Sum all columns whose names start with a pattern, by group

Question

I'm fairly new to R and I'm trying to sum columns by groups based on their names. I have a data frame like this one:

DT <- data.frame(a011=c(0,10,20,0),a012=c(010,10,0,0),a013=c(10,30,0,10),
a021=c(10,20,20,10),a022=c(0,0,0,10),a023=c(20,0,0,0),a031=c(30,0,10,0),
a032=c(0,0,10,0),a033=c(20,0,0,0))

I would like to obtain the sum of all the columns starting with "a01", of all the columns starting with "a02" and all the columns starting with "a03":

a01tot a02tot a03tot
    20     30     50
    50     20      0
    20     20     20
    10     20      0

So far I have used

DT$a01tot <- rowSums(DT[,grep("a01", names(DT))])

and so on, but my real data frame has many more groups and I would like to avoid having to write a line of code for each group. I was wondering if it is possible to include "a01","a02","a03"... in a vector or list and have something that adds the columns "a01tot","a02tot","a03tot"... to the data frame automatically.

I know that my question is very similar to this one: R sum of rows for different group of columns that start with similar string, but the solution pointed out there,

cbind(df, t(rowsum(t(df), sub("_.*", "_t", names(df)))))

does not work in my case because there isn't a common element (like "_") to replace (I cannot change the names of the variables to a01_1, a02_2 etc.).

Switching to the "long" format is not a viable solution in my case either.

Any help will be greatly appreciated.

Answer 1

You can store the patterns in a vector and loop through them. With your example you can use something like this:

patterns <- unique(substr(names(DT), 1, 3))  # store patterns in a vector
new <- sapply(patterns, function(xx) rowSums(DT[,grep(xx, names(DT)), drop=FALSE]))  # loop through
#     a01 a02 a03
#[1,]  20  30  50
#[2,]  50  20   0
#[3,]  20  20  20
#[4,]  10  20   0

You can adjust the names like this:

colnames(new) <- paste0(colnames(new), "tot")  # rename