python：for in变量个数不定

list1=[1,2,3]

list2=['a','b','c']

list3=[4,'e']

combination = [(i,j,k) for i in list1for j in list2for kin list3]

print结果：[(1, 'a', 4), (1, 'a', 'e'), (1, 'b', 4), (1, 'b', 'e'), (1, 'c', 4), (1, 'c', 'e'), (2, 'a', 4), (2, 'a', 'e'), (2, 'b', 4), (2, 'b', 'e'), (2, 'c', 4), (2, 'c', 'e'), (3, 'a', 4), (3, 'a', 'e'), (3, 'b', 4), (3, 'b', 'e'), (3, 'c', 4), (3, 'c', 'e')]

当列表个数不定时，先根据需求生成字符串，然后采用eval()函数将字符串转换为可执行命令。

import string

maxKind =list(string.ascii_letters)   # 26个小写字母和26个大写字母列表，假设最多52种

cc ='a'

dd ='for a in listA[0]'

listA = [[列表1], [列表2], [列表3], [列表4]]

for i in range(1, len(listA)):

    cc = cc +',' + maxKind[i]

    dd = dd +' for ' +str(maxKind[i]) +' in list' +str([i])

print('(' + cc +')' +' ' + dd)   # (a,b,c,d) for a in listA[0] for b in listA[1] for c in listA[2] for d in listA[3]

ee='[(' + cc +')' +' ' + dd +']'

combination =eval(ee)