Re: [R] Fwd: Difficulty in finding some R functions
Hello, Maybe from here ? http://www.ie.boun.edu.tr/~hormannw/BounQuantitiveFinance/Thesis/SilaHALULU.pdf HTH Pascal On Fri, May 2, 2014 at 11:11 AM, prachi jain preciousprachi...@gmail.com wrote: Hey, I am trying to find some of the following functions in R packages: MLEt pt3 cormatrix2vector ParameterEst tCopula riskBT I have checked every package from this link: http://cran.r-project.org/web/packages/available_packages_by_name.html but is unable to find the above functions. These functions belong to a code for calculating Value-at-risk and expected shortfall using copula function and backtesting the model. Could you please help me find these functions by naming the packages they belong to. Its really urgent for me so please reply asap. If required i can send the entire code for better understanding. If this is not the right place to mail a query, please do let me know where can I send my query, its really urgent for me. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Pascal Oettli Project Scientist JAMSTEC Yokohama, Japan __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] substring if a record has a \
Hello togehter, i have a litte problem. I have a data.frame with a view entries like this one: 1 A Marius Muller -\nIT Services B Rockwood\nBrockhues C Microlog Services\nMarcos D Firefox Services I now want only the first description in the column until the \n. How can i do this? The solution look like this one: 1 A Marius Muller - B Rockwood C Microlog Services D Firefox Services Thank you. Best regards. Mat. -- View this message in context: http://r.789695.n4.nabble.com/substring-if-a-record-has-a-tp4689857.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] extracting the first few eigenvectors
I have a symmetric matrix, X, and I just want the first K eigenvectors (those associated with the K largest eigenvalues). Clearly, this works: eigs - eigen( X, symmetric=TRUE ) K_eigenvectors - eigs$vectors[ , 1:K ] K_eigenvalues - eigs$values[ 1:K ] rm(eigs) In order to do that, I have to create the matrix eigs$vectors which is the same size as X. Sometimes X has 100 million elements or more. Usually I want only the first 10 eigenvectors instead of all 10,000. Is there any R function that will allow me to extract just the few that I want? This would be analogous to the [V,d] = eigs(X,K) function in Octave/MATLAB. Best, Mike -- Michael B. Miller, Ph.D. Minnesota Center for Twin and Family Research Department of Psychology University of Minnesota http://scholar.google.com/citations?user=EV_phq4J __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] substring if a record has a \
Hello,
Something like that?
x - c('Marius Muller -\nIT Services','Rockwood\nBrockhues','Microlog
Services\nMarcos','Firefox Services')
x - sapply(strsplit(x, '\n'), '[', 1)
x
[1] Marius Muller - Rockwood Microlog Services
[4] Firefox Services
Regards,
Pascal
On Fri, May 2, 2014 at 4:17 PM, Mat matthias.we...@fnt.de wrote:
Hello togehter,
i have a litte problem. I have a data.frame with a view entries like this
one:
1
A Marius Muller -\nIT Services
B Rockwood\nBrockhues
C Microlog Services\nMarcos
D Firefox Services
I now want only the first description in the column until the \n. How can
i do this?
The solution look like this one:
1
A Marius Muller -
B Rockwood
C Microlog Services
D Firefox Services
Thank you.
Best regards. Mat.
--
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JAMSTEC
Yokohama, Japan
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Re: [R] substring if a record has a \
works perfect, thank you :-) -- View this message in context: http://r.789695.n4.nabble.com/substring-if-a-record-has-a-n-tp4689857p4689860.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Return TRUE only for first match of values between matrix and vector.
I wish to return True in a matrix for only the first match of a value per row where the value equals that in a vector with the same number of values as rosw in the matrix eg: A-matrix(c(2,3,2,1,1,2,NA,NA,NA,5,1,0,5,5,5),5,3) B-c(2,1,NA,1,5) desired result: [,1] [,2] [,3] [1,] TRUE FALSE FALSE [2,] FALSE NA FALSE [3,]NA NANA [4,] TRUE NA FALSE [5,] FALSE TRUE FALSE however A==B returns: [,1] [,2] [,3] [1,] TRUE TRUE FALSE [2,] FALSE NA FALSE [3,]NA NANA [4,] TRUE NA FALSE [5,] FALSE TRUE TRUE and apply(A,1,function(x) match (B,x)) returns [,1] [,2] [,3] [,4] [,5] [1,]1 NA1 NA NA [2,]3 NA NA11 [3,] NA222 NA [4,]3 NA NA11 [5,] NA NA332 thanks [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] extracting the first few eigenvectors
On 02-05-2014, at 09:17, Mike Miller mbmille...@gmail.com wrote: I have a symmetric matrix, X, and I just want the first K eigenvectors (those associated with the K largest eigenvalues). Clearly, this works: eigs - eigen( X, symmetric=TRUE ) K_eigenvectors - eigs$vectors[ , 1:K ] K_eigenvalues - eigs$values[ 1:K ] rm(eigs) In order to do that, I have to create the matrix eigs$vectors which is the same size as X. Sometimes X has 100 million elements or more. Usually I want only the first 10 eigenvectors instead of all 10,000. Is there any R function that will allow me to extract just the few that I want? This would be analogous to the [V,d] = eigs(X,K) function in Octave/MATLAB. You can find possibly relevant stuff with library(sos) findFn(eigenvalue”) which is how I found package rARPACK, which may be exactly what you need/want. Berend Best, Mike -- Michael B. Miller, Ph.D. Minnesota Center for Twin and Family Research Department of Psychology University of Minnesota http://scholar.google.com/citations?user=EV_phq4J __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] R and GML data from the Canadian Goverment
