[R] Rscript silent failures with unmatched brackets
Hi all,
I’ve noticed that a script with unmatched brackets of any sort will fail
silently in Rscript—neither logging nor any output in the shell indicates
that anything went wrong. Example file to run in Rscript:
sink('/tmp/exampleoutfile')
a - 0
{
print(a)
Replacing the ‘{‘ with any other symbol that must be matched (quotes,
parens, etc.) results in the same kind of silent failure. One workaround is
to use
echo source('myscript.R') | R --no-save --no-restore
from the command line but that’s a kludge. My apologies if this has been
brought up before, but doing a quick search I didn’t find anything, and to
me this seems like a significant bug.
Thanks best,
Luke
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Re: [R] integrate with vector arguments
Hi,
The following works.
f2
function(z)
{
f1 - function(t)
{
z*t + z*t^2
}
return(f1)
}
sapply(1:5,function(x)integrate(f2(x),0,1)$value)
[1] 0.83 1.67 2.50 3.33 4.17
--
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[R] lmerTest - difflsmeans: LS means for Post-hoc analysis
Dear fellow R users, I am trying to calculate the difference of the least square means from a lmer model which has a pretty much standard form of Y ~ A*B + (1|C/D) I am using the difflsmeans command from the lmerTest package but it returns the following error line: Error in as.data.frame.default(VarCorr(model)) : cannot coerce class VarCorr.merMod to a data.frame I also tried to do it manually but apparently to no avail. I am using the lme4 v. 1.0-6 package on Linux (if that helps). Any suggestions how to override this? Thanks in advance! Grigorios -- __ Grigorios Georgolopoulos Research Assistant Animal Ecology Department of Ecology and Genetics (IEG) Evolutionary Biology Centre (EBC) Uppsala University Norbyvägen 18d, SE-752 36 Uppsala, Sweden Email: georgolopoulosgg...@hotmail.com, grigorios.georgolopou...@ebc.uu.se __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] mle
Hello,
I am going to estimate the parameter of the count model: pr(N=n)=
integration{B(x, alpha)-C(x,alpha)} by maximum likelihood estimation.
n-c(0,1,2,3) and F- (0,3,4,5) are the vectors of values and observed
frequency respectively. The function C(x,alpha) is not defined for n=0, but
suppose C(x,alpha)=1 when n=0. When I want to insert this exception in the
following loop, I don't receive reasonable estimate.
pari (alpha){
nloglik- function(alpha){
B-function(x,k){}
C-function(x,k){}
A-function(x){
s-rep(0,length(x))
s-s+ C(x,k)
s- s+B(x,k)
}
s
}
d-0
for (n in seq(along=F)){
lik-integrate(A,0,1)$value
d- d - F[n]*log(lik)}}
d }
F- (0,3,4,5)
n-length(F)
mle (nloglik, start=list(alpha=alpha)
}
This program gives the answer when n= 1,2,3. But for n=0 I get error, I have to
consider the exception : C(x,alpha)=1.
Does anybody know where I need to put the exception in the program? ( For 'if'
loops, I don't get reasonable results)
I would appreciate any help
Best Regards,
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Re: [R] Saving Mean Relative Difference from all.equal()
On Feb 26, 2015, at 2:02 PM, Scott Colwell scolw...@uoguelph.ca wrote: I think I have one solution. Not very pretty though. Relies on the text not changing at all. as.numeric(gsub(Mean relative difference: , , all.equal(cov2cor(ITEMCOV),cor(item.data))[2])) Is there a better way? `all.equal` is a generic function and it appears you are interested in `all.equal.numeric`. Not sure if it’s “better” but you could fairly easily hack the all.equal.numeric function to replace this code (near the end of the function code) : if (is.na(xy) || xy tolerance) msg - c(msg, paste(Mean, what, difference:, format(xy))) ,,, with: if (is.na(xy) || xy tolerance) msg - xy (probably best to make it have a different name and not use 'all.equal' to name it.) When it did that I got this as a result: all.EQ( seq(1,2, by=0.1)*10, 10:20) [1] TRUE all.EQ( seq(1,2, by=0.11)*10, 10:20) [1] Numeric: lengths (10, 11) differ all.EQ(seq(1,2, by=0.11)*10, 10:19) [1] 0.03225806 — David Winsemius, MD Alameda, CA, USA __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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] Schedule R function/Code to run at specific time
I have a problem with statistics, I think this forum is not the right option, but I'm running out resolution to my problem, so I thought I expose you here so someone could help me. Realized a certain amount of tests in different groups, this assessment had different weights, and the groups were formed by the different amount of people. I thought of calculating the average to see how much each person contributed to the total amount of the assessment. What I would like to know if there is another way I use the absolute number and the average for quantifying group was better. Table Group Absolute Points Unit Average A 700 35 20 B 500 20 25 C 900 150 6 D 300 10 30 Att.. Mestre. Marcelo Alves Costa Grupo de Estudo e Pesquisa em Desenvolvimento e Aprendizagem Motora (GEPEDAM) Centro Universitário Filadélfia (43) 9646-1071 2015-02-26 16:08 GMT-03:00 Doran, Harold hdo...@air.org: Is there functionality within R proper, without having to revert to the OS, allowing a function or a portion of an R script to be run at a defined time? My google searches haven't provided much other than one at the link below which relies on an OS. Thanks, Harold https://tgmstat.wordpress.com/2013/09/11/schedule-rscript-with-cron/ [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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 -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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] integrate with vector arguments
On Feb 26, 2015, at 1:49 PM, marKo mton...@ffri.hr wrote:
v1-c(1:5)
v2-c(1:5)
f1-function (x) {v1*x+v2*x^2}
The problem is that integrate(f1, 0, 1) does not work.
I does not, even if a pas the arguments (v1, v2)
f1-function (x, v1, v2) {v1*x+v2*x^2}
or if i try to vectorize the function
f1-Vectorize(function(x, v1, v2){v1*x+v2*x^2},
vectorize.args=c(v1, v2))
integrate(f1, 0, 1) gives an error:
f1-function (X, Y) integrate( function(x) {X*x+Y*x^2}, 0, 1)$value
fV-Vectorize(f1)
outer( X=v1,Y=v2, FUN= fV)
[,1] [,2] [,3] [,4] [,5]
[1,] 0.833 1.17 1.5 1.83 2.17
[2,] 1.333 1.67 2.0 2.33 2.67
[3,] 1.833 2.17 2.5 2.83 3.17
[4,] 2.333 2.67 3.0 3.33 3.67
[5,] 2.833 3.17 3.5 3.83 4.17
—
David Winsemius, MD
Alameda, CA, USA
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Re: [R] Summing certain values within columns that satisfy a certain condition
Kate — here is a transparent solution (tested but without NA treatment). Doubtless there are cleverer faster ones, which later posters will present. HTH # example with four columns and 20 rows nrows - 20 A - sample(c(1:100), nrows, replace=T) B - sample(c(1:100), nrows, replace=T) C - sample(c(1:100), nrows, replace=T) D - sample(c(1:100), nrows, replace=T) locs - c(c(1:nrows)[A==max(A)],c(1:nrows)[B==max(B)],c(1:nrows)[C==max(C)],c(1:nrows)[D==max(D)]) mat1 - matrix(rep(0,4*nrows),nrows,4) for (i in 1:4) mat1[,i][locs[i]] - 1 SUM - rowSums(mat1) On Feb 26, 2015, at 12:23 PM, Kate Ignatius kate.ignat...@gmail.com wrote: Hi, Supposed I had a data frame like so: A B C D 0 1 0 7 0 2 0 7 0 3 0 7 0 4 0 7 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 5 0 5 1 5 0 4 1 5 0 8 4 7 0 0 3 0 0 0 3 4 0 0 3 4 0 0 0 5 0 2 0 6 0 0 4 0 0 0 4 0 0 0 4 0 For each row, I want to count how many max column values appear to adventurely get the following outcome, while ignoring zeros and N/As: A B C D Sum 0 1 0 7 1 0 2 0 7 1 0 3 0 7 1 0 4 0 7 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 5 0 0 5 1 5 0 0 4 1 5 0 0 8 4 7 3 0 0 3 0 0 0 0 3 4 0 0 0 3 4 0 0 0 0 5 0 0 2 0 6 0 0 0 4 0 1 0 0 4 0 1 0 0 4 0 1 I've used the following code but it doesn't seem to work (my sum column column is all 1s): (apply(df,1, function(x) (sum(x %in% c(pmax(x)) Is this code too simple? Thanks! K. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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 -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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] Best Mac for R
Dan, FWIW, I have basically the system you describe, except a larger HD — I'm quite happy, but I'm a biased Mac user, although I love my Ubuntu Linux machine as well… One can bring any machine to its knees, so there is the element of expectations. A MacBook Pro stacks up as well or better compared to a similarly configured windows box. The thing is, IMO, there are at least two very good virtual machines to run MS-Windows on if the need arises (as well as Apple's 'Boot Camp') and, I believe, since the core Mac OS is essentially UNIX/Linux you have all that capability natively as well. Tom On Thu, Feb 26, 2015 at 12:50 AM, Dan Murphy chiefmur...@gmail.com wrote: I am possibly in the market for a new laptop. Predominantly a Windows user, I owned a macbook pro 10 years ago and am considering going that route again. Does the standard advice still hold: Get the most powerful processor (i7), most ram (16GB), and largest internal storage (512GB), if affordable? thanks, dan __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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 -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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] aggregating variables (sum within groups)
Even though I was looking in several r-books I could not find a suitable function to this problem Which R books did you look through? Bill Dunlap TIBCO Software wdunlap tibco.com On Thu, Feb 26, 2015 at 4:02 AM, David Studer stude...@gmail.com wrote: Hello everybody! I have a (probabely very easy) problem. Even though I was looking in several r-books I could not find a suitable function to this problem, that's why I hope that someone here could help me: # Sample data: group-c(A,A,A,B,B,C,C,C) var1-c(1,0,0,1,1,0,NA,1) var2-c(0,1,NA,0,1,1,0,0) testdata-data.frame(group, var1, var2) Now, I'd like to generate two aggregated variables: testdata$x- ??? should count the sum of var1 within each group (=4) testdata$y- ??? should count the sum of var2 within each group (=3) Therefore I am looking for a function like ave() which does not calculate the mean value but a sum. Thank you for any hints! David [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] extract file name from a path string
Dear all, what code should I write in order to extract the file name from a give path? Let's say that I want to get the file my file.xls which is in the directory/folder My documents; since I work both with Windows and Linux, the paths I am looking at are in the format: path.windows-\\home$\\lm667\\My Documents\\my file.xls path.linux-/home/My Documents/my file.xls I used two words for the file name because sometimes the file names have multiple words rather than a single one separated by capitals, . or _. The code should now get the file name, which is included between \\ (or /) and .xls but I don't know what regular expression will do the trick. Once the file name has been assigned to a vector, it should be easy to remove it from the path.windows/.linux and obtain a vector with the path on its own. Essentially the output should be as follows: file.name [1] my file.xls path.w [1] \\home$\\lm667\\My Documents\\ path.l [1] /home/My Documents/ Thank you, Luigi __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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] Dummy variable in ARIMA
