Re: [R] R parallel / foreach - aggregation of results
Dear Jim,
Thank you very much for your response. It seems to work now, but the return
value is not the required matrix but a list of matrices (one for each repition
j).
Any idea how it is possible to return only the last matrix and not all?
Thanks and best,
Martin
Gesendet: Freitag, 31. Juli 2015 um 18:22 Uhr
Von: "jim holtman"
An: "Martin Spindler"
Cc: "r-help@r-project.org"
Betreff: Re: [R] R parallel / foreach - aggregation of results
Try this chance to actually return values:
library(doParallel)
Simpar3 <- function(n1) {
L2distance <- matrix(NA, ncol=n1, nrow=n1)
data <- rnorm(n1)
diag(L2distance)=0
cl <- makeCluster(4)
registerDoParallel(cl)
x <- foreach(j=1:n1) %dopar% {
library(np)
datj <- data[j]
for(k in j:n1) {
L2distance[j,k] <- k*datj
}
L2distance # return the value
}
stopCluster(cl)
return(x)
}
Res <- Simpar3(100)
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 Fri, Jul 31, 2015 at 8:39 AM, Martin Spindler
wrote:Dear all,
when I am running the code attached below, it seems that no results are
returned, only the predefined NAs. What mistake do I make?
Any comments and help is highly appreciated.
Thanks and best,
Martin
Simpar3 <- function(n1) {
L2distance <- matrix(NA, ncol=n1, nrow=n1)
data <- rnorm(n1)
diag(L2distance)=0
cl <- makeCluster(4)
registerDoParallel(cl)
foreach(j=1:n1) %dopar% {
library(np)
datj <- data[j]
for(k in j:n1) {
L2distance[j,k] <- k*datj
}
}
stopCluster(cl)
return(L2distance)
}
Res <- Simpar3(100)
__
R-help@r-project.org[R-help@r-project.org] mailing list -- To UNSUBSCRIBE and
more, see
https://stat.ethz.ch/mailman/listinfo/r-help[https://stat.ethz.ch/mailman/listinfo/r-help]
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html[http://www.R-project.org/posting-guide.html]
and provide commented, minimal, self-contained, reproducible code.
__
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] R parallel / foreach - aggregation of results
Dear all,
when I am running the code attached below, it seems that no results are
returned, only the predefined NAs. What mistake do I make?
Any comments and help is highly appreciated.
Thanks and best,
Martin
Simpar3 <- function(n1) {
L2distance <- matrix(NA, ncol=n1, nrow=n1)
data <- rnorm(n1)
diag(L2distance)=0
cl <- makeCluster(4)
registerDoParallel(cl)
foreach(j=1:n1) %dopar% {
library(np)
datj <- data[j]
for(k in j:n1) {
L2distance[j,k] <- k*datj
}
}
stopCluster(cl)
return(L2distance)
}
Res <- Simpar3(100)
__
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] R parallel - slow speed
Thank you very much to you both for your help.
I knew that parallelizing has some additional "overhead" costs, but I was
surprised be the order of magnitude (it was 10 times slower.) Therefore I
thought I made some mistake or that there is a more clever way to do it.
Best,
Martin
Gesendet: Donnerstag, 30. Juli 2015 um 15:28 Uhr
Von: "jim holtman"
An: "Jeff Newmiller"
Cc: "Martin Spindler" , "r-help@r-project.org"
Betreff: Re: [R] R parallel - slow speed
I ran a test on my Windows box with 4 CPUs. THere were 4 RScript processes
started in response to the request for a cluster of 4. Each of these ran for
an elapsed time of around 23 seconds, making the median time around 0.2 seconds
for 100 iterations as reported by microbenchmark. The 'apply' only takes about
0.003 seconds for a single iteration - again what microbenchmark is reporting.
The 4 RScript processes each use about 3 CPU seconds in the 23 seconds of
elapsed time, most of that is probably the communication and startup time for
the processes and reporting results.
So as was pointed out previous there is overhead is running in parallel. You
probably have to have at least several seconds of heavy computation for a
iteration to make trying to parallelize something. You should also investigate
exactly what is happening on your system so that you can account for the time
being spent.
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, Jul 30, 2015 at 8:56 AM, Jeff Newmiller
wrote:Parallelizing comes at a price... and there is no guarantee that you can
afford it. Vectorizing your algorithms is often a better approach.
