使用递归函数删除嵌套字典中的空字典 [英] Remove empty dicts in nested dictionary with recursive function

问题描述

我正在尝试从嵌套字典中删除非值.我的第一次努力工作正常，但不幸的是，指向现在空字典的键仍然存在.

I'm trying to remove non-values from a nested dictionary. My first effort works fine, but unfortunately keys pointing to now empty dicts still persist.

所以如果我这样做:

pass1 = stripper(my_dict)
return stripper(pass1)

这可行，但我认为可能有更优雅的嵌套解决方案?

This works, but i'm thinking a more elegant nested solution might be possible?

def stripper(self, data):
    if isinstance(data, dict):
        d = ({k: stripper(v) for k, v in data.items()
             if v not in [u'', None]})
        if d:
            return d
    else:
        return data

失败的例子，dict 下面返回为 {'foo': 'bar', 'bar': None}:

Failing example, dict below returns as {'foo': 'bar', 'bar': None}:

{
    'foo': 'bar',
    'bar': {
        'foo': None,
        'one': None
    }
}

推荐答案

dict comprehension 固然简洁，但如果你把它展开，解决方案就会变得更加明显:

The dict comprehension is certainly concise but if you expand it out, the solution becomes more obvious:

def stripper(self, data):
    new_data = {}
    for k, v in data.items():
        if isinstance(v, dict):
            v = stripper(v)
        if not v in (u'', None, {}):
            new_data[k] = v
    return new_data

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