python列表组合

发表于

python列表组合

描述

主要记录下两个及多个列表的组合，我们想把两个列表进行组合生成所有可能性。例如：

a = [‘1’,’2’,’3’]

b = [‘a’,’b’,’c’]

期望得到的结果：[‘1’,’a’],[‘1’,’b’],[‘1’,’c’],[‘2’,’a’],[‘2’,’b’]··········

思考

笨办法的话就是for循环，但是写起来不美观。下面学习到几种写法，记录下。

1
2
3
4
a = ["foo","melon"]
b = [True, False]
c = list(itertools.product(a, b))
print(c) --> [("foo", True), ("foo", False), ("melon", True), ("melon", False)]
1
2
3
4
a = ["foo","bar"]
b = [1, 2, 3]
c = [(x,y) for x in a for y in b]
print(c) --> [('foo', 1), ('foo', 2), ('foo', 3), ('bar', 1), ('bar', 2), ('bar', 3)]
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
import itertools

a = [1, 2, 3]
b = ['a', 'b', 'c', 'd']
c = ['@@', '$$']
#简化了参数, 增加了参数类型声明
def combination(*args:list):
    config = [e for e in args]
    res = []
    tmp = []
    dfs(res, tmp, config, 0)
    print(res)
    print('combination num:', len(res))
# dfs没变动
def dfs(res=[], tmp=[], config=[], i=0):
    if i == len(config):
        res.append(tmp.copy())
    else:
        for e in config[i]:
            # change spot
            tmp.append(e)
            # new condition,new recursion
            dfs(res, tmp, config, i + 1)
            # restore spot
            tmp.pop(len(tmp) - 1)
# 执行主体函数         
combination(a,b,c)

Respect

https://www.codenong.com/12935194/
https://blog.csdn.net/littlehaes/article/details/103461622