Anna wrote:
Hi all, I need to calculate a row median for every three columns of a dataframe. I made it work using the following script, but not happy with the script. Is there a simpler way for doing this?
To which Jim L responded:
Hi Anna,
I can't think of a simple way, but this function may make you happier:
step_median<-function(x,window) {
x<-unlist(x)
stop<-length(x)-window+1
xout<-NA
nindx<-1
for(i in seq(1,stop,by=window)) {
xout[nindx]<-do.call("median",list(x[i:(i+window-1)]))
nindx<-nindx+1
}
return(xout)
}
apply(df,1,step_median,3)
This should return a matrix where the columns are the medians
calculated from blocks of "window" width on each row of "df". As Bert
noted, you may want to think about a "rolling" median where the
"windows" overlap. This can be done like so:
library(zoo)
apply(df,1,rollmedian,3)
Jim
Another approach you might try is multiple calls to sapply/lapply. This won't rid you of loops, but it will hide them: # Example data. Some names changed to avoid collisions between # R functions (collisions are in the gap between the headphones, # not i R). dfr <- data.frame(a = c(2,3,4), b = c(3,5,1), c = c(1,3,6), d = c(7,2,1), e = c(2,5,3), f = c(4,5,1)) # Turn each of the three-column groups into their own element # in a list. Note: the subsetting (probably) fails with an # error if ncol(dfr) is not a multiple of 3 dlist <- lapply(seq(1, ncol(dfr), by = 3), function(enn) dfr[ , enn + 0:2]) # Then you can use sapply to calculate the row medians for each # of the elements.. # Both of the following seem to work. I'm not sure which is # more readable? sapply(dlist, function(xx) apply(xx, 1, median)) sapply(dlist, apply, 1, median) # I'm sure the cognoscenti will have a much more elegant way # of doing this. Cheers y'all, DMcP