Hi. First of all, thx. But when using in arima(...xreg=fact,...) then fact should be a vector and not a factor variable? Maybe I should have been more clear in my first mail, sorry. Or else I have to dig deeper into factors. /Mikael On Thu, Feb 26, 2015 at 5:17 PM, Bert Gunter gunter.ber...@gene.com wrote: Inline. Cheers, Bert Bert Gunter Genentech Nonclinical Biostatistics (650) 467-7374 Data is not information. Information is not knowledge. And knowledge is certainly not wisdom. Clifford Stoll On Thu, Feb 26, 2015 at 8:02 AM, Mikael Olai Milhøj mikaelmil...@gmail.com wrote: Hi all I have been searching on the web in vain. I want to include a dummy variable in my ARIMA model. Let's say that I want to make an AR(1) model for X including a dummy variable which should be 1 for observation 4,5,6 and zero otherwise (let's say that there is 50 observations in total). How do I make that? You don't, really. 1. Go through an R tutorial so that you understand the concept of factors and how they are used in R modeling. 2. fact - factor( (1:50) %in% (4:6)) Cheers, Bert This does the trick but seems inefficient: dummy-c(rep(0,3), rep(1,3), rep(0,44)) Thx in advance Best regards /Mikael [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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 -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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] Dummy variable in ARIMA
Inline. Cheers, Bert Bert Gunter Genentech Nonclinical Biostatistics (650) 467-7374 Data is not information. Information is not knowledge. And knowledge is certainly not wisdom. Clifford Stoll On Thu, Feb 26, 2015 at 8:02 AM, Mikael Olai Milhøj mikaelmil...@gmail.com wrote: Hi all I have been searching on the web in vain. I want to include a dummy variable in my ARIMA model. Let's say that I want to make an AR(1) model for X including a dummy variable which should be 1 for observation 4,5,6 and zero otherwise (let's say that there is 50 observations in total). How do I make that? You don't, really. 1. Go through an R tutorial so that you understand the concept of factors and how they are used in R modeling. 2. fact - factor( (1:50) %in% (4:6)) Cheers, Bert This does the trick but seems inefficient: dummy-c(rep(0,3), rep(1,3), rep(0,44)) Thx in advance Best regards /Mikael [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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 -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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] Dummy variable in ARIMA
Hi all I have been searching on the web in vain. I want to include a dummy variable in my ARIMA model. Let's say that I want to make an AR(1) model for X including a dummy variable which should be 1 for observation 4,5,6 and zero otherwise (let's say that there is 50 observations in total). How do I make that? This does the trick but seems inefficient: dummy-c(rep(0,3), rep(1,3), rep(0,44)) Thx in advance Best regards /Mikael [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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] aggregating variables (sum within groups)
For the record, the ave function in R can apply any function you specify, not just mean. The primary feature of ave is that it does not collapse the rows like aggregate does. Choose among them according to how you want the output to be organized. --- 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. On February 26, 2015 4:02:49 AM PST, David Studer stude...@gmail.com wrote: Hello everybody! I have a (probabely very easy) problem. Even though I was looking in several r-books I could not find a suitable function to this problem, that's why I hope that someone here could help me: # Sample data: group-c(A,A,A,B,B,C,C,C) var1-c(1,0,0,1,1,0,NA,1) var2-c(0,1,NA,0,1,1,0,0) testdata-data.frame(group, var1, var2) Now, I'd like to generate two aggregated variables: testdata$x- ??? should count the sum of var1 within each group (=4) testdata$y- ??? should count the sum of var2 within each group (=3) Therefore I am looking for a function like ave() which does not calculate the mean value but a sum. Thank you for any hints! David [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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 -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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] Save a list of list and search for values
You store it as a list of lists and can then use the lapply function
to navigate for values.
result - lapply(1:1, function(x){
mix(param[x]) # whatever your call to 'mix' is with some data
})
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 Thu, Feb 26, 2015 at 9:27 AM, Alaios via R-help r-help@r-project.org wrote:
Dear all,in my code I am using the mix() function that returns results in a
list. The result looks like
List of 10
$ parameters :'data.frame': 2 obs. of 3 variables:
..$ pi : num [1:2] 0.77 0.23
..$ mu : num [1:2] -7034 162783
..$ sigma: num [1:2] 20235 95261
$ se :'data.frame': 2 obs. of 3 variables:
..$ pi.se : num [1:2] 0.0423 0.0423
..$ mu.se : num [1:2] 177 12422
..$ sigma.se: num [1:2] 1067 65551
$ distribution: chr norm
$ constraint :List of 8
..$ conpi : chr NONE
..$ conmu : chr NONE
..$ consigma: chr NONE
..$ fixpi : NULL
..$ fixmu : NULL
..$ fixsigma: NULL
..$ cov : NULL
..$ size: NULL
$ chisq : num 28
$ df : num 5
$ P : num 3.67e-05
$ vmat: num [1:5, 1:5] 1.79e-03 -3.69e-01 -1.17e+02 2.95e+01
-2.63e+03 ...
$ mixdata :Classes ‘mixdata’ and 'data.frame': 11 obs. of 2
variables:
..$ X: num [1:11] 1e+04 2e+04 3e+04 4e+04 5e+04 6e+04 7e+04 8e+04 9e+04
1e+05 ...
..$ count: int [1:11] 993 137 82 30 21 5 7 14 21 2 ...
$ usecondit : logi FALSE
- attr(*, class)= chr mix
In my code I am trying around 10.000 fit (and each of these fits returns the
list above) and I want to keep those in a way that later on I would be able
to search inside all the lists.For example I would like to find inside those
10.000 lists which one has the smallest $chisq value. What would be a
suitable way to implement that in R? Luckily I am working in a computer with
a lot of ram so storing 10.000 lists temporary in memory before saving to
disk would not be a problem.
What would you suggest me?
RegardsAlex
[[alternative HTML version deleted]]
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Re: [R] covert entire dataset to numeric while persuing percentage values
The following data.frame x as one column named Percent. x Percent 1 10% 2 20% 3 30% as.numeric(substr(x$Percent,1,nchar(x$Percent)-1)) [1] 10 20 30 -- View this message in context: http://r.789695.n4.nabble.com/covert-entire-dataset-to-numeric-while-persuing-percentage-values-tp4703862p4703880.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] covert entire dataset to numeric while persuing percentage values
I think you are getting ahead of yourself. You use the term dataset, which is colloquial and not precise. The read.csv function returns a data.frame, in which each column can have its own storage mode (type). Most data.frames do not have all columns of the same type... if they were you might consider converting to a matrix, but with different units in each column that would not be a good idea in this case. From your str output, I think you need to skip loading the second and third lines of your file in the first place, since it looks like they consist of unit strings. Something like: fname - file.choose() xelements - read.csv( fname, header=FALSE, skip=3, stringsAsFactors=FALSE) but this does not get your column names. One way to get those would be: names( xelements ) - names( read.csv( fname ) ) As for the percent signs, you can convert those with something like: xelements$ X..NG.by.Energy - as.numeric( sub( %. , xelements$ X..NG.by.Energy ) ) In the future, please don't post in HTML format, as it just leads to confusion on this plain text mailing list. Read the Posting Guide for other warnings, and let us follow your journey to your problem with a reproducible example. There are various discussions online of what is reproducible.. you might start with [1]. Note that the read.csv function supports a text argument that lets you embed a sample of lines from your file into your example so we could troubleshoot your input process better if that is where your problem is. [1] http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example --- 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. On February 26, 2015 1:06:03 AM PST, Methekar, Pushpa (GE Transportation, Non-GE) pushpa.methe...@ge.com wrote: Hi , I am little confused about how to covert entire dataset to numeric . As I read data like.. Xelements =read.csv(file. Choose(),header = T, stringsAsFactors=FALSE) str(xelements ) str(xelements) 'data.frame': 731 obs. of 4 variables: $ Engine.Speed : chr rpm ES rpm 1049 ... $ X..NG.by.Energy : chr % NG by Energy % 0% ... $ Int.Mfld.Temp: chr Int Mfld Temp �C 49 ... $ Cmd.Advance.Angle: chr Cmd Advance Angle �BTDC 13.8 ... I have second column as in 0%, 10%, In percentage value , Whenever I am going to covert whole dataset its showing NA introduced .second column going to become NA. Converting separately I would be successful . xelements$Engine.Speed - as.numeric(xelements$Engine.Speed) Warning message: NAs introduced by coercion xelements$X..NG.by.Energy- as.numeric(sub(%,,xelements$X..NG.by.Energy))/100 Warning message: NAs introduced by coercion xelements$Int.Mfld.Temp- as.numeric(xelements$Int.Mfld.Temp) Warning message: NAs introduced by coercion xelements$Cmd.Advance.Angle- as.numeric(xelements$Cmd.Advance.Angle) Warning message: NAs introduced by coercion But I want to covert whole dataset at a time. I want to write function which will help me to solve this problem . xelements - data.frame(sapply(xelements, function(x) as.numeric(as.character(x sapply(xelements, class) but it won't be able to covert percentage value like 10%, 20% please do help me if you know the way. Thank you [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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 -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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] format selected columns in dataframe as character
Of course you could have created them as character vectors in the first
place:
dfx - data.frame(
group = c(rep('A', 8), rep('B', 15), rep('C', 6)),
sex = sample(c(M, F), size = 29, replace = TRUE),
age = runif(n = 29, min = 18, max = 54),
stringsAsFactors=FALSE
)
But if that is not possible in your context, then I would suggest this:
for (nm in names(dfx))
if (is.factor(dfx[[nm]])) dfx[[nm]] - as.character(dfx[[nm]])
It's clear and simple.
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062
On 2/26/15, 2:08 AM, Alain D. dialva...@yahoo.de wrote:
Dear R-List,
#I have a df with the first two cols formatted as factor.
dfx - data.frame(
group = c(rep('A', 8), rep('B', 15), rep('C', 6)),
sex = sample(c(M, F), size = 29, replace = TRUE),
age = runif(n = 29, min = 18, max = 54))
# now I want to format both factor VARs as character
# I tried
factor.id-names(dfx[sapply(dfx,is.factor)])
chr.names-which(names(dfx)%in% factor.id)
dfx[ , chr.names]-as.character(dfx[ , chr.names])
# which gives me
str(dfx)
'data.frame': 29 obs. of 3 variables: $ group: chr c(1, 1, 1, 1, 1, 1,
1, 1, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3) c(2, 2, 1,
1, 1,
1, 2, 1, 1, 2, 2, 1, 2, 2, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 2, 2)
c(1,
1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3,
3, 3, 3,
3) c(2, 2, 1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 2, 2, 2, 1, 1, 1, 1, 1, 2, 1,
1, 1,
1, 1, 2, 2, 2) ... $ sex : chr c(2, 2, 1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 2,
2, 2,
1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 2, 2) c(1, 1, 1, 1, 1, 1, 1, 1, 2,
2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3) c(2, 2, 1, 1, 1,
1, 2,
1, 1, 2, 2, 1, 2, 2, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 2, 2) c(1,
1, 1,
1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3,
3, 3)
... $ age : num 38.5 18 33.5 26 22.5 ...
#though I was hoping for something like
'data.frame': 29 obs. of 3 variables:
$ group: chr A A A A ...
$ sex : chr M F F M ...
$ age : num 21.3 35.2 53.8 21 23.6 ...
#What is wrong with my code?
