similar to: Simple programming question

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, 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

R users, I don't know if I can make myself clear but I'll give it a try. I have a data.frame like this x <- "var1,var2,var3,var4 a,b,b,a b,b,c,b c,a,a,a a,b,c,c b,a,c,a c,c,b,b a,c,a,b b,c,a,c c,a,b,c" DF <- read.table(textConnection(x), header=T, sep=",") DF and I would like to sum all the combinations/occurences by a factor (letter in this case) between

Hi R-users, I'm trying to retrieve data from MS SQL database with RODBC's sqlQuery-function: temp <- sqlQuery(channel,"select *, (select text from table1 where koodi='paa' and koodi2=paa) as tempor from table2") str(temp) … $ var0 : num NA NA 1.6 NA NA 1.4 NA 1 NA NA ... $ var1 : Factor w/ 45 levels "

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, How can I extract part of an array? I would like to extract table Supported from this array. If this is not possible, how do I convert array to list? I'm sorry this is not an reproducible example. > spl <- tapply(temp$var1, list(temp$var2, temp$var3, temp$var3), mean) > spl , , Supported 07 08 A 68.38710 71.48387 B 21.67742 20.83871 C

Hello, I'm trying to split my DF into a list using incremental loop. How can I avoid NULL elements in this list? DF <- data.frame(var1 = 1:10, var2 = 11:20, var3 = 21:30, var4 = 31:40) x <- list() i <- 1 while (i <= ncol(DF)-1) { x[[i]] <- DF[, i:c(i+1)] i <- i + 2 } x Many thanks, Lauri

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, 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

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

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 R-users, I have a data frame like this: photographer category picture points Hannu kalat limamikko 1 Teemu kalat verkkovaja 3 Hate kalat munat puoliks padassa 6 Hannu kalat isokala 8 Teemu kasvit, sienet ja muut eliöt harppi 2 Hate kasvit, sienet ja muut eliöt pyynikki 2 Petteri kasvit, sienet ja muut eliöt harmaaleppä 5 Lauri kasvit, sienet ja muut eliöt lumipuu 9 Teemu linnut kainostelua 1

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, 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

Hi R-users, I have a dataset like this: kuvaaja kuva yhteispisteet Hannu isokala 8 Hannu kaapin alta löytynyt 2 Hannu kaapin alta löytynyt 2 8 Hannu limamikko 1 Hannu maukasta marmeladia 8 Hannu skrinnareita 4 Hate madekoukkujen suojelupyhimys 3 Hate matka aikaan joka ei enää palaa 3 Hate munat puoliks padassa 6 Hate pyynikki 2 Hate vailla armeerausta 2