如何生成满足前两位与后两位分别求和条件的itertools.product组合？

解决方案 #

你只需要修改combination列表推导式里的筛选条件即可，把原来的总求和判断，替换为前两位求和、后两位求和分别满足小于等于21的判断：

import itertools
a = [*range(22)], [*range(22)], [*range(22)], [*range(22)]

# 修改筛选条件：前两位求和<=21 且 后两位求和<=21
combination = [seq for seq in itertools.product(*a) if sum(seq[:2]) <= 21 and sum(seq[2:]) <= 21]
print(combination)

lst1 = [item[0] for item in combination]
lst2 = [item[1] for item in combination]
lst3 = [item[2] for item in combination]
lst4 = [item[3] for item in combination]

逻辑说明 #

• seq[:2] 取元组的前两个元素，sum(seq[:2]) 计算前两位的和
• seq[2:] 取元组的后两个元素，sum(seq[2:]) 计算后两位的和
• 两个条件用and关联，同时满足才会被保留，你需要的(0, 21, 1, 20)这类组合就会被正常筛选出来了。

内容的提问来源于stack exchange，提问作者Janani Ethirraj