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使用列表理解的if语句进行字典理解

[英]Dictionary comprehension with if statements using list comprehension

I am trying to filter a large dictionary based on values from another dictionary. 我正在尝试根据另一个字典中的值来过滤大型字典。 I want to store the keys to filter on in a list. 我想存储要在列表中进行过滤的键。 So far I have: 到目前为止,我有:

feature_list = ['a', 'b', 'c']
match_dict = {'a': 1,
              'b': 2,
              'c': 3}

all_dict = {'id1': {'a': 1,
                    'b': 2,
                    'c': 3},
            'id2': {'a': 1,
                    'b': 4,
                    'c': 3},
            'id3': {'a': 2,
                    'b': 5,
                    'c': 3}}

filtered_dict = {k: v for k, v in all_dict.items() for feature in feature_list if
                 v[feature] == match_dict[feature]}

This returns all the ids because I think the if statement is been evaluated as an OR statement when I would like it to be evaluated as an AND statement. 这将返回所有id,因为当我希望if语句被评估为AND语句时,我认为if语句被评估为OR语句。 So I only want back the id1 dictionary. 所以我只想要id1字典。 I would like to get back: 我想回来:

filtered_dict = {'id1': {'a': 1,
                         'b': 2,
                         'c': 3}}

You're right: your test always passes because one condition is true. 您是对的:您的测试始终会通过,因为一个条件成立。 You need all the conditions to be true. 您需要满足所有条件。

You could use all to get the proper behaviour: 您可以使用all来获得适当的行为:

{k: v for k, v in all_dict.items() if all(v[feature] == match_dict[feature] for feature in feature_list)}

note that if match_list keys are the same as feature_list , it's even simpler, just compare dictionaries: 请注意,如果match_list键与feature_list相同,则更为简单,只需比较字典即可:

r = {k: v for k, v in all_dict.items() if v == match_dict}

(or compute a filtered match_dict with the features you require first. Performance will be better) (或使用您首先需要的功能来计算过滤后的match_dict 。性能会更好)

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