用python中列表中的元素替换字典的值

Question

我有一个列表：operation = [5,6]和字典dic = {0: None, 1: None}

我想用运算值替换dic的每个值。

我试过了，但看上去不太好用。

operation = [5,6]

for i in oper and val, key in dic.items():
        dic_op[key] = operation[i]

有人有主意吗？

Answer 1

zip方法将完成这项工作

operation = [5, 6]
dic = {0: None, 1: None}

for key, op in zip(dic, operation):
  dic[key] = op

print(dic)   # {0: 5, 1: 6}  

上面的解决方案假设dic是有序的，以便operation中的元素位置与dic中的键对齐。

Answer 2

其他选择，也许：

operation = [5,6]
dic = {0: None, 1: None}

for idx, val in enumerate(operation):
  dic[idx] = val

dic #=> {0: 5, 1: 6}

此处使用索引的详细信息：Accessing the index in 'for' loops?

Answer 3

在Python zip中使用3.7+，您只需执行以下操作：

operation = [5,6]
dic = {0: None, 1: None}

print(dict(zip(dic, operation)))
# {0: 5, 1: 6}