R 使用tapply按组对多个列求和
我想按组对各个列进行汇总，我的第一个想法是使用R 使用tapply按组对多个列求和,r,tapply,R,Tapply,我想按组对各个列进行汇总，我的第一个想法是使用tapply。 然而，我无法让tapply工作。tapply可以用于多列求和吗？ 若否，原因为何 我在互联网上搜索了很多类似的问题 早在2008年。然而，这些问题都没有得到直接回答。 相反，回答总是建议使用不同的函数 下面是一个示例数据集，我希望将各州的苹果和各州的樱桃相加 和李子。下面，我已经编译了许多备选方案，以取代tapply 干活儿 在底部，我展示了对tapply源代码的一个简单修改，该修改允许 t按执行所需操作 然而，也许我忽略了一个简单的
tapplytapplytapplytapplytapplyt按tapplytapplydf.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)
tapply(X, INDEX, FUN = NULL, ..., simplify = TRUE)
X an atomic object, typically a vector.
弄糊涂了，它通常是一个向量，这让我想知道
可以使用数据帧。我从来都不清楚原子对象是什么意思
这里有几种可行的tapply
替代方案。第一个备选方案是将taply
与apply
相结合的变通方法
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
#
# $BB
# apples cherries plums
# -111 -222 -333
以下是tapply
的源代码，只是我更改了行：
nx <- length(X)
tapply
用于向量，对于data.frame，您可以使用by
（这是tapply
的包装器，请查看代码）：
您正在通过
查找。它使用索引
的方式与您假设的tapply
的方式相同
by(df.1, df.1$state, function(x) colSums(x[,3:5]))
使用tapply
的问题在于，您正在按列索引data.frame
。（因为data.frame
实际上只是列的列表。）因此，tapply
抱怨您的索引与data.frame
的长度不匹配，即5。我按照EDi的建议查看了by的源代码。该代码比我在tapply
中更改为一行的代码要复杂得多。我现在发现，my.tapply
不适用于下面更复杂的场景，苹果
和樱桃
由州
和县
相加。如果我使用my.tapply
处理此案例，我可以稍后在此处发布代码：
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)})
df.2
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
#
# $BB
# apples cherries
# -111 -222
> by(df.1[,c(3:5)], df.1$state, FUN=colSums)
df.1$state: AA
apples cherries plums
111 222 333
-------------------------------------------------------------------------------------
df.1$state: BB
apples cherries plums
-111 -222 -333
by(df.1, df.1$state, function(x) colSums(x[,3:5]))
df.2 <- read.table(text = '
state county apples cherries plums
AA 1 1 2 3
AA 1 1 2 3
AA 2 10 20 30
AA 2 10 20 30
AA 3 100 200 300
AA 3 100 200 300
BB 7 -1 -2 -3
BB 7 -1 -2 -3
BB 8 -10 -20 -30
BB 8 -10 -20 -30
BB 9 -100 -200 -300
BB 9 -100 -200 -300
', header = TRUE, stringsAsFactors = FALSE)
# my function works
tapply(df.2$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)})