R:在j子句中引用cut函数中的data.table字段

Question

基本上,我有以下2 data.tables:

dt- 包含值字段(y)和分组字段(x)
bk- 包含4个" break "字段(bn),用于描述x在中找到的每个组的区间[1,inf]的存储区结构dt.每个bn代表桶的最小值(包括)并延伸到下一个桶(例如:x= 1 的4个桶是[1,3],[3,5],[5,10],[10,inf] ). 请注意,桶结构不一定是唯一的.

> #4 groups (x), each with a bucket structure defined breaks (bn).
> bk<- data.table(x=c(1:4), b1=c(1,1,1,1), b2=c(3,3,4,4), b3=c(5,5,7,8), b4=c(10,10,10,10), key="x")
> bk
   x b1 b2 b3 b4
1: 1  1  3  5 10
2: 2  1  3  5 10
3: 3  1  4  7 10
4: 4  1  4  8 10
> dt<- data.table(x=rep(c(1:4),5), y=rep(c(1:10),2), key="x")
> dt
    x  y
 1: 1  1
 2: 1  5
 3: 1  9
 4: 1  3
 5: 1  7
 6: 2  2
 7: 2  6
 8: 2 10
 9: 2  4
10: 2  8
11: 3  3
12: 3  7
13: 3  1
14: 3  5
15: 3  9
16: 4  4
17: 4  8
18: 4  2
19: 4  6
20: 4 10

我的目标是添加一个字段b,以dt指示哪些桶(1,2,3,或4)在记录落在基于对应于该组的桶结构x.请参阅下面的所需输出:

    x  y  b
 1: 1  1  1  #Buckets for x=1
 2: 1  5  3
 3: 1  9  3
 4: 1  3  2
 5: 1  7  3
 6: 2  2  1  #Buckets for x=2 (same as 1)
 7: 2  6  3
 8: 2 10  4
 9: 2  4  2
10: 2  8  3
11: 3  3  1  #Buckets for x=3
12: 3  7  3
13: 3  1  1
14: 3  5  2
15: 3  9  3
16: 4  4  2  #Buckets for x=4
17: 4  8  3
18: 4  2  1
19: 4  6  2
20: 4 10  4

我最初的想法是加入两个data.tables并使用cut函数返回每个记录的桶号,但是我遇到了这个break问题.首次尝试如下:

> bkt[dt, .(x, y, b=cut(y, breaks=c(b1, b2, b3, b4, "inf"), include.lowest=TRUE, labels=c(1:4)))]
Error in cut.default(y, breaks = c(b1, b2, b3, b4, "inf"), include.lowest = TRUE,  : 
  'breaks' are not unique

如果我创建一个变量a来保存存储桶结构(例如,for x= 1),则以下工作方式与我预期的一样:

> a<- c(1, 3, 5, 10, "inf")
> bkt[dt, .(x, y, b=cut(y, breaks=a, include.lowest=TRUE, labels=c(1:4)))]
    x  y b
 1: 1  1 1
 2: 1  5 2
 3: 1  9 3
 4: 1  3 1
 5: 1  7 3
 6: 2  2 1
 7: 2  6 3
 8: 2 10 3
 9: 2  4 2
10: 2  8 3
11: 3  3 1
12: 3  7 3
13: 3  1 1
14: 3  5 2
15: 3  9 3
16: 4  4 2
17: 4  8 3
18: 4  2 1
19: 4  6 3
20: 4 10 3

对于我的应用程序,这仍然不是一个实用的解决方案,但我希望有人可以帮助我理解如何breaks正确地将桶结构信息传递给参数以获得类似的结果.我已经试过的各种组合c,list,unlist,as.numeric函数传递正确的break说法,但有没有运气.任何帮助/见解将不胜感激.谢谢!

完全披露,我是R的新手,这是我的第一篇文章,请温柔.

Answer 1

您可以使用melt.data.table将bk数据集重组为更简单的形式：

bk_long <- melt.data.table(
  bk,
  id.vars = 'x',
  measure.vars = paste0('b', 1:4),
  value.name = 'y'
)
setkey(bk_long, x)
bk_long[, variable := NULL]
bk_long[, b := seq_len(.N), by = x]
bk_long
#     x  y b
#  1: 1  1 1
#  2: 1  3 2
#  3: 1  5 3
#  4: 1 10 4
#  5: 2  1 1
#  6: 2  3 2
#  7: 2  5 3
#  8: 2 10 4
#  9: 3  1 1
# 10: 3  4 2
# 11: 3  7 3
# 12: 3 10 4
# 13: 4  1 1
# 14: 4  4 2
# 15: 4  8 3
# 16: 4 10 4

然后按照弗兰克的建议进行滚动连接：

bk_long[dt, on = c('x', 'y'), roll = TRUE]
#     x  y b
#  1: 1  1 1
#  2: 1  5 3
#  3: 1  9 3
#  4: 1  3 2
#  5: 1  7 3
#  6: 2  2 1
#  7: 2  6 3
#  8: 2 10 4
#  9: 2  4 2
# 10: 2  8 3
# 11: 3  3 1
# 12: 3  7 3
# 13: 3  1 1
# 14: 3  5 2
# 15: 3  9 3
# 16: 4  4 2
# 17: 4  8 3
# 18: 4  2 1
# 19: 4  6 2
# 20: 4 10 4