similar to: R Help-Implicit loop-lapply

Hi, Try: dat2<- read.csv("BOlValues.csv",header=TRUE,sep="\t",row.names=1) dim(dat2) #[1] 20 28 indx2<-expand.grid(names(dat2),names(dat2),stringsAsFactors=FALSE) nrow(indx2) #[1] 784 indx2New<- indx2[indx2[,1]!=indx2[,2],] nrow(indx2New) #[1] 756 res2<-sapply(seq_len(nrow(indx2New)),function(i) {x1<- indx2New[i,];

Dear all, I am processing a very long and complicated list using lapply through a custom function and I would like to generate some sort of progress report. For instance, print a dot on the screen every time 1000 item have been process. Or even better, reporting the percent of the list that have been process every 10%. However, I can't seem to figure out a way to achieve that. For instance,

HI Irucka, Please check this: temp<- structure(list(`:Bostoncitydata` = structure(list(Month = c(1L, 2L, 3L, NA), Data1 = c(1.5, 12.3, 11.4, NA), Data2 = c(9.1342, 12.31, 3.5, NA)), .Names = c("Month", "Data1", "Data2"), class = "data.frame", row.names = c(NA, -4L)), `:Chicagocitydata` = structure(list(Month = c(1L, 2L, 3L, 4L, 5L, NA), Data1 = c(1.52,

There are a few version of apply() (e.g., lapply(), sapply()). I'm wondering if there is one that does not return anything but just silently apply a function to the list argument. For example, the plot function is applied to each element in 'alist'. It is redundant to return anything from apply. apply(alist,function(x){ plot each element of alist})

Dear R developers, While improving duplicated.array() and friends and developing equivalents for the new ff package for large datasets I came across two questions: 1) is it safe to use duplicated.default(), unique.default() and match() on arbitrary lists? If so, we can speed up duplicated.array and friends considerably by using list() instead of paste(collapse="\r") 2) while

Hi, Try this: #mat1 is the data res<-do.call(cbind,lapply(seq_len(nrow(mat1)),function(i) {new1<-do.call(rbind,lapply(seq_len(nrow(mat1[-i,])),function(j) {x1<-rbind(mat1[i,],mat1[j,]); x2<-(abs(x1[1,1]-x1[2,1])*abs(x1[1,5]-x1[2,5]))+(abs(x1[1,2]-x1[2,2])*abs(x1[1,6]-x1[2,6]))+(abs(x1[1,3]-x1[2,3])*abs(x1[1,7]-x1[2,7]))+(abs(x1[1,4]-x1[2,4])*abs(x1[1,8]-x1[2,8]))}));new1}))

HI, Using c11<- 0.01 c12<- 0.01 c1<- 0.10 c2<- 0.10 One possible problem is that: dim(res5) #[1] 513? 20 res6<-aggregate(.~m1+n1+m+n,data=res5[,c(1:6,9:12,21:24)] ,max) #Error in `[.data.frame`(res5, , c(1:6, 9:12, 21:24)) : ?# undefined columns selected A.K. ________________________________ From: Joanna Zhang <zjoanna2013 at gmail.com> To: arun <smartpink111 at

Dear all, let say I have following list: Dat <- vector("list", length = 26) names(Dat) <- LETTERS My_Function <- function(x) return(rnorm(5)) Dat1 <- lapply(Dat, My_Function) However I want to apply my function 'My_Function' for all elements of 'Dat' except the elements having 'names(Dat) == "P"'. Here I have specified the name

Dear all, # 1. I use tapply to calculate various things based on a grouping variable. # simple example using the warpbreaks dataset: tmp <- with(warpbreaks, tapply(breaks, tension, range)) tmp dimnames(tmp) # 2. I wish to plot the result of each element of tmp, using the names of the elements in tmp as 'main' in each plot # My attempt: # par(mfrow = c(3, 1)) # my preferred

Dear helpers. I often need to make dichotomous variables out of continuous ones (yes, I realize the problems with throwing away much of the information), but I then like to check the min and max of each category. I have the following simple code to do this that cuts each variable (x1,x2,x3) at the 90th percentile, and then prints the min and max of each category:

Hello: Forgive me, this is surely a simple question but I can't figure it out, having consulted the help archives and "Data Manipulation With R" (Spector). I have a list of 11 data frames with one common variable in each (prov). I'd like to use lapply to go through and recode one particular level of that common variable. I can get the recode to work, but it only returns the

Hi Gustav, Try this: lapply(1:length(models),function(i) lapply(models[[i]],function(x) summary(x)$coef[2,]))[[1]] #1st list component [[1]] #??? Estimate?? Std. Error????? z value???? Pr(>|z|) # pm10 #5.999185e-04 1.486195e-04 4.036606e+00 5.423004e-05 #[[2]] #??? Estimate?? Std. Error????? z value???? Pr(>|z|) #ozone #0.0010117294 0.0003792739 2.6675428048 0.0076408155 #[[3]] #???

Hi, Try this: el<- read.csv("el.csv",header=TRUE,sep="\t",stringsAsFactors=FALSE) ?elsplit<- split(el,el$st) ? datetrial<-data.frame(date1=seq.Date(as.Date("1930.1.1",format="%Y.%m.%d"),as.Date("2010.12.31",format="%Y.%m.%d"),by="day")) elsplit1<- lapply(elsplit,function(x)

Hi, directory<- "/home/arunksa111/data.new" #first function filelist<-function(directory,number,list1){ setwd(directory) filelist1<-dir(directory) direct<-dir(directory,pattern = paste("MSMS_",number,"PepInfo.txt",sep=""), full.names = FALSE, recursive = TRUE) list1<-lapply(direct, function(x) read.table(x,header=TRUE, sep =

>I think that the suggestion I made, in response to a posting by Barry >Rowlingson, that the first argument of lapply() be given the name of ".X" rather >than just-plain-X, would be (a) effective, and (b) harmless. It would break any call to *apply() that used X= to name the first argument. There are currently 3020 such calls in the R code in CRAN. One can avoid the problem

Using the input below, can I do something more elegant (and more efficient) than the loop also listed below to pad strings to a width of 5? The true matrix is about 300K rows and 31 columns. ####################### #INPUT ####################### > temp DX1 DX2 DX3 1 13761 8125 49178 2 63371 v75 22237 3 51745 77703 93500 4 64081 32826 v72 5 78477 43828 87645 >

Hi, Try: x<- 1:4 ?y<- c("*","/","-","+") res<-sapply(y,function(i) {x1<-expand.grid(x,x); unlist(lapply(paste0(x1[,1],i,x1[,2]),function(u) eval(parse(text=u))))}) row.names(res)<- as.character(interaction(expand.grid(x,x),sep="_")) head(res) #??? *?? /? - + #1_1 1 1.0? 0 2 #2_1 2 2.0? 1 3 #3_1 3 3.0? 2 4 #4_1 4 4.0? 3 5 #1_2 2 0.5

The documentation is not specific enough on the indented semantics in this situation to consider this a bug. The original R-level implementation of lapply was lapply <- function(X, FUN, ...) { FUN <- match.fun(FUN) if (!is.list(X)) X <- as.list(X) rval <- vector("list", length(X)) for(i in seq(along = X)) rval[i]

Actually, it depends on the number of cores: > fun1 <- function(c){function(i){c*i}} > fun2 <- function(f) f(2) > sapply(mclapply(1:4, fun1, mc.cores=1L), fun2) [1] 8 8 8 8 > sapply(mclapply(1:4, fun1, mc.cores=2L), fun2) [1] 6 8 6 8 > sapply(mclapply(1:4, fun1, mc.cores=4L), fun2) [1] 2 4 6 8 > >/ On Feb 24, 2015, at 10:50 AM,

Looks like I?m bumping a lot into unexpected behaviour lately, but I think I found a bug again, but don?t have access to Bugzilla: Write.table (from core-package utils) doesn?t handle nested data.frames well, the quote arguments only marks top-level character (or-factor columns) for quoting, so this fails: df <- data.frame(a='One;Two;Three',