You want to have a look at the R Spatial Task View for starters... GML files can be read by the rgdal package's readOGR function. But be warned GML is a complex beast. For example, I downloaded canvec_gml_AB_EN.zip from that site, it unzips to a GML (and an xsd) with a lot of layers, some of which aren't spatial. Using the ogrinfo command line tool from the gdal library I can see the layers: $ ogrinfo -so AB_EN.gml Had to open data source read-only. INFO: Open of `AB_EN.gml' using driver `GML' successful. 1: d_y_n (None) 2: d_y_n00 (None) 3: d_y_n01 (None) 4: d_y_n02 (None) 5: d_y_n03 (None) 6: d_y_n04 (None) 7: d_y_n05 (None) 8: d_y_n06 (None) [etc etc] 191: d_y166 (None) 192: d_y167 (None) 193: EN_1360049_0 (Multi Point) 194: EN_1360059_2 (Multi Polygon) 195: EN_1360049_2 (Multi Polygon) 196: EN_1340009_0 (Multi Point) 197: EN_1360059_0 (Multi Point) 198: EN_2170009_0 (Multi Point) 199: EN_1120009_1 (Multi Line String) 200: EN_1180009_1 (Multi Line String) So thats 192 non-spatial layers and a bunch of points n lines n polygons layers. Let's read one: m = readOGR(./AB_EN.gml,EN_1360059_2) plot(m) makes a map. Its a bunch of very small polygons, hard to see really! Further questions will be welcome at the R-sig-geo mailing list, where we whine about file formats on a daily basis. Barry On Thu, May 1, 2014 at 11:26 PM, jcrosbie ja...@crosb.ie wrote: I'm trying to create a map of transmission lines in Alberta. In addition, I'm very new to creating maps. The data can be found at: http://geogratis.gc.ca/site/eng/download http://ftp2.cits.rncan.gc.ca/pub/canvec/doc/CanVec_distribution_formats_en.pdf Would someone be able to point me in the right direction? I haven't been able to find an R package which is able to work with the GML data on the webset. Does anyone know of which package I should use and a good example out there? Thank you -- View this message in context: http://r.789695.n4.nabble.com/R-and-GML-data-from-the-Canadian-Goverment-tp4689843.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Return TRUE only for first match of values between matrix and vector.
Hi,
Try:
indx - A==B
t(apply(indx,1,function(x) {x[duplicated(x) !is.na(x)] - FALSE; x}))
# [,1] [,2] [,3]
#[1,] TRUE FALSE FALSE
#[2,] FALSE NA FALSE
#[3,] NA NA NA
#[4,] TRUE NA FALSE
#[5,] FALSE TRUE FALSE
A.K.
On Friday, May 2, 2014 4:47 AM, nevil amos nevil.a...@gmail.com wrote:
I wish to return True in a matrix for only the first match of a value
per row where the value equals that in a vector with the same number of
values as rosw in the matrix
eg:
A-matrix(c(2,3,2,1,1,2,NA,NA,NA,5,1,0,5,5,5),5,3)
B-c(2,1,NA,1,5)
desired result:
[,1] [,2] [,3]
[1,] TRUE FALSE FALSE
[2,] FALSE NA FALSE
[3,] NA NA NA
[4,] TRUE NA FALSE
[5,] FALSE TRUE FALSE
however A==B returns:
[,1] [,2] [,3]
[1,] TRUE TRUE FALSE
[2,] FALSE NA FALSE
[3,] NA NA NA
[4,] TRUE NA FALSE
[5,] FALSE TRUE TRUE
and
apply(A,1,function(x) match (B,x))
returns
[,1] [,2] [,3] [,4] [,5]
[1,] 1 NA 1 NA NA
[2,] 3 NA NA 1 1
[3,] NA 2 2 2 NA
[4,] 3 NA NA 1 1
[5,] NA NA 3 3 2
thanks
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Re: [R] Return TRUE only for first match of values between matrix and vector.
Hi Nevil, Try apply(A, 2, function(x) x == B) HTH, Jorge.- On Fri, May 2, 2014 at 6:46 PM, nevil amos nevil.a...@gmail.com wrote: I wish to return True in a matrix for only the first match of a value per row where the value equals that in a vector with the same number of values as rosw in the matrix eg: A-matrix(c(2,3,2,1,1,2,NA,NA,NA,5,1,0,5,5,5),5,3) B-c(2,1,NA,1,5) desired result: [,1] [,2] [,3] [1,] TRUE FALSE FALSE [2,] FALSE NA FALSE [3,]NA NANA [4,] TRUE NA FALSE [5,] FALSE TRUE FALSE however A==B returns: [,1] [,2] [,3] [1,] TRUE TRUE FALSE [2,] FALSE NA FALSE [3,]NA NANA [4,] TRUE NA FALSE [5,] FALSE TRUE TRUE and apply(A,1,function(x) match (B,x)) returns [,1] [,2] [,3] [,4] [,5] [1,]1 NA1 NA NA [2,]3 NA NA11 [3,] NA222 NA [4,]3 NA NA11 [5,] NA NA332 thanks [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] substring if a record has a \
Hi, Try: dat - structure(list(`1` = c(Marius Muller -\nIT Services, Rockwood\nBrockhues, Microlog Services\nMarcos, Firefox Services)), .Names = 1, class = data.frame, row.names = c(A, B, C, D)) dat$`1` - gsub(\n.*,,dat$`1`) dat # 1 #A Marius Muller - #B Rockwood #C Microlog Services #D Firefox Services A.K. On Friday, May 2, 2014 3:19 AM, Mat matthias.we...@fnt.de wrote: Hello togehter, i have a litte problem. I have a data.frame with a view entries like this one: 1 A Marius Muller -\nIT Services B Rockwood\nBrockhues C Microlog Services\nMarcos D Firefox Services I now want only the first description in the column until the \n. How can i do this? The solution look like this one: 1 A Marius Muller - B Rockwood C Microlog Services D Firefox Services Thank you. Best regards. Mat. -- View this message in context: http://r.789695.n4.nabble.com/substring-if-a-record-has-a-tp4689857.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] creating multiple orthogonal plans
