Re: [R] Hep with regex! - combination of ^, |, \\, _ and $

> On 17 Sep 2015, at 23:11, Dimitri Liakhovitski > wrote: > > (x <- c("q10_1", "q10_2", "q10_11", "q12_1", "q12_2", "q13_1", "q13_11")) > > # Which strings start with "q10" or "q12? - WORKS > x[grep("^q10|q12", x)] > > # Which strings end with "1"? - WORKS > x[grep("1$", x)] > > # Which stri

[R] Clustering of clients (retail) - Free data sets?

Hi all, I'm learning about how to do clusters of clients. Ç I've founde this nice presentation on the subject, but the data is not avaliable to use. I've contacted the autor, hope he'll answer soon. https://ds4ci.files.wordpress.com/2013/09/user08_jimp_custseg_revnov08.pdf Someone knows similar

Re: [R] Distance in miles btw Zipcodes

On Sep 17, 2015, at 3:36 PM, Farnoosh Sheikhi via R-help wrote: > Hello, > I'm trying to get the distances between two Zipcode variables, but for some > reason I get this error: > "matching was not perfect, returning what was found.Error: no such index at > level 1" > Here is my code: > > lib

Re: [R] best data storage format?

No data. See dput() (?dput) as the preferred way to send data John Kane Kingston ON Canada > -Original Message- > From: alfadia...@mac.com > Sent: Thu, 17 Sep 2015 16:41:46 -0400 > To: r-help@r-project.org > Subject: [R] best data storage format? > > Hello - > > I’m working on dataset

Re: [R] Hep with regex! - combination of ^, |, \\, _ and $

On 17/09/2015 5:46 PM, Dimitri Liakhovitski wrote: > Duncan, > Of course my verbal descriptions and my code don't match my regexp - > otherwise I wouldn't be asking the question, would I? > Please assume my verbal descriptions are correctly describing what I want. Sorry, I interpreted "works" and

[R] Distance in miles btw Zipcodes

 Hello, I'm trying to get the distances between two Zipcode variables, but for some reason I get this error: "matching was not perfect, returning what was found.Error: no such index at level 1" Here is my code: library(ggmap)mapdist(data$Zip.A, data$Zip.B, mode = "driving") The Zip codes are all

Re: [R] Hep with regex! - combination of ^, |, \\, _ and $

On Sep 17, 2015, at 2:11 PM, Dimitri Liakhovitski wrote: > (x <- c("q10_1", "q10_2", "q10_11", "q12_1", "q12_2", "q13_1", "q13_11")) > > # Which strings start with "q10" or "q12? - WORKS > x[grep("^q10|q12", x)] > > # Which strings end with "1"? - WORKS > x[grep("1$", x)] > > # Which strings e

Re: [R] Hep with regex! - combination of ^, |, \\, _ and $

For the last one, looks like this one works: x[grep("^(q10|q12).*\\_1$", x)] On Thu, Sep 17, 2015 at 5:46 PM, Dimitri Liakhovitski wrote: > Duncan, > Of course my verbal descriptions and my code don't match my regexp - > otherwise I wouldn't be asking the question, would I? > Please assume my ver

Re: [R] Hep with regex! - combination of ^, |, \\, _ and $

On 17/09/2015 5:11 PM, Dimitri Liakhovitski wrote: > (x <- c("q10_1", "q10_2", "q10_11", "q12_1", "q12_2", "q13_1", "q13_11")) > > # Which strings start with "q10" or "q12? - WORKS > x[grep("^q10|q12", x)] > > # Which strings end with "1"? - WORKS > x[grep("1$", x)] > > # Which strings end with

Re: [R] Hep with regex! - combination of ^, |, \\, _ and $

Duncan, Of course my verbal descriptions and my code don't match my regexp - otherwise I wouldn't be asking the question, would I? Please assume my verbal descriptions are correctly describing what I want. Thank you! On Thu, Sep 17, 2015 at 5:42 PM, Duncan Murdoch wrote: > On 17/09/2015 5:11 PM,

[R] Rcpp RNG - runif() override

Dear R users, I'm attempting to override the base function runif() with a function, custom_runif(), that I've written and tested in Rcpp - with the aim of using my own RNG in DEoptim. I've attempted setting the name in a namespace changing locking to no avail, e.g. assignInNamespace(runif, cu

Re: [R] aggregate counting variable factors

Thanks everybody! On Thu, Sep 17, 2015 at 6:57 PM, Rui Barradas wrote: > In package reshape2 > > Hope this helps, > > Rui Barradas > > > Em 17-09-2015 17:03, Frank Schwidom escreveu: > >> Hi >> >> where can i find 'melt' and 'dcast' ? >> >> Regards >> >> >> On Thu, Sep 17, 2015 at 08:22:10AM +00

[R] best data storage format?

