我要打印的內(nèi)容如下，為簡單起見，在每個(gè)字符串中使用 2 個(gè)變量。我認(rèn)為使用循環(huán)有問題，我對 Python 有點(diǎn)陌生。提前致謝！com0 <-(a) with (1,)com1 <-(a) with (2,)com2 <-(a) with (1, 2)com3 <-(b) with (1,)com4 <-(b) with (2,)com5 <-(b) with (1, 2)com6 <-(a,b) with (1,)com7 <-(a,b) with (2,)com8 <-(a,b) with (1, 2)這是我試過的：import itertoolsi = 0 #v1j = 0 #v2v1 = [1, 2]v2 = ["a","b"]while j < 2**len(v2): for K in range(0, len(v2)+1): while i < 2**len(v1): for L in range(0, len(v1)+1): for subset2 in itertools.combinations(v1, L): for subset1 in itertools.combinations(v2, K): print("com{0} <-{1} with {2}".format(i,subset1,subset2)) i+=1 j+=1
2 回答
牧羊人nacy
TA貢獻(xiàn)1862條經(jīng)驗(yàn) 獲得超7個(gè)贊
你不需要這么多循環(huán)。只需使用combinations并product從itertoools
>>> from itertools import combinations, product
>>>
>>> v1 = [1, 2]
>>> v2 = ["a","b"]
>>>
>>> all_v1 = [e for i in range(len(v1)) for e in combinations(v1,i+1)]
>>> all_v2 = [e for i in range(len(v1)) for e in combinations(v2,i+1)]
>>>
>>> for i, (x,y) in enumerate(product(all_v2, all_v1)):
... print (f'com{i} <-{x} with {y}')
...
com0 <-('a',) with (1,)
com1 <-('a',) with (2,)
com2 <-('a',) with (1, 2)
com3 <-('b',) with (1,)
com4 <-('b',) with (2,)
com5 <-('b',) with (1, 2)
com6 <-('a', 'b') with (1,)
com7 <-('a', 'b') with (2,)
com8 <-('a', 'b') with (1, 2)
楊魅力
TA貢獻(xiàn)1811條經(jīng)驗(yàn) 獲得超6個(gè)贊
來自 itertools 的結(jié)果product()是兩個(gè)列表的笛卡爾積。通過使用此函數(shù)，您可以避免所有循環(huán)：
import itertools
v1 = [1, 2]
v2 = ["a","b"]
combinations = list(itertools.product(v1, v2))
>> [(1, "a"), (1, "b"), (2, "a"), (2, "b")]
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