Hi, I am trying to create multiple orthogonal designs using the Conjoint package. I have 4 factors with 5 levels in each. library(conjoint) experiment-expand.grid( price-c(a1,a2,a3,a4,a5), tag-c(b1,b2,b3,b4,b5), smell-c(c1,c2,c3,c4,c5), aroma-c(f1,f2,f3,f4,f5)) design-caFactorialDesign(data=experiment,type=fractional) print(design) print(cor(caEncodedDesign(design))) The resulting design from the above code has 22 test combinations.= (listed below). I would like to be able to create multiple designs instead of just having one. Is there a way to do that? Var1 Var2 Var3 Var4 19a4 b4 c1 f1 40a5 b3 c2 f1 96a1 b5 c4 f1 107 a2 b2 c5 f1 148 a3 b5 c1 f2 157 a2 b2 c2 f2 176 a1 b1 c3 f2 195 a5 b4 c3 f2 239 a4 b3 c5 f2 252 a2 b1 c1 f3 299 a4 b5 c2 f3 313 a3 b3 c3 f3 335 a5 b2 c4 f3 366 a1 b4 c5 f3 381 a1 b2 c1 f4 447 a2 b5 c3 f4 469 a4 b4 c4 f4 478 a3 b1 c5 f4 543 a3 b4 c2 f5 559 a4 b2 c3 f5 587 a2 b3 c4 f5 625 a5 b5 c5 f5 [1] R version 3.0.3 (2014-03-06) Thank you [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] extracting the first few eigenvectors
Exactly. The syntax is intended to mimic eigs() in Matlab and Octave. library(rARPACK) eigs(X, 10) # If your X is of class dsyMatrix eigs_sym(X, 10) # If X is of class matrix Best, Yixuan 2014-05-02 4:48 GMT-04:00 Berend Hasselman b...@xs4all.nl: On 02-05-2014, at 09:17, Mike Miller mbmille...@gmail.com wrote: I have a symmetric matrix, X, and I just want the first K eigenvectors (those associated with the K largest eigenvalues). Clearly, this works: eigs - eigen( X, symmetric=TRUE ) K_eigenvectors - eigs$vectors[ , 1:K ] K_eigenvalues - eigs$values[ 1:K ] rm(eigs) In order to do that, I have to create the matrix eigs$vectors which is the same size as X. Sometimes X has 100 million elements or more. Usually I want only the first 10 eigenvectors instead of all 10,000. Is there any R function that will allow me to extract just the few that I want? This would be analogous to the [V,d] = eigs(X,K) function in Octave/MATLAB. You can find possibly relevant stuff with library(sos) findFn(eigenvalue”) which is how I found package rARPACK, which may be exactly what you need/want. Berend Best, Mike -- Michael B. Miller, Ph.D. Minnesota Center for Twin and Family Research Department of Psychology University of Minnesota http://scholar.google.com/citations?user=EV_phq4J __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Yixuan Qiu yixuan@cos.name Department of Statistics, Purdue University __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Adding a column with a count of unique values
Hi, Try: dat - read.table(text=Person Time Change 2 0 3 2 10 5 2 15 7 3 0 4 3 5 2,sep=,header=TRUE) dat1 - transform(dat,Count= ave(rep(1,nrow(dat)), Person, FUN=cumsum)) #or ##If it is ordered by Person dat2 - transform(dat, Count= setNames(sequence(table(Person)),NULL)) #or dat3 - transform(dat,Count= ave(seq_along(Person), Person, FUN=seq_along)) all.equal(dat1,dat2) #[1] TRUE all.equal(dat1,dat3) #[1] TRUE A.K. I have a dataframe that looks like this: Person Time Change 2 0 3 2 10 5 2 15 7 3 0 4 3 5 2 I would like to add a column that counts each row for each person, like this: Person Time Change Count 2 0 3 1 2 10 5 2 2 15 7 3 3 0 4 1 3 5 2 2 Thanks in advance! __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] speeding up applying hist() over rows of a matrix
Your original code, as a function of 'm' and 'bins' is
f0 - function (m, bins) {
t(apply(m, 1, function(x) hist(x, breaks = bins, plot = FALSE)$counts))
}
and the time it takes to run on your m1 is about 5 s. on my machine
system.time(r0 - f0(m1,bins))
user system elapsed
4.950.005.02
hist(x,breaks=bins) is essentially tabulate(cut(x,bins),nbins=length(bins)-1).
See how much it speeds things up by replacing hist() with tabulate(cut()):
f1 - function (m, bins)
{
nbins - length(bins) - 1L
t(apply(m, 1, function(x) tabulate(cut(x, bins), nbins = nbins)))
}
That doesn't help with the time, but it does give the same output
system.time(r1 - f1(m1,bins))
user system elapsed
4.850.105.35
identical(r0, r1)
[1] TRUE
Now try speeding it up by calling cut() on the whole matrix first
and then applying tabulate to each row, as in
f2 - function (m, bins) {
nbins - length(bins) - 1L
m - array(as.integer(cut(m, bins)), dim = dim(m))
t(apply(m, 1, tabulate, nbins = nbins))
}
That saves quite a bit of time and gives the same output
system.time(r2 - f2(m1,bins))
user system elapsed
0.250.000.25
identical(r0, r2)
[1] TRUE
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Thu, May 1, 2014 at 12:48 PM, Ortiz-Bobea, Ariel ortiz-bo...@rff.org wrote:
Hello everyone,
I'm trying to construct bins for each row in a matrix. I'm using apply() in
combination with hist() to do this. Performing this binning for a 10K-by-50
matrix takes about 5 seconds, but only 0.5 seconds for a 1K-by-500 matrix.