Hello - I’m working on dataset that will eventually be used in an xyz-plot. I’m having trouble figuring out the best way to store the data (see an attached .csv sheet exported from Excel). Some information on the data: - Columns B - F are labels that describe the z data points - Rows above x an

Re: [R] fast way to create composite matrix based on mixed indices?

Hi Matt, you could use matrix indexing. Here is a possible solution, which could be optimized further (probably). # The old matrix (old.mat <- matrix(1:30,nrow=3,byrow=TRUE)) # matrix of indices index <- matrix(c(1,1,1,4, 1,3,5,10, 2,2,1,3,

[R] Hep with regex! - combination of ^, |, \\, _ and $

(x <- c("q10_1", "q10_2", "q10_11", "q12_1", "q12_2", "q13_1", "q13_11")) # Which strings start with "q10" or "q12? - WORKS x[grep("^q10|q12", x)] # Which strings end with "1"? - WORKS x[grep("1$", x)] # Which strings end with "_1"? - WORKS x[grep("\\_1$", x)] # Which strings start with "q10" A

[R] fast way to create composite matrix based on mixed indices?

HI all, Sorry for the title here but I find this difficult to describe succinctly. Here's the problem. I want to create a new matrix where each row is a composite of an old matrix, but where the row & column indexes of the old matrix change for different parts of the new matrix. For example, the

Re: [R] c(1:n, 1:(n-1), 1:(n-2), ... , 1)

Note that: >> Also not sure about efficiency but somewhat shorter... >> unlist(lapply(5:1, seq)) >> >>> Peter >>> is almost exactly sequence(5:1) which is unlist(lapply(5:1,seq_len)) which is "must preferred". See ?seq for details Cheers, Bert >>> On Thu, Sep 17, 2015 at 11:19 AM, Dan D

Re: [R] c(1:n, 1:(n-1), 1:(n-2), ... , 1)

If you are interested in speed for long input vectors try the following, which should give the same result as sequence(). mySequence <-function (nvec) { nvec <- as.integer(nvec) seq_len(sum(nvec)) - rep(cumsum(c(0L, nvec[-length(nvec)])), nvec) } E.g., > n <- rpois(1e6, 3) > system

Re: [R] HELP IN GRAPHS - slip screen

Hi Rosa, I don't think the problem is with the split.screen command, for you are getting the eight plots and the screen at the right as you requested. It looks like your margins for each plot need adjusting, and I also think you should have about a 2.2 to 1 width to height ratio in the graphics dev

Re: [R] c(1:n, 1:(n-1), 1:(n-2), ... , 1)

Very nice variety of solutions to create c(1:n, 1:(n-1), 1:(n-2), ... , 1) #Testing the methods with n=1000 (microbenchmark) n<-1000 # by far the nicest-looking, easiest to follow, and fastest is Frank Schwidom's: # it also requires the minimum amount of memory (as do several of the others) # 2.7

Re: [R] Optimization Grid Search Slow

optimx does nothing to speed up optim or the other component optimizers. In fact, it does a lot of checking and extra work to improve reliability and add KKT tests that actually slow things down. The purpose of optimx is to allow comparison of methods and discovery of improved approaches to a p

[R] Optimization Grid Search Slow

R Help - I am trying to use a grid search for a 2 free parameter reinforcement learning model and the grid search is incredibly slow. I've used optimx but can't seem to get reasonable answers. Is there a way to speed up this grid search dramatically? dat <- structure(list(choice = c(0, 1, 1, 1,

Re: [R] c(1:n, 1:(n-1), 1:(n-2), ... , 1)

how abount a more complicated one? outer( 1:5, 1:5, '-')[ outer( 1:5, 1:5, '>')] [1] 1 2 3 4 1 2 3 1 2 1 On Thu, Sep 17, 2015 at 11:52:27AM -0700, David Winsemius wrote: > You can add this to the list of options to be tested, although my bet would > be placed on `sequence(5:1)`: > > > Reduce

Re: [R] c(1:n, 1:(n-1), 1:(n-2), ... , 1)

I'm not too sure this is any better: n<-5 c<-0; # establish result as numeric for(i in seq(n,1,-1)){ c<-c(c,seq(1,i)); str(c); }; #generate array c<-c[2:length(c)]; #remove the leading 0 If you're a fan of recursive programming: > mklist <- function(x) { if (x==1) return(1) else return( c(seq(1,

Re: [R] c(1:n, 1:(n-1), 1:(n-2), ... , 1)

