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I have a text file similar to this (separated by spaces): x <- "DF12 This is an example 1 This DF12 This is an 1232 This is DF14 This is 12334 This is an DF15 This 23 This is an example " and I know the field lengths of each variable (there is 5 variables in this data set), which are: varlength <- c(2, 2, 18, 5, 18) How can I import this kind of data into R, using the varlength

R users, I have a simple lapply question. g <- list(a=1:3, b=4:6, c=7:9) g <- lapply(g, function(x) as.data.frame(x)) lapply(g, function(x) cbind(x, var1 = rep(names(g), each=nrow(x))[1:nrow(x)])) I get $a x var1 1 1 a 2 2 a 3 3 a $b x var1 1 4 a 2 5 a 3 6 a $c x var1 1 7 a 2 8 a 3 9 a And I would like to have $a x var1 1 1 a 2 2 a 3 3 a

Dear list, I tried to find a solution for this problem from the archives but couldn't find any. I would like sample sequence of ten days from vector d d <- seq(as.Date("2007-02-12"), as.Date("2008-08-18"), by="days") so that the days follow each other (sample(d, 10) is not the appropriate solution). Any ideas? Best regards, Lauri

R-users, Is there a function for ceiling to the nearest ten? a <- 1:10*4 a [1] 4 8 12 16 20 24 28 32 36 40 The resulting vector should look like this ("ceiling to the nearest ten") [1] 10 10 20 20 20 30 30 40 40 40 Thanks in advance Lauri

Hello, I have difficulties combining boxes and lines in plot legend. I searched previous R-posts and found this (with no solution): http://tolstoy.newcastle.edu.au/R/help/06/07/30248.html. Is there a way to avoid boxes behind the line legends? x1 <- rnorm(100) x2 <- rnorm(100, 2) hist(x1, main = "", col = "orange",ylab = "density", xlab = "x", freq =

Hi R-users, I have a simple question for R heavy users. If I have a data frame like this dfr <- data.frame(id=1:16, categ=rep(LETTERS[1:4], 4), var3=c(8,7,6,6,5,4,5,4,3,4,3,2,3,2,1,1)) dfr <- dfr[order(dfr$categ),] and I want to score values or points in variable named "var3" following this kind of logic: 1. the highest value of var3 within category (variable named

Hi R-users, What is the best way to achieve a table which contains all days and months between years 2007-2020? I would like to calculate number of days in each month within those years (to data frame). Regards, Lauri [[alternative HTML version deleted]]

R users, I have a vector of dates days <- seq(as.Date("2007/1/1"), as.Date("2008/1/31"), "days") and I would like to have week numbers from 1 to 52 for each year. How do I do that? Now I get 00-53 using format(days, "%W") > range(format(days, "%W")) [1] "00" "53" I have read "Date and Time Classes in R" (R

Hello, I'm producing text from my data.frame using cat function. I would like to use for loop to export each column in my data.frame into separate text files. Here is the example code r <- t(Indometh) for (i in 1:ncol(r)) { cat("Some text,", "\n") cat("\n") cat("More text, More text, More text") cat("\n")

Hi, Suppose I have a data.frame like this Lines <- "var1 var2 var3 var4 var5 var6 0 2 1 2 0 0 2 3 7 6 0 1 1.5 4 9 9 6 0 1.0 6 10 22 3 3 " DF <- read.table(textConnection(Lines), skip=1) names(DF) <- scan(textConnection(Lines), what = "", nlines = 1) How do I find the

Hi, I'm trying to read several Excel sheets from an Excel file into a list. I'm using read.xls from package 'gdata'. I would like to know how I can check the number of sheets before the loop (in the example below) so that I could adjust the loop counter? Any suggestions? DF.list <- list() for (i in 1:4) { DF.list[[i]] <- read.xls("sample_file.xls", sheet=i,

Hi R-users, Following this post http://tolstoy.newcastle.edu.au/R/help/06/06/28965.html , how do I get last two rows (or six or ten) by id group out of the data frame? Here the example gives just the last row. Sincere thanks, Lauri [[alternative HTML version deleted]]

Hi R-users, How do I calculate a number of NA's in a row of every second column in my data frame? As a starting point: dfr <- data.frame(sapply(x, function(x) sample(0:x, 6, replace = TRUE))) dfr[dfr==0] <- NA So, I would like to count the number of NA in row one, two, three etc. of columns X1, X3, X5 etc. Thanks in advance Lauri [[alternative HTML version deleted]]

Hi, I have a data.frame like this: y <- rnorm(60) lev <- gl(3,20, labels=paste("lev", 1:3, sep="")) date1 <- as.Date(seq(ISOdate(2007,9,1), ISOdate(2007,11,5), by=60*60*24)) date1 <- date1[-c(3,4,15,34,38,40)] df <- data.frame(lev=lev, date1=date1, y=y) I would like to produce a new data.frame with missing days in df$date1 in each df$lev, like this: lev

...imes from previous runs of your code that failed). If your code is inside a function, wrap the code in point 1 and 2 on an on.exit() call to ensure that excel is properly closed each time. Hope this helps. Enrique ------------------------------ Date: Wed, 8 Jul 2009 16:06:57 +0300 From: Lauri Nikkinen <lauri.nikkinen at iki.fi> Subject: [R] RDCOMClient: how to close Excel process? To: r-help at stat.math.ethz.ch Message-ID: <ba8c09910907080606n29b8a537uac5ba1788376ff42 at mail.gmail.com> Content-Type: text/plain; charset=windows-1252 Hi, I?m using R package RDCOMClient (http://www...

Hi, I installed R on USB-drive, but when I run Rgui.exe from bin folder, I get this error: ----------------------- R version 2.10.1 (2009-12-14) Copyright (C) 2009 The R Foundation for Statistical Computing ISBN 3-900051-07-0 R is free software and comes with ABSOLUTELY NO WARRANTY. You are welcome to redistribute it under certain conditions. Type 'license()' or 'licence()' for

...pply(tmp1,tmp2,sum)) and I want the levels from tmp2 be shown as a column in the data.frame, not as row name as it now does. To put it in another way, as a result, I want a data frame with two columns: levels and the sums of those levels. Row names can be, for example, numbers from 1 to 10. -Lauri Nikkinen Lahti, Finland [[alternative HTML version deleted]]

Hi R-users, I have a list a <- list(one=matrix(rnorm(20), 5, 4), two=matrix(rnorm(20, 3, 0.5),5,4)) How to add rowSums (calculated using lapply) to corresponding matrix in this list lapply(a, function(x) rowSums(x)) ?? -Lauri

Hello, I have a vector of dates and I would like to grep the year component from this vector (= all digits after the last punctuation character) dates <- c("28.7.08","28.7.2008","28/7/08", "28/7/2008", "28/07/2008", "28-07-2008", "28-07-08") the resulting vector should look like "08" "2008"

R users, I have df like this a <- data.frame( indx = 1:20, var1 = rep(c("I20", "I40", "A50", "B60"), each=5), var1_lab= rep(c("cat", "dog", "mouse", "horse"), each=5), var2 = rep(c("B20", "X40", "D50",