This suggests the bottleneck is accessing rows in apply() rather than the
calculations going on inside hist().
My initial idea is to process as many columns (as make sense for the intended
use) at once. However, I still have many many rows to process and I would
appreciate any feedback on how to speed this up.
Any thoughts?
Thanks,
Ariel
Here is the illustration:
# create data
m1 - matrix(10*rnorm(50*10^4), ncol=50)
m2 - matrix(10*rnorm(50*10^4), ncol=500)
# compute bins
bins - seq(-100,100,1)
system.time({ out1 - t(apply(m1,1, function(x) hist(x,breaks=bins,
plot=FALSE)$counts)) })
system.time({ out2 - t(apply(m2,1, function(x) hist(x,breaks=bins,
plot=FALSE)$counts)) })
---
Ariel Ortiz-Bobea
Fellow
Resources for the Future
1616 P Street, N.W.
Washington, DC 20036
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
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Re: [R] speeding up applying hist() over rows of a matrix
And since as.integer(cut(x,bins)) is essentially findInterval(x,bins)
(since we throw away the labels made by cut()), I tried using
findInterval instead of cut() and it cut the time by more than half,
so your 5.0 s. is now c. 0.1 s.
f3 - function (m, bins)
{
nbins - length(bins) - 1L
m - array(findInterval(m, bins), dim = dim(m))
t(apply(m, 1, tabulate, nbins = nbins))
}
system.time(r3 - f3(m1,bins))
user system elapsed
0.090.000.09
identical(r0,r3)
[1] TRUE
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Fri, May 2, 2014 at 9:23 AM, William Dunlap wdun...@tibco.com wrote:
Your original code, as a function of 'm' and 'bins' is
f0 - function (m, bins) {
t(apply(m, 1, function(x) hist(x, breaks = bins, plot = FALSE)$counts))
}
and the time it takes to run on your m1 is about 5 s. on my machine
system.time(r0 - f0(m1,bins))
user system elapsed
4.950.005.02
hist(x,breaks=bins) is essentially tabulate(cut(x,bins),nbins=length(bins)-1).
See how much it speeds things up by replacing hist() with tabulate(cut()):
f1 - function (m, bins)
{
nbins - length(bins) - 1L
t(apply(m, 1, function(x) tabulate(cut(x, bins), nbins = nbins)))
}
That doesn't help with the time, but it does give the same output
system.time(r1 - f1(m1,bins))
user system elapsed
4.850.105.35
identical(r0, r1)
[1] TRUE
Now try speeding it up by calling cut() on the whole matrix first
and then applying tabulate to each row, as in
f2 - function (m, bins) {
nbins - length(bins) - 1L
m - array(as.integer(cut(m, bins)), dim = dim(m))
t(apply(m, 1, tabulate, nbins = nbins))
}
That saves quite a bit of time and gives the same output
system.time(r2 - f2(m1,bins))
user system elapsed
0.250.000.25
identical(r0, r2)
[1] TRUE
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Thu, May 1, 2014 at 12:48 PM, Ortiz-Bobea, Ariel ortiz-bo...@rff.org
wrote:
Hello everyone,
I'm trying to construct bins for each row in a matrix. I'm using apply() in
combination with hist() to do this. Performing this binning for a 10K-by-50
matrix takes about 5 seconds, but only 0.5 seconds for a 1K-by-500 matrix.
This suggests the bottleneck is accessing rows in apply() rather than the
calculations going on inside hist().
My initial idea is to process as many columns (as make sense for the
intended use) at once. However, I still have many many rows to process and I
would appreciate any feedback on how to speed this up.
Any thoughts?
Thanks,
Ariel
Here is the illustration:
# create data
m1 - matrix(10*rnorm(50*10^4), ncol=50)
m2 - matrix(10*rnorm(50*10^4), ncol=500)
# compute bins
bins - seq(-100,100,1)
system.time({ out1 - t(apply(m1,1, function(x) hist(x,breaks=bins,
plot=FALSE)$counts)) })
system.time({ out2 - t(apply(m2,1, function(x) hist(x,breaks=bins,
plot=FALSE)$counts)) })