Microbenchmarking is usually overkill for evaluating parallelizing.
You assume 4 cores... but many CPUs have 2 cores and use hyperthreading to make
each core look like two.
The operating system can make a difference also... Windows processes are more
expensive to start and communicate between than *nix processes are. In
particular, Windows seems to require duplicated RAM pages while *nix can share
process RAM (at least until they are written to) so you end up needing more
memory and disk paging of virtual memory becomes more likely.
---
Jeff Newmiller The . . Go Live...
DCN: Basics: ##.#.
##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (Solar/Batteries O.O#. #.O#. with
/Software/Embedded Controllers) .OO#. .OO#. rocks...1k
---
Sent from my phone. Please excuse my brevity.
On July 30, 2015 8:26:34 AM EDT, Martin Spindler
wrote:
>Dear all,
>
>I am trying to parallelize the function npnewpar given below. When I am
>comparing an application of "apply" with "parApply" the parallelized
>version seems to be much slower (cf output below). Therefore I would
>like to ask how the function could be parallelized more efficient.
>(With increasing sample size the difference becomes smaller, but I was
>wondering about this big differences and how it could be improved.)
>
>Thank you very much for help in advance!
>
>Best,
>
>Martin
>
>
>library(microbenchmark)
>library(doParallel)
>
>n <- 500
>y <- rnorm(n)
>Xc <- rnorm(n)
>Xd <- sample(c(0,1), replace=TRUE)
>Weights <- diag(n)
>n1 <- 50
>Xeval <- cbind(rnorm(n1), sample(c(0,1), n1, replace=TRUE))
>
>
>detectCores()
>cl <- makeCluster(4)
>registerDoParallel(cl)
>microbenchmark(apply(Xeval, 1, npnewpar, y=y, Xc=Xc, Xd = Xd,
>Weights=Weights, h=0.5), parApply(cl, Xeval, 1, npnewpar, y=y, Xc=Xc,
>Xd = Xd, Weights=Weights, h=0.5), times=100)
>stopCluster(cl)
>
>
>Unit: milliseconds
> expr min lq mean median
>apply(Xeval, 1, npnewpar, y = y, Xc = Xc, Xd = Xd, Weights = Weights,
> h = 0.5) 4.674914 4.726463 5.455323 4.771016
>parApply(cl, Xeval, 1, npnewpar, y = y, Xc = Xc, Xd = Xd, Weights =
>Weights, h = 0.5) 34.168250 35.434829 56.553296 39.438899
> uq max neval
> 4.843324 57.01519 100
> 49.777265 347.77887 100
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>npnewpar <- function(y, Xc, Xd, Weights, h, xeval) {
> xc <- xeval[1]
> xd <- xeval[2]
> l <- function(x,X) {
> w <- Weights[x,X]
> return(w)
> }
> u <- (Xc-xc)/h
> #K <- kernel(u)
> K <- dnorm(u)
> L <- l(xd,Xd)
> nom <- sum(y*K*L)
> denom <- sum(K*L)
> ghat <- nom/denom
> return
Re: [R] R parallel - slow speed
Thank you very much for your help.
I tried it under Unix and then the parallel version was faster than under
Windows (but still slower than the non parall version). This is an important
point to keep in mind. Thanks for this.
Best,
Martin
Gesendet: Donnerstag, 30. Juli 2015 um 14:56 Uhr
Von: "Jeff Newmiller"
An: "Martin Spindler" , "r-help@r-project.org"
Betreff: Re: [R] R parallel - slow speed
Parallelizing comes at a price... and there is no guarantee that you can afford
it. Vectorizing your algorithms is often a better approach. Microbenchmarking
is usually overkill for evaluating parallelizing.
You assume 4 cores... but many CPUs have 2 cores and use hyperthreading to make
each core look like two.
The operating system can make a difference also... Windows processes are more
expensive to start and communicate between than *nix processes are. In
particular, Windows seems to require duplicated RAM pages while *nix can share
process RAM (at least until they are written to) so you end up needing more
memory and disk paging of virtual memory becomes more likely.
---
Jeff Newmiller The . . Go Live...
DCN: Basics: ##.#. ##.#. Live Go...
