sum multiple columns by group with tapply

I wanted to sum individual columns by group and my first thought was to use tapply. However, I cannot get tapply to work. Can tapply be used to sum multiple columns? If not, why not?

I have searched the internet extensively and found numerous similar questions posted as far back as 2008. However, none of those questions have been answered directly. Instead, the responses invariably suggest using a different function.

Below is an example data set for which I wish to sum apples by state, cherries by state and plums by state. Below that I have compiled numerous alternatives to tapply that do work.

At the bottom I show a simple modification to the tapply source code that allows tapply to perform the desired operation.

Nevertheless, perhaps I am overlooking a simple way to perform the desired operation with tapply. I am not looking for alternative functions, although additional alternatives are welcome.

Given the simplicity of my modification to the tapply source code I wonder why it, or something similar, has not already been implemented.

Thank you for any advice. If my question is a duplicate I will be happy to post my question as an answer to that other question.

Here is the example data set:

df.1 <- read.table(text = '

    state   county   apples   cherries   plums
       AA        1        1          2       3
       AA        2       10         20      30
       AA        3      100        200     300
       BB        7       -1         -2      -3
       BB        8      -10        -20     -30
       BB        9     -100       -200    -300

', header = TRUE, stringsAsFactors = FALSE)

This does not work:

tapply(df.1, df.1$state, function(x) {colSums(x[,3:5])})

The help pages says:

tapply(X, INDEX, FUN = NULL, ..., simplify = TRUE)

X       an atomic object, typically a vector.

I was confused by the phrase typically a vector which made me wonder whether a data frame could be used. I have never been clear on what atomic object means.

Here are several alternatives to tapply that do work. The first alternative is a work-around that combines tapply with apply.

apply(df.1[,c(3:5)], 2, function(x) tapply(x, df.1$state, sum))

#    apples cherries plums
# AA    111      222   333
# BB   -111     -222  -333

with(df.1, aggregate(df.1[,3:5], data.frame(state), sum))

#   state apples cherries plums
# 1    AA    111      222   333
# 2    BB   -111     -222  -333

t(sapply(split(df.1[,3:5], df.1$state), colSums))

#    apples cherries plums
# AA    111      222   333
# BB   -111     -222  -333

t(sapply(split(df.1[,3:5], df.1$state), function(x) apply(x, 2, sum)))

#    apples cherries plums
# AA    111      222   333
# BB   -111     -222  -333

aggregate(df.1[,3:5], by=list(df.1$state), sum)

#   Group.1 apples cherries plums
# 1      AA    111      222   333
# 2      BB   -111     -222  -333

by(df.1[,3:5], df.1$state, colSums)

# df.1$state: AA
#   apples cherries    plums 
#      111      222      333 
# ------------------------------------------------------------ 
# df.1$state: BB
#   apples cherries    plums 
#     -111     -222     -333

with(df.1, 
     aggregate(x = list(apples   = apples, 
                        cherries = cherries,
                        plums    = plums), 
               by = list(state   = state), 
               FUN = function(x) sum(x)))

#   state apples cherries plums
# 1    AA    111      222   333
# 2    BB   -111     -222  -333

lapply(split(df.1, df.1$state), function(x) {colSums(x[,3:5])} )

# $AA
#   apples cherries    plums 
#      111      222      333 
#
# $BB
#   apples cherries    plums 
#     -111     -222     -333

Here is the source code for tapply except that I changed the line:

nx <- length(X)

to:

nx <- ifelse(is.vector(X), length(X), dim(X)[1])

This modified version of tapply performs the desired operation:

my.tapply <- function (X, INDEX, FUN = NULL, ..., simplify = TRUE)
{
    FUN <- if (!is.null(FUN)) match.fun(FUN)
    if (!is.list(INDEX)) INDEX <- list(INDEX)
    nI <- length(INDEX)
    if (!nI) stop("'INDEX' is of length zero")
    namelist <- vector("list", nI)
    names(namelist) <- names(INDEX)
    extent <- integer(nI)
    nx     <- ifelse(is.vector(X), length(X), dim(X)[1])  # replaces nx <- length(X)
    one <- 1L
    group <- rep.int(one, nx) #- to contain the splitting vector
    ngroup <- one
    for (i in seq_along(INDEX)) {
    index <- as.factor(INDEX[[i]])
    if (length(index) != nx)
        stop("arguments must have same length")
    namelist[[i]] <- levels(index)#- all of them, yes !
    extent[i] <- nlevels(index)
    group <- group + ngroup * (as.integer(index) - one)
    ngroup <- ngroup * nlevels(index)
    }
    if (is.null(FUN)) return(group)
    ans <- lapply(X = split(X, group), FUN = FUN, ...)
    index <- as.integer(names(ans))
    if (simplify && all(unlist(lapply(ans, length)) == 1L)) {
    ansmat <- array(dim = extent, dimnames = namelist)
    ans <- unlist(ans, recursive = FALSE)
    } else {
    ansmat <- array(vector("list", prod(extent)),
            dim = extent, dimnames = namelist)
    }
    if(length(index)) {
        names(ans) <- NULL
        ansmat[index] <- ans
    }
    ansmat
}

my.tapply(df.1$apples, df.1$state, function(x) {sum(x)})

#  AA   BB 
# 111 -111

my.tapply(df.1[,3:4] , df.1$state, function(x) {colSums(x)})

# $AA
#   apples cherries 
#      111      222 
#
# $BB
#   apples cherries 
#     -111     -222

回答1:

tapply works on a vector, for a data.frame you can use by (which is a wrapper for tapply, take a look at the code):

> by(df.1[,c(3:5)], df.1$state, FUN=colSums)
df.1$state: AA
  apples cherries    plums 
     111      222      333 
------------------------------------------------------------------------------------- 
df.1$state: BB
  apples cherries    plums 
    -111     -222     -333

回答2:

You're looking for by. It uses the INDEX in the way that you assumed tapply would, by row.

by(df.1, df.1$state, function(x) colSums(x[,3:5]))

The problem with your use of tapply is that you were indexing the data.frame by column. (Because data.frame is really just a list of columns.) So, tapply complained that your index didn't match the length of your data.frame which is 5.

回答3:

I looked at the source code for by, as EDi suggested. That code was substantially more complex than my change to the one line in tapply. I have now found that my.tapply does not work with the more complex scenario below where apples and cherries are summed by state and county. If I get my.tapply to work with this case I can post the code here later:

df.2 <- read.table(text = '

    state   county   apples   cherries   plums
       AA        1        1          2       3
       AA        1        1          2       3
       AA        2       10         20      30
       AA        2       10         20      30
       AA        3      100        200     300
       AA        3      100        200     300

       BB        7       -1         -2      -3
       BB        7       -1         -2      -3
       BB        8      -10        -20     -30
       BB        8      -10        -20     -30
       BB        9     -100       -200    -300
       BB        9     -100       -200    -300

', header = TRUE, stringsAsFactors = FALSE)

# my function works

   tapply(df.2$apples  , list(df.2$state, df.2$county), function(x) {sum(x)})
my.tapply(df.2$apples  , list(df.2$state, df.2$county), function(x) {sum(x)})

# my function works

   tapply(df.2$cherries, list(df.2$state, df.2$county), function(x) {sum(x)})
my.tapply(df.2$cherries, list(df.2$state, df.2$county), function(x) {sum(x)})

# my function does not work

my.tapply(df.2[,3:4], list(df.2$state, df.2$county), function(x) {colSums(x)})