similar to: How to recode a factor level (within the list)?

#Hi R-users, #Suppose that I have a data.frame like this: y1 <- rnorm(10) + 6.8 y2 <- rnorm(10) + (1:10*1.7 + 1) y3 <- rnorm(10) + (1:10*6.7 + 3.7) y <- c(y1,y2,y3) x <- rep(1:3,10) f <- gl(2,15, labels=paste("lev", 1:2, sep="")) g <- seq(as.Date("2000/1/1"), by="day", length=30) DF <- data.frame(x=x,y=y, f=f, g=g) DF$g[DF$x == 1]

#Hi R-users, #I have an example DF like this: y1 <- rnorm(10) + 6.8 y2 <- rnorm(10) + (1:10*1.7 + 1) y3 <- rnorm(10) + (1:10*6.7 + 3.7) y <- c(y1,y2,y3) x <- rep(1:3,10) f <- gl(2,15, labels=paste("lev", 1:2, sep="")) g <- seq(as.Date("2000/1/1"), by="day", length=30) DF <- data.frame(x=x,y=y, f=f, g=g) DF$wdays <- weekdays(DF$g)

Hi folks, I would like to trim the trailing spaces in my factor variables using lapply (described in this post by Marc Schwartz: http://tolstoy.newcastle.edu.au/R/e2/help/07/08/22826.html) but the code is not functioning (in this example there is only one factor with trailing spaces): y1 <- rnorm(20) + 6.8 y2 <- rnorm(20) + (1:20*1.7 + 1) y3 <- rnorm(20) + (1:20*6.7 + 3.7) y <-

Hi R-users, I have a data.frame like this (modificated from https://stat.ethz.ch/pipermail/r-help/2007-August/138124.html). y1 <- rnorm(20) + 6.8 y2 <- rnorm(20) + (1:20*1.7 + 1) y3 <- rnorm(20) + (1:20*6.7 + 3.7) y <- c(y1,y2,y3) x <- rep(1:5,12) f <- gl(3,20, labels=paste("lev", 1:3, sep="")) d <- data.frame(x=x,y=y, f=f) and this is how I can

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

Dear list members, I have problems to interpret the coefficients from a lm model involving the interaction of a numeric and factor variable compared to separate lm models for each level of the factor variable. ## data: y1 <- rnorm(20) + 6.8 y2 <- rnorm(20) + (1:20*1.7 + 1) y3 <- rnorm(20) + (1:20*6.7 + 3.7) y <- c(y1,y2,y3) x <- rep(1:20,3) f <- gl(3,20,

Hi R-experts, suppose I have a list with containing data frame elements: [[1]] (Intercept) y1 y2 y3 y4 -6.64 0.761 0.383 0.775 0.163 [[2]] (Intercept) y2 y3 -3.858 0.854 0.834 Now I want to put them into ONE dataframe like this: (Intercept) y1

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 intention is to calculate some summary statistics across factor levels. I know that in Hmisc package there is a summary function which produces neat summary statistics when using "cross" option. I would like to produce similar output with N and Missing columns but produce a data.frame. Is there any built-in function for that? #example data install.packages("Hmisc")

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

Dear R-users; Previously I posted a question about the problem of rank deficiency in summary.manova. As somebody suggested, I'm attaching a small part of the data set. #*************************************************** "test" <- structure(.Data = list(structure(.Data = c(rep(1,3),rep(2,18),rep(3,10)), levels = c("1", "2", "3"), class =

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]]

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]]

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, 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

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")