#Thank you for any help
Best wishes
Alain
[[alternative HTML version deleted]]
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Re: [R] library(multcomp) does not work for loading Tukey
Dear Xavier See below for comments On 26/02/2015 11:20, CHIRIBOGA Xavier wrote: Dear colleagues, For Tukey, I tried to load the function with library(multcomp) but again a message says: Error in library(multcomp) : any package called ‘multcomp’ has been found I suspect you translated this? I would have thought the message in English would have been there is no package called 'multcomp' which gives you the clue that you should install it first ?install.packages is your friend Michael Thanks for ur help, Xavier __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. - No virus found in this message. Checked by AVG - www.avg.com Version: 2015.0.5645 / Virus Database: 4299/9183 - Release Date: 02/26/15 -- Michael http://www.dewey.myzen.co.uk __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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] Processing key_column, begin_date, end_date in R
Here is another way. Have not tested for large scale efficiency, but if you
convert dta to a data.table that might improve things.
library(dplyr)
dta - read.csv( text=
key_column,begin_date,end_date
123456,2013-01-01,2014-01-01
123456,2013-07-01,2014-07-01
789102,2012-03-01,2014-03-01
789102,2015-02-01,2016-02-01
789102,2015-02-06,2016-02-06
789102,2015-02-28,2015-03-31
789102,2015-04-30,2015-05-31
, as.is=TRUE)
( dta
%% mutate( begin_date = as.Date( begin_date ),
end_date = as.Date( end_date ) )
%% arrange( key_column, begin_date )
) - dta
mkgp - function( begin_date, cend ) {
ix - c( TRUE, cend[ -length( begin_date ) ] begin_date[ -1 ] )
cumsum( ix )
}
result - ( dta
%% group_by( key_column )
%% mutate( cend = as.Date( cummax( as.numeric( end_date ) )
, origin=1970-01-01 )
, gp = mkgp( begin_date, cend )
)
%% ungroup
%% group_by( key_column, gp )
%% summarise( begin_date = begin_date[ 1 ]
, end_date = cend[ length( cend ) ]
)
%% ungroup
%% select( -gp )
%% as.data.frame
)
---
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.
On February 25, 2015 1:18:58 PM PST, Matt Gross gro...@gmail.com wrote:
Hi,
I am trying to process a large dataset in R. The dataset contains the
following three columns:
key_column - a unique key identifier
begin_date - the start date of the active period
end_date - the end date of the active period
Example data is here:
key_column,begin_date,end_date
123456,2013-01-01,2014-01-01
123456,2013-07-01,2014-07-01
789102,2012-03-01,2014-03-01
789102,2015-02-01,2016-02-01
789102,2015-02-06,2016-02-06
I want to build a condensed table of key_column and begin_date's and
end_date's. As you can see in the example data above, some begin and
end
date periods overlap with begin_date and end_date pairs for the same
key_column. In situations where overlap exists I want to have one
record
for the key_column with the min(begin_date) and the max(end_date).
Can anyone help me build the commands to process this data in R?
Thanks,
Matt
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[R] retrieving protein for swissport
It's possible to retrieve protein for swissport by protein name?? I try using seqinr and query but i didnt find a way to get all protein that named Delta 9 acyl CoA desaturase. If some one have an idea i be glad :P thank you all ᐧ [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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] Save a list of list and search for values
On 26/02/2015 9:27 AM, Alaios via R-help wrote:
Dear all,in my code I am using the mix() function that returns results in a
list. The result looks like
List of 10
$ parameters :'data.frame': 2 obs. of 3 variables:
..$ pi : num [1:2] 0.77 0.23
..$ mu : num [1:2] -7034 162783
..$ sigma: num [1:2] 20235 95261
$ se :'data.frame': 2 obs. of 3 variables:
..$ pi.se : num [1:2] 0.0423 0.0423
..$ mu.se : num [1:2] 177 12422
..$ sigma.se: num [1:2] 1067 65551
$ distribution: chr norm
$ constraint :List of 8
..$ conpi : chr NONE
..$ conmu : chr NONE
..$ consigma: chr NONE
..$ fixpi : NULL
..$ fixmu : NULL
..$ fixsigma: NULL
..$ cov : NULL
..$ size: NULL
$ chisq : num 28
$ df : num 5
$ P : num 3.67e-05
$ vmat: num [1:5, 1:5] 1.79e-03 -3.69e-01 -1.17e+02 2.95e+01
-2.63e+03 ...
$ mixdata :Classes ‘mixdata’ and 'data.frame': 11 obs. of 2
variables:
..$ X: num [1:11] 1e+04 2e+04 3e+04 4e+04 5e+04 6e+04 7e+04 8e+04 9e+04
1e+05 ...
..$ count: int [1:11] 993 137 82 30 21 5 7 14 21 2 ...
$ usecondit : logi FALSE
- attr(*, class)= chr mix
In my code I am trying around 10.000 fit (and each of these fits returns the
list above) and I want to keep those in a way that later on I would be able to
search inside all the lists.For example I would like to find inside those
10.000 lists which one has the smallest $chisq value. What would be a suitable
way to implement that in R? Luckily I am working in a computer with a lot of
ram so storing 10.000 lists temporary in memory before saving to disk would not
be a problem.
What would you suggest me?
If all of the lists have the same components, then it would be
convenient to convert them into a big matrix or dataframe, with one row
per fit. It would need to be a dataframe if you include character data
along with the numbers, but a matrix would be faster, if it's only
numbers that you need. You'd use code like this to produce the matrix:
results - matrix(NA_real_, 1, ncols = however many you keep )
for (i in 1:1) {
fit - code to get the fit object
results[i,] - with(fit, c(parameters$pi, parameters$mu,
parameters$sigma, .. fill in the rest ..)
}
Duncan Murdoch
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Re: [R] format selected columns in dataframe as character
Try as.character like the following shows.
dfx - data.frame(
+ group = c(rep('A', 8), rep('B', 15), rep('C', 6)),
+ sex = sample(c(M, F), size = 29, replace = TRUE),
+ age = runif(n = 29, min = 18, max = 54))
dfx
group sex age
1 A M 41.35554346
2 A F 47.73245469
3 A F 42.97870796
4 A M 52.51180396
5 A F 46.72228944
6 A M 48.64668630
7 A M 36.07894452
8 A M 26.96805121
9 B M 30.67208692
10 B M 45.09322672
11 B M 31.86601692
12 B F 53.28861780
13 B M 27.74271305
14 B F 52.05845066
15 B F 18.94612430
16 B M 48.66673378
17 B F 53.07004762
18 B F 48.15222416
19 B M 32.17737802
20 B M 37.02122907
21 B M 39.31442046
22 B M 27.90392578
23 B M 44.70562356
24 C F 53.43127126
25 C M 49.85362283
26 C M 40.40779822
27 C F 31.41189728
28 C M 47.49351314
29 C M 46.34333618
summary(dfx)
group sex age
A: 8 F:10 Min. :18.94612
B:15 M:19 1st Qu.:32.17738
C: 6 Median :44.70562
Mean :41.46947
3rd Qu.:48.64669
Max. :53.43127
dfx$group - *as.character*(dfx$group)
summary(dfx)
group sex age
Length:29 F:10 Min. :18.94612
Class :character M:19 1st Qu.:32.17738
Mode :character Median :44.70562
Mean :41.46947
3rd Qu.:48.64669
Max. :53.43127
dfx
group sex age
1 A M 41.35554346
2 A F 47.73245469
3 A F 42.97870796
4 A M 52.51180396
5 A F 46.72228944
6 A M 48.64668630
7 A M 36.07894452
8 A M 26.96805121
9 B M 30.67208692
10 B M 45.09322672
11 B M 31.86601692
12 B F 53.28861780
13 B M 27.74271305
14 B F 52.05845066
15 B F 18.94612430
16 B M 48.66673378
17 B F 53.07004762
18 B F 48.15222416
19 B M 32.17737802
20 B M 37.02122907
21 B M 39.31442046
22 B M 27.90392578
23 B M 44.70562356
24 C F 53.43127126
25 C M 49.85362283
26 C M 40.40779822
27 C F 31.41189728
28 C M 47.49351314
29 C M 46.34333618
dfx$sex - *as.character*(dfx$sex)
summary(dfx)
group sex age
Length:29 Length:29 Min. :18.94612
Class :character Class :character 1st Qu.:32.17738
Mode :character Mode :character Median :44.70562
Mean :41.46947
3rd Qu.:48.64669
Max. :53.43127
class(dfx$group)
[1] character
dfx$group
[1] A A A A A A A A B B B B B B B B B B
B B B B B C
[25] C C C C C
class(dfx$sex)
[1] character
dfx$sex
[1] M F F M F M M M M M M F M F F M F F
M M M M M F
[25] M M F M M
--
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Sent from the R help mailing list archive at Nabble.com.
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Re: [R-es] DUDA LLENAR MATRIZ CREADA
Hola David,
Puedes hacer como ya comentaron en las otras respuestas. Pero por
si sirve de utilidad, pego al final parte de un código que empleo para
ilustrar la repetición de contrastes en simulación.
Está con un bucle y guardando en vectores, si se quisiera modificar
para obtener una matriz y emplear lapply (o sapply), tendrías que crear
una función del tipo:
fun.test - function(muestra)
unlist(test(muestra)[c('statistic','p.value')])
reemplazando test por el contraste que te interesa (y suponiendo que
devuelve el objeto estándar de clase htest).
sapply(muestras, fun.test) podría ser lo que buscas (suponiendo que
muestras es una lista con los datos en cada componente). Para que
conste, para ciertas cosas yo no soy reacio a emplear bucles en lugar de
applies...
Un saludo, Rubén.
# --
# Repetición de contrastes
# --
# --
# Valores iniciales
set.seed(1) # Fijar semilla para reproductibilidad
n - 500
nsim - 1000
estadistico - numeric(nsim)
pvalor - numeric(nsim)
# --
# Realizar contrastes
for(isim in 1:nsim) {
u - runif(n)
tmp - ks.test(u, punif, min=0, max=1)
estadistico[isim] - tmp$statistic
pvalor[isim] - tmp$p.value
}
#Probar a cambiar por u - rnorm(n) y tmp - ks.test(u, pnorm, mean =
mean(u), sd = sd(u))
# Mantengo algo del resto del código por si resulta de interés...
# --
# Proporción de rechazos
cat(\nProporción de rechazos al 1% =, mean(pvalor 0.01), \n)
cat(Proporción de rechazos al 5% =, mean(pvalor 0.01), \n)
cat(Proporción de rechazos al 10% =, mean(pvalor 0.01), \n)
# --
# Análisis del estadístico contraste
# Histograma
hist(estadistico, freq=FALSE)
#...