---
Ariel Ortiz-Bobea
Fellow
Resources for the Future
1616 P Street, N.W.
Washington, DC 20036
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[R] Parsing XML file to data frame
Hi all - I am trying to parse out the attached XML file into a data frame. The file is extracted from Hadoop File Services (HFS). I am new in using the XML package so need some help in parsing out the data. Below is some code that I explore to get the attribute into a data frame. Any help is appreciated. library(XML) temp - xmlParseDoc(sample.xml) temp.root - xmlRoot(temp) xmlName(temp.root) xmlSize(temp.root) #21 child nodes temp.root[[2]] #headers temp.root[[2]][[2]] #extracts just the revision temp.2 - xmlToList(temp.root[[2]]) #extracts the info in temp.root[[2]] into a list temp.2 temp.2.df - xmlToDataFrame(temp.root[[2]]) #data frame of the list temp.2.df xmlValue(temp.root[[2]]) #string the values of the node inside [[2]] temp.revision - xmlValue(temp.root[[2]][[Revision]]) temp.revision test - xmlTreeParse(sample.xml) test -- View this message in context: http://r.789695.n4.nabble.com/Parsing-XML-file-to-data-frame-tp4689883.html Sent from the R help mailing list archive at Nabble.com. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] matafor package: adding additional information under a forest plot
Dear R-Users, i have to questions about plotting a forest plot with the metafor package: 1) Is there a (text()) function to add additional information (e.g. heterogeneity statistics) under the forest plot? 2) I want to add various columns of additional information about each study (e.g. quality, time point, measure). I thought i could do that with the ilab-argument, but i keep getting in the way with the line that restricts the plot and although i tweaked everything around, I do not succeed. Basecally, i want the format of the forest plot to be rather wide than tall, because i have little studies but many columns with additional study information. Does someone know how to solve that (over xlim, ylim???)? Thanks a lot for your help Linde __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How to fix the format in write.csv
By dafault, write.csv will change the characters such as 5/38 as a date May-38. How can I not change the format? Thanks. ChangJiang [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Writing to gzcon with rawConnection
I would like to encode/decode some text with deflate/gzip, but I'm having trouble. Decoding values from a rawConnection is fairly straightforward, but I'm having trouble encoding values and then retrieving them. gz_out - gzcon(raw_out - rawConnection(raw(0),open=wb)) writeLines(test,con=gz_out) flush(gz_out) rawConnectionValue(raw_out) Error: cannot allocate vector of size 131069.2 Gb raw_out - rawConnection(raw(0),open=wb) writeLines(test,con=raw_out) rawConnectionValue(raw_out) [1] 74 65 73 74 0a Has anyone had success doing this? --Jamie __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to fix the format in write.csv
Read your Excel documentation. AFAIK, R just writes text files -- you need to tell Excel how to read them in. Cheers, Bert Bert Gunter Genentech Nonclinical Biostatistics (650) 467-7374 Data is not information. Information is not knowledge. And knowledge is certainly not wisdom. H. Gilbert Welch On Fri, May 2, 2014 at 11:34 AM, ChangJiang Xu changjiang.h...@gmail.com wrote: By dafault, write.csv will change the characters such as 5/38 as a date May-38. How can I not change the format? Thanks. ChangJiang [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to fix the format in write.csv
ChangJiang: Open the .csv file with Notepad or other plain vanilla text processor, NOT EXCEL. You will find that the columns are text. Excel automatically converts them to dates. Read Excel's docs or get help from someone to learn how to convert the dates back to text. -- Bert Bert Gunter Genentech Nonclinical Biostatistics (650) 467-7374 Data is not information. Information is not knowledge. And knowledge is certainly not wisdom. H. Gilbert Welch On Fri, May 2, 2014 at 12:35 PM, ChangJiang Xu changjiang.h...@gmail.com wrote: Thanks. I still didn't get the solution. For example, I have a data frame, called temp temp Chr Ref Var AFFUNAFF AFF.test pvalue 1 10 A G 2/1/240/0/1905/49 2.429e-09 2 18 G A 1/9/17 0/23/167 11/43 2.484e-04 3 1 G A 2/2/220/8/176 6/46 4.293e-04 4 11 T G 1/1/250/2/188 3/51 1.193e-03 5 2 A T 1/10/16 38/90/60 12/42 2.220e-03 6 3 G A 1/16/10 8/49/13318/36 4.549e-03 Then I want to write a csv file using R function write.csv, as follows: write.csv(temp, file=temp.csv, row.names = FALSE) and the csv file, temp.csv, looks like the below, not same as original data ChrRefVarAFF UNAFFAFF.testpvalue 10AG02/01/20240/0/190May-492.43E-09 18GA01/09/20170/23/167 Nov-430.0002484 1 GA02/02/20220/8/176Jun-460.0004293 11 TG01/01/20250/2/188Mar-510.001193 2 AT01/10/201638/90/60 Dec-420.00222 3 GA1/16/108/49/133 18/36 0.004549 On Fri, May 2, 2014 at 3:06 PM, Bert Gunter gunter.ber...@gene.com wrote: Read your Excel documentation. AFAIK, R just writes text files -- you need to tell Excel how to read them in. Cheers, Bert Bert Gunter Genentech Nonclinical Biostatistics (650) 467-7374 Data is not information. Information is not knowledge. And knowledge is certainly not wisdom. H. Gilbert Welch On Fri, May 2, 2014 at 11:34 AM, ChangJiang Xu changjiang.h...@gmail.com wrote: By dafault, write.csv will change the characters such as 5/38 as a date May-38. How can I not change the format? Thanks. ChangJiang [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to fix the format in write.csv
