word is to use
> compressed high-resolution tiff, like this:
>
> word.tif = function(filename="Word_Figure_%03d.tif", zoom=4, width=17,
> height=10, pointsize=10, ...) {
> if (!grepl("[.]ti[f]+$", filename, ignore.case=TRUE))
> filename = paste0(filename,".tif")
> tiff(filename=filename, compression="lzw", res=96*zoom,
>width=width, height=height, units='cm', pointsize=pointsize, ...)
> }
> word.tif('test')
> plot(rnorm(100))
> dev.off()
>
> Now drag the file test.tif into your Word document.
>
> Sure, it's a bitmap format rather than a vector format, but the quality is
> excellent and the file sizes are still quite small. None of the vector
> formats works as well as this.
>
> cheers,
> Steve
>
> __
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
Kevin Wright
[[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] plot() function: color transparency
As Duncan suggested, this will probably get you what you want. You can set the transparency using alpha.f in adjustcolor(). x <- c(1:5) color <- c(2,2,3,4,5) color_transparent <- adjustcolor(color, alpha.f = 0.3) plot(x, col = color, pch = 20, cex = 4) plot(x, col = color_transparent, pch = 20, cex = 4) ?adjustcolor Richard On Wed, Dec 18, 2013 at 2:35 PM, Christoph Scherber wrote: > you will need to specify colours as RGB values and then set transparency > via the "alpha" argument. > > e.g.: color=rgb(0,0,0,alpha=0.3) > > # will give black (0,0,0) and a transparency of 30%. > > Best wishes > Christoph > > > > On 18/12/2013 23:23, capricy gao wrote: > >> I checked as you suggested. However, I found that the number in those >> functions are the number of colors. In contrast, my number here means a >> specific color, for example, 2 in my code means "red", 3 in my code means >> "green" >> >> >> >> >> >> On Wednesday, December 18, 2013 1:18 PM, Duncan Murdoch < >> murdoch.dun...@gmail.com> wrote: >> On 13-12-18 2:08 PM, capricy gao wrote: >> >> I found all the color transparency was defined with character color, or >>> rgb color. What if I have number code and still try to modify the >>> transparency? >>> >>> For example: >>> >>> x=c(1:5) color=c(2,2,3,4,5) plot(x, col=color) plot(x, col=color,pch=20) >>> here I defined color by numbers, how can I modify the transparency? >>> >> See the examples for ?palette. >> >> Duncan Murdoch >> [[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. > [[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] plot() function: color transparency
On 13-12-18 5:23 PM, capricy gao wrote: I checked as you suggested. However, I found that the number in those functions are the number of colors. In contrast, my number here means a specific color, for example, 2 in my code means "red", 3 in my code means "green" You didn't read very carefully. Duncan Murdoch On Wednesday, December 18, 2013 1:18 PM, Duncan Murdoch wrote: On 13-12-18 2:08 PM, capricy gao wrote: > I found all the color transparency was defined with character color, or rgb color. What if I have number code and still try to modify the transparency? > > For example: > >> x=c(1:5) >> color=c(2,2,3,4,5) >> plot(x, col=color) >> plot(x, col=color,pch=20) > > here I defined color by numbers, how can I modify the transparency? See the examples for ?palette. 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] binary symmetric matrix combination
Hi,
You could try:
Either:
dat1 <- read.table("Test.txt",header=TRUE)
dim(dat1)
#[1] 4735 4735
dat2 <- read.table("1991res.txt",header=TRUE)
dim(dat2)
#[1] 574 574
m1 <- as.matrix(dat1)
m2 <- as.matrix(dat2)
library(data.table)
d1 <-
data.table(Name1=as.vector(outer(rownames(m1),colnames(m1),paste0)),value1=as.vector(m1),key='Name1')
d2 <-
data.table(Name1=as.vector(outer(rownames(m2),colnames(m2),paste0)),value2=as.vector(m2),key='Name1')
res <- d2[d1]
res1 <- as.data.frame(res)
res1[,3][!is.na(res1[,2])] <- res1[,2][!is.na(res1[,2])]
vec1 <- as.vector(outer(rownames(m1),colnames(m1),paste0))
res2 <- res1[match(vec1,res1[,1]),-2]
res3 <- matrix(res2[,2],dimnames=list(rownames(m1),colnames(m1)),ncol=ncol(m1))
#or
vec1 <- paste0(rownames(m1)[row(m1)],colnames(m1)[col(m1)])
vec2 <- paste0(rownames(m2)[row(m2)],colnames(m2)[col(m2)])
indx <- match(vec1,vec2)
indx1 <- indx[!is.na(indx)]
indx2 <- match(vec2,vec1)
indx2N <- indx2[!is.na(indx2)]
m1[indx2N] <- m2[indx1]
A.K.