# --
# Análisis de los p-valores
# (si todo 'correcto' con distribución uniforme)
# Histograma
hist(pvalor, freq=FALSE)
curve(dunif(x,0,1), add=TRUE) #abline(h=1)
# Test de Kolmogorov-Smirnov
ks.test(pvalor, punif, min=0, max=1)
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Re: [R] Help with nonlinear least squares regression curve fitting
Andrew's suggestion for Year is a help, but package nlmrt shows the problem you are trying to solve is truly one where there is a Jacobian singularity. (nlmrt produces the Jacobian singular values -- but read the output carefully because these are placed for compact output as if they correspond to parameters, which they do not). Unfortunately, nlmrt tries to use analytic derivatives, and sign() is not in the derivatives table for the double sigmoid. BTW, your function has a typo. Do provide reproducible results. Here is what I did using callaghan.csv: Area,Year 104.7181283,1984 32.88026974,1985 56.07395863,1986 191.3422143,1987 233.4661392,1988 57.28317116,1989 201.1273404,1990 34.42570796,1991 165.8962342,1992 58.21905274,1993 114.6643724,1994 342.3461986,1995 184.8877994,1996 94.90509356,1997 45.2026941,1998 68.6196393,1999 575.2440229,2000 519.7557581,2001 904.157509,2002 1107.357517,2003 1682.876061,2004 40.55667824,2005 740.5032604,2006 885.7243469,2007 395.4190968,2008 1031.314519,2009 2597.544987,2010 1316.968695,2011 848.7093901,2012 5076.675075,2013 6132.975491,2014 code: library(nlmrt) df - read.csv(callaghan.csv) fitmodeliq - nlxb(Area ~ (-a*Year)*(Year + b), data = df, start=list(a=1,b=1, c=1)) fitmodelsig - nlxb(Area~a/(1+exp(-(b+c*Year))), data=df, start=list(a=1,b=1, c=1)) fitmodelds - nlxb(Area ~ a+2*b*(1/(1+exp(-abs(-c*Year+d)))-1/2)*sign(-c*Year+d), data=df, start=list(a=1, b=1, c=1)) For information of readers, Duncan Murdoch and I have been working on nls14 to replace/augment nls(), but we've a way to go yet before this is ready for CRAN. Collaborators welcome. John Nash On 15-02-26 06:00 AM, r-help-requ...@r-project.org wrote: Message: 24 Date: Thu, 26 Feb 2015 07:26:50 +1100 From: Andrew Robinson a.robin...@ms.unimelb.edu.au To: Corey Callaghan ccallaghan2...@fau.edu Cc: R help \(r-help@r-project.org\) r-help@r-project.org Subject: Re: [R] Help with nonlinear least squares regression curve fitting Message-ID: cahygmd6rruc_aobhrhw7babxnmzrsbi4b7zjt0vn5lrwvw2...@mail.gmail.com Content-Type: text/plain; charset=UTF-8 Finding starting values is a bit of a dark art. That said, there are steps you can take, but it may take time. First, I would scale Year so that it's not in the thousands! Experiment with subtracting 1980 or so. For specific advice, see inline. On Thu, Feb 26, 2015 at 3:03 AM, Corey Callaghan ccallaghan2...@fau.edu wrote: The curves' functions that I want to test are in the code here (hopefully correctly): Inverse Quadratic Curve: fitmodel - nls(Area ~ (-a*Year)*(Year + b), data = df, start=list(a=??, b=??, c=??)) I would plot the data and a smooth spline, differentiate the curve function, identify some parameter values somewhere stable, and estimate some values by eye, or even predict them from the first derivative of the spline - spline.smooth will do this. Sigmodial Curve: fitmodel - nls(Area~a/(1+exp(-(b+c*Year))), data=df, start=list(a=???, b=???, c=??)) I'd use the highest value as a, fit spline as above then invert area at two times to get b and c. Double sigmoidal Curve: fitmodel - nls(Area~a+2b(1/(1+exp(-abs(-c*Year+d)))-1/2)*sign(-c*Year+d), data=df, start=list(a=???, b=???, c=???) I'd use min(Area) as a, figure out b from the maximum (I guess 2b+a is the asymptote), and experiment with two values for year to retrieve c and d uniroot might help? Cheers Andrew -- Andrew Robinson Deputy Director, CEBRA, School of Biosciences Reader Associate Professor in Applied Statistics Tel: (+61) 0403 138 955 School of Mathematics and Statistics Fax: +61-3-8344 4599 University of Melbourne, VIC 3010 Australia Email: a.robin...@ms.unimelb.edu.au Website: http://www.ms.unimelb.edu.au/~andrewpr MSME: http://www.crcpress.com/product/isbn/9781439858028 FAwR: http://www.ms.unimelb.edu.au/~andrewpr/FAwR/ SPuR: http://www.ms.unimelb.edu.au/spuRs/ [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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 many digits are there in left of dot of 0.0001 ?
Hi, Some modification to work for both positive and negative number: nchar(format(*abs*(a),scientific=FALSE))-(trunc(log10(max(1,trunc(abs(a)+1) -1. -- View this message in context: http://r.789695.n4.nabble.com/How-many-digits-are-there-in-left-of-dot-of-0-0001-tp4703842p4703872.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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] covert entire dataset to numeric while persuing percentage values
It would help a lot if you posted a subset of your data using 'dput'
so that we know what it actually looks like. You have character data
mixed with numerics, so you will be NAs in some cases.
Conversion of percent to numeric is accomplished with something like this:
x - c('12%', '6%', '3.75%')
# convert to a number
as.numeric(gsub(%, , x)) / 100
[1] 0.1200 0.0600 0.0375
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 Thu, Feb 26, 2015 at 4:06 AM, Methekar, Pushpa (GE Transportation,
Non-GE) pushpa.methe...@ge.com wrote:
Hi ,
I am little confused about how to covert entire dataset to numeric .
As I read data like..
Xelements =read.csv(file. Choose(),header = T, stringsAsFactors=FALSE)
str(xelements )
str(xelements)
'data.frame': 731 obs. of 4 variables:
$ Engine.Speed : chr rpm ES rpm 1049 ...
$ X..NG.by.Energy : chr % NG by Energy % 0% ...
$ Int.Mfld.Temp: chr Int Mfld Temp °C 49 ...
$ Cmd.Advance.Angle: chr Cmd Advance Angle °BTDC 13.8 ...
I have second column as in 0%, 10%, In percentage value ,
Whenever I am going to covert whole dataset its showing NA introduced .second
column going to become NA.
Converting separately I would be successful .
xelements$Engine.Speed - as.numeric(xelements$Engine.Speed)
Warning message:
NAs introduced by coercion
xelements$X..NG.by.Energy-
as.numeric(sub(%,,xelements$X..NG.by.Energy))/100
Warning message:
NAs introduced by coercion
xelements$Int.Mfld.Temp- as.numeric(xelements$Int.Mfld.Temp)
Warning message:
NAs introduced by coercion
xelements$Cmd.Advance.Angle- as.numeric(xelements$Cmd.Advance.Angle)
Warning message:
NAs introduced by coercion
But I want to covert whole dataset at a time. I want to write function which
will help me to solve this problem .
xelements - data.frame(sapply(xelements, function(x)
as.numeric(as.character(x
sapply(xelements, class)
but it won't be able to covert percentage value like 10%, 20%
please do help me if you know the way. Thank you
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[R] twitteR
Hello, i need help, I'm trying to get oauth authorization to get rcredentials .RData the code that i use is: library(twitteR) library(tm) library(wordcloud) library(RColorBrewer) library(RCurl) library(ROAuth) options(RCurlOptions = list(cainfo = system.file(CurlSSL, cacert.pem, package = RCurl))) u = https://raw.github.com/tonybreyal/Blog-Reference-Functions/master/R/bingSearchXScraper/bingSearchXScraper . x = getURL(u, cainfo = system.file(CurlSSL, cacert.pem, package = RCurl)) download.file(url=http://curl.haxx.se/ca/cacert.pem;, destfile=cacert.pem requestURL - https://api.twitter.com/oauth/request_token; accessURL - https://api.twitter.com/oauth/access_token; authURL - https://api.twitter.com/oauth/authorize; consumerKey - eED. consumerSecret - qo. twitCred - OAuthFactory$new(consumerKey=consumerKey, consumerSecret=consumerSecret, requestURL=reqURL, accessURL=accessURL, authURL=authURL) ) when Asking for access twitCred$handshake(cainfo=cacert.pem) i receive this message: ERROR: AUTHORIZATION NO FOUND Please help me, thanks a lot!! [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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] Save a list of list and search for values
Dear all,in my code I am using the mix() function that returns results in a list. The result looks like List of 10 $ parameters :'data.frame': 2 obs. of 3 variables: ..$ pi : num [1:2] 0.77 0.23 ..$ mu : num [1:2] -7034 162783 ..$ sigma: num [1:2] 20235 95261 $ se :'data.frame': 2 obs. of 3 variables: ..$ pi.se : num [1:2] 0.0423 0.0423 ..$ mu.se : num [1:2] 177 12422 ..$ sigma.se: num [1:2] 1067 65551 $ distribution: chr norm $ constraint :List of 8 ..$ conpi : chr NONE ..$ conmu : chr NONE ..$ consigma: chr NONE ..$ fixpi : NULL ..$ fixmu : NULL ..$ fixsigma: NULL ..$ cov : NULL ..$ size : NULL $ chisq : num 28 $ df : num 5 $ P : num 3.67e-05 $ vmat : num [1:5, 1:5] 1.79e-03 -3.69e-01 -1.17e+02 2.95e+01 -2.63e+03 ... $ mixdata :Classes ‘mixdata’ and 'data.frame': 11 obs. of 2 variables: ..$ X : num [1:11] 1e+04 2e+04 3e+04 4e+04 5e+04 6e+04 7e+04 8e+04 9e+04 1e+05 ... ..$ count: int [1:11] 993 137 82 30 21 5 7 14 21 2 ... $ usecondit : logi FALSE - attr(*, class)= chr mix In my code I am trying around 10.000 fit (and each of these fits returns the list above) and I want to keep those in a way that later on I would be able to search inside all the lists.For example I would like to find inside those 10.000 lists which one has the smallest $chisq value. What would be a suitable way to implement that in R? Luckily I am working in a computer with a lot of ram so storing 10.000 lists temporary in memory before saving to disk would not be a problem. What would you suggest me? RegardsAlex [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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] Dummy variable in ARIMA