Hi ChangJiang, Date conversion is one of the biggest headaches with Excel. Even if you import those data into Excel and then specify that the column should be text format, it won't convert the values back into what they originally were. Be aware that Excel may, when encountering dates from a different locale, _silently_ convert the ones that aren't valid in its current locale, usually by swapping the day and month values. That one has bitten me when I have had to use Excel. The solution I found was to always use international format (-mm-dd) as that didn't seem to be altered. Jim On 05/03/2014 06:56 AM, Bert Gunter wrote: ChangJiang: Open the .csv file with Notepad or other plain vanilla text processor, NOT EXCEL. You will find that the columns are text. Excel automatically converts them to dates. Read Excel's docs or get help from someone to learn how to convert the dates back to text. -- Bert Bert Gunter Genentech Nonclinical Biostatistics (650) 467-7374 Data is not information. Information is not knowledge. And knowledge is certainly not wisdom. H. Gilbert Welch On Fri, May 2, 2014 at 12:35 PM, ChangJiang Xu changjiang.h...@gmail.com wrote: Thanks. I still didn't get the solution. For example, I have a data frame, called temp temp Chr Ref Var AFFUNAFF AFF.test pvalue 1 10 A G 2/1/240/0/1905/49 2.429e-09 2 18 G A 1/9/17 0/23/167 11/43 2.484e-04 3 1 G A 2/2/220/8/176 6/46 4.293e-04 4 11 T G 1/1/250/2/188 3/51 1.193e-03 5 2 A T 1/10/16 38/90/60 12/42 2.220e-03 6 3 G A 1/16/10 8/49/13318/36 4.549e-03 Then I want to write a csv file using R function write.csv, as follows: write.csv(temp, file=temp.csv, row.names = FALSE) and the csv file, temp.csv, looks like the below, not same as original data ChrRefVarAFF UNAFFAFF.testpvalue 10AG02/01/20240/0/190May-492.43E-09 18GA01/09/20170/23/167 Nov-430.0002484 1 GA02/02/20220/8/176Jun-460.0004293 11 TG01/01/20250/2/188Mar-510.001193 2 AT01/10/201638/90/60 Dec-420.00222 3 GA1/16/108/49/133 18/36 0.004549 On Fri, May 2, 2014 at 3:06 PM, Bert Guntergunter.ber...@gene.com wrote: Read your Excel documentation. AFAIK, R just writes text files -- you need to tell Excel how to read them in. Cheers, Bert Bert Gunter Genentech Nonclinical Biostatistics (650) 467-7374 Data is not information. Information is not knowledge. And knowledge is certainly not wisdom. H. Gilbert Welch On Fri, May 2, 2014 at 11:34 AM, ChangJiang Xu changjiang.h...@gmail.com wrote: By dafault, write.csv will change the characters such as 5/38 as a date May-38. How can I not change the format? Thanks. ChangJiang [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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 the first few eigenvectors
Thank you, and thanks for writing that code! That is the perfect answer to my question. I hope the R development team will consider expanding the functionality of eigen() to include the option to retain only the first few eigenvectors and/or eigenvalues. To me it seems very common in statistics to work with a few eigenvectors instead of with all of them. Mike On Fri, 2 May 2014, Yixuan Qiu wrote: Exactly. The syntax is intended to mimic eigs() in Matlab and Octave. library(rARPACK) eigs(X, 10) # If your X is of class dsyMatrix eigs_sym(X, 10) # If X is of class matrix Best, Yixuan 2014-05-02 4:48 GMT-04:00 Berend Hasselman b...@xs4all.nl: On 02-05-2014, at 09:17, Mike Miller mbmille...@gmail.com wrote: I have a symmetric matrix, X, and I just want the first K eigenvectors (those associated with the K largest eigenvalues). Clearly, this works: eigs - eigen( X, symmetric=TRUE ) K_eigenvectors - eigs$vectors[ , 1:K ] K_eigenvalues - eigs$values[ 1:K ] rm(eigs) In order to do that, I have to create the matrix eigs$vectors which is the same size as X. Sometimes X has 100 million elements or more. Usually I want only the first 10 eigenvectors instead of all 10,000. Is there any R function that will allow me to extract just the few that I want? This would be analogous to the [V,d] = eigs(X,K) function in Octave/MATLAB. You can find possibly relevant stuff with library(sos) findFn(eigenvalue”) which is how I found package rARPACK, which may be exactly what you need/want. Berend Best, Mike__ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to fix the format in write.csv
Thanks. I still didn't get the solution. For example, I have a data frame, called temp temp Chr Ref Var AFFUNAFF AFF.test pvalue 1 10 A G 2/1/240/0/1905/49 2.429e-09 2 18 G A 1/9/17 0/23/167 11/43 2.484e-04 3 1 G A 2/2/220/8/176 6/46 4.293e-04 4 11 T G 1/1/250/2/188 3/51 1.193e-03 5 2 A T 1/10/16 38/90/60 12/42 2.220e-03 6 3 G A 1/16/10 8/49/13318/36 4.549e-03 Then I want to write a csv file using R function write.csv, as follows: write.csv(temp, file=temp.csv, row.names = FALSE) and the csv file, temp.csv, looks like the below, not same as original data ChrRefVarAFF UNAFFAFF.testpvalue 10AG02/01/20240/0/190May-492.43E-09 18GA01/09/20170/23/167 Nov-430.0002484 1 GA02/02/20220/8/176Jun-460.0004293 11 TG01/01/20250/2/188Mar-510.001193 2 AT01/10/201638/90/60 Dec-420.00222 3 GA1/16/108/49/133 18/36 0.004549 On Fri, May 2, 2014 at 3:06 PM, Bert Gunter gunter.ber...@gene.com wrote: Read your Excel documentation. AFAIK, R just writes text files -- you need to tell Excel how to read them in. Cheers, Bert Bert Gunter Genentech Nonclinical Biostatistics (650) 467-7374 Data is not information. Information is not knowledge. And knowledge is certainly not wisdom. H. Gilbert Welch On Fri, May 2, 2014 at 11:34 AM, ChangJiang Xu changjiang.h...@gmail.com wrote: By dafault, write.csv will change the characters such as 5/38 as a date May-38. How can I not change the format? Thanks. ChangJiang [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] SQL vs R
Hi, How do I do something like this without using sqldf? a - sqldf(SELECT COUNT(*) FROM b WHERE c = 'd') or e - sqldf(SELECT f, COUNT(*) FROM b GROUP BY f ORDER BY f) greetings, el __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] speeding up applying hist() over rows of a matrix
This works great, thanks a lot!