Live: OO#.. Dead: OO#.. Playing
Research Engineer (Solar/Batteries O.O#. #.O#. with
/Software/Embedded Controllers) .OO#. .OO#. rocks...1k
---
Sent from my phone. Please excuse my brevity.
On July 30, 2015 8:26:34 AM EDT, Martin Spindler wrote:
>Dear all,
>
>I am trying to parallelize the function npnewpar given below. When I am
>comparing an application of "apply" with "parApply" the parallelized
>version seems to be much slower (cf output below). Therefore I would
>like to ask how the function could be parallelized more efficient.
>(With increasing sample size the difference becomes smaller, but I was
>wondering about this big differences and how it could be improved.)
>
>Thank you very much for help in advance!
>
>Best,
>
>Martin
>
>
>library(microbenchmark)
>library(doParallel)
>
>n <- 500
>y <- rnorm(n)
>Xc <- rnorm(n)
>Xd <- sample(c(0,1), replace=TRUE)
>Weights <- diag(n)
>n1 <- 50
>Xeval <- cbind(rnorm(n1), sample(c(0,1), n1, replace=TRUE))
>
>
>detectCores()
>cl <- makeCluster(4)
>registerDoParallel(cl)
>microbenchmark(apply(Xeval, 1, npnewpar, y=y, Xc=Xc, Xd = Xd,
>Weights=Weights, h=0.5), parApply(cl, Xeval, 1, npnewpar, y=y, Xc=Xc,
>Xd = Xd, Weights=Weights, h=0.5), times=100)
>stopCluster(cl)
>
>
>Unit: milliseconds
> expr min lq mean median
>apply(Xeval, 1, npnewpar, y = y, Xc = Xc, Xd = Xd, Weights = Weights,
> h = 0.5) 4.674914 4.726463 5.455323 4.771016
>parApply(cl, Xeval, 1, npnewpar, y = y, Xc = Xc, Xd = Xd, Weights =
>Weights, h = 0.5) 34.168250 35.434829 56.553296 39.438899
> uq max neval
> 4.843324 57.01519 100
> 49.777265 347.77887 100
>
>
>
>
>
>
>
>
>
>
>
>
>
>
>npnewpar <- function(y, Xc, Xd, Weights, h, xeval) {
> xc <- xeval[1]
> xd <- xeval[2]
> l <- function(x,X) {
> w <- Weights[x,X]
> return(w)
> }
> u <- (Xc-xc)/h
> #K <- kernel(u)
> K <- dnorm(u)
> L <- l(xd,Xd)
> nom <- sum(y*K*L)
> denom <- sum(K*L)
> ghat <- nom/denom
> return(ghat)
>}
>
>__
>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[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] R parallel - slow speed
Dear all,
I am trying to parallelize the function npnewpar given below. When I am
comparing an application of "apply" with "parApply" the parallelized version
seems to be much slower (cf output below). Therefore I would like to ask how
the function could be parallelized more efficient. (With increasing sample size
the difference becomes smaller, but I was wondering about this big differences
and how it could be improved.)
Thank you very much for help in advance!
