问题描述
我想做一个字符串列表的组合,在组合结果中没有重复的字符串,但我不知道如何做到这一点。
例如:
# Input
pickup_list = [["a","b","c"],["d","e"],["a","c","f"]]
print(combinaison_function(pickup_list))
# Output
> [["a","d","f"],"e",["b","a"],...]
在示例中 ["a",”a"] 没有返回,因为 "a" 是重复的并且我想要这样做。
我想解决方案是使用 itertools 函数,但我不知道该怎么做。
感谢您提前回复。
解决方法
这可以通过 itertools.product()
完成,然后过滤重复项。检查重复项的一种快速方法是强制转换为 set
。
comb_list = []
for comb in itertools.product(*pickup_list):
if len(set(comb))==len(comb):
comb_list.append(comb)
用你的 pickup_list
我得到:
[('a','d','c'),('a','f'),'e',('b','a'),('c','f')]
,
您可以制作自己的递归生成器:
def uCombo(L,used=None):
if used is None: used = set() # track used values
if not L: yield [];return # end of recursion
for n in L[0]: # combine each unused value
if n in used: continue # with remaining sub-lists
yield from ( [n]+c for c in uCombo(L[1:],used | {n}) )
输出:
pickup_list = [["a","b","c"],["d","e"],["a","c","f"]]
for combo in uCombo(pickup_list): print(combo)
['a','c']
['a','f']
['a','f']
['b','a']
['b','c']
['b','f']
['c','a']
['c','f']