基于键值在python中过滤嵌套字典 [英] Filter nested dictionary in python based on key values

问题描述

如何根据键值过滤python中的嵌套字典：

How do I filter a nested dictionary in python based on key values:

d = {'data': {'country': 'US', 'city': 'New York', 'state': None},
     'tags': ['US', 'New York'],
     'type': 'country_info',
     'growth_rate': None
     }

我想过滤此字典以消除NoneType值，因此得到的dict应该是：

I want to filter this dictionary to eliminate NoneType values so the resulting dict should be:

d = {'data': {'country': 'US', 'city': 'New York'},
     'tags': ['US', 'New York'],
     'type': 'country_info',
     }

此外，dict可以有多个级别的嵌套。我想从dict中删除所有的NoneType值。

Also, the dict can have multiple levels of nesting. I want to remove all NoneType values from the dict.

推荐答案

您可以使用 dict comprehension 。

def remove_keys_with_none_values(item):
    if not hasattr(item, 'items'):
        return item
    else:
        return {key: remove_keys_with_none_values(value) for key, value in item.items() if value is not None}

递归不是过于优化Python，但是考虑到可能的嵌套数量相对较少，我不用担心。

Recursion isn't too optimised in Python, but given the relatively small number of nestings that are likely, I wouldn't worry.

在我们飞跃之前看起来不是太诡异，我认为这是一个比捕获异常更好的选择 - 大部分时间（可能我们比分支机构有更多的叶子），这个值可能不会是 dict 。

Looking before we leap isn't too Pythonic, I think it is a better option than catching the exception - as it's likely that the value will not be a dict most of the time (it is likely we have more leaves than branches).

还要注意，在Python 2.x中，你可能想要在 iteritems（）中交换 items（）。

Also note that in Python 2.x, you probably want to swap in iteritems() for items().

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