similar to: splitting character vectors into multiple vectors using strsplit

Context: I'm relatively new to R and am working with very large datasets. General problem: If working on a dataset requires that I produce more than two objects of roughly the size of the dataset, R quickly uses up its available memory and slows to a virtual halt. My tentative solution: To save and remove objects as they're created, and load them when I need them. To do this I'm

Hi, I'm turning my wheels on this and keep coming around to the same wrong solution - please have a look and give a hand ... The premise is: a DF like so > loremIpsum <- "Lorem ipsum dolor sit amet, consectetur adipiscing elit. Quisque leo ipsum, ultricies scelerisque volutpat non, volutpat et nulla. Curabitur consequat ullamcorper tellus id imperdiet. Duis semper malesuada

On Tuesday, March 07, 2000 5:40 PM, Richard Bilonick wrote: >I need to merge several data.frames into one data.frame. In S-Plus I would >use >"merge" but I don't see a merge command in R. What is the best way to >accomplish >this? The easiest way to to this, I think, is as follows: if you have several data frames, dataframe1, dataframe2, . . . , dataframen, you

Hello fellow R people, I can not figure out a pretty way to use strplit with vectors Imagine that I got the following data from someone with ID's representing several factors ID data A1-B1-t1 0 A1-B1-t2 1 A1-B2-t1 5 A1-B2-t2 10 A1-B10-t1 0 A1-B10-t2 1 A1-B20-t1 5 A1-B20-t2 10 ... I would like to turn this dataframe to station substation time data A1 B1 t1 0 A1 B1 t2 1 A1 B2 t1 5

Hello, Sorry if anyone gets this message twice, as my mailserver may not be working. Thanks for your response. Your idea makes a lot of sense to me, but I've been unable to get seconds to work. I ended up with this format finally: "2007-10-31_16:20:22" Problem is I am unable to get it recognized as a date using timeDate(): R>

Hello, I am new to R. I often collect data at multiple sites and need to create separate graphs (such as scatterplots or histograms) of specific variables for each site. I have tried to do this by splitting the data frame and then using lapply, but it seems that the graphing commands cannot be called as functions. Here is a sample of my data, called "seeddist2": site

Dear R-Devel List, **TL;DR:** The function **parLapplyLB** of the parallel package has [reportedly][1] (see also attached RRD output) not been doing its job, i.e. not actually balancing the load. My colleague Dirk Sarpe and I found the cause of the problem and we also have a patch to fix it (attached). A similar fix has also been provided [here][2]. [1]:

Hi, I'm trying to understand the rationale for your proposed amount of splitting and more precisely why that one is THE one. If I put labels on your example numbers in one of your previous post: nbrOfElements <- 97 nbrOfWorkers <- 5 With these, there are two extremes in how you can split up the processing in chunks such that all workers are utilized: (A) Each worker, called

Dear Christian and Henrik, thank you for spotting the problem and suggestions for a fix. We'll probably add a chunk.size argument to parLapplyLB and parLapply to follow OpenMP terminology, which has already been an inspiration for the present code (parLapply already implements static scheduling via internal function staticClusterApply, yet with a fixed chunk size; parLapplyLB already

I wasn't even aware I was using midnightStandard. You won't find it in my script. Here is the relevant loop: date1 = timeDate(charvec = Sys.Date(), format = "%Y-%m-%d") date1 dow = 3; for (i in 1:length(V4) ) { x = read.csv(as.character(V4[[i]]), header = FALSE, na.strings=""); y = x[,1]; year = V2[[i]]; week = V3[[i]]; dtstr =

Hi folks. This set of code used to work, but after upgrading to the latest version of Rmetrics it no longer does. Any ideas? SP500<-read.table("SP500.csv",header=TRUE,sep=",") > head(SP500) Date Open High Low Close Volume Close2 1 8/4/2006 1280.26 1292.92 1273.82 1279.40 2530970112 1279.40 2 8/3/2006 1278.22 1283.96 1271.25 1280.27

I need to multiply to very large, nonsparse matrices, and so get the error "allocMatrix: too many elements specified". Is there a way to set the limit for allocMatrix? In my case, the two matrices, A and B, are nxm and mxp where m is small, so I could subdivide each into blocks of submatrices A=rbind(A1,A2,...) and B=cbind(B1,B2,...) then multiply each pair of submatrices, but I was

Dear R users, I have a dataframe where column is has countries, column 2 is dates (monthly) for each countrly, the next 10 columns are my factors where I have measurements for each country and for each date. I have attached a sample of the data in csv format with the data for 3 countries. I would like to convert my monthly data into quarterly data, finding the mean over 3 month periods for

For a model I am working on, I have samples organized by year and week of the year. For this model, the data (year and week) comes from the basic sample data, but I require a value representing the amount of time since the sample was taken (actually, for the purpose of the model, it is sufficient to use the number of weeks from the middle of the sample week to the present). What I have found so

Thank you for your answer! I agree with you except for the 3 (Error) example and I realize now I should have started with that in the explanation. >From my point of view parLapply(cl = clu, X = 1:2, fun = fun, c = 1) shouldn't give an error. This could be easily avoided by using all the argument names in the custerApply call of parLapply which means changing, parLapply <-

Dear R users, I have a dataframe with one character field, and I would like to create two new fields (columns) in my dataset, by spliting the existing character field into two using an existing substring. ... something that in SAS I could solve e.g. combining substr(which I am aware exist in R) and "index" for determining the position of the pattern within the string. e.g. if my

Not paying close attention to detail, I entered the equivalent of pstr<-c("b1=200", "b2=50", "b3=0.3") when what I wanted was pnum<-c(b1=200, b2=50, b3=0.3) There was a list thread in 2010 that shows how to deal with un-named vectors, but the same lapply solution doesn't seem to work here i.e., pnum<-lapply(pstr, as.numeric) or similar vapply

On 03/15/2018 05:25 PM, Henrik Bengtsson wrote: > On Thu, Mar 15, 2018 at 3:39 AM, <FlorianSchwendinger at gmx.at> wrote: >> Thank you for your answer! >> I agree with you except for the 3 (Error) example and >> I realize now I should have started with that in the explanation. >> >> From my point of view >> parLapply(cl = clu, X = 1:2, fun = fun, c =

Hi, I have a vector of strings like: c("a1b1","a2b2","a1b2") which I want to spilt into two parts like: c("a1","a2","a2") and c("b1","b2,"b2"). So there is always a first part with a+number and a second part with b+number. Unfortunately there is no separator I could use to directly split the vectors.. Any idea

Greetings! For some reason I am not managing to work out how to do this (in principle) simple task! As a result of applying strsplit() to a vector of character strings, I have a long list L (N elements), where each element is a vector of two character strings, like: L[1] = c("A1","B1") L[2] = c("A2","B2") L[3] = c("A3","B3")