Python:如何从NESTED数据结构(列表和字典)中删除None值? [英] Python: How RECURSIVELY remove None values from a NESTED data structure (lists and dictionaries)?

问题描述

以下是一些嵌套数据，其中包括列表，元组和字典:

Here is some nested data, that includes lists, tuples, and dictionaries:

data1 = ( 501, (None, 999), None, (None), 504 )
data2 = { 1:601, 2:None, None:603, 'four':'sixty' }
data3 = OrderedDict( [(None, 401), (12, 402), (13, None), (14, data2)] )
data = [ [None, 22, tuple([None]), (None,None), None], ( (None, 202), {None:301, 32:302, 33:data1}, data3 ) ]

目标:删除所有无键或值(从数据"中删除).如果列表或字典包含一个值，该值本身就是列表，元组或字典，则请RECURSE删除嵌套无".

Goal: Remove any keys or values (from "data") that are None. If a list or dictionary contains a value, that is itself a list, tuple, or dictionary, then RECURSE, to remove NESTED Nones.

所需的输出:

[[22, (), ()], ((202,), {32: 302, 33: (501, (999,), 504)}, OrderedDict([(12, 402), (14, {'four': 'sixty', 1: 601})]))]

或更容易理解的是，这里是格式化输出:

Or more readably, here is formatted output:

StripNones(data)= list:
. [22, (), ()]
. tuple:
. . (202,)
. . {32: 302, 33: (501, (999,), 504)}
. . OrderedDict([(12, 402), (14, {'four': 'sixty', 1: 601})])

我将提出一个可能的答案，因为我还没有找到解决该问题的方法.我感谢任何替代方法，或对已有解决方案的指导.

I will propose a possible answer, as I have not found an existing solution to this. I appreciate any alternatives, or pointers to pre-existing solutions.

编辑 我忘了提到它必须在Python 2.7中起作用.我目前无法使用Python 3.

EDIT I forgot to mention that this has to work in Python 2.7. I can't use Python 3 at this time.

尽管值得 IS 值得为其他人发布Python 3解决方案.因此，请指明您要回答的python.

Though it IS worth posting Python 3 solutions, for others. So please indicate which python you are answering for.

推荐答案

如果可以假定各个子类的__init__方法具有与典型基类相同的签名:

If you can assume that the __init__ methods of the various subclasses have the same signature as the typical base class:

def remove_none(obj):
  if isinstance(obj, (list, tuple, set)):
    return type(obj)(remove_none(x) for x in obj if x is not None)
  elif isinstance(obj, dict):
    return type(obj)((remove_none(k), remove_none(v))
      for k, v in obj.items() if k is not None and v is not None)
  else:
    return obj

from collections import OrderedDict
data1 = ( 501, (None, 999), None, (None), 504 )
data2 = { 1:601, 2:None, None:603, 'four':'sixty' }
data3 = OrderedDict( [(None, 401), (12, 402), (13, None), (14, data2)] )
data = [ [None, 22, tuple([None]), (None,None), None], ( (None, 202), {None:301, 32:302, 33:data1}, data3 ) ]
print remove_none(data)

请注意，例如，该不会与defaultdict一起使用，因为defaultdict带有__init__的附加参数.要使其与defaultdict一起使用，将需要另一种特殊情况elif(在用于常规命令的情况下).

Note that this won't work with a defaultdict for example since the defaultdict takes and additional argument to __init__. To make it work with defaultdict would require another special case elif (before the one for regular dicts).

还要注意，我实际上已经构造了 new 对象.我没有修改旧的.如果不需要支持修改tuple之类的不可变对象，则可以修改旧对象.

Also note that I've actually constructed new objects. I haven't modified the old ones. It would be possible to modify the old objects if you didn't need to support modifying immutable objects like tuple.

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