calculate row median of every three columns for a dataframe

Hi

As usual in R, things could be done by different ways.

idx <- (0:(ncol(dfr)-1))%/%3

aggregate(t(dfr), list(idx), median)
  Group.1 V1 V2 V3
1       0  2  3  4
2       1  4  5  1

Results should be OK although its structure is different, performance is
not tested.

Cheers
Petr

-----Original Message-----
From: R-help <r-help-bounces at r-project.org> On Behalf Of David McPearson
Sent: Friday, April 17, 2020 7:50 AM
To: r-help at r-project.org
Cc: dcmcp at telstra.com
Subject: Re: [R] calculate row median of every three columns for a

dataframe

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