计算唯一的值组合

Question

我的数据框看起来像这样:

ID | value 1 | value 2 | value 3 | value 4
1  |    M    |    D    |    F    |   A
2  |    F    |    M    |    G    |   B
3  |    M    |    D    |    F    |   A
4  |    L    |    D    |    E    |   B

我想得到这样的东西.

value 1 | value 2 | value 3 | value 4|  Number of combinations
  M     |    D    |    F    |   A    |     2
  F     |    M    |    G    |   B    |     1
  L     |    D    |    E    |   B    |     1

例如,计算列值1 - 值4的唯一组合的数量.

Answer 1

count在plyr包中将完成该任务.

> df
  ID   value.1   value.2   value.3 value.4
1  1     M         D         F           A
2  2     F         M         G           B
3  3     M         D         F           A
4  4     L         D         E           B
> library(plyr)
> count(df[, -1])
    value.1   value.2   value.3 value.4 freq
1     F         M         G           B    1
2     L         D         E           B    1
3     M         D         F           A    2

Answer 2

N <- 10000

d <- data.frame(
  ID=seq(1, N), 
  v1=sample(c("M","F", "M", "L"), N, replace = TRUE), 
  v2=sample(c("D","M","D","D"), N, replace = TRUE), 
  v3=sample(c("F","G","F","E"), N, replace = TRUE),
  v4=sample(c("A","B","A","B"), N, replace = TRUE)
)

使用data.table(最快)

dt <- data.table::as.data.table(d)
dt[, .N, by = c('v1','v2','v3','v4')]

用dplyr

dplyr::count_(d, vars = c('v1','v2','v3','v4'))

随着plyr

plyr::count(d, vars = c('v1','v2','v3','v4'))
plyr::ddply(d, .variables = c('v1','v2','v3','v4'), nrow)

使用聚合(最慢)

aggregate(ID ~ ., d, FUN = length)

基准

microbenchmark::microbenchmark(dt[, .N, by = c('v1','v2','v3','v4')],
                               plyr::count(d, vars = c('v1','v2','v3','v4')),
                               plyr::ddply(d, .variables = c('v1','v2','v3','v4'), nrow),
                               dplyr::count_(d, vars = c('v1','v2','v3','v4')),
                               aggregate(ID ~ ., d, FUN = length), 
                               times = 1000)

Unit: microseconds
                                                         expr      min       lq      mean   median        uq        max neval  cld
                     dt[, .N, by = c("v1", "v2", "v3", "v4")]  887.807 1107.543  1263.777 1174.258  1289.724   4263.156  1000 a   
             plyr::count(d, vars = c("v1", "v2", "v3", "v4")) 3912.791 4270.387  5379.080 4498.053  5791.743 157146.103  1000   c 
 plyr::ddply(d, .variables = c("v1", "v2", "v3", "v4"), nrow) 7737.874 8553.370 10630.849 9018.266 11126.517 187301.696  1000    d
           dplyr::count_(d, vars = c("v1", "v2", "v3", "v4")) 2126.913 2432.957  2763.499 2568.251  2789.386  12549.669  1000  b  
                           aggregate(ID ~ ., d, FUN = length) 7395.440 8121.828 10546.659 8776.371 10858.263 210139.759  1000    d

最好简单地使用data.table而不是data.frame最快,并且不需要其他函数或库来计算.另请注意,aggregate函数在大型数据集上的执行速度要慢得多.

最后注意事项:随意使用新方法进行更新.

Answer 3

没有plyr.

aggregate(ID ~ ., d, FUN=length)# . means all variables in d except ID