On 26 Feb 2015, at 17:29 , Mikael Olai Milhøj mikaelmil...@gmail.com wrote: Hi. First of all, thx. But when using in arima(...xreg=fact,...) then fact should be a vector and not a factor variable? Maybe I should have been more clear in my first mail, sorry. Or else I have to dig deeper into factors. You can always do things like as.numeric((1:50) %in% (4:6)), but longer term I think it is more generalizable to play with model.matrix(), i.e. M - model.matrix(~fact). As far as I recall, arima() will automatically include a constant so you need to say xreg=M[,-1] to get rid of the column of ones. /Mikael On Thu, Feb 26, 2015 at 5:17 PM, Bert Gunter gunter.ber...@gene.com wrote: Inline. Cheers, Bert Bert Gunter Genentech Nonclinical Biostatistics (650) 467-7374 Data is not information. Information is not knowledge. And knowledge is certainly not wisdom. Clifford Stoll On Thu, Feb 26, 2015 at 8:02 AM, Mikael Olai Milhøj mikaelmil...@gmail.com wrote: Hi all I have been searching on the web in vain. I want to include a dummy variable in my ARIMA model. Let's say that I want to make an AR(1) model for X including a dummy variable which should be 1 for observation 4,5,6 and zero otherwise (let's say that there is 50 observations in total). How do I make that? You don't, really. 1. Go through an R tutorial so that you understand the concept of factors and how they are used in R modeling. 2. fact - factor( (1:50) %in% (4:6)) Cheers, Bert This does the trick but seems inefficient: dummy-c(rep(0,3), rep(1,3), rep(0,44)) Thx in advance Best regards /Mikael [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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 -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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 -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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] extract file name from a path string
Look at dirname() and basename(). The first would be what you call the path. The second is the file.name without the path. On Thu, Feb 26, 2015 at 10:58 AM, Luigi Marongiu marongiu.lu...@gmail.com wrote: Dear all, what code should I write in order to extract the file name from a give path? Let's say that I want to get the file my file.xls which is in the directory/folder My documents; since I work both with Windows and Linux, the paths I am looking at are in the format: path.windows-\\home$\\lm667\\My Documents\\my file.xls path.linux-/home/My Documents/my file.xls I used two words for the file name because sometimes the file names have multiple words rather than a single one separated by capitals, . or _. The code should now get the file name, which is included between \\ (or /) and .xls but I don't know what regular expression will do the trick. Once the file name has been assigned to a vector, it should be easy to remove it from the path.windows/.linux and obtain a vector with the path on its own. Essentially the output should be as follows: file.name [1] my file.xls path.w [1] \\home$\\lm667\\My Documents\\ path.l [1] /home/My Documents/ Thank you, Luigi __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- He's about as useful as a wax frying pan. 10 to the 12th power microphones = 1 Megaphone Maranatha! John McKown __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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] Dummy variable in ARIMA
Dig deeper. -- Bert Bert Gunter Genentech Nonclinical Biostatistics (650) 467-7374 Data is not information. Information is not knowledge. And knowledge is certainly not wisdom. Clifford Stoll On Thu, Feb 26, 2015 at 8:29 AM, Mikael Olai Milhøj mikaelmil...@gmail.com wrote: Hi. First of all, thx. But when using in arima(...xreg=fact,...) then fact should be a vector and not a factor variable? Maybe I should have been more clear in my first mail, sorry. Or else I have to dig deeper into factors. /Mikael On Thu, Feb 26, 2015 at 5:17 PM, Bert Gunter gunter.ber...@gene.com wrote: Inline. Cheers, Bert Bert Gunter Genentech Nonclinical Biostatistics (650) 467-7374 Data is not information. Information is not knowledge. And knowledge is certainly not wisdom. Clifford Stoll On Thu, Feb 26, 2015 at 8:02 AM, Mikael Olai Milhøj mikaelmil...@gmail.com wrote: Hi all I have been searching on the web in vain. I want to include a dummy variable in my ARIMA model. Let's say that I want to make an AR(1) model for X including a dummy variable which should be 1 for observation 4,5,6 and zero otherwise (let's say that there is 50 observations in total). How do I make that? You don't, really. 1. Go through an R tutorial so that you understand the concept of factors and how they are used in R modeling. 2. fact - factor( (1:50) %in% (4:6)) Cheers, Bert This does the trick but seems inefficient: dummy-c(rep(0,3), rep(1,3), rep(0,44)) Thx in advance Best regards /Mikael [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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 -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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] Processing key_column, begin_date, end_date in R
here is yet another way:
dta - read.csv( text=
+ key_column,begin_date,end_date
+ 123456,2013-01-01,2014-01-01
+ 123456,2013-07-01,2014-07-01
+ 789102,2012-03-01,2014-03-01
+ 789102,2015-02-01,2016-02-01
+ 789102,2015-02-06,2016-02-06
+ 789102,2015-02-28,2015-03-31
+ 789102,2015-04-30,2015-05-31
+ , as.is=TRUE)
# check for overlap by sorting into time order and the adding 1 for
# begin and -1 for end and create cumsum
# select only resulting entries with begin @ 1 and end @ 0
dta - dta %%
+ mutate(begin_date = as.Date(begin_date) # convert the times
+ , end_date = as.Date(end_date)
+ ) %%
+ gather(time, value, -key_column) %% # create 'long' data
+ mutate(oper = ifelse(grepl('^b', time), 1, -1)) %% # value for begin/end
+ arrange(value) %% # sort by time
+ group_by(key_column) %% # separate into groups
+ mutate(depth = cumsum(oper)) %%
+ filter((grepl(^b, time) depth == 1) | # filter on begin@1 and end@0
+ (grepl(^e, time) depth == 0)
+ ) %%
+ ungroup() %%
+ arrange(key_column, value)
# now have pairs of lines for the times
indx - seq(1, nrow(dta), 2)
result - data.frame(key_column = dta$key_column[indx]
+ , begin_time = dta$value[indx]
+ , end_time = dta$value[indx + 1L]
+ , stringsAsFactors = FALSE
+ )
result
key_column begin_time end_time
1 123456 2013-01-01 2014-07-01
2 789102 2012-03-01 2014-03-01
3 789102 2015-02-01 2016-02-06
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, Feb 25, 2015 at 4:18 PM, Matt Gross gro...@gmail.com wrote:
Hi,
I am trying to process a large dataset in R. The dataset contains the
following three columns:
key_column - a unique key identifier
begin_date - the start date of the active period
end_date - the end date of the active period
Example data is here:
key_column,begin_date,end_date
123456,2013-01-01,2014-01-01
123456,2013-07-01,2014-07-01
789102,2012-03-01,2014-03-01
789102,2015-02-01,2016-02-01
789102,2015-02-06,2016-02-06
I want to build a condensed table of key_column and begin_date's and
end_date's. As you can see in the example data above, some begin and end
date periods overlap with begin_date and end_date pairs for the same
key_column. In situations where overlap exists I want to have one record
for the key_column with the min(begin_date) and the max(end_date).
Can anyone help me build the commands to process this data in R?
Thanks,
Matt
--
Matt Gross
gro...@gmail.com
503.329.4545
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Re: [R] Saving Mean Relative Difference from all.equal()
I think I have one solution. Not very pretty though. Relies on the text not changing at all. as.numeric(gsub(Mean relative difference: , , all.equal(cov2cor(ITEMCOV),cor(item.data))[2])) Is there a better way? -- View this message in context: http://r.789695.n4.nabble.com/Saving-Mean-Relative-Difference-from-all-equal-tp4703905p4703908.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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] Schedule R function/Code to run at specific time
Everything Duncan said, plus:
A construction like this might do the job
run.at - as.POSIXct('2015-02-26 13:05')
while(TRUE) {
if ( trunc(Sys.time(),'min') == run.at) source('whatever-it-is.r')
Sys.sleep(60)
}
but I wouldn't count on it to be as reliable as cron (or Windows
equivalent).
--
Don MacQueen
Lawrence Livermore National Laboratory
7000 East Ave., L-627
Livermore, CA 94550
925-423-1062
On 2/26/15, 11:20 AM, Duncan Murdoch murdoch.dun...@gmail.com wrote:
On 26/02/2015 2:08 PM, Doran, Harold wrote:
Is there functionality within R proper, without having to revert to the
OS, allowing a function or a portion of an R script to be run at a
defined time? My google searches haven't provided much other than one at
the link below which relies on an OS.
If you want it to start 1000 seconds from now, use Sys.sleep(1000) as
your first statement. You'll have a process sitting there using no CPU
(but perhaps lots of virtual memory) until the sleeping is done. Using
cron is better.
Duncan Murdoch
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[R] Saving Mean Relative Difference from all.equal()
Hello, Does anyone know how to save the numeric value of the mean relative difference when using the all.equal() command? For example this: all.equal(cov2cor(ITEMCOV),cor(item.data)) Gives: [1] Attributes: Length mismatch: comparison on first 1 components [2] Mean relative difference: 0.01523708 I'd like to save the value 0.01523708 in a numeric format. Thanks, -- View this message in context: http://r.789695.n4.nabble.com/Saving-Mean-Relative-Difference-from-all-equal-tp4703905.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Extracting Factor Pattern Matrix Similar to Proc Factor
Thanks everyone -- View this message in context: http://r.789695.n4.nabble.com/Extracting-Factor-Pattern-Matrix-Similar-to-Proc-Factor-tp4703704p4703904.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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] Schedule R function/Code to run at specific time
On 26/02/2015 2:08 PM, Doran, Harold wrote: Is there functionality within R proper, without having to revert to the OS, allowing a function or a portion of an R script to be run at a defined time? My google searches haven't provided much other than one at the link below which relies on an OS. If you want it to start 1000 seconds from now, use Sys.sleep(1000) as your first statement. You'll have a process sitting there using no CPU (but perhaps lots of virtual memory) until the sleeping is done. Using cron is better. Duncan Murdoch __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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] integrate with vector arguments
-BEGIN PGP SIGNED MESSAGE-
Hash: SHA1
I'm a bit stuck.
I have to integrate a series of polynomial functions with vector
arguments.
v1-c(1:5)
v2-c(1:5)
f1-function (x) {v1*x+v2*x^2}
The problem is that integrate(f1, 0, 1) does not work.
I does not, even if a pas the arguments (v1, v2)
f1-function (x, v1, v2) {v1*x+v2*x^2}
or if i try to vectorize the function
f1-Vectorize(function(x, v1, v2){v1*x+v2*x^2},
vectorize.args=c(v1, v2))
integrate(f1, 0, 1) gives an error:
Error in integrate(f1, 0, 1) :
evaluation of function gave a result of wrong length
Any help will be greatly appreciated.
