similar to: A problem with text manipulation

Hello again, Let say I have following vector: Vec <- c("0.0365780769", "(1.09738648244378)", "(0.812507787221523)", "0.5778069963", "(0.452456601362355)", "-1.8900812605", "-1.8716093762", "0.0055217041", "-0.4769192333", "-2.4133018880") Now I want to convert this vector to numeric vector. I

Hello again, Let say I have following string: Vec <- c("sada", "asdsa", "sa") Now I want to make each element of this vector with equal length. Basically I want following vector: c("sada ", "asdsa", "sa ") Therefore we can get: > nchar(c("sada ", "asdsa", "sa ")) [1] 5 5 5 Is there any

Hello again, Let say I have an non-negative integer vector (which may be random): Vec <- c(0, 13, 10, 4) And I have a date: > Date <- as.Date(Sys.time()) > Date [1] "2013-03-09" Using these 2 information, I want to get following date-vector: New_Vec <- c("2013-03-01", "2014-04-01", "2014-01-01", "2013-07-01") Basically the

Hello again, Let say I have a matrix: Mat <- matrix(1:12, 4, 3) And a vector: Vec <- 5:8 Now I want to do following: Each element of row-i in 'Mat' will be divided by i-th element of Vec Is there any direct way to doing that? Thanks for your help

Hi all, let say I have following matrix: > Dat <- matrix(1:30, 5, 6); colnames(Dat) <- rep(c("Name1", "Names2"), 3) > Dat Name1 Names2 Name1 Names2 Name1 Names2 [1,] 1 6 11 16 21 26 [2,] 2 7 12 17 22 27 [3,] 3 8 13 18 23 28 [4,] 4 9 14 19 24 29 [5,] 5

Dear all: I have this: A1 B1 C1 D1 E1 A2 B2 C2 D2 E2 A3 B3 C3 D3 E3 And I want this A1 E1 B1 E1 C1 E1 D1 E1 A2 E2 B2 E2 C2 E2 D2 E2 A3 E3 B3 E3 C3 E3 D3 E3 Example: m<- matrix(1:15,nrow=3,byrow=T) m v<- unlist(list(t(m[,1:4]))) u<- rep(c(5,10,15),c(4,4,4)) data.frame(v,u) This is the result I want but I would like to learn a simpler way to do it. Any clue?

I happened to see these: > round(.5, 0) [1] 0 > round(1.5, 0) [1] 2 > round(2.5, 0) [1] 2 > round(3.5, 0) [1] 4 > round(4.5, 0) [1] 4 What is the rule here? Should not round(.5, 0) = 1, round(2.5, 0) = 3 etc? Thanks and regards,

Dear all, Let say I have following data-frame: Dat <- structure(list(dat = c(-0.387795842956327, -0.23270882099043, -0.89528973290562, 0.95857175595512, 1.61680582493783, -1.17738110289352, 0.210601060411423, -0.827369747447338, -0.36896112964414, 0.440288648776096, 1.28018410608809, -0.897113649961341, 0.342216546981718, -1.17288066266219, -1.57994101992621, -0.913655547602414,

#I have a data set that looks like this. A bit more complicated actually with # three factor levels but these calculations need to be done on one factor at a #I then have a set of different rates that are applied #to it. #dataset cata <- c( 1,1,6,1,1,2) catb <- c( 1,2,3,4,5,6) doga <- c(3,5,3,6,4, 0) data1 <- data.frame(cata, catb, doga) rm(cata,catb,doga) data1 # start rates #

Hi, In ?shapiro.test, there seems to be no option to pass mean and sd information of the Normal distribution which I want to compare my sample data to. For example in the code below, I want to test my sample to N(0, 10). shapiro.test(rnorm(100, mean = 5, sd = 3)) Is there any way to pass the information of the benchmark normal distribution?

Dear all, Let say I have following data: RawData <- matrix(1:101, nr = 1); colnames(RawData) <- c("ASD", as.character(as.yearmon(seq(as.Date("2012-03-01"), length.out = 100, by = "1 month")))); rownames(RawData) <- "XYZ" CutOffDate <- as.Date("2012-09-01") NewDateSeries <- as.character(as.yearmon(seq(CutOffDate, to =

Dear all, let say I have following data: dat <- structure(list(V1 = structure(c(1L, 4L, 5L, 3L, 3L, 5L, 6L, 6L, 4L, 3L, 5L, 6L, 5L, 5L, 4L, 4L, 6L, 2L, 3L, 4L, 3L, 3L, 2L, 5L, 3L, 6L, 3L, 3L, 6L, 3L, 6L, 1L, 6L, 5L, 2L, 2L), .Label = c("C", "G", "I", "O", "R", "T"), class = "factor"), V2 = c(0L, 0L, 0L, 1L, 1L, 1L, 1L, 0L,

See if the following helps: > m <- outer(letters[1:5], 1:4, paste, sep="") > m [,1] [,2] [,3] [,4] [1,] "a1" "a2" "a3" "a4" [2,] "b1" "b2" "b3" "b4" [3,] "c1" "c2" "c3" "c4" [4,] "d1" "d2" "d3" "d4" [5,]

Dear all:   I have this:  A1 B1 C1 D1 E1 A2 B2 C2 D2 E2 A3 B3 C3 D3 E3   And I want this A1 E1 B1 E1 C1 E1 D1 E1 A2 E2 B2 E2 C2 E2 D2 E2 A3 E3 B3 E3 C3 E3 D3 E3   Example:   m<- matrix(1:15,nrow=3,byrow=T) m v<- unlist(list(t(m[,1:4]))) u<-

Hi, Suppose, if I create 15 files in my working directory. set.seed(48) lapply(1:15,function(i) {m1 <- matrix(sample(1:20,1686*2,replace=TRUE),nrow=1686,ncol=2); write.table(m1,paste0("file_",i,".txt"),row.names=FALSE,quote=FALSE)}) ?D <-dir() D1 <- D[order(as.numeric(gsub("\\D+","",D)))] D1 ?res <- t(sapply(D1,function(x) {x1<-

Dear all, I am trying to define "+" method for my newly defined s4 class which is as follows: setClass("Me", sealed=F,representation(x1 = "numeric", x2 = "character")) new1 <- new("Me", x1=2, x2="comment1") new2 <- new("Me", x1=3, x2="comment1") setMethod("+", "Me",

Below is my full implementation (tried to make it simple as for demonstration) Lapply_me = function(X = X, FUN = FUN, Apply_MC = FALSE, ...) { if (Apply_MC) { return(mclapply(X, FUN, ...)) } else { if (any(names(list(...)) == 'mc.cores')) { myList = list(...)[!names(list(...)) %in% 'mc.cores'] } return(lapply(X, FUN, myList)) } } Lapply_me(as.list(1:4), function(xx) { if (xx ==

Hi all, please see my code: > library(zoo) > a <- as.yearmon("March-2010", "%B-%Y") > b <- as.yearmon("May-2010", "%B-%Y") > > nn <- (b-a)*12 # number of months in between them > nn [1] 2 > as.integer(nn) [1] 1 What is the correct way to find the number of months between "a" and "b", still

The reason that it works for Apply_MC=TRUE is that in that case you call mclapply(X,FUN,...) and the mclapply() function strips off the mc.cores argument from the "..." list before calling FUN, so FUN is being called with zero arguments, exactly as it is declared. A quick workaround is to change the line Lapply_me(as.list(1:4), function(xx) { to Lapply_me(as.list(1:4),

My modified function looks below : Lapply_me = function(X = X, FUN = FUN, Apply_MC = FALSE, ...) { if (Apply_MC) { return(mclapply(X, FUN, ...)) } else { if (any(names(list(...)) == 'mc.cores')) { myList = list(...)[!names(list(...)) %in% 'mc.cores'] } return(lapply(X, FUN, myList)) } } Here, I am not passing ... anymore rather passing myList On Sun, Mar 4, 2018 at 10:37 PM,