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]和一个字典dic = {0: None, 1: None}

我想用运算的值替换dic的每个值。

我尝试了此操作，但它似乎没有运行。

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

其他选项，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 2

zip方法将完成工作

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

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

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

以上解决方案假设dic的顺序是为了使operation中的元素位置与dic中的键对齐。

Answer 3

在Python 3.7+中使用zip，您可以这样做：

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

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