On Monday, December 16, 2013 1:54 PM, Elio Shijaku wrote:
Hi Arun,
I hope you remember me. You helped me build several symmetrical matrices in R.
I have one question though. I have now build all the matrices and I want to
combine each of them to a larger one. To make the matter simpler to understand,
I am attaching the text version of both files. What I would like is to merge
the contents of 1991res (574x574 symmetric matrix) into the Test file
(4703x4703, symmetric matrix which includes the row names of the 1991res.
Could you help me by showing the steps I should take to merge the matrices?
Thank you in advance and sorry for disturbing.
Best,
Elio
On Mon, Sep 23, 2013 at 2:19 AM, arun wrote:
Hi Elio,
>
>There was only one matrix with that error. Glad you were able to correct it.
>
>I sent an linkedin request to you.
>Regards
>Arun
>
>
>
>
>
>
>
>
>From: Elio Shijaku
>To: arun
>Sent: Sunday, September 22, 2013 7:21 PM
>
>Subject: Re: binary symmetric matrix combination
>
>
>
>Hi Arun,
>
>You are very right, that mistake which i corrected by removal fixed the issue,
>now the matrix works perfectly.
>
>Many thanks for all your help, please if you are in LinkedIn, I would be
>delighted to add you as a friend.
>
>Here is my profile in case you're interested:
>es.linkedin.com/pub/elio-shijaku/11/b7b/147/
>
>Hopefully I can "disturb" you in case I have further questions, I am just
>learning R and everything is new to me.
>
>All the best,
>
>Elio
>
>
>
>
>On Mon, Sep 23, 2013 at 12:05 AM, arun wrote:
>
>Hi,
>>Actually, you have duplicate names.
>>
>>
>>"p226 p226 s112"
>>
>>What do you need to do in those situations? Looks like it is a mistake in
>>the file.
>>
>>
>>
>>
>>
>>From: Elio Shijaku
>>To: arun
>>Sent: Sunday, September 22, 2013 5:41 PM
>>
>>Subject: Re: binary symmetric matrix combination
>>
>>
>>
>>Hi Arun,
>>
>>Here is the file, I tried many options, once I enter the command
>>
>>lst2<- lapply(lst1[lapply(lst1,length)>0],function(x)
>>as.matrix(read.table(text=x,row.name=1)))
>>
>>
>>I get:
>>
>>Error in read.table(text = x, row.name = 1) : duplicate 'row.names' are not
>>allowed
>>
>>
>>Any idea? Thanks a lot for the help.
>>
>>
>>Elio
>>
>>
>>
>>
>>
>>On Sun, Sep 22, 2013 at 5:02 PM, arun wrote:
>>
>>Hi Elio,
>>>Check the new text file. I used the first line:
>>>
>>>lines1<-str_trim(gsub("\t"," ",readLines("file.txt")))
>>>because the file was "\t" separated
>>>
>>>In the new file, it could be just space separated. So, you may only need:
>>>lines1<- readLines("file.txt")
>>>
>>>If possible, could you email me the file. I can take a look into it.
>>>A.K.
>>>
>>>
>>>
>>>
>>>
>>>
>>>
>>>
>>>
>>>
>>>
>>>From: Elio Shijaku
>>>To: arun
>>>Sent: Sunday, September 22, 2013 7:55 AM
>>>
>>>Subject: Re: binary symmetric matrix combination
>>>
>>>
>>>
>>>
>>>Hi Arun,
>>>
>>>I am replying to this e-mail address to not break the forum's rules.
>>>Everything worked great. However, when I use another text file called
>>>"file", I use the commands you gave me but I get this problem:
>>>
>>>library(stringr)
>>>
>>>lines1<-str_trim(gsub("\t"," ",readLines("file.txt")))
>>> lst1<-lapply(split(lines1,
>>>cumsum(lines1=="")),function(x) x[x!=""])
>>>
>>>
>>>lst2<- lapply(lst1[lapply(lst1,length)>0],function(x)
>>>as.matrix(read.table(text=x,row.names=1)))
>>>
>>>Error in read.table(text = x, row.names = 1) : duplicate 'row.names' are
>>>not allowed
>>>
>>>#I cannot then continue with the rest:
>>>
>>>
>>>names(lst2)<- paste0("m",seq_along(lst2))
>>>dat<- do.call(rbind,lapply(names(lst2),function(x) {x1<- lst2[[x]];
>>>cbind(expand.grid(rep(list(colnames(x1)),2),stringsAsFactors=FALSE),value=as.vector(x1))}))
>>>library(reshape2)
>>>res<- dcast(dat,Var1~Var2,value.var="value",sum)
>>> row.names(res)<- res[,1]
>>> res<- as.matrix(res[,-1])
>>> dim(res)
>>>
>>>
>>>Any idea on why such problem?
