# 使用data.table的嵌套分组中的前n个

``````2019 Q1  Klaus 2
2019 Q2   Karl 3
``````

As this is just a toy example going forward I also want to have the top 4, 5 etc by count per quarter and name. Do you have any good ideas how to implement this with `data.table` (no `dplyr` please). Many thanks!

``````library(data.table)

dt <- data.table(x = c("2019 Q1", "2019 Q1", "2019 Q1", "2019 Q2", "2019 Q2", "2019 Q2", "2019 Q2"),
y = c("Klaus", "Gustav", "Klaus", "Karl", "Karl", "Karl", "Stefan"))

# Structure of dt
# x      y
# 1: 2019 Q1  Klaus
# 2: 2019 Q1 Gustav
# 3: 2019 Q1  Klaus
# 4: 2019 Q2   Karl
# 5: 2019 Q2   Karl
# 6: 2019 Q2   Karl
# 7: 2019 Q2 Stefan

dt[, .N, by = .(x, y)]

# Output:
# x      y N
# 1: 2019 Q1  Klaus 2
# 2: 2019 Q1 Gustav 1
# 3: 2019 Q2   Karl 3
# 4: 2019 Q2 Stefan 1
``````

• ~签名 回复

Here is a base R solution using `aggregate`

``````> aggregate(y~x,dt,function(v) as.matrix(head(data.frame(sort(table(v),decreasing = TRUE)),1)))
x   y.1 y.2
1 2019 Q1 Klaus   2
2 2019 Q2  Karl   3
``````
• 龙舌兰 回复

here is another `data.table` approach, almost the same as Gilean's answer, but without `head()`.

``````dt[, .N, by = .(x,y) ][ order(-N), .SD[1:1], by = x ]

#          x     y N
# 1: 2019 Q2  Karl 3
# 2: 2019 Q1 Klaus 2
``````
• 笙歌i 回复

您可以首先计算每个名称和每个季度的N个，然后对data.table进行排序，然后选择每个季度的前n行：

``````dt[, .N, by = .(x, y)][order(-N), head(.SD, 1), by = x]
``````