python递归列表组合而不使用生成器 [英] python recursion list combinations without using generator

问题描述

我正在学习python3.为了更多地考虑递归，我想实现一个函数comb(n，k)，该函数返回一个列表，其中包含集合{1,2，…，n}中kk元素的所有组合.

I am learning python3. To think more about recursion, I want to implement a function comb(n, k) that returns a list consisting of all the combinations of kk elements out of a set {1,2,…,n}.

我认为使用循环是不明智的，因为嵌套循环的数量取决于k.因此，我以递归的方式考虑它.我尝试编写受此问题启发的函数虽然我找不到正确的答案.

I think it's not wise to use the loop since the number of the nested loop depends on k. So I consider it with recursion. I try to write the function inspired by This question while I can't get the right answer.

def combinations(sub, data_set, index, still_needed):
    if still_needed == 0:
        return sub

    for i in range(index, len(data_set)):
        sub.append(data_set[i])
        still_needed = still_needed - 1
        return combinations(sub, data_set, index+1, still_needed)

def comb(n, k):
    data_set = list(range(1, n+1))
    print (combinations([], data_set, 0, k))

如果我测试Comb(6,3)，我只会得到[1,2,3].我想获得所有组合.我的代码有什么问题?还是错过了重要的事情?我只想学习python的递归，这不是一项家庭作业，谢谢.

If I test Comb(6,3), I only get [1,2,3]. I want to get all combinations. What's the problem in my code? or something important missed? I just want to learn the recursion of python and it is not a homework, thanks.

预期结果如下:

[[1, 5, 6],
[2, 5, 6],
[3, 5, 6],
[4, 5, 6],
[1, 4, 6],
[2, 4, 6],
[3, 4, 6],
[1, 3, 6],
[2, 3, 6],
[1, 2, 6],
[1, 4, 5],
[2, 4, 5],
[3, 4, 5],
[1, 3, 5],
[2, 3, 5],
[1, 2, 5],
[1, 3, 4],
[2, 3, 4],
[1, 2, 4],
[1, 2, 3]]

虽然顺序并不重要.我将很感激，如果有任何pythonic的方法来解决这个问题，例如.嵌套的 [可迭代的项目表达式] (因为我尝试过但失败了).

While the order is not important. And I will appreciate that if there is any pythonic way to solve this question eg. nested [expression for item in iterable](since I have tried it but failed).

再次感谢.

推荐答案

函数中的问题是，在 for 循环中有一个 return 语句:它将停止在第一次迭代中执行该功能.

The problem in your function is that you have a return statement inside your for loop: it stops execution of the function during the first iteration.

这是您可以用于递归的基本结构:

Here's the basic structure you could use for recursion:

def combinations(n, k, min_n=0, accumulator=None):
    if accumulator is None:
        accumulator = []
    if k == 0:
        return [accumulator]
    else:
        return [l for x in range(min_n, n)
                for l in combinations(n, k - 1, x + 1, accumulator + [x + 1])]

print(combinations(6, 3))
# [[1, 2, 3], [1, 2, 4], [1, 2, 5], [1, 2, 6], [1, 3, 4], [1, 3, 5], [1, 3, 6], [1, 4, 5], [1, 4, 6], [1, 5, 6], [2, 3, 4], [2, 3, 5], [2, 3, 6], [2, 4, 5], [2, 4, 6], [2, 5, 6], [3, 4, 5], [3, 4, 6], [3, 5, 6], [4, 5, 6]]

要检查结果是否正确，可以针对 itertools 进行测试:

To check if the result's correct, you can test it against itertools:

import itertools
print(list(itertools.combinations(range(1,7),3)))
# [(1, 2, 3), (1, 2, 4), (1, 2, 5), (1, 2, 6), (1, 3, 4), (1, 3, 5), (1, 3, 6), (1, 4, 5), (1, 4, 6), (1, 5, 6), (2, 3, 4), (2, 3, 5), (2, 3, 6), (2, 4, 5), (2, 4, 6), (2, 5, 6), (3, 4, 5), (3, 4, 6), (3, 5, 6), (4, 5, 6)]
print(
        list(itertools.combinations(range(1, 7), 3))
          ==
        [tuple(comb) for comb in combinations(6, 3)]
     )
# True

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