similar to: lapply - function with arguments

Dear R helpers: I wonder how to pass more than one argument to the function called by lapply. For example, #R code below --------------------------- indf <- data.frame(id=I(c('a','b')),y=c(1,10)) #I want to add an addition argument cutoff into the function called by lapply. outside.fun <- function(indf, cutoff) { unlist(lapply(split(indf, indf[,'id']),

I have split my original dataframe to generate a list of dataframes each of which has 3 columns of factors and a 4th column of numeric data. I would like to use lapply to apply the fitdistr() function to only the 4th column (x$isi) of the dataframes in the list. Is there a way to do this or am I misusing lapply? As a second solution I tried splitting only the numeric data column to yield a list

Dear all, I am using lapply to generate plots by applying a plot function to a list of dataframes. e.g. lapply(dataSet, FUN = plotFunction) Is there a way to call the list item number inside the plot function so that I can identify each graph? Thanks Chibisi [[alternative HTML version deleted]]

In the following code, I'd like to be able to create a new variable containing the value of the names of the list. a <- data.frame(var.1 = 1:5) b <- data.frame(var.1 = 11:15) test.list <- list(a=a, b=b) # in this case, names(test.list) is "a" and "b" # and I'd like to use lapply() so that # I get something that looks like # var.1 var.2 # 1 a # 2

Hello everybody, I have problem with a lapply command, which rather proves that I don't fully understand it. I want to extract from a list that consists of dataframes, the length of the first sequences from a given variable (its part of a simulation exercises). Below is code which does the job, but I think it should be possible to make it more compact. ### Example Data dat <-list()

Hi All, I'm trying to understand the difference between do.call and lapply for applying a function to a list. Below is one of the variations of programs (by Marc Schwartz) discussed here recently to select the first and last n observations per group. I've looked in several books, the R FAQ and searched the archives, but I can't find enough to figure out why lapply doesn't do what

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 >

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:

After a year my R programming style is still very "C like". I am still writing a lot of "for loops" and finding it difficult to recognize where, in place of loops, I could just do the same with one line of code, using "sapply", "lapply", or the like. On-line examples for such high level function do not help me. Even if, sooner or later, I am getting my R

So I have a function that does lapply's for me based on dimension. Currently only works for length(pivotColumns)=2 because I haven't fixed the rbinds. I have two versions. One runs WAYYY faster than the other. And I'm not sure why. Fast Version: fedb.ddplyWrapper2Fast <- function(data, pivotColumns, listNameFunctions, ...){ lapplyFunctionRecurse <- function(cdata, level=1,

Hi all, I am trying to find a way to rename R objects with names pulled from a vector of names. For example, I have a data frame, my.data.frame, and a list of names, my.names. My.names is simply the column names of my.data.frame. I want save the histogram with the column name as the name of the object. for (i in 1:ncol(my.data.frame) { tmp<-hist(my.data.frame[,i])

Suppose I have a nx2 matrix of data, X, the following code generate density estimation for each column and plot them denlist <- apply(X, 2, density) par(mfrow=c(1,2)) lapply(denlist, plot) Does anyone know how to change the main title of each density plot to "var 1", "var 2" by passing optional argument "main"? I've tried lapply(denlist, plot,

If anyone's interested, especially client developers. It's been a bit quiet there after the initial rush. Begin forwarded message: > From: Randall Gellens <randy at qualcomm.com> > Date: August 1, 2008 5:08:39 AM EDT > To: ietf-imapext at imc.org, morg at ietf.org > Subject: [MORG] IMAP5 List > > At the MORG BOF, a discussion as to if the proposed IMAP

Hi all, I have a data frame called "rst", see below: ------------------------------------------------------------------------------------------ # This is a paste able example # In case you don't have "KernSmooth" package installed, please uncomment below line. # install.packages("KernSmooth") library(KernSmooth) rst <- data.frame(hsp = rnorm(23), dal =

Hi, I've tried to figure this out using Intro to R and help(), to no avail - I am new at this. I'm trying to write a script that will read multiple files from a directory and then merge them into a single new data frame. The original data are in a tree-ring specific format, and so I've first used a function (read.rwl) from the dplR package to read each file, translate each into a

Hello, I am facing a problem with lapply which I ''''think''' may be a bug. This is the most basic function in which I can reproduce it: myfun <- function() { foo = data.frame(1:10,10:1) foos = list(foo) fooCollumn=2 cFoo = lapply(foos,subset,select=fooCollumn) return(cFoo) } I am building a list of dataframes, in each of which I want to keep only column

Greetings, I am not familiar with processing text in R. Can someone tell me how to read each line of words as separate elements in a list? FE, I would like to turn: word1 word2 word3 word2 word4 into a list of length two with three character elements in the first list and two elements in the second. I know that this should be easy, but I am a little confused by the text functions. Thanks in

Hi, anyone interested in this: I tried to simply this loop with lapply or something but haven't figured it out: mapt = c("203929_s_at", "203930_s_at", "203928_x_at", "206401_s_at") mapt.combn <- lapply(1:4, function(i) combn(mapt, i)) out = list() k = 1 for (i in 1:length(mapt.combn)){ for (j in 1:ncol(mapt.combn[[i]])){ out[[k]] =

Dear R users: Is there a way to append selectively to components of a list (if possible, loops are to be avoided)? To illustrate the point, in the example below, I would like to append 99 to vector b of the list l. > l <- list(a=c(1), b=c(2,3), c=c(4,5,6)) > l $a [1] 1 $b [1] 2 3 $c [1] 4 5 6 As you may expect, the result should look like: > l $a [1] 1 $b [1] 2 3 99 $c [1] 4 5

Suppose I have a list of logicals, such as returned by lapply: Theoph$Dose[1] <- NA Theoph$Time[2] <- NA Theoph$conc[3] <- NA lapply(Theoph,is.na) Is there a direct way to execute logical "or" across all vectors? The following gives the desired result, but seems unnecessarily complex. as.logical(apply(do.call("rbind",lapply(Theoph,is.na)),2,"sum"))