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(labels=c("a","a","b","c","c"),data=matrix(1:20,nrow=5))
> z
labels data.1 data.2 data.3 data.4
1 a 1 6 11 16
2 a 2 7 12 17
3 b 3 8 13 18
4 c 4 9 14 19
5 c 5 10 15 20
I can use R's by
command to get the column means according to the labels column:
> by(z[,2:5],z$labels,colMeans)
z[, 1]: a
data.1 data.2 data.3 data.4
1.5 6.5 11.5 16.5
------------------------------------------------------------
z[, 1]: b
data.1 data.2 data.3 data.4
3 8 13 18
------------------------------------------------------------
z[, 1]: c
data.1 data.2 data.3 data.4
4.5 9.5 14.5 19.5
But how do I coerce the output back to a data frame? as.data.frame
doesn't work...
> as.data.frame(by(z[,2:5],z$labels,colMeans))
Error in as.data.frame.default(by(z[, 2:5], z$labels, colMeans)) :
cannot coerce class '"by"' into a data.frame
You can use ddply
from plyr
package
library(plyr)
ddply(z, .(labels), numcolwise(mean))
labels data.1 data.2 data.3 data.4
1 a 1.5 6.5 11.5 16.5
2 b 3.0 8.0 13.0 18.0
3 c 4.5 9.5 14.5 19.5
Or aggregate
from stats
aggregate(z[,-1], by=list(z$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$labels), colMeans))
data.1 data.2 data.3 data.4
a 1.5 6.5 11.5 16.5
b 3.0 8.0 13.0 18.0
c 4.5 9.5 14.5 19.5
The output of by
is a list
so you can use do.call
to rbind
them and then convert this:
as.data.frame(do.call("rbind",by(z[,2:5],z$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
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With