You can add this to the list of options to be tested, although my bet would be placed on `sequence(5:1)`: > Reduce( function(x,y){c( 1:y, x)}, 1:5) [1] 1 2 3 4 5 1 2 3 4 1 2 3 1 2 1 On Sep 17, 2015, at 11:40 AM, Achim Zeileis wrote: > On Thu, 17 Sep 2015, Peter Langfelder wrote: > >> Not sur

Re: [R] c(1:n, 1:(n-1), 1:(n-2), ... , 1)

On Thu, 17 Sep 2015, Peter Langfelder wrote: Not sure if this is slicker or easier to follow than your solution, but it is shorter :) do.call(c, lapply(n:1, function(n1) 1:n1)) Also not sure about efficiency but somewhat shorter... unlist(lapply(5:1, seq)) Peter On Thu, Sep 17, 2015 at 11:

Re: [R] c(1:n, 1:(n-1), 1:(n-2), ... , 1)

sequence( 5:1) Regards. On Thu, Sep 17, 2015 at 11:19:05AM -0700, Dan D wrote: > Can anyone think of a slick way to create an array that looks like c(1:n, > 1:(n-1), 1:(n-2), ... , 1)? > > The following works, but it's inefficient and a little hard to follow: > n<-5 > junk<-array(1:n,dim=c(n,n)

Re: [R] c(1:n, 1:(n-1), 1:(n-2), ... , 1)

Not sure if this is slicker or easier to follow than your solution, but it is shorter :) do.call(c, lapply(n:1, function(n1) 1:n1)) Peter On Thu, Sep 17, 2015 at 11:19 AM, Dan D wrote: > Can anyone think of a slick way to create an array that looks like c(1:n, > 1:(n-1), 1:(n-2), ... , 1)? > >

[R] c(1:n, 1:(n-1), 1:(n-2), ... , 1)

Can anyone think of a slick way to create an array that looks like c(1:n, 1:(n-1), 1:(n-2), ... , 1)? The following works, but it's inefficient and a little hard to follow: n<-5 junk<-array(1:n,dim=c(n,n)) junk[((lower.tri(t(junk),diag=T)))[n:1,]] Any help would be greatly appreciated! -Dan -

Re: [R] finding those elements of listA that are found in listB

John, The intersect() function may help you. For example: listA <- sort(sample(10, 5)) listB <- sort(sample(10, 5)) both <- intersect(listA, listB) > listA [1] 2 4 7 8 9 > listB [1] 1 2 3 8 10 > both [1] 2 8 Jean On Wed, Sep 16, 2015 at 9:43 PM, John Sorkin wrote: > I have two structure

Re: [R] aggregate counting variable factors

In package reshape2 Hope this helps, Rui Barradas Em 17-09-2015 17:03, Frank Schwidom escreveu: Hi where can i find 'melt' and 'dcast' ? Regards On Thu, Sep 17, 2015 at 08:22:10AM +, PIKAL Petr wrote: Hi -Original Message- From: R-help [mailto:r-help-boun...@r-project.org] O

Re: [R] Factor response and explanatory variables.

Best advice I can give: Find a local statistical expert to work with. You appear to be asking for help understanding statistical methodology when you do not have the necessary background to do so. Cheers, Bert Bert Gunter "Data is not information. Information is not knowledge. And knowledge is c

Re: [R] Multiple if function

On Thu, 17 Sep 2015, Berend Hasselman wrote: On 17 Sep 2015, at 01:42, Dénes Tóth wrote: On 09/16/2015 04:41 PM, Bert Gunter wrote: Yes! Chuck's use of mapply is exactly the split/combine strategy I was looking for. In retrospect, exactly how one should think about it. Many thanks to all

Re: [R] aggregate counting variable factors

Hi where can i find 'melt' and 'dcast' ? Regards On Thu, Sep 17, 2015 at 08:22:10AM +, PIKAL Petr wrote: > Hi > > > -Original Message- > > From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Kai Mx > > Sent: Wednesday, September 16, 2015 10:43 PM > > To: r-help mailing

Re: [R] aggregate counting variable factors

Hi res <- sapply( df1[ , -1], function( x) table(x)[as.character( 0:5)]) rownames( res) <- paste( sep='', 'result', 0:5) res[ is.na( res)] <- 0 res item1 item2 item3 item4 item5 result0 1 0 1 1 0 result1 1 2 0 0 0 result2 1 2 1 1

[R] Need data labels to jitter with datapoints in boxplot

Hello, I have created a boxplot with the data points overlayed on top using the below code. I am happy with the way the datapoints are jittered, however I cannot figure out how to get the labels to jitter along with the datapoints. The labels remain in the center and are unreadable. I have tried a

[R] Factor response and explanatory variables.

