similar to: ave gives unexpected NA's

library(plyr) n<-100000 grp1<-sample(1:750, n, replace=T) grp2<-sample(1:750, n, replace=T) d<-data.frame(x=rnorm(n), y=rnorm(n), grp1=grp1, grp2=grp2) system.time({ d$avx1 <- ave(d$x, list(d$grp1, d$grp2)) d$avy1 <- ave(d$y, list(d$grp1, d$grp2)) }) # user system elapsed # 39.300 0.279 40.809 system.time({ d$avx2 <- ave(d$x, interaction(d$grp1, d$grp2, drop =

Hi, I have a data.frame that has a categorical variable, for which I would like to look at the distribution of levels of this variable, based on a grouping of two other variables. As an example: x <- data.frame(obs=sample(c('low', 'high'),100, replace=TRUE), grp1=sample(1:10, 100, replace=TRUE), grp2=runif(100)) cut.grp1 <- cut(x$grp1, 3) cut.grp2 <- cut(x$grp2, 3)

x.fun <- function( formula, data ) dotplot( formula, data ) x.grp <- function( formula, groups, data ) dotplot( formula, groups, data ) data( barley ) > x.fun( variety ~ yield | site, data=barley ) # no problem > dotplot( variety ~ yield | site, groups=year, data=barley ) # no problem > x.grp( variety ~ yield | site, groups=year, data=barley ) object "year" not found

Hi, I have a vector of data, that I group based on two factors via tapply. For each such grouping I would like to plot a pie chart. I can layout these pie charts in a matrix layout, correpsonding to the levels of the two factors. But I am getting stuck on how to label the rows and colums. My current approach looks like this: x <- data.frame(obs=sample(c('low', 'high'),100,

Dear Helpful R Users, I am graphing some data using the beanplot, but I am having trouble getting the output I desire. I have five tanks (A-E) and 2 groups for each tank grp1 or grp2, except tank C where there is only grp1. (I only changed the grouprep to "C grp1" for the example) When I plot them, I would like A B C(only grp1 - half of the bean plot) then D and E (as full beans).

I'd like to summarize several variables in a data frame, for multiple groups, and store the results in a data.frame. To do so, I'm using by(). For example: df<-data.frame(a=1:10,b=11:20,c=21:30,grp1=c("x","y"),grp2=c("x","y"),grp3=c("x","y")) dfsum<-by(df[c("a","b","c")],

Hi folks, has anybody a simple solution for the following request? I have a group alias (all at company.com). (1) Only company.com accounts should be able to send an email to everybody in that company via all at company.com. (2) - rather optional: refine the restrictions, e.g. two groups, grp1 at company.com and grp2 at company.com. Grp1 members should be able to send mails to grp2 but not vice

Dear all Is there a clean plyr version of the following by() and do.call(rbind, ...) construct: > df<-data.frame(a=1:10,b=11:20,c=21:30,grp1=c("x","y"),grp2=c("x","y"),grp3=c("x","y")) > dfsum<-by(df[c("a","b","c")], df[c("grp1","grp2","grp3")], range) >

Hi experts I have a trouble in access rights I am running Samba 3.0.31 on Solaris 10 x86 64 bits as member server of an Active Directory 2003 R2 domain (MYDOMAIN) using Identity Management for Unix I set rights to access a sub folder of a Samba share. On Solaris the user "toto" jdoe can write a new file. From Windows, the same user can't. Itlooks like OK when the primary group

Hi, I wanted to generate a plot which is almost like the plot generated by the following codes. category <- paste("Geographical Category", 1:10) grp1 <- rnorm(10, mean=10, sd=10) grp2 <- rnorm(10, mean=20, sd=10) grp3 <- rnorm(10, mean=15, sd=10) grp4 <- rnorm(10, mean=12, sd=10) mydat <- data.frame(category,grp1,grp2,grp3,grp4) dat.m <- melt(mydat) p <-

Hello, I am trying to add a less than equal to symbol in a ggplot2 legend text. See sample code below. I have tried using the expression function and \u2264. I also tried adding labels to legend.text under theme. Neither of these 3 options work. Please help, Mahesh ++++++++++++++ Extra.column=ifelse(data[,covariate]>cutpoint,1,0) Grp1 <- "\u2264 1.5" Grp2 <-

On Wed, 5 Dec 2018, Alexander Dalloz wrote: >> I have a group alias (all at company.com). >> (1) Only company.com accounts should be able to send an email to everybody >> in that company via all at company.com. >> (2) - rather optional: refine the restrictions, e.g. two groups, >> grp1 at company.com and grp2 at company.com. Grp1 members should be able to send

Is there a simple possibility to become directly a matrix from a call of sub() on a matrix? --------- START OF LOGFILE ---------------- # R 1.4.1 > a <- matrix( letters[1:6], 2, 3 ) # a is a matrix > print( a ) [,1] [,2] [,3] [1,] "a" "c" "e" [2,] "b" "d" "f" > b <- sub( '(.)', '-\\1-', a )

I have: d <- sample(10:100, 9) o <- order(d) r <- d[o] How I can get d (in the original order), knowing only r and o? Thanks - Wolfram

What is the easiest way to change within vector of strings each letter after a space into a capital letter? E.g.: c( "this is an element of the vector of strings", "second element" ) becomes: c( "This Is An Element Of The Vector Of Strings", "Second Element" ) My reason to try to do this is to get more readable abbreviations. (A suggestion would be to

What is the reason, that the levels of the factor returned by cut() are not marked as ordered levels? > is.ordered( cut( breaks=3, sample(10 ) ) ) FALSE > help(factor) ... If 'ordered' is 'TRUE', the factor levels are assumed to be ordered. ... Wolfram

I propose to add a function that allows to display colors selected by a text pattern or by color vectors in a plot. Wolfram Fischer #--- showcolors.R showcolors <- function( col = "red" , index = NULL , pie = TRUE , lwd = 6 , cex = 1.0 , main = NULL , sub = NULL , ... ){ n.colors <- length( col ) if( n.colors > 1 ){ main <- deparse( substitute( col ) )

Hi everybody, I assume this is no real problem but only some setup thing but I don't get through it. How can I mount only shares that a specified group is supposed to see? Meaning if I have somebody member of grp1 I want him to see dir A. If some one is member of grp2 I want him to see dir B. So a section of our company can have his own shared directory for their own.. At the moment I

I would propose to add " See also: `nchar' for counting the number of character in character vectors. " to the helpfile of length(), because it is rather difficult to find nchar() if one has only search terms as "length", "len", "strlen" in mind. Sincerly Wolfram Fischer

I wanted to test if there exists already a name (which is incidentally a substring of another name) in a dataframe. I did e.g.: > data(swiss) > names(swiss) [1] "Fertility" "Agriculture" "Examination" "Education" [5] "Catholic" "Infant.Mortality" > ! is.null(swiss$EduX) [1] FALSE > !