试试itertools

import itertools for x in itertools.product([1,2,3,4],[2,3,4],[1,1,2,2,3,3]): print x

哈哈，其实就是一个求笛卡尔积的问题
要学会组织关键词，善用搜索

我來表演重複發明輪子:

一個用 iteration 的版本:

def product(*iterables): START = 0 END = 1 se_idxs = [(0, len(iterable)) for iterable in iterables] curr_idxs = [0 for iterable in iterables] def next(): curr_idxs[-1] += 1 for k in reversed(range(len(iterables))): if curr_idxs[k]==se_idxs[k][END]: if k==0: return False curr_idxs[k] = 0 curr_idxs[k-1] += 1 else: break return True while True: yield [iterable[idx] for idx, iterable in zip(curr_idxs, iterables)] if not next(): return

一個 recursive 的版本:

def product_core(result, *iterables): if len(iterables)==0: yield result else: for item in iterables[0]: yield from product_core(result+[item], *iterables[1:]) def product(*iterables): return product_core([], *iterables)

測試:

a = [1, 2, 3] b = [4, 5, 6] c = [7, 8, 9] for p in product(a, b, c): print(p)

結果:

[1, 4, 7] [1, 4, 8] [1, 4, 9] [1, 5, 7] [1, 5, 8] [1, 5, 9] [1, 6, 7] [1, 6, 8] [1, 6, 9] [2, 4, 7] [2, 4, 8] [2, 4, 9] [2, 5, 7] [2, 5, 8] ...

人生苦短 別傻傻的自己做重複的事情啊...

我回答過的問題: Python-QA

类似于全排列生成算法
用递归或者递推都可以

递归实现起来比较简单，递推稍微麻烦一点

列表生成式可以吗？

>>> [m + n for m in 'ABC' for n in 'XYZ'] ['AX', 'AY', 'AZ', 'BX', 'BY', 'BZ', 'CX', 'CY', 'CZ']

写递归确实可以，难道不会写dfs深度优先搜索嘛？