How to run tapply() on multiple columns of data frame using R?

Question

I have a data frame like the following:

a   b1  b2  b3  b4  b5  b6  b7  b8  b9
D   4   6   9   5   3   9   7   9   8
F   7   3   8   1   3   1   4   4   3
R   2   5   5   1   4   2   3   1   6
D   9   2   1   4   3   3   8   2   5
D   5   4   3   1   6   4   1   8   3
R   3   7   9   1   8   5   3   4   2
D   4   1   8   2   6   3   2   7   5
F   7   1   7   2   7   1   6   2   4
D   6   3   9   3   9   9   7   1   2

The function tapply(df[,2], INDEX = df$a, sum) works fine to produce a table that sums everything in df[,2] by df$a, but when I try tapply(df[,2:10], INDEX = df$a, sum) to get a similar table, except with a sum for each column (2, 3, 4,..., 10), I get an error message reading:

Error in tapply(df[, 2:10], INDEX = df$a, sum) : arguments must have same length

Additionally, I would like the row names of the table to be the column names of df[,2:10], such that row 1 is b1, row 2 is b2, and row 9 is b9.

Answer 1

That's because tapply works on vectors, and transforms df[,2:10] to a vector. Next to that, sum will give you the total sum, not the sum per column. Use aggregate(), eg :

aggregate(df[,2:10],by=list(df$a), sum)

If you want a list returned, you could use by() for that. Make sure to specify colSums instead of sum, as by works on a splitted dataframe :

by(df[,2:10],df$a,FUN=colSums)

Answer 2

Another possibility is to combine apply and tapply.

apply(df[,-1], 2, function(x) tapply(x, df$a, sum))

Will produce the output (which is a matrix)

    b1  ...   b9
D   sD1 ...  sD9
F   sF1 ...  sF9
R   sR1 ...  sR9

You can then use as.data.frame() to get a data frame as output.

Answer 3

Here is a way to apply data.table to this problem.

library(data.table)
DT <- data.table(df)
DT[, lapply(.SD, sum), by=a]

And here is a dplyr approach

library(dplyr)
df %>% group_by(a) %>% summarise_all(funs(sum))