converting output of R's “by” command to data frame

Question

I\'m trying to use R\'s by command to get column means for subsets of a data frame. For example, consider this data frame:

> z = data.frame(         


        

Answer 1

Dealing with the by output can be really annoying. I just found a way to withdraw what you want in a format of a data frame and you won't need extra packages.

So, if you do this:

aux <- by(z[,2:5],z$labels,colMeans)

You can then transform it in a data frame by doing this:

  aux_df <- as.data.frame(t(aux[seq(nrow(aux)),seq(ncol(aux))]))

I'm just getting all the rows and columns from aux, transposing it and using as.data.frame.

I hope that helps.

Answer 2

You can use ddply from plyr package

library(plyr)
ddply(z, .(labels), numcolwise(mean))
  labels data.1 data.2 data.3 data.4
1      a    1.5    6.5   11.5   16.5
2      b    3.0    8.0   13.0   18.0
3      c    4.5    9.5   14.5   19.5

Or aggregate from stats

aggregate(z[,-1], by=list(z$labels), mean)
  Group.1 data.1 data.2 data.3 data.4
1       a    1.5    6.5   11.5   16.5
2       b    3.0    8.0   13.0   18.0
3       c    4.5    9.5   14.5   19.5

Or dcast from reshape2 package

library(reshape2)
dcast( melt(z), labels ~ variable, mean)

Using sapply :

 t(sapply(split(z[,-1], z$labels), colMeans))
  data.1 data.2 data.3 data.4
a    1.5    6.5   11.5   16.5
b    3.0    8.0   13.0   18.0
c    4.5    9.5   14.5   19.5

Answer 3

The output of by is a list so you can use do.call to rbind them and then convert this:

as.data.frame(do.call("rbind",by(z[,2:5],z$labels,colMeans)))
  data.1 data.2 data.3 data.4
a    1.5    6.5   11.5   16.5
b    3.0    8.0   13.0   18.0
c    4.5    9.5   14.5   19.5