>>>
>>>Thanks
[R] plot different groups as factors
dear all, i would like to plot the value of different response groups. when i simply use plot(y ~ x) i obtain a series of boxplots. i would rather use dots. i also tried with stripchart(y ~ x), which gives better results but does not place properly the labels since place them alphabetically. in addition i actually have 6 response groups: 3 classes (low, medium, high) and 2 sampling time (12 and 18 months). how can i generate these individual groups and plot them in the correct order (low, medium, high and 12, 18)? i believe is something to do with the factors but i don't know how to implement them. what i am looking for is to generate a figure such as the one i sketched in the attached file. i also attached a dataframe version of my data. the vectors containing the same data are: time <-c( 18,18,18,18,18,18,18,18,18, 18,18,18,18,18,18,18,18,18,18, 18,18,18,18,18,18,18,18,18,12, 12,12,12,12,12,12,12,12,12,12, 12,12,12,12,12,12,12,12,12,12, 12,12,12,12,12,12,12,12,12,12, 12,12,12,12,12,12,12) class <-c(medium,medium,medium,medium,medium, medium,medium,medium,medium,medium,medium, medium,medium,medium,high,high,high,high, high,high,high,low,low,low,low,low,low, low,medium,medium,medium,medium,medium,medium, medium,medium,medium,medium,medium,medium, medium,medium,medium,medium,medium,medium,high, high,high,high,high,high,high,high,high, high,low,low,low,low,low,low,low,low, low,low) value <-c(2.92,0.01,0.36,3.16,0.99,0.38, 0.01,5.1,0.04,0.01,1.33,4.13,0.15,0.15, 14.18,4290.14,26.8,5.33,17.58,14.29,248.5, 0.01,0,0,0,0,0,0,0.151395382, 5.327863403,5.10096383,1.32567787,4.352404124, 0.458606982,2.915908912,0.011996374,0.364710382, 0.033016026,3.161701212,0.381564497,0.010971385, 0.035646472,0.014781805,4.129708296,0.153094117, 0.018497847,15.09178491,17.58393041,14.17643928, 4290.143561,26.79730719,294.6367065,14.2888441, 248.495231,209.3131795,2014.506722,0.010751273, 0.002325138,0.000637473,0.003984336,0.006018154, 0.003620907,0.745936,0.000142311,0.002460417, 0.001280189) Thank you very much for any help you could provide! regards Luigi timeclass value 1 18 medium 2.92 2 18 medium 0.01 3 18 medium 0.36 4 18 medium 3.16 5 18 medium 0.99 6 18 medium 0.38 7 18 medium 0.01 8 18 medium 5.1 9 18 medium 0.04 10 18 medium 0.01 11 18 medium 1.33 12 18 medium 4.13 13 18 medium 0.15 14 18 medium 0.15 15 18 high14.18 16 18 high"4,290.14" 17 18 high26.8 18 18 high5.33 19 18 high17.58 20 18 high14.29 21 18 high248.5 22 18 low 0.01 23 18 low 0 24 18 low 0 25 18 low 0 26 18 low 0 27 18 low 0 28 18 low 0 29 12 medium 0.151395382 30 12 medium 5.327863403 31 12 medium 5.10096383 32 12 medium 1.32567787 33 12 medium 4.352404124 34 12 medium 0.458606982 35 12 medium 2.915908912 36 12 medium 0.011996374 37 12 medium 0.364710382 38 12 medium 0.033016026 39 12 medium 3.161701212 40 12 medium 0.381564497 41 12 medium 0.010971385 42 12 medium 0.035646472 43 12 medium 0.014781805 44 12 medium 4.129708296 45 12 medium 0.153094117 46 12 medium 0.018497847 47 12 high15.09178491 48 12 high17.58393041 49 12 high14.17643928 50 12 high4290.143561 51 12 high26.79730719 52 12 high294.6367065 53 12 high14.2888441 54 12 high248.495231 55 12 high209.3131795 56 12 high2014.506722 57 12 low 0.010751273 58 12 low 0.002325138 59 12 low 0.000637473 60 12 low 0.003984336 61 12 low 0.006018154 62 12 low 0.003620907 63 12 low 7.46E-05 64 12 low 0.000142311 65 12 low 0.002460417 66 12 low 0.001280189 __ R-hel
[R] assessment of validity of PLS predictions
