根据列中的值为每个组选择前 N 行

Selecting top N rows for each group based on value in column

我有如下数据框:-

x<-c(3,2,1,8,7,11,10,9,7,5,4)
y<-c("a","a","a", "b","b","c","c","c","c","c","c")
z<-c(2,2,2,1,1,3,3,3,3,3,3)
df<-data.frame(x,y,z)

df
    x y z
1   3 a 2
2   2 a 2
3   1 a 2
4   8 b 1
5   7 b 1
6  11 c 3
7  10 c 3
8   9 c 3
9   7 c 3
10  5 c 3
11  4 c 3

我想select每组按第 y 列排在前 n 行,其中 n 在第 z 列中提供。 所以输出应该是这样的:

output:
       x   y  z
     1 3   a  2
     2 2   a  2
     3 8   b  1
     4 11  c  3
     5 10  c  3
     6 9   c  3

data.table的一种方法:

library(data.table)
setDT(df)
df[,.(inc=seq_len(.N)<=z,x,z),by=.(y)][inc==T ,-2]
#   y  x z
#1: a  3 2
#2: a  2 2
#3: b  8 1
#4: c 11 3
#5: c 10 3
#6: c  9 3

基于 R 的解决方案:

# df is split according to y, then we keep only the top "z" value (after ordering x) 
# and rbind everything back together:
do.call(rbind, 
        lapply(split(df, df$y), 
               function(df1) df1[order(df1$x, decreasing=TRUE), ][1:unique(df1$z), ]))
#     x y z
#a.1  3 a 2
#a.2  2 a 2
#b    8 b 1
#c.6 11 c 3
#c.7 10 c 3
#c.8  9 c 3

编辑:
@mt1022:

的评论中提供了一种更直接的方式(仍在基础 R 中)
df[ave(1:nrow(df), df$y, FUN = seq_along) <= df$z, ]
#   x y z
#1  3 a 2
#2  2 a 2
#4  8 b 1
#6 11 c 3
#7 10 c 3
#8  9 c 3

dplyr 的解决方案使用 do:

df %>%
   group_by(y) %>%
   do(head(.,as.numeric(unique(.$z))))

我正在发布我使用 dplyr 寻找的解决方案。它基于@HNSKD:

library(dplyr)
x<-c(3,2,1,8,7,11,10,9,7,5,4)
y<-c("a","a","a", "b","b","c","c","c","c","c","c")
z<-c(2,2,2,1,1,3,3,3,3,3,3)

df<-data.frame(x,y,z)

df %>% group_by(y) %>% slice(1:2)

其中returns前两个元素各y:

# A tibble: 6 x 3
# Groups:   y [3]
      x y         z
  <dbl> <fct> <dbl>
1     3 a         2
2     2 a         2
3     8 b         1
4     7 b         1
5    11 c         3
6    10 c         3