长度为 k 的子集？

Question

以下代码以优化的方式工作，我想理解它但失败了。我理解第一行，但第二行我迷路了。如果有人可以用 C 或 Python 来解释它，那将是很棒的，甚至是一般的想法。谢谢

combinations :: Int -> [a] -> [[a]]
combinations k xs = combinations' (length xs) k xs
  where combinations' n k' l@(y:ys) 
    | k' == 0   = [[]] 
    | k' >= n   = [l] 
    | null l    = [] 
    | otherwise = map (y :) (combinations' (n - 1) (k' - 1) ys) 

Answer 1

Python 中的等价物。请注意，这在 CPython 中非常糟糕，因为递归在那里实现得非常糟糕，并且在所有 Python 风格中都非常糟糕，因为列表的构造方式不同（1: [2, 3, 4]) 在 Haskell 中很简单，但在 Python 中做同样的事情需要一些 @ 987654323@ 或丑陋的[1] + [2,3,4] 东西）

from typing import TypeVar, List

A = TypeVar('A')
def combinations(k: int, xs: List[A]) -> List[List[A]]:
    def helper(n: int, k2: int, lst: List[A]) -> List[List[A]]:
        # base cases
        if k2 == 0: return [[]]     # if the requested subsequence length is zero
        elif k2 >= n: return [lst]  # if the requested subsequence length is longer than the whole sequence
        elif not lst: return []     # if the sequence itself is empty

        y, ys = lst[0], lst[1:]
        return [[y] + rest for rest in helper(n-1, k2-1, ys)]

    return helper(len(xs), k, xs)

请注意，Haskell 中过度使用守卫让我觉得有些丑陋。我可能会改写为：

combinations :: Int -> [a] -> [[a]]
combinations k xs = combinations' (length xs) k xs
  where
    combinations' _ 0  _  = [[]]
    combinations' _ _  [] = []
    combinations' n k' l@(y:ys)
      | k' >= n           = [l]
      | otherwise         = map (y :) (combinations' (n - 1) (k' - 1) ys) 

Answer 2

Khuldraeseth na'Barya in this question 提供了一种与模式匹配等效的简单替代方法：

subsetsOfLength 0 _  = [[]]
subsetsOfLength _ [] = []
subsetsOfLength n (x:xs) = map (x:) (subsetsOfLength (n-1) xs) ++ subsetsOfLength n xs