英文:
purrr - how to apply recursively a function with changing arguments
问题
理想情况下，我想要利用purrr的accumulate函数或类似的功能。
假设我想要迭代地使用utils::combn函数，并获取所有中间结果（最好放在一个列表的列表中）。
在下面的示例中，最初，参数x = 4，因此m也将是4（但(x, m)可以是(5, 5)，(6, 6)，...）。然后，在第一个循环之后，x将成为先前的结果，而m将逐渐减少，直到m = 2。
n1 <- combn(x = 4, m = 4, simplify = FALSE)
n2 <- map(n1, ~ combn(.x, 3, simplify = FALSE))
n3 <- map(n2, ~ map(., ~ combn(.x, 2, simplify = FALSE)))
你可以想象，我想要获取所有可能的组合，例如：
choose(4, 4) -> choose(result, 3) -> choose(result, 2)。
我该如何实现这一点？
英文:
Ideally I would like to make use of purrr's accumulate function or similar.
Let's say I want to make use of utils::combn function iteratively, and get all the intermediate results (ideally put inside a list of lists).
In example below, initially, parameter x = 4, thus m will be also 4 (but (x, m) could be (5, 5), (6, 6), ...). Then, after first loop, x will be previous result, whilst m goes down by one, iteratively until m = 2.
n1 <- combn(x = 4, m = 4, simplify = FALSE)
n2 <- map(n1, ~ combn(.x, 3, simplify = FALSE))
n3 <- map(n2, ~ map(., ~ combn(.x, 2, simplify = FALSE)))
> n1
[[1]]
[1] 1 2 3 4
> n2
[[1]]
[[1]][[1]]
[1] 1 2 3
[[1]][[2]]
[1] 1 2 4
[[1]][[3]]
[1] 1 3 4
[[1]][[4]]
[1] 2 3 4
> n3
[[1]]
[[1]][[1]]
[[1]][[1]][[1]]
[1] 1 2
[[1]][[1]][[2]]
[1] 1 3
[[1]][[1]][[3]]
[1] 2 3
[[1]][[2]]
[[1]][[2]][[1]]
[1] 1 2
[[1]][[2]][[2]]
[1] 1 4
[[1]][[2]][[3]]
[1] 2 4
[[1]][[3]]
[[1]][[3]][[1]]
[1] 1 3
[[1]][[3]][[2]]
[1] 1 4
[[1]][[3]][[3]]
[1] 3 4
[[1]][[4]]
[[1]][[4]][[1]]
[1] 2 3
[[1]][[4]][[2]]
[1] 2 4
[[1]][[4]][[3]]
[1] 3 4
As you can imagine, I want to get all possible combinations, e.g.:
> choose(4, 4) -> choose(result, 3) -> choose(result, 2).
How can I do this?
答案1
得分: 4
你可以使用accumulate + map_depth：
combn_recur <- function(n) {
accumulate(c(n, 0:(n-2)),
~ map_depth(.x, .y, combn, m = n-.y, simplify = FALSE))[-1]
}
all.equal(combn_recur(4), c(n1, n2, n3))
# TRUE
combn_recur(3)
# [[1]]
# [1] 1 2 3
#
# [[2]]
# [[2]][[1]]
# [1] 1 2
#
# [[2]][[2]]
# [1] 1 3
#
# [[2]][[3]]
# [1] 2 3
combn_recur(2)
# [[1]]
# [1] 1 2
combn_recur(1)
# Error in .f(.x[[i]], ...) : n < m
(Note: The code remains in its original language as requested.)
英文:
You can use accumulate + map_depth:
combn_recur <- function(n) {
accumulate(c(n, 0:(n-2)),
~ map_depth(.x, .y, combn, m = n-.y, simplify = FALSE))[-1]
}
all.equal(combn_recur(4), c(n1, n2, n3))
# TRUE
combn_recur(3)
# [[1]]
# [1] 1 2 3
#
# [[2]]
# [[2]][[1]]
# [1] 1 2
#
# [[2]][[2]]
# [1] 1 3
#
# [[2]][[3]]
# [1] 2 3
combn_recur(2)
# [[1]]
# [1] 1 2
combn_recur(1)
# Error in .f(.x[[i]], ...) : n < m
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