前置条件:
res = [['616', '2712354', '1905885'], ['616', '2712357', '1905888'],
['616', '2712360', '1905891'], ['616', '2712363', '1905894'],
['616', '2712360', '1905891'], ['616', '2712363', '1905894'],
['616', '2712360', '1905891'], ['616', '2712363', '1905894'],
['616', '2712360', '1905891'], ['616', '2712363', '1905894'],
['616', '2712360', '1905891'], ['616', '2712363', '1905894'],
['616', '2712366', '1905897'], ['616', '2712369', '1905900']]
value = ['user1', 'user2', 'user3']
需求结果:
将value里的元素循环遍历添加到res里面的嵌套列表中,一个嵌套列表只加一个user
代码输出:
count = 0
while count < len(res):
"""
※取余规律:
1.value长度是n的时候;
2.当b小于n时,count % n的余数就是count本身,当count = n-1时候,恰好是列表value的最大循环角标;
3.当b = n时,count % n的余数又从0开始,那么value[count % len(value)]就又从第一个元素开始遍历;
4.如此循环,即使value里的元素个数和嵌套列表中的元素个数不能整除,也不影响结果
"""
i = value[count % len(value)]
# print(i)
res[count].append(i)
count += 1
print(res)
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