在一次调用中按组对多个变量应用多个汇总函数

Question

我有以下数据框架

x <- read.table(text = "  id1 id2 val1 val2
1   a   x    1    9
2   a   x    2    4
3   a   y    3    5
4   a   y    4    9
5   b   x    1    7
6   b   y    4    4
7   b   x    3    9
8   b   y    2    8", header = TRUE)

我想计算按id1和id2分组的val1和val2的平均值,并同时计算每个id1-id2组合的行数.我可以单独执行每个计算:

# calculate mean
aggregate(. ~ id1 + id2, data = x, FUN = mean)

# count rows
aggregate(. ~ id1 + id2, data = x, FUN = length)

为了在一次通话中进行两次计算,我试过了

do.call("rbind", aggregate(. ~ id1 + id2, data = x, FUN = function(x) data.frame(m = mean(x), n = length(x))))

但是,我收到一个乱码输出和警告:

#     m   n
# id1 1   2
# id2 1   1
#     1.5 2
#     2   2
#     3.5 2
#     3   2
#     6.5 2
#     8   2
#     7   2
#     6   2
# Warning message:
#   In rbind(id1 = c(1L, 2L, 1L, 2L), id2 = c(1L, 1L, 2L, 2L), val1 = list( :
#   number of columns of result is not a multiple of vector length (arg 1)

我可以使用plyr包,但是当数据集的大小增加时,我的数据集非常大并且plyr非常慢(几乎不可用).

如何aggregate在一次通话中使用或其他功能执行多项计算？

Answer 1

您可以一步完成所有操作并获得适当的标签:

> aggregate(. ~ id1+id2, data = x, FUN = function(x) c(mn = mean(x), n = length(x) ) )
#   id1 id2 val1.mn val1.n val2.mn val2.n
# 1   a   x     1.5    2.0     6.5    2.0
# 2   b   x     2.0    2.0     8.0    2.0
# 3   a   y     3.5    2.0     7.0    2.0
# 4   b   y     3.0    2.0     6.0    2.0

这将创建一个包含两个id列和两个矩阵列的数据框:

str( aggregate(. ~ id1+id2, data = x, FUN = function(x) c(mn = mean(x), n = length(x) ) ) )
'data.frame':   4 obs. of  4 variables:
 $ id1 : Factor w/ 2 levels "a","b": 1 2 1 2
 $ id2 : Factor w/ 2 levels "x","y": 1 1 2 2
 $ val1: num [1:4, 1:2] 1.5 2 3.5 3 2 2 2 2
  ..- attr(*, "dimnames")=List of 2
  .. ..$ : NULL
  .. ..$ : chr  "mn" "n"
 $ val2: num [1:4, 1:2] 6.5 8 7 6 2 2 2 2
  ..- attr(*, "dimnames")=List of 2
  .. ..$ : NULL
  .. ..$ : chr  "mn" "n"

正如下面的@ lord.garbage所指出的那样,可以通过使用将其转换为带有"简单"列的数据框 do.call(data.frame, ...)

str( do.call(data.frame, aggregate(. ~ id1+id2, data = x, FUN = function(x) c(mn = mean(x), n = length(x) ) ) ) 
    )
'data.frame':   4 obs. of  6 variables:
 $ id1    : Factor w/ 2 levels "a","b": 1 2 1 2
 $ id2    : Factor w/ 2 levels "x","y": 1 1 2 2
 $ val1.mn: num  1.5 2 3.5 3
 $ val1.n : num  2 2 2 2
 $ val2.mn: num  6.5 8 7 6
 $ val2.n : num  2 2 2 2

这是LHS上多个变量的语法:

aggregate(cbind(val1, val2) ~ id1 + id2, data = x, FUN = function(x) c(mn = mean(x), n = length(x) ) )

Answer 2

鉴于此问题:

我可以使用plyr包,但是当数据集的大小增加时,我的数据集非常大并且plyr非常慢(几乎不可用).

然后在data.table(1.9.4+)你可以尝试:

> DT
   id1 id2 val1 val2
1:   a   x    1    9
2:   a   x    2    4
3:   a   y    3    5
4:   a   y    4    9
5:   b   x    1    7
6:   b   y    4    4
7:   b   x    3    9
8:   b   y    2    8

> DT[ , .(mean(val1), mean(val2), .N), by = .(id1, id2)]   # simplest
   id1 id2  V1  V2 N
1:   a   x 1.5 6.5 2
2:   a   y 3.5 7.0 2
3:   b   x 2.0 8.0 2
4:   b   y 3.0 6.0 2

> DT[ , .(val1.m = mean(val1), val2.m = mean(val2), count = .N), by = .(id1, id2)]  # named
   id1 id2 val1.m val2.m count
1:   a   x    1.5    6.5     2
2:   a   y    3.5    7.0     2
3:   b   x    2.0    8.0     2
4:   b   y    3.0    6.0     2

> DT[ , c(lapply(.SD, mean), count = .N), by = .(id1, id2)]   # mean over all columns
   id1 id2 val1 val2 count
1:   a   x  1.5  6.5     2
2:   a   y  3.5  7.0     2
3:   b   x  2.0  8.0     2
4:   b   y  3.0  6.0     2

对于时间比较aggregate(用于问题和所有其他3个答案),data.table看看 这个基准(agg和agg.x案例).

Answer 3

你可以添加一个count列,聚合sum,然后缩小以获得mean:

x$count <- 1
agg <- aggregate(. ~ id1 + id2, data = x,FUN = sum)
agg
#   id1 id2 val1 val2 count
# 1   a   x    3   13     2
# 2   b   x    4   16     2
# 3   a   y    7   14     2
# 4   b   y    6   12     2

agg[c("val1", "val2")] <- agg[c("val1", "val2")] / agg$count
agg
#   id1 id2 val1 val2 count
# 1   a   x  1.5  6.5     2
# 2   b   x  2.0  8.0     2
# 3   a   y  3.5  7.0     2
# 4   b   y  3.0  6.0     2

它具有保留列名和创建单个count列的优点.

Answer 4

也许你想合并？

x.mean <- aggregate(. ~ id1+id2, p, mean)
x.len  <- aggregate(. ~ id1+id2, p, length)

merge(x.mean, x.len, by = c("id1", "id2"))

  id1 id2 val1.x val2.x val1.y val2.y
1   a   x    1.5    6.5      2      2
2   a   y    3.5    7.0      2      2
3   b   x    2.0    8.0      2      2
4   b   y    3.0    6.0      2      2

Answer 5

使用dplyr包你可以通过使用来实现这一点summarise_all.有了这个汇总功能可以应用其他功能(在这种情况下mean和n())到每个非分组列:

x %>%
  group_by(id1, id2) %>%
  summarise_all(funs(mean, n()))

这使:

     id1    id2 val1_mean val2_mean val1_n val2_n
1      a      x       1.5       6.5      2      2
2      a      y       3.5       7.0      2      2
3      b      x       2.0       8.0      2      2
4      b      y       3.0       6.0      2      2

如果您不想将函数应用于所有非分组列,请指定应应用它们的列,或者使用summarise_at()函数排除非想要的减号:

# inclusion
x %>%
  group_by(id1, id2) %>%
  summarise_at(vars(val1, val2), funs(mean, n()))

# exclusion
x %>%
  group_by(id1, id2) %>%
  summarise_at(vars(-val2), funs(mean, n()))