Hello, I've built a PLSR model to predict the concentrations of mixture components from experimental data using the 'pls' library. I calculated the Q residual (or lack of fit) and T squared value for each of the samples used to build the model in order to assess how well each sample is described by the model. This is straightforward to do for these data because their X scores are returned by the 'plsr' function in addition to the X loadings of the model- both of these are needed to calculate the Q residual and T squared value. I'd like to calculate values for Q residual and T squared for predictions for samples for which the concentrations aren't known. However, the 'plsr' function doesn't calculate the X scores for predictions. I can solve for them if I solve the equation (in R code): X = T%*%trans(P) for T: trans(T) = (P*)%*%trans(X) where X is the data matrix from prediction samples, T is X scores matrix for prediction samples, P is X loadings matrix from the model and P* is the pseudo-inverse of this matrix. From these calculated X scores of the prediction samples and the X loadings of the model, I can calculate T squared and the Q residual values. My main question: Is my approach a reasonable one to identify samples that may not be well described by a given model? If not, can anyone direct me to a resource that describes better methods? Thanks [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] read ".slk" file
I suspect this could be SYLK format, a very old spreadsheet exchange format.
The only hit that I get from RSiteSearch("sylk") is read.gnumeric.sheet, which
depends on an external program "ssconvert" to extract CSV. In the plus
department, it is a text-based format that has been implemented numerous times
so you could roll your own reader based on open source code. In the minus
department, there is no published specification, so if your files have quirks
then you will have to work around them one by one, and it really is like
Latin... a dead language.
---
Jeff NewmillerThe . . Go Live...
DCN:Basics: ##.#. ##.#. Live Go...
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---
Sent from my phone. Please excuse my brevity.
PIKAL Petr wrote:
>Hi
>
>Extension does not specify file format. You can rename any file with
>any extension without changing its nature. However slk stays for
>symbolic link and therefore it just brings actual file to Excel.
>
>Maybe you could start to play with
>
>http://stat.ethz.ch/R-manual/R-devel/library/base/html/files.html
>
>Regards
>Petr
>
>> -Original Message-
>> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
>> project.org] On Behalf Of Santosh
>> Sent: Tuesday, December 17, 2013 12:11 AM
>> To: r-help
>> Subject: [R] read ".slk" file
>>
>> Dear Rxperts..
>>
>> I recently received a data file with the extension ".slk". If I save
>> the file as MS Excel file, I am able to read in R without issues. Is
>> it possible to read this ".slk" file without converting into another
>R-
>> readable data format?
>>
>> Regards,
>> Santosh
>>
>> [[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.
__
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] plot different groups as factors
On 12/19/2013 11:18 AM, Luigi Marongiu wrote:
dear all,
i would like to plot the value of different response groups. when i simply
use plot(y ~ x) i obtain a series of boxplots. i would rather use dots. i
also tried with stripchart(y ~ x), which gives better results but does not
place properly the labels since place them alphabetically.
in addition i actually have 6 response groups: 3 classes (low, medium,
high) and 2 sampling time (12 and 18 months).
how can i generate these individual groups and plot them in the correct
order (low, medium, high and 12, 18)? i believe is something to do with the
factors but i don't know how to implement them.