Best,
Martin
library(microbenchmark)
library(doParallel)
n <- 500
y <- rnorm(n)
Xc <- rnorm(n)
Xd <- sample(c(0,1), replace=TRUE)
Weights <- diag(n)
n1 <- 50
Xeval <- cbind(rnorm(n1), sample(c(0,1), n1, replace=TRUE))
detectCores()
cl <- makeCluster(4)
registerDoParallel(cl)
microbenchmark(apply(Xeval, 1, npnewpar, y=y, Xc=Xc, Xd = Xd, Weights=Weights,
h=0.5), parApply(cl, Xeval, 1, npnewpar, y=y, Xc=Xc, Xd = Xd, Weights=Weights,
h=0.5), times=100)
stopCluster(cl)
Unit: milliseconds
expr minlq meanmedian
apply(Xeval, 1, npnewpar, y = y, Xc = Xc, Xd = Xd, Weights = Weights,
h = 0.5) 4.674914 4.726463 5.455323 4.771016
parApply(cl, Xeval, 1, npnewpar, y = y, Xc = Xc, Xd = Xd, Weights = Weights,
h = 0.5) 34.168250 35.434829 56.553296 39.438899
uq max neval
4.843324 57.01519 100
49.777265 347.77887 100
npnewpar <- function(y, Xc, Xd, Weights, h, xeval) {
xc <- xeval[1]
xd <- xeval[2]
l <- function(x,X) {
w <- Weights[x,X]
return(w)
}
u <- (Xc-xc)/h
#K <- kernel(u)
K <- dnorm(u)
L <- l(xd,Xd)
nom <- sum(y*K*L)
denom <- sum(K*L)
ghat <- nom/denom
return(ghat)
}
__
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] Problem with predict.lm()
Thank you! I think I now understand where the problem was. Best, Martin Gesendet: Mittwoch, 29. April 2015 um 16:50 Uhr Von: "David L Carlson" An: "Martin Spindler" , "r-help@r-project.org" Betreff: RE: [R] Problem with predict.lm() Since you passed a matrix to lm() and then a data.frame to predict(), predict can't match up what variables to use for the prediction so it falls back on the original data. This seems to work: > set.seed(42) > y <- rnorm(100) > X <- matrix(rnorm(100*10), ncol=10) > Xd <- data.frame(X) > lm <- lm(y~., Xd) > Xnew <- matrix(rnorm(100*20), ncol=10) > Xnewd <- data.frame(Xnew) > ynew <- predict(lm, newdata=Xnewd) > head(ynew) 1 2 3 4 5 6 0.35404067 0.14073495 -0.45442499 0.31065562 -0.02091366 0.25358175 > head(predict(lm)) 1 2 3 4 5 6 0.75474817 0.06024122 -0.27221466 -0.20344713 0.20218135 -0.24045859 > - David L Carlson Department of Anthropology Texas A&M University College Station, TX 77840-4352 -Original Message- From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Martin Spindler Sent: Wednesday, April 29, 2015 9:21 AM To: r-help@r-project.org Subject: [R] Problem with predict.lm() Dear all, the following example somehow uses the "old data" (X) to make the predictions, but not the new data Xnew as intended. y <- rnorm(100) X <- matrix(rnorm(100*10), ncol=10) lm <- lm(y~X) Xnew <- matrix(rnorm(100*20), ncol=10) ynew <- predict(lm, newdata=as.data.frame(Xnew)) #prediction in not made for Xnew How can I foce predict.lm to use use the new data? Thank you very much for your efforts in advance! Best, Martin __ 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[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] Problem with predict.lm()
Dear Arnab, Thank you very much for your reply. It does not give an error message. The problem is that predict does "work and predict" on the old data and does not make the predictions with the provided new data. Best, Martin Gesendet: Mittwoch, 29. April 2015 um 16:51 Uhr Von: "ARNAB KR MAITY" An: "Martin Spindler" , "r-help@r-project.org" Betreff: Re: [R] Problem with predict.lm() Hi, It seems to be working in my R. Although it is throwing the warning message Warning message: 'newdata' had 200 rows but variables found have 100 rows y [1] -1.071307580 0.102414204 -0.965046207 1.386057875 0.726835339 [6] -0.186549950 -0.777144258 1.137210314 -1.069446945 -0.696084338 [11] -0.467686285 0.997316781 0.776265490 -1.385720997 -0.007438381 [16] 0.302821728 0.024075173 -0.590401970 0.877104292 0.652724314 [21] 2.142135078 2.023051454 -0.547221960 0.342600702 0.080848203 [26] 0.074609232 0.255946197 -0.191242759 1.036445108 0.895068954 [31] 0.589477883 0.123230750 1.467210735 -1.636657283 -0.131504288 [36] -0.665000122 0.390977868 0.546802014 0.445498091 1.063872749 [41] 1.406788635 -0.037147550 -1.047190960 -0.189105987 0.069617165 [46] -0.049760285 -1.454279226 0.358351554 0.246587937 -0.060735329 [51] 1.664530111 -0.475931484 0.405480604 1.560446941 -0.030537155 [56] -1.060319583 -1.828624216 -0.429391165 0.301697744 -0.029593593 [61] -1.696307754 0.342678986 -0.433965195 -0.947338037 0.318186677 [66] 0.539630789 -1.354555193 0.086168702 0.002950100 1.783486665 [71] -1.182419158 -0.930524123 0.376579158 -1.085035387 1.186125702 [76] 0.719738391 -0.486692820 -2.105396602 0.531238276 1.302812739 [81] 0.347851244 0.016452693 0.417535566 0.277705766 2.286275977 [86] 1.610183518 2.032037030 1.319074179 1.129375593 0.176684807 [91] -0.630517144 1.302785450 0.994275267 -0.060116993 -0.655966924 [96] 1.628197169 1.935532651 -1.635783346 -1.172511179 1.238336597 > ynew 1 2 3 4 5 6 -0.270916637 0.169149841 0.191348061 -0.009541999 0.112027155 0.016242323 7 8 9 10 11 12 -0.062178365 0.275322344 0.397030485 0.565078468 0.301230303 0.305405674 13 14 15 16 17 18 0.552136794 -0.151275710 0.470280882 0.349631748 0.022005869 0.181384646 19 20 21 22 23 24 0.143719339 0.478791323 0.518731127 0.229860133 -0.199433324 0.310576455 25 26 27 28 29 30 0.127612633 -0.157347145 0.413807523 0.007961485 -0.288867750 0.208759771 31 32 33 34 35 36 0.286165027 0.299492579 0.197312294 0.135601904 0.452828662 0.187191405 37 38 39 40 41 42 0.335596502 -0.109960231 -0.303770506 -0.276385255 0.429700474 0.003930969 43 44 45 46 47 48 0.184186301 0.140858190 0.479882236 0.182523553 -0.133845870 0.443940376 49 50 51 52 53 54 0.070571673 -0.383780163 0.362153269 0.202527841 0.164299813 0.327998904 55 56 57 58 59 60 0.047612361 -0.032167295 0.060976285 0.231929803 -0.449532973 0.109925656 61 62 63 64 65 66 0.468842330 0.108507841 0.158697337 -0.125813680 0.501159861 0.101646132 67 68 69 70 71 72 0.194383106 -0.006185569 0.354467348 0.340013811 0.088757961 0.439984356 73 74 75 76 77 78 0.330976669 0.449337326 0.081841142 -0.190123754 0.337794560 -0.111895039 79 80 81 82 83 84 0.598231564 0.444399789 0.388313945 0.244270482 0.200026237 0.009025077 85 86 87 88 89 90 0.341093767 -0.164196034 0.825849472 0.325975911 0.494473323 0.270037159 91 92 93 94 95 96 0.369787280 0.247455471 0.282701738 -0.541688411 -0.145796547 0.073172268 97 98 99 100 0.685833173 -0.079174316 -0.193161949 -0.137517175 Arnab Kumar Maity Graduate Teaching Assistant Division of Statistics Northern Illinois University DeKalb, Illinois 60115 U.S.A -------- From: Martin Spindler To: r-help@r-project.org Sent: Wednesday, April 29, 2015 9:21 AM Subject: [R] Problem with p
[R] Problem with predict.lm()
Dear all, the following example somehow uses the "old data" (X) to make the predictions, but not the new data Xnew as intended. y <- rnorm(100) X <- matrix(rnorm(100*10), ncol=10) lm <- lm(y~X) Xnew <- matrix(rnorm(100*20), ncol=10) ynew <- predict(lm, newdata=as.data.frame(Xnew)) #prediction in not made for Xnew How can I foce predict.lm to use use the new data? Thank you very much for your efforts in advance! Best, Martin __ 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] Using openBLAS in R under Unix / Linux
Dear all, I would like to us openBLAS in R under Linux / Unix. Which steps do I have to undertake? Does someone know a detailed description? (I found some sources on the web, but none was really helpful for me.) Thanks and best, Martin [[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] Standardisation of variables with Lasso (glmnet)
Dear all, I am working with glmnet but the problem arises also in all other Lasso implementations: It is ususally recommended to standardize the variables / use intercept and this works well with the implemented options: x <- matrix(rnorm(1), ncol=50) y <- rnorm(200) cv.out =cv.glmnet(x,y, alpha =1, intercept=T , standardize=T) coef <- coef(cv.out, s = "lambda.min") ind1 <- which(coef>0) coef[ind1,] but when I would like to do this by hand: xs <- apply(x,2, function(x) (x-mean(x))/sqrt(var(x))) ys <- y - mean(y) cv.out =cv.glmnet(xs,ys, alpha =1, intercept=F , standardize=F) coef <- coef(cv.out, s = "lambda.min") ind1 <- which(coef>0) coef[ind1,] The following error appears: > cv.out =cv.glmnet(xs,ys, alpha =1, intercept=F , standardize=F) Error in elnet(x, is.sparse, ix, jx, y, weights, offset, type.gaussian, : NA/NaN/Inf in foreign function call (arg 5) Therefore my question is what am I doing wrong and what is the "best practice" with Lasso (intercept yes / no, standardisation by hand, ...) Thank you very much for your efforts and replies in advance! Best, Martin __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Negative Binomial Regression - glm.nb
