R数据表：将条件值与行值比较 [英] R data table: compare row value to group values, with condition

问题描述

这是问题的延续：

R数据表：将行值与组值比较

现在：

x = data.table( id=c(1,1,1,1,1,1,1,1), price = c(10, 10, 12, 12, 12, 15, 
8, 11), subgroup = c(1, 1, 1, 1, 1, 1, 2, 2))

   id price subgroup
1:  1    10        1
2:  1    10        1
3:  1    12        1
4:  1    12        1
5:  1    12        1
6:  1    15        1
7:  1     8        2
8:  1    11        2

$ b b

并希望计算每个ID的价格较低的行数，但只计算子组1中的行数

and would like to calculate the number of rows with lower prices per id, but only counting the ones in subgroup 1.

如果我使用：

x[,cheaper := rank(price, ties.method="min")-1, by=id]

结果是：

> x
   id price subgroup cheaper
1:  1    10        1       1   # only 1 is cheaper (row 7)
2:  1    10        1       1   # only 1 is cheaper (row 7)
3:  1    12        1       4   # 4 frows are cheaper (row 1,2,7,8)
4:  1    12        1       4   # etc
5:  1    12        1       4
6:  1    15        1       7
7:  1     8        2       0
8:  1    11        2       3

但我想要的结果是：

> x
   id price subgroup cheaper_in_subgroup_1
1:  1    10        1       0    # nobody in subgroup 1 is cheaper
2:  1    10        1       0    # nobody in subgroup 1 is cheaper
3:  1    12        1       2    # only row 1 and 2 are cheaper in subgroup 1
4:  1    12        1       2
5:  1    12        1       2
6:  1    15        1       5
7:  1     8        2       0    # nobody in subgroup 1 is cheaper
8:  1    11        2       2    # only row 1 and 2 are cheaper in subgroup 1

推荐答案

可能有更多的 data.table ish方式实现，在每个 id

There's probably a more data.tableish way achieving this, but here an attempt using vapply within each id

x[, cheaper := vapply(price, 
                      function(x) sum(price[subgroup == 1L] < x),
                      FUN.VALUE = integer(1L)), 
               by = id]
x
#    id price subgroup cheaper
# 1:  1    10        1       0
# 2:  1    10        1       0
# 3:  1    12        1       2
# 4:  1    12        1       2
# 5:  1    12        1       2
# 6:  1    15        1       5
# 7:  1     8        2       0
# 8:  1    11        2       2

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