r - 金额(仅限整数)在不同桶中的分布组合

Question

假设有 5 个桶 (1 - 5)，并且为每个桶分配一个(整数)值。例如

> bucket = 1:5
> value = c(14, 12, 9, 20, 7)
> data.frame(bucket, value)
  bucket value
1      1    14
2      2    12
3      3     9
4      4    20
5      5     7

然后，要求将桶中的值增加到 3(仅在允许的情况下增加)。有多种方法可以在这些存储桶上分配总共 3 个(仅随着允许的整数而增加)。

R 中是否有一个函数可以给出 5 个桶中总计 3 的所有可能分布？

更具体地说，类似于:

  distr1 distr2 distr3 distr4 distr5 distr6 ....
1      3      2      2      2      2      1 ....
2      0      1      0      0      0      2 ....
3      0      0      1      0      0      0 ....
4      0      0      0      1      0      0 ....
5      0      0      0      0      1      0 ....

我研究了 combn() 和 expand.grid()，但那些似乎不是合适的函数......

Answer 1

这是一种可能性，即 expand.grid ，但可能不是最优雅的:

n_buckets <- 5
increase <- 3

foo <- do.call(
  expand.grid, 
  replicate(increase, seq_len(n_buckets), simplify = FALSE)
)

res <- apply(foo, 1, function(x) {
  sapply(seq_len(n_buckets), function(y) sum(y == x))
})


#      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11]
# [1,]    3    2    2    2    2    2    1    1    1     1     2
# [2,]    0    1    0    0    0    1    2    1    1     1     0
# [3,]    0    0    1    0    0    0    0    1    0     0     1
# [4,]    0    0    0    1    0    0    0    0    1     0     0
# [5,]    0    0    0    0    1    0    0    0    0     1     0
#      [,12] [,13] [,14] [,15] [,16] [,17] [,18] [,19] [,20]
# [1,]     1     1     1     1     2     1     1     1     1
# [2,]     1     0     0     0     0     1     0     0     0
# [3,]     1     2     1     1     0     0     1     0     0
# [4,]     0     0     1     0     1     1     1     2     1
# [5,]     0     0     0     1     0     0     0     0     1
#
# ...

注意:刚刚注意到我的答案中有冗余，因为我区分了 a在桶里1和b在桶里2 ，和b在桶里1和a在桶里2 。您仍然可以获得所有可能性...

编辑:删除重复项:

res[, !duplicated(res, MARGIN = 2)]

#      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11]
# [1,]    3    2    2    2    2    1    1    1    1     1     1
# [2,]    0    1    0    0    0    2    1    1    1     0     0
# [3,]    0    0    1    0    0    0    1    0    0     2     1
# [4,]    0    0    0    1    0    0    0    1    0     0     1
# [5,]    0    0    0    0    1    0    0    0    1     0     0
#      [,12] [,13] [,14] [,15] [,16] [,17] [,18] [,19] [,20] [,21]
# [1,]     1     1     1     1     0     0     0     0     0     0
# [2,]     0     0     0     0     3     2     2     2     1     1
# [3,]     1     0     0     0     0     1     0     0     2     1
# [4,]     0     2     1     0     0     0     1     0     0     1
# [5,]     1     0     1     2     0     0     0     1     0     0
#      [,22] [,23] [,24] [,25] [,26] [,27] [,28] [,29] [,30] [,31]
# [1,]     0     0     0     0     0     0     0     0     0     0
# [2,]     1     1     1     1     0     0     0     0     0     0
# [3,]     1     0     0     0     3     2     2     1     1     1
# [4,]     0     2     1     0     0     1     0     2     1     0
# [5,]     1     0     1     2     0     0     1     0     1     2
#      [,32] [,33] [,34] [,35]
# [1,]     0     0     0     0
# [2,]     0     0     0     0
# [3,]     0     0     0     0
# [4,]     3     2     1     0
# [5,]     0     1     2     3