how do I search for columns with same name, add the column values and replace these columns with same name by their sum? Using R

Question

I have a data frame where some consecutive columns have the same name. I need to search for these, add their values in for each row, drop one column and replace the other wi

Answer 1

> dfrm <- data.frame(a = 1:10, b= 1:10, cc= 1:10, dd=1:10, ee=1:10)
> names(dfrm) <- c("a", "a", "b", "b", "b")
> sapply(unique(names(dfrm)[duplicated(names(dfrm))]), 
      function(x) Reduce("+", dfrm[ , grep(x, names(dfrm))]) )
       a  b
 [1,]  2  3
 [2,]  4  6
 [3,]  6  9
 [4,]  8 12
 [5,] 10 15
 [6,] 12 18
 [7,] 14 21
 [8,] 16 24
 [9,] 18 27
[10,] 20 30

EDIT 2: Using rowSums allows simplification of the first sapply argumentto just unique(names(dfrm)) at the expense of needing to remember to include drop=FALSE in "[":

sapply(unique(names(dfrm)), 
       function(x) rowSums( dfrm[ , grep(x, names(dfrm)), drop=FALSE]) )

To deal with NA's:

sapply(unique(names(dfrm)), 
      function(x) apply(dfrm[grep(x, names(dfrm))], 1, 
              function(y) if ( all(is.na(y)) ) {NA} else { sum(y, na.rm=TRUE) }
       )               )

(Edit note: addressed Tommy counter-example by putting unique around the names(.)[.] construction. The erroneous code was:

sapply(names(dfrm)[unique(duplicated(names(dfrm)))], 
     function(x) Reduce("+", dfrm[ , grep(x, names(dfrm))]) )

Answer 2

Some sample data.

dfr <- data.frame(
  foo = rnorm(20),
  bar = 1:20,
  bar = runif(20),
  check.names = FALSE
)

Method: Loop over unique column names; if there is only one of that name, then selecting all columns with that nme will return a vector, but if there are duplicates it will also be a data frame. Use rowSums to sum over rows. (Duh. EDIT: Not quite as 'duh' as previously thought!) lapply returns a list, which we need to reform into a data frame, and finally we fix the names. EDIT: sapply avoids the need for the last step.

unique_col_names <- unique(colnames(dfr))
new_dfr <- sapply(unique_col_names, function(name)
{
  subs <- dfr[, colnames(dfr) == name]
  if(is.data.frame(subs))
    rowSums(subs)
  else
    subs
})

Answer 3

Here is my one liner

# transpose data frame, sum by group = rowname, transpose back.
t(rowsum(t(dfrm), group = rownames(t(dfrm))))

Answer 4

One way is to identify duplcates using (surprise) the duplicated function, and then loop through them to calculate the sums. Here is an example:

dat.dup <- data.frame(x=1:10, x=1:10, x=1:10, y=1:10, y=1:10, z=1:10, check.names=FALSE)
dups <- unique(names(dat.dup)[duplicated(names(dat.dup))])
for (i in dups) {
dat.dup[[i]] <- rowSums(dat.dup[names(dat.dup) == i])
}
dat <- dat.dup[!duplicated(names(dat.dup))]