Dear all, I would like to ask, if there is a way to make the variance / dispersion parameter $\theta$ (referring to MASS, 4th edition, p. 206) in the function glm.nb dependent on the data, e.g. $1/ \theta = exp(x \beta)$ and to estimate the parameter vector $\beta$ additionally. If this is not possible with glm.nb, is there another function / package which might do that? Thank you very much for your answer in advance! Best, Martin __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Splitting up of a dataframe according to the type of variables
Dear R users, I have a dataframe which consists of variables of type numeric and factor. What is the easiest way to split up the dataframe to two dataframe which contain all variables of the type numeric resp. factors? Thank you very much for your efforts in advance! Best, Martin __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] Function for Generating all Permutations with Repetition
Dear all, I am looking for a function in R which returns all possible permutations of an object x with r number of repitions. For example If x <- c(0,1) and r <-3 the result should be 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 1 1 1 0 1 1 1 and consist of 2^3=8 elements. Unfortunately, I have found only functions which return the combinations for "choose n over k" but not the case described above. I would appreciate hints and help highly. Best, Martin -- __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] non-finite finite-difference value
Dear all, when applying the optim function the following error occured "non-finite finite-difference value" Therefore I would like to ask how one can try to handle such a problem and which strategies have proven useful. (There is only litte guidance on the help list for this kind of problem.) Thank you in advance for your efforts! Best, Martin -- __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] problem with as.Date
Dear all, I would like to convert the first column of a dataframe to a date (original format: year (4 digits) and month (last 2 digits)) >str(dat_FF) 'data.frame': 1022 obs. of 4 variables: $ date : int 192607 192608 192609 192610 192611 192612 192701 192702 192703 192704 ... $ Rm.Rf: num 2.69 2.52 0 -3.06 2.42 2.66 0 4.29 0.51 0.57 ... $ SMB : num -2.49 -1.25 -1.38 -0.2 -0.34 -0.07 -0.11 0.35 -1.87 0.44 ... $ HML : num -2.91 4.25 0.22 0.71 -0.4 -0.11 4.92 3.17 -2.92 1.33 ... But >dat_FF$date <- as.Date(as.character(dat_FF$date), format="%Y%m") delievers NAs: >str(dat_FF) 'data.frame': 1022 obs. of 4 variables: $ date : Date, format: NA NA ... I am very grateful for hints! Thanks in advance! Best, Martin __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] package VGAM: vglm.fit()
Dear all, I am estimating a bivariate probit model using the package VGAM: fit1 = vglm(cbind(y1, y2) ~ x1+x2, binom2.rho, data=dat, trace=TRUE) I would like to estimate this via the method vglm.fit (in an analogous way as lm.fit for lm or glm.fit for glm). Unfortunately my trials did not succeed. Therefore I would like to ask if this is possible at all and - if yes - how one can access this function. Thank you very much for your efforts in advance! Best, Martin [[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] conversion of matrix into list
Dear all, I have a matrix X which consists of 2 columns. I would like to convert this matrix into a list where every entry of the list consists of a single row of the matrix. Does anyone have a suggestions how to manage this? Thank you for your efforts in advance! Best, Martin [[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] partial evaluation of a function with several arguments
Dear all,
I have the following problem:
add <- function(x,y) {x+y}
What is the easiest / most elegant way to create a new function (e.g. with the
name "addev") that sets the second argument of the function "add" to a fixed
value (e.g. y=3), i.e. addev <- add(x,3). But this does not work.
Thank you for your efforts in advance!
