[R] calculate row median of every three columns for a dataframe

[David McPearson]

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