Question

我试图用i和y更新字典。但是y仅作为变量值的最后一个列表出现

multiples_6 = []
multiples_7 = []
multiples_8 = []
multiples_9 = []
multiples_10 = []

for n in range(1,5+1):
  multiples_6.append(6*n)
  multiples_7.append(7*n)
  multiples_8.append(8*n)
  multiples_9.append(9*n)
  multiples_10.append(10*n)

keys = [6,7,8,9,10]
values = [multiples_6,multiples_7,multiples_8,multiples_9,multiples_10]

new_multiples = {}

for i in keys:
  for y in values:
    new_multiples.update({i:y})

当我尝试这样做时，它只给了我值列表中的最后一个元素。

Output: {6: [10, 20, 30, 40, 50], 7: [10, 20, 30, 40, 50], 8: [10, 20, 30, 40, 50], 9: [10, 20, 30, 40, 50], 10: [10, 20, 30, 40, 50]}

Answer 1

您好，@ Pranav Hosangadi的评论说明了一切。您只应有一个循环。您可以使用Python的许多公式来做到这一点。我最喜欢的是：

for key, value in zip(keys, values):
    new_multiples.update({key: value})

将列表与字典的值相乘

1 个答案: