similar to: Google Custom Search Engine for R

Hi, I'm having a weird result with the length() function: >a [... omited ...] [9994] NA "2003-12-03 16:37:00" "2002-06-26 18:43:00" [9997] "2005-07-04 04:00:00" "2007-02-16 22:09:00" "2007-02-24 15:49:00" [10000] NA > length(LastModified) [1] 9 > length(c(LastModified)) [1] 9 I was expecting to get

Hi, I have several files with data in this format: 20070102 20070102 20070106 20070201 ... The data is sorted and each line represents a date (YYYYMMDD). I would like to analyze this data using R. For instance, I would like to have a histogram by year, month or day. I've already made a simple Perl script that aggregates this data but I believe that R can be much more powerful and easy on

Hi, I'm trying to aggregate date values using the aggregate function. For example: aggregate(data,by=list(weekdays(LM),months(LM)),FUN=length) I would also like to aggregate by year but there seems to be no years() function. Should there be one? Is there any alternative choice? Also, a hours() function would be great. Any tip on this? Thanks in advance! S?rgio Nunes

Hi, I'm looking for a function to measure the dispersion of a set of values ranging from 0 to 1. This function should be 0 if all the values are evenly spaced within the interval and it should be > 0 if values are clustered. The more clustered the values are, the higher should the function be. An example: [0; 0.2; 0.4; 0.6; 0.8; 1] - function should be ~ 0 [0; 0.1; 0.1; 0.15; 1] -

Hi, What's the best way to average dates? I though mean.POISXct would work fine but... > a [1] "2007-04-02 19:22:00 WEST" > b [1] "2007-03-17 16:23:00 WET" > class(a) [1] "POSIXt" "POSIXct" > class(b) [1] "POSIXt" "POSIXct" > mean(a,b) [1] "2007-04-02 19:22:00 WEST" > mean(b,a) [1] "2007-03-17

Hi, I'm trying to draw a 2D plot using multiple tints of red. The (simplified) setup is the following: || year | x | y || My idea is that each year is plotted with a different tint of red. Older year (lightest) -> Later year (darkest). I've managed to plot this with different scales of grays simply by doing: palette(gray(length(years):0/length(years))) before the plot and for each

Hello all, I wanted to announce a new R search engine I made that covers all the major R mailing lists, CRAN, r-project.org, and more. Results are tabbed, so you can refine the search. Current refinements include searching just the mailing lists, searching just introductions, and searching the web for source files ending in .R. Please send comments and suggestions. If you want to add sites to

gReetings: Is there any way to download a (or a sample of a) crossection of stock market prices? Or is it possible to use get.hist.quote with a *wild card*? Thanks, mihai Mihai Nica 170 East Griffith St. G5 Jackson, MS 39201 601-914-0361 ____________________________________________________________________________________ 8:00? 8:25? 8:40? Find a flick in no time [[alternative HTML

Just for the record, here are my steps for producing a date based histogram. Data is stored in a file where each line only has a date - 2007/02/16 >d<-readLines("filename.dat") >d<-as.Date(d, format="%Y/%m/%d") >pdf(yearly.pdf) >hist(d, "years") >dev.off() Instead of "years" you can also use "days", "weeks",

Hi, I'm pretty new to R and I've been having some problems filtering data in matrices. I have the following initial dataset: || year | name | varA || I have multiple values for "varA" for the same "year" and the same "name". Having this as the input I would like to obtain the following: || year | name | {varA mean} || Where I only have one line for each

Hi all, This question involves using a "for" loop to make a "decision" in a script. I've written a rather intricate script, and near the start of it, I want it either to do a loop (if a variable called "number.runs" > 1) or not do a loop (if "number.runs" is 1). This is probably trivial but I can't figure it out. Here's a self-contained

Hi all, [this is a bit hard to describe, so if my initial description is confusing, please try running my code below] #WHAT I'M TRYING TO DO I'd appreciate any help in trying to speed up some code. I've written a script that converts a matrix of integers (usually between 1-10,000 - these represent allele names) into two new matrices of normally distributed values (representing

Hi, It seems that hist() has a buggy behavior when breaking over "days". The bug can be reproduced in a few steps: > d=data.frame(date=c("2009-01-01", "2009-01-02", "2009-01-02")) > d$date=as.Date(d$date) > d$date [1] "2009-01-01" "2009-01-02" "2009-01-02" > h=hist(d$date, "days") > h$count [1] 3

Hi all, Probably a rudimentary question. I have a flat file that looks like this (the real one has ~10e6 elements): 10110100101001011101011 and I want to pull that into R as a vector, but with each digit being it's own element. There are no separators between the digits. How can I accomplish this? Thanks in advance! Matt -- Matthew C Keller Asst. Professor of Psychology University of

Hi, I'm pretty new to R, I've just installed version 2.1.1 on Windows. I have a simple (it seems) doubt - how do I change the Console default Language ? It's set to Portuguese but I would like to view it in English. Thanks in advance, S??rgio Nunes

Hi all, Perhaps this is more of a unix question, but I'll give it a try here. I am running 9 different R processes at the same time (called from a shell script using R CMD BATCH). When I use the top program to monitor how they are doing, it is impossible to tell which R process is related to which R script. Is there a way to rename a specific instantiation of an R process in top with

Hi all, Is there a way to skip non-sequential lines using the "skip" argument in the scan function? E.g., I have a matrix with 100 rows and 1e7 columns. I open a connection and want to read only lines 5, 7, 9, etc [i.e., seq(5,99,2)] It might seem that the syntax to do this would be something like this (if only the "skip" allowed vectors in the same way colClasses does in

hi all, I'm full of questions today :). Thanks in advance for your help! Here's the problem: x <- c('18x.6','12x.9','302x.3') I want to get a vector that is c('18x','12x','302x') This is easily done using this code: unlist(lapply(strsplit(x,".",fixed=TRUE),function(x) x[1])) So far so good. The problem is that x is a vector

Hi all, I used to use fwrite() function in data.table but I cannot get it to work now. The function is not in the data.table package, even though a help page exists for it. My session info is below. Any ideas on how to get fwrite() to work would be much appreciated. Thanks! > sessionInfo() R version 3.2.0 (2015-04-16) Platform: x86_64-unknown-linux-gnu (64-bit) Running under: Red Hat

Hi all, I've had one of my most miserable R weeks in memory. I'm trying to deal with huge datasets (>1GB each) but am running up against those pesky memory limits. The libraries filehash and g.data are not very suitable for what I need. I haven't gotten into the sql thing yet. Most recently I've been trying to install the new package ff (not yet on the CRAN repository). I