Hi Luigi,
Perhaps you want something like this:
lmdf<-read.table(text="time class value
18 medium 2.92
18 medium 0.01
18 medium 0.36
18 medium 3.16
18 medium 0.99
18 medium 0.38
18 medium 0.01
18 medium 5.1
18 medium 0.04
18 medium 0.01
18 medium 1.33
18 medium 4.13
18 medium 0.15
18 medium 0.15
18 high 14.18
18 high 4290.14
18 high 26.8
18 high 5.33
18 high 17.58
18 high 14.29
18 high 248.5
18 low 0.01
18 low 0
18 low 0
18 low 0
18 low 0
18 low 0
18 low 0
12 medium 0.151395382
12 medium 5.327863403
12 medium 5.10096383
12 medium 1.32567787
12 medium 4.352404124
12 medium 0.458606982
12 medium 2.915908912
12 medium 0.011996374
12 medium 0.364710382
12 medium 0.033016026
12 medium 3.161701212
12 medium 0.381564497
12 medium 0.010971385
12 medium 0.035646472
12 medium 0.014781805
12 medium 4.129708296
12 medium 0.153094117
12 medium 0.018497847
12 high 15.09178491
12 high 17.58393041
12 high 14.17643928
12 high 4290.143561
12 high 26.79730719
12 high 294.6367065
12 high 14.2888441
12 high 248.495231
12 high 209.3131795
12 high 2014.506722
12 low 0.010751273
12 low 0.002325138
12 low 0.000637473
12 low 0.003984336
12 low 0.006018154
12 low 0.003620907
12 low 7.46E-05
12 low 0.000142311
12 low 0.002460417
12 low 0.001280189",header=TRUE)
# get the factor levels in the correct order
lmdf$class<-factor(lmdf$class,levels=c("low","medium","high"))
library(plotrix)
brkdn.plot("value","class","time",lmdf,col=2:4)
legend(14,1200,c("low","medium","high"),col=2:4,lty=1)
Jim
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and provide commented, minimal, self-contained, reproducible code.
Re: [R] Welcome to the "R-help" mailing list
Hi there, I am new to R and programming in general and am looking for help with writing a function with dates and times. I have checked around but am still a bit stuck. Basically, I have data in the format "dd/mm/ HH:MM" and I have to calculate how much time has passed between various events. I have given the following command ( where "date" is the column in my data frame that indicates the date and time) date=as.Date.factor(date,format="%d/%m/%Y %H:%M") However, this displays only the date, without the time. I also have tried: date=substr(argo1$date,1,907) And it shows the date and time. However, when I try to find the difference between two dates i.e.the time that has passed with the command: difftime(date[2],date[3],unit="secs"), it returns that 0 seconds have passed. When I try to find the difference with the command: date[3]-date[2] it tells me Error in date[3] - date[2] : non-numeric argument to binary operator The class(date) is "character". Any idea what I am doing wrong? Thank you very much in advance! On 19 December 2013 15:22, wrote: > Welcome to the R-help@r-project.org mailing list! > > To post to this list, send your message to: > > r-help@r-project.org > > General information about the mailing list is at: > > https://stat.ethz.ch/mailman/listinfo/r-help > > If you ever want to unsubscribe or change your options (eg, switch to > or from digest mode, change your password, etc.), visit your > subscription page at: > > https://stat.ethz.ch/mailman/options/r-help/lydiakeppler%40gmail.com > > You can also make such adjustments via email by sending a message to: > > r-help-requ...@r-project.org > > with the word `help' in the subject or body (don't include the > quotes), and you will get back a message with instructions. > > You must know your password to change your options (including changing > the password, itself) or to unsubscribe without confirmation. It is: > > whaletail27 > > Normally, Mailman will remind you of your r-project.org mailing list > passwords once every month, although you can disable this if you > prefer. This reminder will also include instructions on how to > unsubscribe or change your account options. There is also a button on > your options page that will email your current password to you. > -- *Protect the environment: think before you print* ><º> [[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] Welcome to the "R-help" mailing list
use POSIXct instead of Date:
> x <- c("2013-12-12 12:00:00", "2013-12-15 03:15:23")
> # convert using as.POSIXct
>
> times <- as.POSIXct(x, format = "%Y-%m-%d %H:%M:%S")
> times
[1] "2013-12-12 12:00:00 EST" "2013-12-15 03:15:23 EST"
>
> difftime(times[2], times[1], units = 'secs')
Time difference of 227723 secs
>
>
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.
On Wed, Dec 18, 2013 at 9:24 PM, Lydia Keppler wrote:
> Hi there,
>
> I am new to R and programming in general and am looking for help with
> writing a function with dates and times. I have checked around but am still
> a bit stuck.
>
> Basically, I have data in the format "dd/mm/ HH:MM" and I have to
> calculate how much time has passed between various events.
>
> I have given the following command ( where "date" is the column in my data
> frame that indicates the date and time)
> date=as.Date.factor(date,format="%d/%m/%Y %H:%M")
> However, this displays only the date, without the time.
>
> I also have tried:
> date=substr(argo1$date,1,907)
> And it shows the date and time.
>
> However, when I try to find the difference between two dates i.e.the time
> that has passed with the command: difftime(date[2],date[3],unit="secs"), it
> returns that 0 seconds have passed.
