Python字典:如何根据键更新字典值-使用单独的字典键 [英] Python Dictionary: How to update dictionary value, base on key - using separate dictionary keys

问题描述

我有两个长度不等的字典，例如:

I've got two dictionaries of unequal length, e.g.:

people = {"john" : "carpenter", "jill": "locksmith", "bob":"carpenter", "jane": "pilot", "dan": "locksmith"}

jobcode = {"carpenter": 1, "locksmith": 2, "pilot": 3}

我想要做的是将people中的值替换为jobcode值. 因此，您最终会得到: n

What I'm wanting to do is replace the values in people with the jobcode value. So youd end up with: n

people = {"john": 1, "jill": 2, "bob": 1, "jane": 3, "dan":2} 

我很乐意制作另一个也封装了这些新数据的新dict，但到目前为止，我认为最接近的是

I'd be happy to make another new dict that encapsulates this new data as well but so far the closest I think I've come is this... I think...

任何帮助将不胜感激.

推荐答案

您可以通过dict理解轻松实现这一目标

You can easily achieve this with dict comprehension

{k: jobcode[v] for k, v in people.items()}

但是您应该小心，因为它可以提高KeyError.

However you should be careful since it can raise KeyError.

使用默认jobcode和dict 方法:

default_jobcode = 1000
final_dict = {k: jobcode.get(v, default_jobcode) for k, v in people.items()}

更新

如 @Graipher 所述，如果jobcode dict缺少键值对，则可以保留项目保持原样:

As @Graipher kindly noted, if jobcode dict is lack of key-value pair, you can leave item untouched as such:

final_dict = {k: jobcode.get(v, v) for k, v in people.items()}

使用默认的jobcode可能是更好的解决方案.

Which is probably better solution that having default jobcode.

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