rowsum: Give Column Sums of a Matrix or Data Frame, Based on a...
In robertzk/monadicbase: The R Base Package

Description Usage Arguments Details Value See Also Examples

Compute column sums across rows of a numeric matrix-like object for each level of a grouping variable. rowsum is generic, with a method for data frames and a default method for vectors and matrices.

rowsum(x, group, reorder = TRUE, ...)

## S3 method for class 'data.frame'
rowsum(x, group, reorder = TRUE, na.rm = FALSE, ...)

## Default S3 method:
rowsum(x, group, reorder = TRUE, na.rm = FALSE, ...)

`x`	a matrix, data frame or vector of numeric data. Missing values are allowed. A numeric vector will be treated as a column vector.
`group`	a vector or factor giving the grouping, with one element per row of `x`. Missing values will be treated as another group and a warning will be given.
`reorder`	if `TRUE`, then the result will be in order of `sort(unique(group))`, if `FALSE`, it will be in the order that groups were encountered.
`na.rm`	logical (`TRUE` or `FALSE`). Should `NA` (including `NaN`) values be discarded?
`...`	other arguments to be passed to or from methods

The default is to reorder the rows to agree with tapply as in the example below. Reordering should not add noticeably to the time except when there are very many distinct values of group and x has few columns.

The original function was written by Terry Therneau, but this is a new implementation using hashing that is much faster for large matrices.

To sum over all the rows of a matrix (ie, a single group) use colSums, which should be even faster.

For integer arguments, over/underflow in forming the sum results in NA.

A matrix or data frame containing the sums. There will be one row per unique value of group.

tapply, aggregate, rowSums

require(stats)

x <- matrix(runif(100), ncol = 5)
group <- sample(1:8, 20, TRUE)
(xsum <- rowsum(x, group))
## Slower versions
tapply(x, list(group[row(x)], col(x)), sum)
t(sapply(split(as.data.frame(x), group), colSums))
aggregate(x, list(group), sum)[-1]