>
> When I try to find the difference with the command:
> date[3]-date[2]
> it tells me Error in date[3] - date[2] : non-numeric argument to binary
> operator
>
> The class(date) is "character".
>
> Any idea what I am doing wrong?
> Thank you very much in advance!
>
>
> On 19 December 2013 15:22, wrote:
>
>> Welcome to the R-help@r-project.org mailing list!
>>
>> To post to this list, send your message to:
>>
>> r-help@r-project.org
>>
>> General information about the mailing list is at:
>>
>> https://stat.ethz.ch/mailman/listinfo/r-help
>>
>> If you ever want to unsubscribe or change your options (eg, switch to
>> or from digest mode, change your password, etc.), visit your
>> subscription page at:
>>
>> https://stat.ethz.ch/mailman/options/r-help/lydiakeppler%40gmail.com
>>
>> You can also make such adjustments via email by sending a message to:
>>
>> r-help-requ...@r-project.org
>>
>> with the word `help' in the subject or body (don't include the
>> quotes), and you will get back a message with instructions.
>>
>> You must know your password to change your options (including changing
>> the password, itself) or to unsubscribe without confirmation. It is:
>>
>> whaletail27
>>
>> Normally, Mailman will remind you of your r-project.org mailing list
>> passwords once every month, although you can disable this if you
>> prefer. This reminder will also include instructions on how to
>> unsubscribe or change your account options. There is also a button on
>> your options page that will email your current password to you.
>>
>
>
>
> --
> *Protect the environment: think before you print*
>><º>
>
> [[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
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
[R] GLMM parameter estimates giving opposite trends
I apologize if this is a simple question. I am running GLMM's using glmmML and model averaging with MuMIn. One of the parameter estimates for a parameter (firefreq) in the best model is giving a positive number, where in reality I know this to be a negative correlation. I have checked and double checked the data that has gone in and this is not the issue. This is occurring for numerous variables in my models. As far as I was aware the parameter estimate is indicative of the direction of the relationship? Is there any reason why this model would give me opposite trends? Let me know if any other information would be useful in answering this question. Thank you in advance for any input. This is the best model: M17<- glmmML(ldeli~bare+firefreq+canopy+treatment,cluster=season,family=poisson,data=ldeli) Model-averaged coefficients: Estimate Std. Error z value Pr(>|z|) (Intercept) -84.248439 30.376197 2.774 0.00555 ** bare -0.102111 0.023231 4.396 1.11e-05 *** firefreq3.370089 1.183091 2.849 0.00439 ** canopy -0.013598 0.007420 1.832 0.06688 . treatmentLU87.939276 30.376750 2.895 0.00379 ** treatmentSM01 67.595184 24.477350 2.762 0.00575 ** treatmentSMWF 64.612540 23.322285 2.770 0.00560 ** treatmentT01 80.142838 28.030787 2.859 0.00425 ** treatmentT03 77.088813 26.836430 2.873 0.00407 ** treatmentTB56.163472 19.744217 2.845 0.00445 ** treatmentWF84.313036 29.214505 2.886 0.00390 ** -- _ _ _ _ Diana [[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 with Installing R ('failure to expand shell folder constant "userdocs")
You didn't say what gave that message (and there is no 'R 3.02': did you mean 3.0.2?). If this came from the installer, that is not part of R: look for info on 'Inno Setup'. On 18/12/2013 21:43, Barry DeCicco wrote:  Hello,  I’m trying to install R 3.02 for windows on a Windows 7 machine (we were just upgraded from Windows XP). It’s a networked machine in a work environment. I have administrator rights. Program files go on C; data is stored on D (it's a partitioned drive).  I get the message 'failure to expand shell folder constant "userdocs"  I searched, and found some stuff on the Microsoft website, which made me believe that I needed to unlock a folder with my user name. There are two, one in ‘C:\Users’, which has a padlock symbol on it, and another in ‘D:\’m which doesn’t have a padlock icon. Both had been set to read only; I removed that.  I’ve not been able to find anything on the CRAN website about this.  Has anybody run into this problem before?   Thank you very much,  Barry DeCicco [[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 do. -- Brian D. Ripley, rip...@stats.ox.ac.uk Professor of Applied Statistics, http://www.stats.ox.ac.uk/~ripley/ University of Oxford, Tel: +44 1865 272861 (self) 1 South Parks Road, +44 1865 272866 (PA) Oxford OX1 3TG, UKFax: +44 1865 272595 __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
