# rowsum: Give Column Sums of a Matrix or Data Frame, Based on a...

## Description

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.

## Usage

 ```1 2 3 4 5 6 7``` ```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, ...) ```

## Arguments

 `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

## Details

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 (i.e., a single `group`) use `colSums`, which should be even faster.

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

## Value

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

`tapply`, `aggregate`, `rowSums`
 ```1 2 3 4 5 6 7 8 9``` ```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] ```