Thanks,
Marko
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[R] Schedule R function/Code to run at specific time
Is there functionality within R proper, without having to revert to the OS, allowing a function or a portion of an R script to be run at a defined time? My google searches haven't provided much other than one at the link below which relies on an OS. Thanks, Harold https://tgmstat.wordpress.com/2013/09/11/schedule-rscript-with-cron/ [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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-es] como eliminar los nombres de las series en el grafico
Funciono perfecto !! tan facil que era, jajaja ... el que sabe sabe, como dicen. Muchas gracias Jorge, Eric. On 26/02/15 03:02, Jorge I Velez wrote: Eric, Creo que necesitas Hmisc:::xYplot(..., label.curves = FALSE) Saludos, Jorge.- 2015-02-26 11:15 GMT+11:00 eric ericconchamu...@gmail.com mailto:ericconchamu...@gmail.com: Estimada comunidad, estoy haciendo un grafico con la funcion xYplot (paquete Hmisc) para agregar barras de error a los puntos del grafico. Pero pasa algo que no me gusta: aparece el nombre de las series en el grafico. Alguien sabe como puedo eliminarlas ? no quiero usar ggplot porque ya tengo hechos todos los graficos con lattice. Aqui el codigo que uso y los datos (estan separados por tab, no por ,) y el grafico los envio adjuntos. xYplot(Cbind(ave,ul,ll) ~ con , groups=sol,data=ca.med.sincon ,xlab=Solvent concentration (mM) , ylab=Contact Angle (°), method=lower bars, cex=1.1, type=b , scales=list(x=list(log=10), equispaced.log=FALSE) , abline=list(h=ca.med[ca.med$sol==con,3]) ) Espero que alguien sepa porque ya he gastado mucho tiempo buscando y probando cosas. Gracias, Eric. -- Forest Engineer Master in Environmental and Natural Resource Economics Ph.D. student in Sciences of Natural Resources at La Frontera University Member in AguaDeTemu2030, citizen movement for Temuco with green city standards for living Nota: Las tildes se han omitido para asegurar compatibilidad con algunos lectores de correo. ___ R-help-es mailing list R-help-es@r-project.org mailto:R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es -- Forest Engineer Master in Environmental and Natural Resource Economics Ph.D. student in Sciences of Natural Resources at La Frontera University Member in AguaDeTemu2030, citizen movement for Temuco with green city standards for living Nota: Las tildes se han omitido para asegurar compatibilidad con algunos lectores de correo. ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es
[R] Summing certain values within columns that satisfy a certain condition
Hi, Supposed I had a data frame like so: A B C D 0 1 0 7 0 2 0 7 0 3 0 7 0 4 0 7 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 5 0 5 1 5 0 4 1 5 0 8 4 7 0 0 3 0 0 0 3 4 0 0 3 4 0 0 0 5 0 2 0 6 0 0 4 0 0 0 4 0 0 0 4 0 For each row, I want to count how many max column values appear to adventurely get the following outcome, while ignoring zeros and N/As: A B C D Sum 0 1 0 7 1 0 2 0 7 1 0 3 0 7 1 0 4 0 7 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 5 0 0 5 1 5 0 0 4 1 5 0 0 8 4 7 3 0 0 3 0 0 0 0 3 4 0 0 0 3 4 0 0 0 0 5 0 0 2 0 6 0 0 0 4 0 1 0 0 4 0 1 0 0 4 0 1 I've used the following code but it doesn't seem to work (my sum column column is all 1s): (apply(df,1, function(x) (sum(x %in% c(pmax(x)) Is this code too simple? Thanks! K. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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] PEA and APE Tobit
Dear Annelies On 26 February 2015 at 09:12, hnlki annelies.hoebe...@ugent.be wrote: I estimated a tobit model tobit.fit-tobit(y~x,left=0, right=Inf) (library AER) or tobit2.fit-censReg(y~x, left=0, right=Inf) (librarycensReg) I' have estimated the partial effect at the average as: pea-(pnorm((colMeans(x)%*%tobit.fit$coef[-1])/tobit.fit$scale))%*%tobit.fitt$coef[-1] and the average partial effect as: ape- (length(x[,1]))^(-1)*sum(pnorm((x%*%tobit.fit$coef[-1])/tobit.fit$scale))*tobit.fit$coef[-1] I guess I did something wrong as margEff( tobit2.fit) (library(censReg) gives a different result than my partial effect at the average. Any ideas about what I did wrong? I did not find the underlying code of margEff. [...] PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Please follow the posting guide and provide a self-contained reproducible example. Best regards, Arne -- Arne Henningsen http://www.arne-henningsen.name __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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] Schedule R function/Code to run at specific time
On 26 Feb 2015, at 20:20 , Duncan Murdoch murdoch.dun...@gmail.com wrote: On 26/02/2015 2:08 PM, Doran, Harold wrote: Is there functionality within R proper, without having to revert to the OS, allowing a function or a portion of an R script to be run at a defined time? My google searches haven't provided much other than one at the link below which relies on an OS. If you want it to start 1000 seconds from now, use Sys.sleep(1000) as your first statement. You'll have a process sitting there using no CPU (but perhaps lots of virtual memory) until the sleeping is done. Using cron is better. There are also some asynchronous possibilities using tcltk: tcl(after, 1, quote(print(boo))) Peter D. Duncan Murdoch __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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 -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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] Summing certain values within columns that satisfy a certain condition
I guess the answer to your question is yes. dta - read.table( text= A B C D 0 1 0 7 0 2 0 7 0 3 0 7 0 4 0 7 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 5 0 5 1 5 0 4 1 5 0 8 4 7 0 0 3 0 0 0 3 4 0 0 3 4 0 0 0 5 0 2 0 6 0 0 4 0 0 0 4 0 0 0 4 0 , header=TRUE ) dtacmax - sapply( dta, max ) followed by dta$Sum - apply(dta,1, function(x) (sum(0!=x x == dtacmax,na.rm=TRUE))) or dtam - as.matrix( dta[,1:4] ) dta$Sum2 - rowSums( !is.na(dtam) 0!=dtam dtam == matrix( dtacmax, ncol=ncol( dtam ), nrow=nrow( dtam ), byrow=TRUE ) ) --- 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. On February 26, 2015 12:23:48 PM PST, Kate Ignatius kate.ignat...@gmail.com wrote: Hi, Supposed I had a data frame like so: A B C D 0 1 0 7 0 2 0 7 0 3 0 7 0 4 0 7 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 5 0 5 1 5 0 4 1 5 0 8 4 7 0 0 3 0 0 0 3 4 0 0 3 4 0 0 0 5 0 2 0 6 0 0 4 0 0 0 4 0 0 0 4 0 For each row, I want to count how many max column values appear to adventurely get the following outcome, while ignoring zeros and N/As: A B C D Sum 0 1 0 7 1 0 2 0 7 1 0 3 0 7 1 0 4 0 7 1 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 5 0 0 5 1 5 0 0 4 1 5 0 0 8 4 7 3 0 0 3 0 0 0 0 3 4 0 0 0 3 4 0 0 0 0 5 0 0 2 0 6 0 0 0 4 0 1 0 0 4 0 1 0 0 4 0 1 I've used the following code but it doesn't seem to work (my sum column column is all 1s): (apply(df,1, function(x) (sum(x %in% c(pmax(x)) Is this code too simple? Thanks! K. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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 -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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 many digits are there in left of dot of 0.0001 ?
Hi, I assume you want to know the digit count to the left of decimal point. If this is the case, then you may use trunc(log10(max(1,trunc(abs(a)+1 for a numerical variable a. Count 0.12 as having one digit to the left of decimal point. trunc(log10(max(1,trunc(abs(-10.99)+1 [1] 6 trunc(log10(max(1,trunc(abs(0)+1 [1] 1 trunc(log10(max(1,trunc(abs(9.999)+1 [1] 1 trunc(log10(max(1,trunc(abs(19.999)+1 [1] 2 trunc(log10(max(1,trunc(abs(-1999.999)+1 [1] 4 -- View this message in context: http://r.789695.n4.nabble.com/How-many-digits-are-there-in-left-of-dot-of-0-0001-tp4703842p4703847.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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] Processing key_column, begin_date, end_date in R
Hi, Here is an implemenation: date key_column begin_dateend_date 1 123456 2013-01-01 2014-01-01 2 123456 2013-07-01 2014-07-01 3 789102 2012-03-01 2014-03-01 4 789102 2015-02-01 2016-02-01 5 789102 2015-02-06 2016-02-06 y - t(sapply(unique(date$key_column),function(x) c(x,min(as.character(date[date$key_column==x,begin_date])),max(as.character(date[date$key_column==x,end_date]) y [,1] [,2] [,3] [1,] 123456 2013-01-01 2014-07-01 [2,] 789102 2012-03-01 2016-02-06 colnames(y) NULL colnames(y) - c(key_column,begin_date,end_date) y key_column begin_date end_date [1,] 123456 2013-01-01 2014-07-01 [2,] 789102 2012-03-01 2016-02-06 -- View this message in context: http://r.789695.n4.nabble.com/Processing-key-column-begin-date-end-date-in-R-tp4703835p4703850.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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] Solved: Re: text miner error: Error in UseMethod(meta, x)
Hi list Closing this one off myself, this is what I did: The error seems to concern the update of tm to version 0.6: the conversion to lower case text should now be: docs - tm_map(docs, content_transformer(tolower)) Everything else seems to work fine thereafter. The issue in the tutorial concerns section 3.1. wherein Graham creates a function toSpace. This seems to introduce an additional term that tm_map and later DocumentTermMatrix do not seem to know how to handle. This is probably an incorrect interpretation of what's going on, but the fix appears to be to use the above line earlier in the preparation stage. If anyone has more informed insight, please share. Cheers Sun On 25/02/15 17:33, Sun Shine wrote: Hi list I've been working my way through a tutorial on text mining ( http://onepager.togaware.com/TextMiningO.pdf ) and all was well until I came across this problem using tm (text miner): ++code+++ docs - tm_map(docs, content_transformer(tolower)) Warning messages: 1: In mclapply(x$content[i], function(d) tm_reduce(d, x$lazy$maps)) : all scheduled cores encountered errors in user code 2: In mclapply(content(x), FUN, ...) : all scheduled cores encountered errors in user code ++end-code After some searching, it appears the best fix for this problem was to pass an explicit lazy=TRUE argument to tm, like this: docs - tm_map(docs, content_transformer(tolower), lazy=TRUE) However, a little further on in the tutorial to set up the text matrix, a related (?) error was returned: ++code+++ dtm - DocumentTermMatrix(docs) Error in UseMethod(meta, x) : no applicable method for 'meta' applied to an object of class try-error In addition: Warning message: In mclapply(unname(content(x)), termFreq, control) : all scheduled cores encountered errors in user code ++end-code I tried applying the explicit lazy=TRUE again, but doesn't change things. I have gone over the tutorial again and have followed all of the steps (including loading the requisite libraries). Moreover, searching on the web seems to return several contradictory suggestions and I'm no wiser than I was before. The closest I came to an answer was at Stack Overflow http://stackoverflow.com/questions/24771165/r-project-no-applicable-method-for-meta-applied-to-an-object-of-class-charact and that answer suggested using the latest tm (v 0.6) and claimed that the earlier tolower step was wrong. However, my code used the recommended: corpus - tm_map(corpus, content_transformer(tolower)) Is there anyone on the list who could either sign-post me to a solution or assist in debugging this please? I'm running R version 3.1.2 and tm is 0.6 Many thanks Sun __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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] covert entire dataset to numeric while persuing percentage values
Hi , I am little confused about how to covert entire dataset to numeric . As I read data like.. Xelements =read.csv(file. Choose(),header = T, stringsAsFactors=FALSE) str(xelements ) str(xelements) 'data.frame': 731 obs. of 4 variables: $ Engine.Speed : chr rpm ES rpm 1049 ... $ X..NG.by.Energy : chr % NG by Energy % 0% ... $ Int.Mfld.Temp: chr Int Mfld Temp �C 49 ... $ Cmd.Advance.Angle: chr Cmd Advance Angle �BTDC 13.8 ... I have second column as in 0%, 10%, In percentage value , Whenever I am going to covert whole dataset its showing NA introduced .second column going to become NA. Converting separately I would be successful . xelements$Engine.Speed - as.numeric(xelements$Engine.Speed) Warning message: NAs introduced by coercion xelements$X..NG.by.Energy- as.numeric(sub(%,,xelements$X..NG.by.Energy))/100 Warning message: NAs introduced by coercion xelements$Int.Mfld.Temp- as.numeric(xelements$Int.Mfld.Temp) Warning message: NAs introduced by coercion xelements$Cmd.Advance.Angle- as.numeric(xelements$Cmd.Advance.Angle) Warning message: NAs introduced by coercion But I want to covert whole dataset at a time. I want to write function which will help me to solve this problem . xelements - data.frame(sapply(xelements, function(x) as.numeric(as.character(x sapply(xelements, class) but it won't be able to covert percentage value like 10%, 20% please do help me if you know the way. Thank you [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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] format selected columns in dataframe as character
Dear R-List,
#I have a df with the first two cols formatted as factor.
dfx - data.frame(
group = c(rep('A', 8), rep('B', 15), rep('C', 6)),
sex = sample(c(M, F), size = 29, replace = TRUE),
age = runif(n = 29, min = 18, max = 54))
# now I want to format both factor VARs as character
# I tried
factor.id-names(dfx[sapply(dfx,is.factor)])
chr.names-which(names(dfx)%in% factor.id)
dfx[ , chr.names]-as.character(dfx[ , chr.names])
# which gives me
str(dfx)
'data.frame': 29 obs. of 3 variables: $ group: chr c(1, 1, 1, 1, 1, 1, 1, 1, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3) c(2, 2, 1, 1, 1,
1, 2, 1, 1, 2, 2, 1, 2, 2, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 2, 2) c(1,
1, 1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3,
3) c(2, 2, 1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 2, 2, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1,
1, 1, 2, 2, 2) ... $ sex : chr c(2, 2, 1, 1, 1, 1, 2, 1, 1, 2, 2, 1, 2, 2, 2,
1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 2, 2) c(1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2,
2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3) c(2, 2, 1, 1, 1, 1, 2,
1, 1, 2, 2, 1, 2, 2, 2, 1, 1, 1, 1, 1, 2, 1, 1, 1, 1, 1, 2, 2, 2) c(1, 1, 1,
1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3)
... $ age : num 38.5 18 33.5 26 22.5 ...
