similar to: sapply and matrix command

I have a table like this: Date TIME Q A a 1 A b 2 A c 3 B a 4 B b 5 B c 6 C a 7 C b 8 C c 9 I want use R language to turn it to a matrix like : a b c A 1 2 3 B 4 5 6 C 7 8 9 I am new to R , anyone can help? Thanks in advance! have a table like this: Date TIME Q A a 1 A b 2 A c 3 B a

Hi, I'd like to add "alpha" in latex code to my image. Any suggestions? library(UsingR) BagA = c(rep(10,6),rep(20,5), rep(30,4),rep(40,3),rep(50,2),60,70) BagA BagB = c(10, 20,rep(30,2),rep(40,3),rep(50,4),rep(60,5),rep(70,6)) BagB par(mfrow = c(2, 1)) DOTplot(BagA) abline(v=60,lwd=2,lty=2,col="red") text(65,3, "alpha-->", col=2) abline(v=50, lwd=2, lty=2,

runif appears to give 31 bits of precision, but this isn't mentioned in the documentation page. The R numeric type supports 53 digits of precision, and other numeric functions (sin, etc.) give full-precision results. So I'd think that either runif should give full precision or its documentation should mention this limitation. #integers table(runif(10000,-2^30,2^30) %% 1) 0 0.5 4972

Hi all, I've got the following problem. I have a vector containing file names. I want to read these files as csv and calculate the density-function for each file (has just one column with data). Then, I'd like to plot all density functions into one window. I did the following to calculate the density data: s <- sapply(filelist, function(x) { if(file.exists(x)) { file <-

I am puzzled by the scope rules that apply with sapply. If I want to modify a vector with sapply I tried... N <- 10 vec <- vector(mode="numeric", length=N) test <- function(i){ vec[i] <- i } sapply(1:N, test) vec but it not work. How can this be done? Worik [[alternative HTML version deleted]]

Hi, sapply((1:NCOL(hermes)),function(x) hist(hermes[,x],main=names(hermes)[x])) .......this works , but i would like use it with apply to generate many plots in one step! #################################################################### >>apply((1:ncol(hermes)),2,function(x) hist(hermes[,x],main=names(hermes)[x])) Error in apply(1:ncol(hermes), 2, function(x) hist(hermes[, x], main =

I have to compare four different grape varieties proteome in two different years. I don't know what test would be more suitable for my data. I think that an anova two way can be usefull also if someone suggested me to perform a manova. In addiction, I can perform each test on a single protein a time, but I can't loose my whole life carrying out anova (I have more than 1000 protein to

colnames(dd) #[1] "col1" "colb" null_vector<- colnames(dd) sapply(null_vector,makeNull,dd) #???? col1 colb #[1,]?? NA??? 4 #[2,]??? 2?? NA #[3,]??? 3??? 2 #[4,]??? 4?? NA #[5,]??? 1??? 4 #[6,]?? NA??? 5 #[7,]??? 1??? 6 A.K. >I am trying to make a column value in a dataframe = NA if there is a 0 or high value in that column. I need to do this process repeatedly, hence

Sorry if this has been answered elsewhere, but I can't find any discussion of it. Wondering why the following situation occurs (duplicated on 3.2.2 CentOS6 and 3.0.1 Win2k, so I don't think it is a bug): > sapply(1, identical, 1) [1] TRUE > sapply(1:2, identical, 1) [1] FALSE FALSE > sapply(1:2, function(i) identical(as.numeric(i),1) ) [1] TRUE FALSE > sapply(1:2,

I highly recommend making friends with the str function. Try str( 1 ) str( 1:2 ) for the clue you need, and then sapply( 1:2, identical, 1L ) -- Sent from my phone. Please excuse my brevity. On April 8, 2016 3:24:31 PM PDT, "Paulson, Ariel" <apa at stowers.org> wrote: >Sorry if this has been answered elsewhere, but I can't find any >discussion of it. >

This is just a warning/hint to others; package/function writers in particular : The `cluster' package (originally S-plus from Rousseeuw et al) currently has code such as ## check type of input matrix if((!is.data.frame(x) && !is.numeric(x)) || !all(sapply(x, data.class) == "numeric")) stop("x is not a numeric dataframe or matrix.") x <-

Full_Name: Drouilhet R?my Version: 1.8.1 OS: Linux Submission from: (NULL) (195.221.43.136) Hi!!! Is there some mistake? This is some part of one of my R programs. I created this function to show you the trouble!!! bugTest<-function() { paramnames <- c("mean","sd") param <- list(expression(1),expression(2)) paramtmp <- list(mean=0,sd=1) if(length(param))

sapply calls lapply as answer <- lapply(as.list(X), FUN, ...) which, when X is a list, causes X to be duplicated unnecessarily. The coercion is unnecessary for other mode(X) because in lapply we have if (!is.list(X)) X <- as.list(X) More generally, perhaps as.vector might not duplicate when mode(x) == mode ? Martin R version 2.4.0 Under development (unstable) (2006-07-05

Full_Name: Robert McGehee Version: 2.7.1 OS: Windows Submission from: (NULL) (192.223.226.6) R-developers, The results below seem inconsistent. From the documentation for is.numeric, I expect FALSE in both cases. > x <- data.frame(dt=Sys.Date()) > is.numeric(x$dt) [1] FALSE > sapply(x, is.numeric) dt TRUE ## Yet, sapply seems aware of the Date class > sapply(x, class) dt

FYI,=20 I've tried posting the below message twice to the bug tracking system, once by email (below), and the second time 5 days later directly to the bugs.r-project.org website. As far as I can tell, the bug tracking system hasn't picked this up. Also it looks like the latest "incoming" bug is dated 25 May 2008, so perhaps others are having difficulty as well. (cc: r-bugs)

>>>>> "RobMcG" == McGehee, Robert <Robert.McGehee at geodecapital.com> >>>>> on Tue, 29 Jul 2008 15:40:37 -0400 writes: RobMcG> FYI, RobMcG> I've tried posting the below message twice to the bug tracking system, [....... r-bugs problems discussed in a separate thread ....] RobMcG> R-developers, RobMcG> The

What about "Rtips" at http://lark.cc.ukans.edu/~pauljohn/R/statsRus.html ? Regards, Heinrich. > -----Urspr?ngliche Nachricht----- > Von: rossini at blindglobe.net [mailto:rossini at blindglobe.net] > Gesendet: Mittwoch, 26. Juni 2002 14:48 > An: r.hankin at auckland.ac.nz > Cc: r-help at stat.math.ethz.ch > Betreff: Re: [R] sapply() and Monte Carlo > > >

Ajay, thank you very much for picking up that age-old habit of posting summaries. It existed years ago on s-help and I find it is still a great thing: I would not have bothered to read your original question nor the answers you got, but I did read the summary -- and I learned something quite interesting! Maybe some others who receive multiple non-elementary answers to their questions could

Hi, All How does one remove relevant information from a regression output besides just the coefficients? I've been able to modify the example given under "help(by)" to give me some additional information, but not everything I need. If you adjust the call statement from what is listed by adding the summary statement like so: tmp <- by(warpbreaks, tension, function(x)

I have the following two mapping data frames (r) and (h). I want to fill teh value of r$seid with the value of r$seid where r$cid==h$cid. I can do it with sapply as such: > r$seid = sapply(r$cid, function(cid) h[h$cid==cid,]$seid) Is ther a better (faster) way to do this? > r <- data.frame(seid=NA, cid= c(2181,2221,2222)) > r seid cid 1 NA 2181 2 NA 2221 3 NA