-AOB
-Original Message-
From: William Dunlap [mailto:wdun...@tibco.com]
Sent: Friday, May 02, 2014 12:31 PM
To: Ortiz-Bobea, Ariel
Cc: r-help@r-project.org
Subject: Re: [R] speeding up applying hist() over rows of a matrix
And since as.integer(cut(x,bins)) is essentially findInterval(x,bins) (since we
throw away the labels made by cut()), I tried using findInterval instead of
cut() and it cut the time by more than half, so your 5.0 s. is now c. 0.1 s.
f3 - function (m, bins)
{
nbins - length(bins) - 1L
m - array(findInterval(m, bins), dim = dim(m))
t(apply(m, 1, tabulate, nbins = nbins)) }
system.time(r3 - f3(m1,bins))
user system elapsed
0.090.000.09
identical(r0,r3)
[1] TRUE
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Fri, May 2, 2014 at 9:23 AM, William Dunlap wdun...@tibco.com wrote:
Your original code, as a function of 'm' and 'bins' is
f0 - function (m, bins) {
t(apply(m, 1, function(x) hist(x, breaks = bins, plot =
FALSE)$counts)) } and the time it takes to run on your m1 is about 5
s. on my machine
system.time(r0 - f0(m1,bins))
user system elapsed
4.950.005.02
hist(x,breaks=bins) is essentially tabulate(cut(x,bins),nbins=length(bins)-1).
See how much it speeds things up by replacing hist() with tabulate(cut()):
f1 - function (m, bins)
{
nbins - length(bins) - 1L
t(apply(m, 1, function(x) tabulate(cut(x, bins), nbins = nbins)))
} That doesn't help with the time, but it does give the same output
system.time(r1 - f1(m1,bins))
user system elapsed
4.850.105.35
identical(r0, r1)
[1] TRUE
Now try speeding it up by calling cut() on the whole matrix first and
then applying tabulate to each row, as in
f2 - function (m, bins) {
nbins - length(bins) - 1L
m - array(as.integer(cut(m, bins)), dim = dim(m))
t(apply(m, 1, tabulate, nbins = nbins)) } That saves quite a bit
of time and gives the same output
system.time(r2 - f2(m1,bins))
user system elapsed
0.250.000.25
identical(r0, r2)
[1] TRUE
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Thu, May 1, 2014 at 12:48 PM, Ortiz-Bobea, Ariel ortiz-bo...@rff.org
wrote:
Hello everyone,
I'm trying to construct bins for each row in a matrix. I'm using apply() in
combination with hist() to do this. Performing this binning for a 10K-by-50
matrix takes about 5 seconds, but only 0.5 seconds for a 1K-by-500 matrix.
This suggests the bottleneck is accessing rows in apply() rather than the
calculations going on inside hist().
My initial idea is to process as many columns (as make sense for the
intended use) at once. However, I still have many many rows to process and I
would appreciate any feedback on how to speed this up.
Any thoughts?
Thanks,
Ariel
Here is the illustration:
# create data
m1 - matrix(10*rnorm(50*10^4), ncol=50)
m2 - matrix(10*rnorm(50*10^4), ncol=500)
# compute bins
bins - seq(-100,100,1)
system.time({ out1 - t(apply(m1,1, function(x) hist(x,breaks=bins,
plot=FALSE)$counts)) })
system.time({ out2 - t(apply(m2,1, function(x) hist(x,breaks=bins,
plot=FALSE)$counts)) })
---
Ariel Ortiz-Bobea
Fellow
Resources for the Future
1616 P Street, N.W.
Washington, DC 20036
[[alternative HTML version deleted]]
__
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PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
__
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[R] Making a panel of scatter plots in R
Hello, I'm trying to figure out how to panel scatter plots of participant data in R using the ggplot2 package. I do not want to display separate variables, I simply want to create a 4 by 4 or 5 by 5 grid of scatter plots that represent the data of each individual participant since I am looking at longitudinal data and wish to examine it visually. I am currently generating basic plots with the function: p+geom_point()+facet_grid(~ID). This generates all of the plots, but in a single row that is not easy to visualize. Any help you can provide on the appropriate command syntax is greatly appreciated! Sincerely, Charlie __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Making a panel of scatter plots in R
Are you looking for facet_wrap? facet_grid usually has two grouping variables... --- 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 May 2, 2014 4:50:08 PM PDT, Charlie Nagle cna...@me.com wrote: Hello, I'm trying to figure out how to panel scatter plots of participant data in R using the ggplot2 package. I do not want to display separate variables, I simply want to create a 4 by 4 or 5 by 5 grid of scatter plots that represent the data of each individual participant since I am looking at longitudinal data and wish to examine it visually. I am currently generating basic plots with the function: p+geom_point()+facet_grid(~ID). This generates all of the plots, but in a single row that is not easy to visualize. Any help you can provide on the appropriate command syntax is greatly appreciated! Sincerely, Charlie __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Seeking well-commented versions of mgcv source code
Dear List, I'm looking for well-commented versions of various functions comprising mgcv, so that I can modify a piece of it for a project I'm working on. In particular I'm looking for * testStat * summary.gam * liu2 * simf Obviously I can find these by typing mgcv:::whatever. But there are a lot of nested if statements, making it difficult to follow. Comments in the code describing exactly what is happening at each step would make my life a lot easier. Where can I find more-detailed versions of the code? Thanks, Andrew [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] SQL vs R
Hi, With the new package dplyr you can create equivalent SQL sintaxt queries like the one you need. You can find examples of how to apply it here: http://martinsbioblogg.wordpress.com/2014/03/26/using-r-quickly-calculating-summary-statistics-with-dplyr/ http://martinsbioblogg.wordpress.com/2014/03/27/more-fun-with-and/ Regards, Carlos. 2014-05-02 23:23 GMT+02:00 Dr Eberhard Lisse nos...@lisse.na: Hi, How do I do something like this without using sqldf? a - sqldf(SELECT COUNT(*) FROM b WHERE c = 'd') or e - sqldf(SELECT f, COUNT(*) FROM b GROUP BY f ORDER BY f) greetings, el __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. -- Saludos, Carlos Ortega www.qualityexcellence.es [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Seeking well-commented versions of mgcv source code
