列数 [英] R ave by columns
问题描述
我想在数据框中的许多列（十个）上使用 ave 函数：
I want to use the ave function on many columns (tens) on the data frame:
ave(df[,the_cols], df[,c('site', 'month')], FUN = mean)
问题是 ave 在平均值函数上运行所有 the_cols 列在一起。有没有办法分别对每个 the_cols 列运行它？
The problem is that ave runs the mean function on all the_cols columns together. Is there any way to run it for each of the_cols columns separately?
我试图查看其他功能。 tapply 和 aggregate 不同，它们每组仅返回一行。我需要 ave 行为，即返回与原始 df 相同的行数。还有一个 by 函数，但是使用它非常笨拙，因为它返回了一个复杂的列表结构，必须以某种方式进行转换。
I tried to look at the other functions. tapply and aggregate are different, they return only one row per group. I need the ave behaviour, i.e. to return the same number of rows as given in the original df. There is also a by function, but using it would be very clumsy as it returns a complicated list structure that would have to be converted somehow.
肯定存在许多笨拙和丑陋的解决方案（通过& do.call，多个* apply函数调用等），但是真的有一些简单而优雅的方法吗？
Certainly many clumsy and ugly (by & do.call, multiple *apply function calls etc.) solutions exist but is there some really easy and elegant?
推荐答案
也许我遗漏了一些东西，但是这里的 apply（）方法会很好地工作，并且不会丑陋或需要任何丑陋的骇客。一些伪数据：
Perhaps I'm missing something, but an apply() approach here would work very well and wouldn't be ugly or require any ugly hacks. Some dummy data:
df <- data.frame(A = rnorm(20), B = rnorm(20), site = gl(5,4), month = gl(10, 2))
出了什么问题
sapply(df[, c("A","B")], ave, df$site, df$month)
？如果确实要通过 data.frame（）将其强制为数据框。
? Coerce that to a data frame via data.frame() if you really want that.
R> sapply(df[, c("A","B")], ave, df$site, df$month)
A B
[1,] 0.0775 0.04845
[2,] 0.0775 0.04845
[3,] -1.5563 0.43443
[4,] -1.5563 0.43443
[5,] 0.7193 0.01151
[6,] 0.7193 0.01151
[7,] -0.9243 -0.28483
[8,] -0.9243 -0.28483
[9,] 0.3316 0.14473
[10,] 0.3316 0.14473
[11,] -0.2539 0.20384
[12,] -0.2539 0.20384
[13,] 0.5558 -0.37239
[14,] 0.5558 -0.37239
[15,] 0.1976 -0.22693
[16,] 0.1976 -0.22693
[17,] 0.2031 1.11041
[18,] 0.2031 1.11041
[19,] 0.3229 -0.53818
[20,] 0.3229 -0.53818
将其放在一起，怎么样
AVE <- function(df, cols, ...) {
dots <- list(...)
out <- sapply(df[, cols], ave, ...)
out <- data.frame(as.data.frame(dots), out)
names(out) <- c(paste0("Fac", seq_along(dots)), cols)
out
}
R> AVE(df, c("A","B"), df$site, df$month)
Fac1 Fac2 A B
1 1 1 0.0775 0.04845
2 1 1 0.0775 0.04845
3 1 2 -1.5563 0.43443
4 1 2 -1.5563 0.43443
5 2 3 0.7193 0.01151
6 2 3 0.7193 0.01151
7 2 4 -0.9243 -0.28483
8 2 4 -0.9243 -0.28483
9 3 5 0.3316 0.14473
10 3 5 0.3316 0.14473
11 3 6 -0.2539 0.20384
12 3 6 -0.2539 0.20384
13 4 7 0.5558 -0.37239
14 4 7 0.5558 -0.37239
15 4 8 0.1976 -0.22693
16 4 8 0.1976 -0.22693
17 5 9 0.2031 1.11041
18 5 9 0.2031 1.11041
19 5 10 0.3229 -0.53818
20 5 10 0.3229 -0.53818
与 ... 一起工作的细节此刻让我迷惑了，但您应该能够为 Fac1取个更好的名字等等。
The details of working with ... escape me at the moment, but you should be able to get better names for the Fac1 etc that I used here.
我会扔一个替代表示形式： aggregate（），但使用 ave（）函数而不是 mean（）：
I'll throw an alternative representation out there for you: aggregate() but use the ave() function instead of mean():
R> aggregate(cbind(A, B) ~ site + month, data = df, ave)
site month A.1 A.2 B.1 B.2
1 1 1 0.0775 0.0775 0.04845 0.04845
2 1 2 -1.5563 -1.5563 0.43443 0.43443
3 2 3 0.7193 0.7193 0.01151 0.01151
4 2 4 -0.9243 -0.9243 -0.28483 -0.28483
5 3 5 0.3316 0.3316 0.14473 0.14473
6 3 6 -0.2539 -0.2539 0.20384 0.20384
7 4 7 0.5558 0.5558 -0.37239 -0.37239
8 4 8 0.1976 0.1976 -0.22693 -0.22693
9 5 9 0.2031 0.2031 1.11041 1.11041
10 5 10 0.3229 0.3229 -0.53818 -0.53818
请注意相当清楚的输出，但这是可以根据需要轻松地重塑形状。
Note quite the stated output, but it is something that is simple to reshape if needed.
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