Question

我们说我有data.frame

DF = structure(list(AE = c(148, 1789, 1223, 260, 1825, 37, 1442, 484, 
10, 163, 1834, 254, 445, 837, 721, 1904, 1261, 382, 139, 213), 
    FW = structure(c(1L, 3L, 2L, 3L, 3L, 1L, 2L, 3L, 2L, 2L, 
    3L, 2L, 3L, 2L, 1L, 3L, 1L, 1L, 1L, 3L), .Label = c("LYLR", 
    "OCXG", "BIYX"), class = "factor"), CP = c("WYB/NXO", "HUK/NXO", 
    "HUK/WYB", "HUK/NXO", "WYB/NXO", "HUK/WYB", "HUK/NXO", "HUK/NXO", 
    "WYB/NXO", "HUK/NXO", "WYB/NXO", "HUK/NXO", "HUK/WYB", "WYB/NXO", 
    "HUK/WYB", "WYB/NXO", "WYB/NXO", "HUK/WYB", "WYB/NXO", "WYB/NXO"
    ), SD = c(1, 1, -1, 1, 1, 1, 1, -1, 1, 1, -1, -1, 1, -1, 
    -1, 1, -1, 1, 1, 1)), .Names = c("AE", "FW", "CP", "SD"), row.names = c(NA, -20L), class = "data.frame")

或以人类可读的格式：

     AE   FW      CP SD
1   148 LYLR WYB/NXO  1
2  1789 BIYX HUK/NXO  1
3  1223 OCXG HUK/WYB -1
4   260 BIYX HUK/NXO  1
5  1825 BIYX WYB/NXO  1
6    37 LYLR HUK/WYB  1
7  1442 OCXG HUK/NXO  1
8   484 BIYX HUK/NXO -1
9    10 OCXG WYB/NXO  1
10  163 OCXG HUK/NXO  1
11 1834 BIYX WYB/NXO -1
12  254 OCXG HUK/NXO -1
13  445 BIYX HUK/WYB  1
14  837 OCXG WYB/NXO -1
15  721 LYLR HUK/WYB -1
16 1904 BIYX WYB/NXO  1
17 1261 LYLR WYB/NXO -1
18  382 LYLR HUK/WYB  1
19  139 LYLR WYB/NXO  1
20  213 BIYX WYB/NXO  1

作为data.table，

DT = data.table(DF)
setkey(DT, CP)

现在考虑以下两个操作：

  DT[, amount_sum_fh   :=  DT[.(CP = CP), 
                     on = .(CP), mean(AE * SD), by=.EACHI]$V1]
  DT[,  amount_sum_sh   :=  DT[.(CP = CP), 
                     on = .(CP), mean(AE), by=.EACHI]$V1]

有没有办法一举完成？

Answer 1

问题，评论和答案到目前为止已经提出了三种方法：

涉及联接的OP posted代码。
Frank suggested data.table方法，无需加入。
Zelazny7 gave基础R答案。

要确定替代品的最快，可以使用microbenchmark包：

library(data.table)
DT = data.table(DF)
setkey(DT, CP)

mb <- microbenchmark::microbenchmark(
  OP = {
    DT[, amount_sum_fh   :=  DT[.(CP = CP), 
                                on = .(CP), mean(AE * SD), by=.EACHI]$V1]
    DT[,  amount_sum_sh   :=  DT[.(CP = CP), 
                                 on = .(CP), mean(AE), by=.EACHI]$V1]
  },
  Frank = DT[, `:=`(amount_sum_fh = mean(AE*SD), amount_sum_sh = mean(AE)), by = CP],
  DF = transform(DF, 
                 amount_sum_fh = ave(AE * SD, CP, FUN = mean),
                 amount_sum_sh = ave(AE, CP, FUN = mean)),
  times = 100L
)

mb
#Unit: microseconds
#  expr      min        lq      mean    median        uq      max neval cld
#    OP 4090.271 4288.2800 4614.9625 4417.2700 4633.7880 7470.179   100   c
# Frank  548.833  612.9355  687.6306  643.5160  711.5745 1142.041   100 a  
#    DF  725.649  769.8660  840.5960  811.9315  870.3365 1376.425   100  b

即使样本量相当小，Frank的data.table版本也比基本R版本快25％左右。

Answer 2

这是使用基数R的一种方法：

DF2 <- transform(DF, 
  amount_sum_fh=ave(AE * SD, CP, FUN = mean),
  amount_sum_sh=ave(AE, CP, FUN = mean))

但data.frame并未按CP排序。

> DF2
     AE   FW      CP SD amount_sum_fh amount_sum_sh
1   148 LYLR WYB/NXO  1      34.11111      907.8889
2  1789 BIYX HUK/NXO  1     486.00000      732.0000
3  1223 OCXG HUK/WYB -1    -216.00000      561.6000
4   260 BIYX HUK/NXO  1     486.00000      732.0000
5  1825 BIYX WYB/NXO  1      34.11111      907.8889
6    37 LYLR HUK/WYB  1    -216.00000      561.6000
7  1442 OCXG HUK/NXO  1     486.00000      732.0000
8   484 BIYX HUK/NXO -1     486.00000      732.0000
9    10 OCXG WYB/NXO  1      34.11111      907.8889
10  163 OCXG HUK/NXO  1     486.00000      732.0000
11 1834 BIYX WYB/NXO -1      34.11111      907.8889
12  254 OCXG HUK/NXO -1     486.00000      732.0000
13  445 BIYX HUK/WYB  1    -216.00000      561.6000
14  837 OCXG WYB/NXO -1      34.11111      907.8889
15  721 LYLR HUK/WYB -1    -216.00000      561.6000
16 1904 BIYX WYB/NXO  1      34.11111      907.8889
17 1261 LYLR WYB/NXO -1      34.11111      907.8889
18  382 LYLR HUK/WYB  1    -216.00000      561.6000
19  139 LYLR WYB/NXO  1      34.11111      907.8889
20  213 BIYX WYB/NXO  1      34.11111      907.8889

一举创建多列？

2 个答案: