我有一个包含15个数字的列表,我需要编写一些代码来生成这些数字的所有32,768种组合。

我已经找到了一些代码(通过谷歌搜索),显然做了我正在寻找的,但我发现代码相当不透明,并谨慎使用它。而且我觉得肯定有更优雅的解决方案。

我唯一想到的就是循环十进制整数1-32768并将其转换为二进制,并使用二进制表示作为筛选器来挑选适当的数字。

有人知道更好的办法吗?使用map(),也许?


当前回答

不使用itertools:

def combine(inp):
    return combine_helper(inp, [], [])


def combine_helper(inp, temp, ans):
    for i in range(len(inp)):
        current = inp[i]
        remaining = inp[i + 1:]
        temp.append(current)
        ans.append(tuple(temp))
        combine_helper(remaining, temp, ans)
        temp.pop()
    return ans


print(combine(['a', 'b', 'c', 'd']))

其他回答

如文档中所述

def combinations(iterable, r):
    # combinations('ABCD', 2) --> AB AC AD BC BD CD
    # combinations(range(4), 3) --> 012 013 023 123
    pool = tuple(iterable)
    n = len(pool)
    if r > n:
        return
    indices = list(range(r))
    yield tuple(pool[i] for i in indices)
    while True:
        for i in reversed(range(r)):
            if indices[i] != i + n - r:
                break
        else:
            return
        indices[i] += 1
        for j in range(i+1, r):
            indices[j] = indices[j-1] + 1
        yield tuple(pool[i] for i in indices)


x = [2, 3, 4, 5, 1, 6, 4, 7, 8, 3, 9]
for i in combinations(x, 2):
    print i

我知道使用itertools来获得所有的组合要实际得多,但是如果你碰巧想要,假设你想要编写很多代码,你可以只使用列表理解来部分实现这一点

对于两对组合:

lambda l: [(a, b) for i, a in enumerate(l) for b in l[i+1:]]

而且,对于三对组合,它是这样简单的:

lambda l: [(a, b, c) for i, a in enumerate(l) for ii, b in enumerate(l[i+1:]) for c in l[i+ii+2:]]

结果和使用itertools.combination是一样的:

import itertools
combs_3 = lambda l: [
    (a, b, c) for i, a in enumerate(l) 
    for ii, b in enumerate(l[i+1:]) 
    for c in l[i+ii+2:]
]
data = ((1, 2), 5, "a", None)
print("A:", list(itertools.combinations(data, 3)))
print("B:", combs_3(data))
# A: [((1, 2), 5, 'a'), ((1, 2), 5, None), ((1, 2), 'a', None), (5, 'a', None)]
# B: [((1, 2), 5, 'a'), ((1, 2), 5, None), ((1, 2), 'a', None), (5, 'a', None)]

3个功能:

列出n个元素的所有组合 列出n个元素的所有组合,其中顺序不明确 所有的排列

import sys

def permutations(a):
    return combinations(a, len(a))

def combinations(a, n):
    if n == 1:
        for x in a:
            yield [x]
    else:
        for i in range(len(a)):
            for x in combinations(a[:i] + a[i+1:], n-1):
                yield [a[i]] + x

def combinationsNoOrder(a, n):
    if n == 1:
        for x in a:
            yield [x]
    else:
        for i in range(len(a)):
            for x in combinationsNoOrder(a[:i], n-1):
                yield [a[i]] + x
    
if __name__ == "__main__":
    for s in combinations(list(map(int, sys.argv[2:])), int(sys.argv[1])):
        print(s)
flag = 0
requiredCals =12
from itertools import chain, combinations

def powerset(iterable):
    s = list(iterable)  # allows duplicate elements
    return chain.from_iterable(combinations(s, r) for r in range(len(s)+1))

stuff = [2,9,5,1,6]
for i, combo in enumerate(powerset(stuff), 1):
    if(len(combo)>0):
        #print(combo , sum(combo))
        if(sum(combo)== requiredCals):
            flag = 1
            break
if(flag==1):
    print('True')
else:
    print('else')

来自itertools的组合

import itertools
col_names = ["aa","bb", "cc", "dd"]
all_combinations = itertools.chain(*[itertools.combinations(col_names,i+1) for i,_ in enumerate(col_names)])
print(list(all_combinations))