aL3*_*3xa 6 r aggregate-functions aggregation plyr reshape
我用来做那种-的melt和cast所有的时间,而这个时候,我正在寻找一个整齐的衬垫.
require(reshape)
# first I melt some data:
m <- melt(mtcars, id.vars = c("cyl", "am"), measure.vars = "hp")
# then cast it:
cast(m, cyl + am ~ ., each(min, mean, sd, max))
cyl am min mean sd max
1 4 0 62 84.66667 19.65536 97
2 4 1 52 81.87500 22.65542 113
3 6 0 105 115.25000 9.17878 123
4 6 1 110 131.66667 37.52777 175
5 8 0 150 194.16667 33.35984 245
6 8 1 264 299.50000 50.20458 335
这是可能的ddply还是smth？我非常渴望一个班轮.我试过了:
ddply(mtcars, cyl + am ~ hp, each(min, max))
cyl am hp min max
1 4 0 62 0 146.7
2 4 0 95 0 140.8
3 4 0 97 0 120.1
4 4 1 52 1 75.7
5 4 1 65 1 71.1
6 4 1 66 1 79.0
7 4 1 91 0 120.3
8 4 1 93 1 108.0
9 4 1 109 1 121.0
10 4 1 113 1 113.0
11 6 0 105 0 225.0
12 6 0 110 0 258.0
13 6 0 123 0 167.6
14 6 1 110 0 160.0
15 6 1 175 0 175.0
16 8 0 150 0 318.0
17 8 0 175 0 400.0
18 8 0 180 0 275.8
19 8 0 205 0 472.0
20 8 0 215 0 460.0
21 8 0 230 0 440.0
22 8 0 245 0 360.0
23 8 1 264 0 351.0
24 8 1 335 0 335.0
当然,这是有效的,但不是hp通过cyl和总结am.我用了一段时间plyr,reshape所以我有点失去了我的肌肉...所以... 借口 - moi pour une question triviale ... = /
Cha*_*ase 12
summmarize 可能是你的朋友:
ddply(m, c("cyl", "am"), summarize
, min = min(value)
, mean = mean(value)
, sd = sd(value)
, max = max(value)
)
cyl am min mean sd max
1 4 0 62 84.66667 19.65536 97
2 4 1 52 81.87500 22.65542 113
3 6 0 105 115.25000 9.17878 123
4 6 1 110 131.66667 37.52777 175
5 8 0 150 194.16667 33.35984 245
6 8 1 264 299.50000 50.20458 335
使用plyr:
> require(plyr)
> ddply(mtcars,c("cyl","am"),summarise, min=min(hp), mean=mean(hp), sd=sd(hp), max=max(hp))
cyl am min mean sd max
1 4 0 62 84.66667 19.65536 97
2 4 1 52 81.87500 22.65542 113
3 6 0 105 115.25000 9.17878 123
4 6 1 110 131.66667 37.52777 175
5 8 0 150 194.16667 33.35984 245
6 8 1 264 299.50000 50.20458 335