问题描述
我想按组对各个列进行求和，而我的第一个想法是使用tapply.但是，我无法使tapply正常工作.可以使用tapply求和多列吗?如果没有，为什么不呢?
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?
我在互联网上进行了广泛搜索，发现张贴了许多类似的问题最早可以追溯到2008年.但是，这些问题都没有直接得到回答.相反，响应总是建议使用其他函数.
I have searched the internet extensively and found numerous similar questions postedas far back as 2008. However, none of those questions have been answered directly.Instead, the responses invariably suggest using a different function.
以下是我希望按州对苹果，按州对樱桃求和的示例数据集和李子按州.在此之下，我为tapply编写了许多替代方案，工作.
Below is an example data set for which I wish to sum apples by state, cherries by stateand plums by state. Below that I have compiled numerous alternatives to tapply thatdo work.
在底部，我显示了对tapply源代码的简单修改，该修改允许tapply执行所需的操作.
At the bottom I show a simple modification to the tapply source code that allowstapply to perform the desired operation.
尽管如此，也许我忽略了一种执行所需操作的简单方法用tapply.我没有在寻找替代功能，尽管欢迎其他替代功能.
Nevertheless, perhaps I am overlooking a simple way to perform the desired operationwith tapply. I am not looking for alternative functions, although additional alternatives are welcome.
鉴于对tapply源代码的修改很简单，所以我想知道为什么，或者类似的东西尚未实施.
Given the simplicity of my modification to the tapply source code I wonder why it, orsomething similar, has not already been implemented.
谢谢您的任何建议.如果我的问题是重复的，我将很乐意张贴我的问题作为对其他问题的答案.
Thank you for any advice. If my question is a duplicate I will be happy to post myquestion as an answer to that other question.
这是示例数据集:
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)
这不起作用:
tapply(df.1, df.1$state, function(x) {colSums(x[,3:5])})
帮助页面上显示:
tapply(X, INDEX, FUN = NULL, ..., simplify = TRUE)
X an atomic object, typically a vector.
我对短语typically a vector感到困惑，这让我怀疑可以使用一个数据帧.我还不清楚atomic object是什么意思.
I was confused by the phrase typically a vector which made me wonder whethera data frame could be used. I have never been clear on what atomic object means.
这是tapply的几种可行的替代方法.第一种选择是将tapply与apply组合在一起的解决方法.
Here are several alternatives to tapply that do work. The first alternative is a work-around that combines tapply with apply.
apply(df.1[,c(3:5)], 2, function(x) tapply(x, df.1$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
#
# $BB
# apples cherries plums
# -111 -222 -333
这是tapply的源代码，除了我更改了该行:
Here is the source code for tapply except that I changed the line:
nx <- length(X)
收件人:
nx <- ifelse(is.vector(X), length(X), dim(X)[1])
此修改后的tapply版本执行所需的操作:
This modified version of tapply performs the desired operation:
my.tapply <- function (X, INDEX, FUN = NULL, ..., simplify = TRUE)
{
FUN <- if (!is.null(FUN)) match.fun(FUN)
if (!is.list(INDEX)) INDEX <- list(INDEX)
nI <- length(INDEX)
if (!nI) stop("'INDEX' is of length zero")
namelist <- vector("list", nI)
names(namelist) <- names(INDEX)
extent <- integer(nI)
nx <- ifelse(is.vector(X), length(X), dim(X)[1]) # replaces nx <- length(X)
one <- 1L
group <- rep.int(one, nx) #- to contain the splitting vector
ngroup <- one
for (i in seq_along(INDEX)) {
index <- as.factor(INDEX[[i]])
if (length(index) != nx)
stop("arguments must have same length")
namelist[[i]] <- levels(index)#- all of them, yes !
extent[i] <- nlevels(index)
group <- group + ngroup * (as.integer(index) - one)
ngroup <- ngroup * nlevels(index)
}
if (is.null(FUN)) return(group)
ans <- lapply(X = split(X, group), FUN = FUN, ...)
index <- as.integer(names(ans))
if (simplify && all(unlist(lapply(ans, length)) == 1L)) {
ansmat <- array(dim = extent, dimnames = namelist)
ans <- unlist(ans, recursive = FALSE)
} else {
ansmat <- array(vector("list", prod(extent)),
dim = extent, dimnames = namelist)
}
if(length(index)) {
names(ans) <- NULL
ansmat[index] <- ans
}
ansmat
}
my.tapply(df.1$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
#
# $BB
# apples cherries
# -111 -222
推荐答案
tapply适用于矢量，对于data.frame，您可以使用by(它是tapply的包装，请看一下代码):
tapply works on a vector, for a data.frame you can use by (which is a wrapper for tapply, take a look at the code):
> by(df.1[,c(3:5)], df.1$state, FUN=colSums)
df.1$state: AA
apples cherries plums
111 222 333
-------------------------------------------------------------------------------------
df.1$state: BB
apples cherries plums
-111 -222 -333
这篇关于用tapply按组求和多列的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！