#though I was hoping for something like
'data.frame': 29 obs. of 3 variables:
$ group: chr A A A A ...
$ sex : chr M F F M ...
$ age : num 21.3 35.2 53.8 21 23.6 ...
#What is wrong with my code?
#Thank you for any help
Best wishes
Alain
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[R] Rgraphviz and NA indices error
Hi list Can someone help me debug the following please: Having downloaded and installed the bioconductor packages and Rgraphviz, I am attempting to plot a network graph showing the relation among chosen words in the corpus of text data. I first did this: plot(dtm, terms=findFreqTerms(dtm, lowfreq=100) [1:30], corThreshold=0.75) and received the error message: Error in `[.simple_triplet_matrix`(m, , terms) : NA indices not allowed. My next step was to remove any NA indices (although to be honest, this is more of a stab in the dark because there shouldn't be any NA values in the corpus): docsNA - (docs[!is.na(docs)]) Then redid the DTM with the NA values removed dtmNA - DocumentTermMatrix(docsNA) Then re-ran Rgraphviv with the new set plot(dtmNA, terms=findFreqTerms(dtmNA, lowfreq=100) [1:10], corThreshold=0.5) But, still get an error: Error in `[.simple_triplet_matrix`(m, , terms) : NA indices not allowed. I have not been successful in finding out why this error persists nor what to do about it. Anyone have any ideas to progress past this issue? Thanks Sun __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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 Tukey
Dear all , I am trying to do a Tukey comparison, but a message appears: tuk-glht(mod0,linfct=mcp(Soil=Tukey));summary(tuk) Error: could not find function glht Anyone knows how to fix it? Thanks a lot! Xavier __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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 Tukey
??glht would tell you that glht is a function from the multcomp package. You need to load a package before you can use its functions. library(multcomp) ir. Thierry Onkelinx Instituut voor natuur- en bosonderzoek / Research Institute for Nature and Forest team Biometrie Kwaliteitszorg / team Biometrics Quality Assurance Kliniekstraat 25 1070 Anderlecht Belgium To call in the statistician after the experiment is done may be no more than asking him to perform a post-mortem examination: he may be able to say what the experiment died of. ~ Sir Ronald Aylmer Fisher The plural of anecdote is not data. ~ Roger Brinner The combination of some data and an aching desire for an answer does not ensure that a reasonable answer can be extracted from a given body of data. ~ John Tukey 2015-02-26 11:21 GMT+01:00 CHIRIBOGA Xavier xavier.chirib...@unine.ch: Dear all , I am trying to do a Tukey comparison, but a message appears: tuk-glht(mod0,linfct=mcp(Soil=Tukey));summary(tuk) Error: could not find function glht Anyone knows how to fix it? Thanks a lot! Xavier __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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 -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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-es] nueva en R
hola!soy nueva aqui, estoy haciendo un trabajo sobre R y toda la informacion que encuentro esta en ingles. me gustaria saber algun libro sobre R o donde podria conseguir buena informacion sobre el programa en español, ya sea de su funcionamiento, finalidad del programa, caracteristicas etc. Nunca he utilizado este programa. gracias. [[alternative HTML version deleted]] ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es
[R] library(multcomp) does not work for loading Tukey
Dear colleagues, For Tukey, I tried to load the function with library(multcomp) but again a message says: Error in library(multcomp) : any package called ‘multcomp’ has been found Thanks for ur help, Xavier __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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 many digits are there in left of dot of 0.0001 ?
yes this is exactly what I want and it works. thanks. -Original Message- From: JS Huang [js.hu...@protective.com] Date: 02/26/2015 03:22 AM To: r-help@r-project.org Subject: Re: [R] How many digits are there in left of dot of 0.0001 ? Hi, To get the number of digits to the right of decimal point: nchar(format(a,scientific=FALSE))-(trunc(log10(max(1,trunc(abs(a)+1) -1. The part (trunc(log10(max(1,trunc(abs(a)+1) is the number of digits to the left of decimal. At the end, subtract 1 for the decimal point. Negative number needs more work. options(digits=10) a - 0.0001 nchar(format(a,scientific=FALSE))-(trunc(log10(max(1,trunc(abs(a)+1) -1 [1] 4 a - 999.123456 nchar(format(a,scientific=FALSE))-(trunc(log10(max(1,trunc(abs(a)+1) -1 [1] 6 -- View this message in context: http://r.789695.n4.nabble.com/How-many-digits-are-there-in-left-of-dot-of-0-0001-tp4703842p4703849.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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 -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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] Replace the value with 1 and 0
On 2015-02-26 00:33, JS Huang wrote: Hi, Here is an implementation. More data are added. An extra column hasRain is added instead of replacing column Amount. rain Year Month Day Amount 1 1950 1 10.0 2 1950 1 2 35.5 3 1950 1 3 17.8 4 1950 1 4 24.5 5 1950 1 5 12.3 6 1950 1 6 11.5 7 1950 1 75.7 8 1950 1 8 13.2 9 1950 1 9 11.3 10 1950 1 10 14.7 11 1950 1 11 11.9 12 1950 1 12 17.5 13 1950 1 138.1 14 1950 1 140.4 15 1950 1 150.0 16 1950 1 16 19.5 17 1950 1 17 10.7 18 1950 1 180.5 19 1950 1 19 12.7 20 1950 1 206.3 21 1950 2 10.0 22 1950 2 2 35.5 23 1950 2 3 17.8 24 1950 2 4 24.5 25 1950 2 5 12.3 26 1950 2 6 11.5 27 1950 2 75.7 28 1950 2 8 13.2 29 1950 2 9 11.3 30 1950 2 10 14.7 31 1950 2 11 11.9 32 1950 2 12 17.5 33 1950 2 138.1 34 1950 2 140.4 35 1950 2 150.0 36 1950 2 16 19.5 37 1950 2 17 10.7 38 1950 2 180.0 39 1950 2 190.0 40 1950 2 200.0 rain$hasRain - ifelse(rain$Amount0,1,0) No! Better is rain$hasRain - as.numeric(rain$Amount 0) See previous discussions about the use of 'ifelse'. Göran rain Year Month Day Amount hasRain 1 1950 1 10.0 0 2 1950 1 2 35.5 1 3 1950 1 3 17.8 1 4 1950 1 4 24.5 1 5 1950 1 5 12.3 1 6 1950 1 6 11.5 1 7 1950 1 75.7 1 8 1950 1 8 13.2 1 9 1950 1 9 11.3 1 10 1950 1 10 14.7 1 11 1950 1 11 11.9 1 12 1950 1 12 17.5 1 13 1950 1 138.1 1 14 1950 1 140.4 1 15 1950 1 150.0 0 16 1950 1 16 19.5 1 17 1950 1 17 10.7 1 18 1950 1 180.5 1 19 1950 1 19 12.7 1 20 1950 1 206.3 1 21 1950 2 10.0 0 22 1950 2 2 35.5 1 23 1950 2 3 17.8 1 24 1950 2 4 24.5 1 25 1950 2 5 12.3 1 26 1950 2 6 11.5 1 27 1950 2 75.7 1 28 1950 2 8 13.2 1 29 1950 2 9 11.3 1 30 1950 2 10 14.7 1 31 1950 2 11 11.9 1 32 1950 2 12 17.5 1 33 1950 2 138.1 1 34 1950 2 140.4 1 35 1950 2 150.0 0 36 1950 2 16 19.5 1 37 1950 2 17 10.7 1 38 1950 2 180.0 0 39 1950 2 190.0 0 40 1950 2 200.0 0 tapply(rain$hasRain,list(rain$Year,rain$Month),sum) 1 2 1950 18 15 -- View this message in context: http://r.789695.n4.nabble.com/Replace-the-value-with-1-and-0-tp4703838p4703840.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Processing key_column, begin_date, end_date in R
Hi,
It's not as easy as I originally thought. Here is a revision with the
function beginEnd to get it done.
date
key_column begin_dateend_date
1 123456 2013-01-01 2014-01-01
2 123456 2013-07-01 2014-07-01
3 789102 2012-03-01 2014-03-01
4 789102 2015-02-01 2016-02-01
5 789102 2015-02-06 2016-02-06
beginEnd
function()
{
date[order(date$key_column,date$begin_date),]
key - numeric(0)
begin - character(0)
end - character(0)
currentKey - as.numeric(date$key_column[1])
key - c(key, currentKey)
currentBegin - as.character(date$begin_date[1])
begin - c(begin, currentBegin)
currentEnd - as.character(date$end_date[1])
for (i in 2:length(date$key_column))
{
if (currentKey == as.numeric(date$key_column[i]))
{
if (currentEnd = as.character(date$begin_date[i]))
{
currentEnd - max(currentEnd, as.character(date$end_date[i]))
}
else
{
end - c(end, currentEnd)
currentKey - as.numeric(date$key_column[i])
key - c(key, currentKey)
currentBegin - as.character(date$begin_date[i])
begin - c(begin, currentBegin)
currentEnd - as.character(date$end_date[i])
}
if (i == length(date$key_column))
{
end - c(end, currentEnd)
}
}
else
{
end - c(end, currentEnd)
currentKey - as.numeric(date$key_column[i])
key - c(key, currentKey)
currentBegin - as.character(date$begin_date[i])
begin - c(begin, currentBegin)
currentEnd - as.character(date$end_date[i])
if (i == length(date$key_column))
{
end - list(end, currentEnd)
}
}
}
result - cbind(key, begin, end)
colnames(result) - c(key.column,begin.date,end.date)
return(result)
}
beginEnd()
key.column begin.date end.date
[1,] 123456 2013-01-01 2014-07-01
[2,] 789102 2012-03-01 2014-03-01
[3,] 789102 2015-02-01 2016-02-06
--
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[R] PEA and APE Tobit
Hi, I estimated a tobit model tobit.fit-tobit(y~x,left=0, right=Inf) (library AER) or tobit2.fit-censReg(y~x, left=0, right=Inf) (librarycensReg) I' have estimated the partial effect at the average as: pea-(pnorm((colMeans(x)%*%tobit.fit$coef[-1])/tobit.fit$scale))%*%tobit.fitt$coef[-1] and the average partial effect as: ape- (length(x[,1]))^(-1)*sum(pnorm((x%*%tobit.fit$coef[-1])/tobit.fit$scale))*tobit.fit$coef[-1] I guess I did something wrong as margEff( tobit2.fit) (library(censReg) gives a different result than my partial effect at the average. Any ideas about what I did wrong? I did not find the underlying code of margEff. Kind regards, -- View this message in context: http://r.789695.n4.nabble.com/PEA-and-APE-Tobit-tp4703856.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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 many digits are there in left of dot of 0.0001 ?
