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python - 从具有“nan”值的字典中删除键

问题描述

我有源字典，请参考下面的示例。它具有深层嵌套的子字典和列表，我需要删除值为“nan”的所有键。

data = {"col1":"val1","col2":"val2","col3":"val3","col4":"val3","list1":[{"l1":"v1","l2":"nan"},{"K1":"Kv1","K2":"nan"},{"M1":"Mv1","M2":"nan","sublist1":[{"SL1":"SV1","SL2":"nan"}]}],"list2":[{"l1":"v1","l2":"nan"},{"K1":"Kv1","K2":"nan"},{"M1":"Mv1","M2":"nan","sublist2":[{"SL1":"SV1","SL2":"nan"}]}]}

我通过创建一个函数尝试了以下代码，但它没有按预期工作：

def cleanNullTerms(d): 
   clean = {}
   for k, v in d.items():
      if isinstance(v, list):
         for values in v:
            nested = cleanNullTerms(values)
            if values =='nan':
                clean[k] = nested
      elif v is not 'nan':
         clean[k] = v
   return clean

标签： python

解决方案

您的代码将列表的值放在列表所在的位置。如果有很多值，每个值都会覆盖最后一个，所以只有最后一个被保存。它并没有被卷入字典。如果它们与字典和列表嵌套，它仍然只删除“nan”。例如，它不进入集合。如果你想让它进入集合，写评论，我可以添加这个。

def cleanNullTerms(d):
   clean={}
   for k, v in d.items():
      if isinstance(v, list):
         v2=[]
         for value0 in v:
            nested=cleanNullTerms(value0)
            if value0!='nan':
                v2.append(nested)
         clean[k]=v2
      elif isinstance(v, dict):
          clean[k]=cleanNullTerms(v)
      elif v!='nan':
         clean[k]=v
   return clean

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