Read the Posting Guide, which points out that this list is for plain text emails so you should set your email program appropriately. Read the Writing R Extensions document, which tells you how packages are constructed, and from which it will become clear that you want the tar.gz package file for whatever package you are interested in. --- 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 May 2, 2014 7:15:27 PM PDT, Andrew Crane-Droesch andre...@gmail.com wrote: Dear List, I'm looking for well-commented versions of various functions comprising mgcv, so that I can modify a piece of it for a project I'm working on. In particular I'm looking for * testStat * summary.gam * liu2 * simf Obviously I can find these by typing mgcv:::whatever. But there are a lot of nested if statements, making it difficult to follow. Comments in the code describing exactly what is happening at each step would make my life a lot easier. Where can I find more-detailed versions of the code? Thanks, Andrew [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] How to fix the format in write.csv
Hi I second Jim's comments. I have also been bitten by British being converted to American format as well as ANSI numeric to a date format which I did not want . And if you want to take a daily series from the 1890's to the 2000's caveat emptor. If you are dealing with dates use something that will give you a proper output date format. Even beware of old versions of Access exporting excel files they are really html files. Regards Duncan Duncan Mackay Department of Agronomy and Soil Science University of New England Armidale NSW 2351 Email: home: mac...@northnet.com.au -Original Message- From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On Behalf Of Jim Lemon Sent: Saturday, 3 May 2014 07:21 To: Bert Gunter Cc: r-help@r-project.org Subject: Re: [R] How to fix the format in write.csv Hi ChangJiang, Date conversion is one of the biggest headaches with Excel. Even if you import those data into Excel and then specify that the column should be text format, it won't convert the values back into what they originally were. Be aware that Excel may, when encountering dates from a different locale, _silently_ convert the ones that aren't valid in its current locale, usually by swapping the day and month values. That one has bitten me when I have had to use Excel. The solution I found was to always use international format (-mm-dd) as that didn't seem to be altered. Jim On 05/03/2014 06:56 AM, Bert Gunter wrote: ChangJiang: Open the .csv file with Notepad or other plain vanilla text processor, NOT EXCEL. You will find that the columns are text. Excel automatically converts them to dates. Read Excel's docs or get help from someone to learn how to convert the dates back to text. -- Bert Bert Gunter Genentech Nonclinical Biostatistics (650) 467-7374 Data is not information. Information is not knowledge. And knowledge is certainly not wisdom. H. Gilbert Welch On Fri, May 2, 2014 at 12:35 PM, ChangJiang Xu changjiang.h...@gmail.com wrote: Thanks. I still didn't get the solution. For example, I have a data frame, called temp temp Chr Ref Var AFFUNAFF AFF.test pvalue 1 10 A G 2/1/240/0/1905/49 2.429e-09 2 18 G A 1/9/17 0/23/167 11/43 2.484e-04 3 1 G A 2/2/220/8/176 6/46 4.293e-04 4 11 T G 1/1/250/2/188 3/51 1.193e-03 5 2 A T 1/10/16 38/90/60 12/42 2.220e-03 6 3 G A 1/16/10 8/49/13318/36 4.549e-03 Then I want to write a csv file using R function write.csv, as follows: write.csv(temp, file=temp.csv, row.names = FALSE) and the csv file, temp.csv, looks like the below, not same as original data ChrRefVarAFF UNAFFAFF.testpvalue 10AG02/01/20240/0/190May-492.43E-09 18GA01/09/20170/23/167 Nov-430.0002484 1 GA02/02/20220/8/176Jun-460.0004293 11 TG01/01/20250/2/188Mar-510.001193 2 AT01/10/201638/90/60 Dec-420.00222 3 GA1/16/108/49/133 18/36 0.004549 On Fri, May 2, 2014 at 3:06 PM, Bert Guntergunter.ber...@gene.com wrote: Read your Excel documentation. AFAIK, R just writes text files -- you need to tell Excel how to read them in. Cheers, Bert Bert Gunter Genentech Nonclinical Biostatistics (650) 467-7374 Data is not information. Information is not knowledge. And knowledge is certainly not wisdom. H. Gilbert Welch On Fri, May 2, 2014 at 11:34 AM, ChangJiang Xu changjiang.h...@gmail.com wrote: By dafault, write.csv will change the characters such as 5/38 as a date May-38. How can I not change the format? Thanks. ChangJiang [[alternative HTML version deleted]] __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] SQL vs R
By making the effort to learn R? See e.g. the Introduction to R tutorial that ships with R. -- Bert Bert Gunter Genentech Nonclinical Biostatistics (650) 467-7374 Data is not information. Information is not knowledge. And knowledge is certainly not wisdom. H. Gilbert Welch On Fri, May 2, 2014 at 2:23 PM, Dr Eberhard Lisse nos...@lisse.na wrote: Hi, How do I do something like this without using sqldf? a - sqldf(SELECT COUNT(*) FROM b WHERE c = 'd') or e - sqldf(SELECT f, COUNT(*) FROM b GROUP BY f ORDER BY f) greetings, el __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R-es] gracias y pregunta off topic
Lo primero, muchisimas gracias a los que respondisteis a mis problemas con mapas :-D Luego, aunque la lista sea de R, tengo una duda matematica que quizas podriais responderme. Hay algun palabro en castellano que se utilice para designar los state process y observational process en modelos jerarquicos? El viejo truco de ir a la wiki en ingles y pasarse al castellano no me ha funcionado y no tengo ningun diccionario tecnico que me resuelva la duda, y estoy haciendo un resumen en castellano de mi tesis de master y no se como traducirlos o adaptarlos. Quizas proceso real para state process y proceso observacional para observational process? Suenan un poco cutres... Gracias por adelantado Fer ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es