Best,
Martin
--
--
Empfehlen Sie GMX DSL Ihren Freunden und Bekannten und wir
belohnen Sie mit bis zu 50,- Euro! https://freundschaftswerbung.gmx.de
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/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] subset with two factors
Hey Michael, Thank you very much. It works! Best, Martin Original-Nachricht > Datum: Fri, 10 Dec 2010 22:35:56 +1100 > Von: Michael Bedward > An: Martin Spindler > CC: r-help@r-project.org > Betreff: Re: [R] subset with two factors > Hello Martin, > > You were almost there :) > > T1 <- subset(daten1, Geschlecht=="M" & GG=="A") > > Hope this helps. > > Michael > > On 10 December 2010 22:25, Martin Spindler wrote: > > Dear all, > > > > I have a dataframe of the following strucutre > > > > numacc_b coverage_b Geschlecht GG > > 1 0 1 W A > > 2 0 1 M A > > 3 0 1 M B > > 4 0 1 M B > > 5 0 1 W A > > 6 0 1 M B > > > > I would like to form a subset consisting of all entries with > Geschlecht=M and GG=A. > > > > Using > > > >>T1 <- subset(daten1, Geschlecht=="M", GG=="A") > > > > delievers > > > > data frame with 0 columns and 6 rows > > > >> T1 <- subset(daten1, Geschlecht=="M") > > > > delievers > > > > numacc_b coverage_b Geschlecht GG > > 2 0 1 M A > > 3 0 1 M B > > 4 0 1 M B > > 6 0 1 M B > > 9 0 1 M B > > 10 0 1 M B > > > > But I want to select the dataframe according to both factos. > > > > What can I do? > > > > Thank you answers in advance! > > > > Best, > > > > Martin > > -- > > GMX DSL Doppel-Flat ab 19,99 €/mtl.! Jetzt auch mit > > gratis Notebook-Flat! http://portal.gmx.net/de/go/dsl > > > > __ > > R-help@r-project.org mailing list > > https://stat.ethz.ch/mailman/listinfo/r-help > > PLEASE do read the posting guide > http://www.R-project.org/posting-guide.html > > and provide commented, minimal, self-contained, reproducible code. > > -- GRATIS! Movie-FLAT mit über 300 Videos. -- GMX DSL Doppel-Flat ab 19,99 €/mtl.! Jetzt auch mit gratis Notebook-Flat! http://portal.gmx.net/de/go/dsl __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] subset with two factors
Dear all, I have a dataframe of the following strucutre numacc_b coverage_b Geschlecht GG 10 1 W A 20 1 M A 30 1 M B 40 1 M B 50 1 W A 60 1 M B I would like to form a subset consisting of all entries with Geschlecht=M and GG=A. Using >T1 <- subset(daten1, Geschlecht=="M", GG=="A") delievers data frame with 0 columns and 6 rows > T1 <- subset(daten1, Geschlecht=="M") delievers numacc_b coverage_b Geschlecht GG 2 0 1 M A 3 0 1 M B 4 0 1 M B 6 0 1 M B 9 0 1 M B 100 1 M B But I want to select the dataframe according to both factos. What can I do? Thank you answers in advance! Best, Martin -- GMX DSL Doppel-Flat ab 19,99 €/mtl.! Jetzt auch mit gratis Notebook-Flat! http://portal.gmx.net/de/go/dsl __ R-help@r-project.org mailing list https://stat.ethz.ch/mailman/listinfo/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] question concerning VGAM
Hello everyone, using the VGAM package and the following code library(VGAM) bp1 <- vglm(cbind(daten$anzahl_b, daten$deckung_b) ~ ., binom2.rho, data=daten1) summary(bp1) coef(bp1, matrix=TRUE) produced this error message: error in object$coefficients : $ operator not defined for this S4 class I am bit confused because some day ago this error message did not show up and everything was fine. Thank you very much in advance for your help. Best, Martin [[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] linear predicted values of the index function in an ordered probit model
Hello,
currently I am estimating an ordered probit model with the function polr
(MASS package).
Is there a simple way to obtain values for the prediction of the index
function ($X*\hat{\beta}$)?
(E..g. in the GLM function there is the linear.prediction value for this
purpose).
If not, is there another function / package where this feature is
implemented?
Thank you very much for your answer in advance!
Best,
Martin
[[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.
