Hi. I searched the list and didn't found nothing similar to this. I simplified my example like below:
#I need calculate correlation (for example) between 2 columns classified by a third one at a data.frame, like below: #number of rows nr = 10 #the third column is to enforce that I need correlation on two variables only dataf = as.data.frame(matrix(c(rnorm(nr),rnorm(nr)*2,runif(nr),sort(c(1,1,2,2,3,3,sample(1:3,nr-6,replace=TRUE)))),ncol=4)) names(dataf)[4] = "class" #> dataf # V1 V2 V3 class #1 0.56933020 1.2529931 0.30774422 1 #2 0.41702299 -1.6441547 0.76140046 1 #3 -1.07671647 -4.8747575 0.43706944 1 #4 -1.97701167 1.3015196 0.04390175 2 #5 0.56501325 1.8597720 0.08174124 2 #6 0.70068638 1.7922641 0.74730126 2 #7 -1.39956177 -1.9918904 0.64521918 3 #8 0.27086664 0.3745362 0.61026133 3 #9 0.04282347 3.7360407 0.48696109 3 #10 -0.34262654 0.7933674 0.09824913 3 #I tried: tapply(dataf$V1, dataf$class, cor, dataf$V2) #Error FUN(X[[1L]], ...) : incompatible dimensions tapply(dataf$V1, dataf$class, cor, tapply(dataf$V2, dataf$class)) #Error FUN(X[[1L]], ...) : incompatible dimensions #But using "by" I obtain: by(dataf[,c("V1","V2")], dataf$class, cor) #dataf$class: 1 # V1 V2 #V1 1.00000 0.91777 #V2 0.91777 1.00000 #-------------------------------------------------------------------------------------------------- #dataf$class: 2 # V1 V2 #V1 1.000000 0.987857 #V2 0.987857 1.000000 #-------------------------------------------------------------------------------------------------- #dataf$class: 3 # V1 V2 #V1 1.0000000 0.7318938 #V2 0.7318938 1.0000000 #My interest is on cor(V1,V2)[1,2], so I can take 0.91777, 0.987857 and 0.7318938, but I think that tapply can works better, if I can solve the problem. Thanks, Cezar Novos endereços, o Yahoo! que você conhece. Crie um email novo com a sua cara @ymail.com ou @rocketmail.com. http://br.new.mail.yahoo.com/addresses [[alternative HTML version deleted]]
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