similar to: simple problem

hi everybody, i try to extract a part of name to a date : like : "VGT1_VGT2_CONTR_B020030401.H0V0.MIR" to "20030401" but the beginning of the files changes i have a list of files: [,1] [1,] "VGT1_CONTR_B020020301.H0V0.MIR" [2,] "VGT1_VGT2_CONTR_B020020611.H0V0.MIR" [3,]

Hi everybody, it's a technical question . i have a matrix with 30 columns and 40000 rows, i start my program and i found this error: "error in solve.default(sBB + Z[,,i] %*% Gamma %*% t(Z[,,i])): system is computationally singular: reciprocal condition number = 7.6377e-018 " is it R limit? thanks a lot. Jonathan Charrier

Hi all, I would like to iterate through a list with named elements and access the names within an lapply / sapply call. One way to do this is iterate through the names and index the list with the name. Is there a way to iterate through the list elements themselves and access the element names within in the function? For example, mylist <-

Perhaps this toy example will help: ## example data output <- list(1:5, 1:7, 1:4) lens <- lapply(output, length) maxlen <- max(unlist(lens)) outputmod <- lapply(output, function(x, maxl) c(x, rep(NA, maxl-length(x))), maxl=maxlen) outputmat <- do.call(cbind, outputmod) write.csv(outputmat, na='') The idea is to pad the shorter vectors with NA (missing) before converting

if I have.... QQQQ<-priceIts("QQQQ",quote="Close") QQQQ<-priceIts("QQQQ",quote="Close");plot(QQQQ) and then i want to do the same thing but say with IBM instead of QQQQ is there an easy way like replace qqqq/ibm Thanks in advance./Jonathan

Within a function I'm assigning dynamic variable names and values to them using the "assign" function. I want to pass back the results but am uncertain how to do this. Basically, my function reads a number of data files and uses the filename of each file as the variable name for a list-to-become-dataframe. I want then to pass all these lists back, but again, the names of the

Hello, I have no previous experience with R, but had to learn on the fly in the past couple of weeks. Basically, what I am trying to do is read a certain variable from a series of files and save it as csv-table. The variable has an hourly value for each month in a year for the past 20 years and has to be read for different geographical locations. So there will be 12 files per year (1 for each

Dear R helpers, I wish to use the "sum" operator for each row of a data frame. However, it appears that the operator acts on the entire data frame, over all columns. What is the best way to obtain row- wise operation? The following code shows my attempts so far, and their problems:- test1=array(rbinom(120,1,0.5),c(20,3)) test1[,3]=NA sum(test1[,1:2]) test1[,3][sum(test1[,1:2])>=2]=1

Quick question -- if I have a vector of strings that I'd like to split into two new vectors based on a substring that is inside of each string, what is the most efficient way to do this? The substring that I want to split on is multiple characters, if that matters, and it is contained in every element of the character vector. --j -- Jonathan A. Greenberg, PhD Postdoctoral Scholar

I have a list of suffixes I want to turn into file names with extensions. suff<- c("C1", "C2", "C3") paste("filename_", suff[[1]], ".ext", sep="") [1] "filename_C1.ext" How do I use lapply() on that call to paste()? What's the right way to do this: filenames <- lapply(suff, paste, ...) ? Can I have lapply()

I'd appreciate some tips, I'm making a simple mistake trying to extract elements from a list of fit objects. The following command works, coef( cph.list[[1]] ) ,.... coef( cph.list[[3]] ) and so forth. These extract regression coefficients from the appropriate list element. When I try to run this inside a for loop, or an lapply (lapply(cph.list,coef)) I get this error

Rers: What is the preferred library/function for doing stratified random sampling from a dataset, given I want to control the number of samples (rather than the proportion of samples) per strata? Thanks! --j -- Jonathan A. Greenberg, PhD Postdoctoral Scholar Center for Spatial Technologies and Remote Sensing (CSTARS) University of California, Davis One Shields Avenue The Barn, Room

Hi Consider the following example: f <- function(expr) g(expr) g <- function(expr) { ? h(expr) } h <- function(expr) { ? expr # evaluation happens here ? i(expr) } i <- function(expr) { ? expr # already evaluated, no costs here ? invisible() } rprof <- tempfile() Rprof(rprof) f(replicate(1e2, sample.int(1e4))) Rprof(NULL) cat(readLines(rprof), sep = "\n") #>

Hi, I have a data frame df and a list of names of columns that I want to turn into factors: df.names <- attr(df,"names") sapply(factors, function (name) { pos <- match(name,df.names) if (is.na(pos)) stop(paste(name,": no such column\n")) df[[pos]] <- factor(df[[pos]]) cat(name,"(",pos,"):",is.factor(df[[pos]]),"\n")

> On Feb 24, 2015, at 10:50 AM, <luke-tierney at uiowa.edu> wrote: > > The documentation is not specific enough on the indented semantics in > this situation to consider this a bug. The original R-level > implementation of lapply was > > lapply <- function(X, FUN, ...) { > FUN <- match.fun(FUN) > if (!is.list(X)) > X <-

Hi, I bet this one has be asked before, but doing sapply(x, FUN=as.character) where 'x' is a vector, then the result "should [] be simplified to a vector" according to ?sapply, correct? However, > x <- 1:10 > sapply(x, FUN=as.character) [1] "1" "2" "3" "4" "5" "6" "7" "8"

On 28/03/17 15:26, Charles C. Berry wrote: > On Mon, 27 Mar 2017, Rolf Turner wrote: > >> >> From time to time I get myself into a state of bewilderment when using >> apply() by calling it with FUN equal to a function which has an >> "optional" argument named "X". >> >> E.g. >> >> xxx <-

I have a numerical vector which contains a (poorly) formatted time column, which, in theory, should be HHMM, but was distributed as an integer, so, for 12:15 am, it is saved as "15" (e.g. HHMM = 0015 with the zeroes stripped). I'm trying to use this in conjunction with strptime, but I'm thinking because each time is an integer ranging from 1 to 4 digits, I probably need to

Yesterday I posted the following question (my apologies for not putting a subject line): =================question====================== Hello -- I would like to know of a more efficient way of writing the following piece of code. Thanks. options(stringsAsFactors=FALSE) orig <- c(rep('11111111',100000),rep('22222222',200000),rep('33333333'

I am tying myself in knots over subscripts when applied to lists I have a list along the lines of: lis<-list(c("a","b","next","want1","c"),c("d", "next", "want2", "a")) >From which I want to extract the values following "next" in each member of the list, i.e. something along the lines of