Hi, To get the number of digits to the right of decimal point: nchar(format(a,scientific=FALSE))-(trunc(log10(max(1,trunc(abs(a)+1) -1. The part (trunc(log10(max(1,trunc(abs(a)+1) is the number of digits to the left of decimal. At the end, subtract 1 for the decimal point. Negative number needs more work. options(digits=10) a - 0.0001 nchar(format(a,scientific=FALSE))-(trunc(log10(max(1,trunc(abs(a)+1) -1 [1] 4 a - 999.123456 nchar(format(a,scientific=FALSE))-(trunc(log10(max(1,trunc(abs(a)+1) -1 [1] 6 -- View this message in context: http://r.789695.n4.nabble.com/How-many-digits-are-there-in-left-of-dot-of-0-0001-tp4703842p4703849.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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] Best Mac for R
On 26 Feb 2015, at 06:26 , Dan Murphy chiefmur...@gmail.com wrote: Quick responses as usual. Can always count on R-Help! Bert's point that it depends is key, of course. Mark and Karim reminded me that R does not use all cores natively. Putting those together, an expensive quad core machine is not necessary for simple package development, documentation, etc. And for hard core (no pun intended) analysis, a multi-core machine won't be fully utilized without parallel implementation of some type. Thanks all for your advice. Just what I was looking for. Dan Notice though, that parallel features _can_ be exploited fairly easily on the multi-CPU Macs (it depends somewhat on whether you need fine-grained parallelism as in fast matrix operations or just embarrassingly parallel tasks like simulation studies - the former needs R to be linked against the Accelerate framework). Also notice that the real Mac experts live on the R-SIG-Mac list and not so much on R-help. -pd On Wed, Feb 25, 2015 at 2:53 PM, Karim Mezhoud kmezh...@gmail.com wrote: Hi, It is not so efficient to have the most speed processor or biggest RAM. In general One processor is working at the time. It is more interesting to work with Linux for multiple multi_thread package and 64 bit. I am not sure if turbo boost is working with R. http://stackoverflow.com/questions/1395309/how-to-make-r-use-all-processors On Wed, Feb 25, 2015 at 9:12 PM, Mark Sharp msh...@txbiomed.org wrote: For what I do, which does not require a lot of parallel work, the high end iMac was faster and much less expensive than the Mac Pro. Mark R. Mark Sharp, Ph.D. msh...@txbiomed.org On Feb 25, 2015, at 1:50 PM, Dan Murphy chiefmur...@gmail.com wrote: I am possibly in the market for a new laptop. Predominantly a Windows user, I owned a macbook pro 10 years ago and am considering going that route again. Does the standard advice still hold: Get the most powerful processor (i7), most ram (16GB), and largest internal storage (512GB), if affordable? thanks, dan __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. NOTICE: This E-Mail (including attachments) is confidential and may be legally privileged. It is covered by the Electronic Communications Privacy Act, 18 U.S.C.2510-2521. If you are not the intended recipient, you are hereby notified that any retention, dissemination, distribution or copying of this communication is strictly prohibited. Please reply to the sender that you have received this message in error, then delete it. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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 -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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 -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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-es] nueva en R
Hola, Gloria. Bienvenida! Hay algunos manuales en la parte inferior de http://cran.r-project.org/other-docs.html Saludos cordiales, Jorge.- 2015-02-26 22:36 GMT+11:00 Gloria Perez Fuentes fuper...@gmail.com: hola!soy nueva aqui, estoy haciendo un trabajo sobre R y toda la informacion que encuentro esta en ingles. me gustaria saber algun libro sobre R o donde podria conseguir buena informacion sobre el programa en espa�ol, ya sea de su funcionamiento, finalidad del programa, caracteristicas etc. Nunca he utilizado este programa. gracias. [[alternative HTML version deleted]] ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es [[alternative HTML version deleted]] ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es
[R] aggregating variables (sum within groups)
Hello everybody! I have a (probabely very easy) problem. Even though I was looking in several r-books I could not find a suitable function to this problem, that's why I hope that someone here could help me: # Sample data: group-c(A,A,A,B,B,C,C,C) var1-c(1,0,0,1,1,0,NA,1) var2-c(0,1,NA,0,1,1,0,0) testdata-data.frame(group, var1, var2) Now, I'd like to generate two aggregated variables: testdata$x- ??? should count the sum of var1 within each group (=4) testdata$y- ??? should count the sum of var2 within each group (=3) Therefore I am looking for a function like ave() which does not calculate the mean value but a sum. Thank you for any hints! David [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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] aggregating variables (sum within groups)
Hi David, You have your answer in the question: aggregate() aggregate(cbind(var1,var2)~group, data=testdata, FUN=sum) Although I am not sure what you intended to do with testdata$x- as the result cannot have the same number of rows than testdata HTH, Ivan -- Ivan Calandra, ATER University of Reims Champagne-Ardenne GEGENAA - EA 3795 CREA - 2 esplanade Roland Garros 51100 Reims, France +33(0)3 26 77 36 89 ivan.calan...@univ-reims.fr https://www.researchgate.net/profile/Ivan_Calandra Le 26/02/15 13:02, David Studer a écrit : Hello everybody! I have a (probabely very easy) problem. Even though I was looking in several r-books I could not find a suitable function to this problem, that's why I hope that someone here could help me: # Sample data: group-c(A,A,A,B,B,C,C,C) var1-c(1,0,0,1,1,0,NA,1) var2-c(0,1,NA,0,1,1,0,0) testdata-data.frame(group, var1, var2) Now, I'd like to generate two aggregated variables: testdata$x- ??? should count the sum of var1 within each group (=4) testdata$y- ??? should count the sum of var2 within each group (=3) Therefore I am looking for a function like ave() which does not calculate the mean value but a sum. Thank you for any hints! David [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How many digits are there in left of dot of 0.0001 ?
On 25/02/2015 8:55 PM, ce wrote: Dear all, I would like to count how many digits are there on the left of a the dot of a numeric variable a=0.0001 This will depend on the formatting used. If default formatting used by as.character() is fine, then nchar(sub(^[[:digit:]]*[.], , a)) should work. (Note that default formatting is scientific for 0.0001.) If you want some other formatting, then format first, and pass a character object, e.g. chars - format(a, scientific = FALSE) nchar(sub(^[[:digit:]]*[.], , chars)) Duncan Murdoch __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] aggregating variables (sum within groups)
Dear David, your email is quite confusing. Do you want to get the sum for each group (A,B,C) or each variable as would be indicated by your result? sum by group: aggregate(data=testdata,var1~group,sum) count by group: aggregate(data=testdata,var1~group,length) sum by variable: sum(na.omit(testdata$var1)) But homework shouldn't be posted on this list. Best regards Christian 2015-02-26 13:02 GMT+01:00 David Studer stude...@gmail.com: Hello everybody! I have a (probabely very easy) problem. Even though I was looking in several r-books I could not find a suitable function to this problem, that's why I hope that someone here could help me: # Sample data: group-c(A,A,A,B,B,C,C,C) var1-c(1,0,0,1,1,0,NA,1) var2-c(0,1,NA,0,1,1,0,0) testdata-data.frame(group, var1, var2) Now, I'd like to generate two aggregated variables: testdata$x- ??? should count the sum of var1 within each group (=4) testdata$y- ??? should count the sum of var2 within each group (=3) Therefore I am looking for a function like ave() which does not calculate the mean value but a sum. Thank you for any hints! David [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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 -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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] PEA and APE Tobit
Dear Mr Henningsen, I have read the posting guide but apparently not well enough. I didn't find how to include R code in my post. I'll read it again and I'll try to give a clearer example. Sorry for the inconvenience. Kind regards, Annelies Op 26-feb.-2015 om 21:59 heeft Arne Henningsen-3 [via R] ml-node+s789695n4703915...@n4.nabble.com het volgende geschreven: Dear Annelies On 26 February 2015 at 09:12, hnlki [hidden email] wrote: I estimated a tobit model tobit.fit-tobit(y~x,left=0, right=Inf) (library AER) or tobit2.fit-censReg(y~x, left=0, right=Inf) (librarycensReg) I' have estimated the partial effect at the average as: pea-(pnorm((colMeans(x)%*%tobit.fit$coef[-1])/tobit.fit$scale))%*%tobit.fitt$coef[-1] and the average partial effect as: ape- (length(x[,1]))^(-1)*sum(pnorm((x%*%tobit.fit$coef[-1])/tobit.fit$scale))*tobit.fit$coef[-1] I guess I did something wrong as margEff( tobit2.fit) (library(censReg) gives a different result than my partial effect at the average. Any ideas about what I did wrong? I did not find the underlying code of margEff. [...] PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Please follow the posting guide and provide a self-contained reproducible example. Best regards, Arne -- Arne Henningsen http://www.arne-henningsen.name __ [hidden email] mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. If you reply to this email, your message will be added to the discussion below: http://r.789695.n4.nabble.com/PEA-and-APE-Tobit-tp4703856p4703915.html To unsubscribe from PEA and APE Tobit, click here. NAML -- View this message in context: http://r.789695.n4.nabble.com/PEA-and-APE-Tobit-tp4703856p4703919.html Sent from the R help mailing list archive at Nabble.com. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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] get_map in ggplot doesn't allow the exact specification of my box's corners?
Hello! get_map help says: location: an address, longitude/latitude pair (in that order), or left/bottom/right/top bounding box My code: library(ggmap) library(mapproj) lat_bottom = 52.33 # bottom latitude of Berlin lat_top= 52.5 # top latitude of Berlin lon_left = 13.0 # left longitude of Berlin lon_rigth = 13.95 # right longitude of Berlin mymap - get_map(location = c(lon_left,lat_bottom,lon_rigth,lat_top), source=google) ggmap(mymap) Why is it giving me a warning: Warning: bounding box given to google - spatial extent only approximate. converting bounding box to center/zoom specification. (experimental) Does it mean that there is no way for me to create a map with these exact corners? Thank you! -- Dimitri Liakhovitski __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/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] integrate with vector arguments
marKo mtoncic at ffri.hr writes:
I'm a bit stuck.
I have to integrate a series of polynomial functions with vector
arguments.
v1-c(1:5)
v2-c(1:5)
f1-function (x) {v1*x+v2*x^2}
The problem is that integrate(f1, 0, 1) does not work.
The point is not that there are vector arguments, but that your function
is vector-valued and so the generated error message below rightly says
evaluation of function gave a result of wrong length.
You could integrate each dimension separately or, e.g., you use quadv()
from package 'pracma' which handles vector-valued functions:
v1 - v2 - 1:5
f1-function (x) {v1*x+v2*x^2}
library(pracma)
quadv(f1, 0, 1, tol=1e-10)
$Q
[1] 0.833 1.667 2.500 3.333 4.167
$fcnt
[1] 13
$estim.prec
[1] 0.03
quadv() employs an adaptive Simpson quadrature where the recursion is
applied to all components at once.
I does not, even if a pas the arguments (v1, v2)
f1-function (x, v1, v2) {v1*x+v2*x^2}
or if i try to vectorize the function
f1-Vectorize(function(x, v1, v2){v1*x+v2*x^2},
vectorize.args=c(v1, v2))
integrate(f1, 0, 1) gives an error:
Error in integrate(f1, 0, 1) :
evaluation of function gave a result of wrong length
Any help will be greatly appreciated.
Thanks,
Marko
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