Try this:
fil <- sapply(ran, '<', e1 = dat[,1]) & sapply(ran[2:(length(ran) +
1)], '>=', e1 = dat[,1])
mm <- apply(fil, 2, function(idx)mean(dat[idx, 2]))
On Wed, Apr 6, 2011 at 5:48 AM, Fabrice Tourre <fabrice.c...@gmail.com> wrote:
> Dear list,
>
> I have a dataframe with two column as fellow.
>
>> head(dat)
> V1 V2
> 0.15624 0.94567
> 0.26039 0.66442
> 0.16629 0.97822
> 0.23474 0.72079
> 0.11037 0.83760
> 0.14969 0.91312
>
> I want to get the column V2 mean value based on the bin of column of
> V1. I write the code as fellow. It works, but I think this is not the
> elegant way. Any suggestions?
>
> dat<-read.table("dat.txt",head=F)
> ran<-seq(0,0.5,0.05)
> mm<-NULL
> for (i in c(1:(length(ran)-1)))
> {
> fil<- dat[,1] > ran[i] & dat[,1]<=ran[i+1]
> m<-mean(dat[fil,2])
> mm<-c(mm,m)
> }
> mm
>
> Here is the first 20 lines of my data.
>
>> dput(head(dat,20))
> structure(list(V1 = c(0.15624, 0.26039, 0.16629, 0.23474, 0.11037,
> 0.14969, 0.16166, 0.09785, 0.36417, 0.08005, 0.29597, 0.14856,
> 0.17307, 0.36718, 0.11621, 0.23281, 0.10415, 0.1025, 0.04238,
> 0.13525), V2 = c(0.94567, 0.66442, 0.97822, 0.72079, 0.8376,
> 0.91312, 0.88463, 0.82432, 0.55582, 0.9429, 0.78956, 0.93424,
> 0.87692, 0.83996, 0.74552, 0.9779, 0.9958, 0.9783, 0.92523, 0.99022
> )), .Names = c("V1", "V2"), row.names = c(NA, 20L), class = "data.frame")
>
> ______________________________________________
> R-help@r-project.org mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
--
Henrique Dallazuanna
Curitiba-Paraná-Brasil
25° 25' 40" S 49° 16' 22" O
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