   Hello everyone,    I am going to ask this certainly tricky question here not (yet) with the intention of getting a definitive answer, as I need to deepen my questions much more, but just to have an approximate idea of which direction taking next.    I have a dataset where the potential respo

Re: [R] (no subject)

Hello, Try th following. Q1 <- matrix(c(sample(4, 200, replace = TRUE), rbinom(200,1,0.7)), ncol = 2) Q1 Hope this helps, Rui Barradas Em 17-09-2015 15:37, thanoon younis escreveu: Dear R - Users I have a small problem when i generated two column with 200 rows and as follows Q1[i,1]=sample

[R] (no subject)

Dear R - Users I have a small problem when i generated two column with 200 rows and as follows Q1[i,1]=sample(4, 200, replace = TRUE); Q1[i,2]=rbinom(200,1,0.7) the first vector is ordered categorical variable and the second vector is dichotomous variable but when i run this code i found this err

Re: [R] Fwd: generate ordered categorical variable in R

Vectors have no columns or rows. rep( NA, 200 ) If you need a matrix, you have to turn it into one: matrix( rep( NA, 200 ), ncol=1 ) --- Jeff NewmillerThe . . Go Live... DCN:

Re: [R] Fwd: generate ordered categorical variable in R

... and, less explicitly, but more compactly: y <- array(dim=c(200,1)) B. On Sep 17, 2015, at 10:07 AM, Boris Steipe wrote: > x <- rep(NA, 200) > > For all cases I can think of, that is enough. If you MUST have a matrix with > one column and two hundred rows, set: > > dim(x) <- c(200,1) >

Re: [R] Fwd: generate ordered categorical variable in R

x <- rep(NA, 200) For all cases I can think of, that is enough. If you MUST have a matrix with one column and two hundred rows, set: dim(x) <- c(200,1) B. On Sep 17, 2015, at 9:40 AM, thanoon younis wrote: > Dear all users > > I want to write a vector with one column and just NA values and

[R] Fwd: generate ordered categorical variable in R

Dear all users I want to write a vector with one column and just NA values and nrow=200 when i write X=numeric(NA) is not correct how can i do this please? Regards [[alternative HTML version deleted]] __ R-help@r-project.org mailing list --

[R] short course: Bayesian Data Analysis with R and WinBUGS

Dear list members, Apologies for cross-posting. Please, find below the information of a course: "Bayesian Data Analysis with R and WinBUGS". The course takes place in a nice area nearby the city of Essen. In the course’s price is included teaching material (The BUGS Book and my slides), an ex

Re: [R] getting means by group within time point for data on multiple lines (long rather than wide file)

Hi John, This will not be the complete answer, but it can probably help you in the right direction. First, I would subset your data.frame to include only subjects with one observation at each time point (and I'm not sure how to do that easily). But then, the aggregate() function is what you

Re: [R] getting means by group within time point for data on multiple lines (long rather than wide file)

On 17/09/2015 7:06 AM, John Sorkin wrote: > I have a long (rather than wide file), i.e. the data for each subject is on > multiple lines rather than one line. Each line has the following layout: > subject group time value > I have two groups, multiple subjects, each subject can be seen up to three

Re: [R] HELP IN GRAPHS - slip screen

Hi Rosa, Try this: # do the first split, to get the rightmost screen for the legend split.screen(figs=matrix(c(0,0.84,0,1,0.84,1,0,1),nrow=2,byrow=TRUE)) # now split the first screen to get your eight screens (numbered 3 to 10) for the plots split.screen(figs=matrix(c(0,0.25,0.5,1,

[R] getting means by group within time point for data on multiple lines (long rather than wide file)

I have a long (rather than wide file), i.e. the data for each subject is on multiple lines rather than one line. Each line has the following layout: subject group time value I have two groups, multiple subjects, each subject can be seen up to three times a time 0, and at most once at times 4 and

Re: [R] aggregate counting variable factors

Hi > -Original Message- > From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of Kai Mx > Sent: Wednesday, September 16, 2015 10:43 PM > To: r-help mailing list > Subject: [R] aggregate counting variable factors > > Hi everybody, > > >From a questionnaire, I have a dataset like t

Re: [R] Multiple if function

> On 17 Sep 2015, at 01:42, Dénes Tóth wrote: > > > > On 09/16/2015 04:41 PM, Bert Gunter wrote: >> Yes! Chuck's use of mapply is exactly the split/combine strategy I was >> looking for. In retrospect, exactly how one should think about it. >> Many thanks to all for a constructive discussion .