从嵌套字典中删除字段的优雅方法

Question

我不得不从字典中删除一些字段，这些字段的键位于列表中。所以，我写这篇文章的功能：

def delete_keys_from_dict(dict_del, lst_keys): 
    """ 
    Delete the keys present in the lst_keys from the dictionary. 
    Loops recursively over nested dictionaries. 
    """ 
    dict_foo = dict_del.copy()#Used as iterator to avoid the 'DictionaryHasChanged' error 
    for field in dict_foo.keys(): 
     if field in lst_keys: 
      del dict_del[field] 
     if type(dict_foo[field]) == dict: 
      delete_keys_from_dict(dict_del[field], lst_keys) 
    return dict_del 

此代码的工作，但它不是很优雅，我敢肯定，你可以编写一个更好的解决方案。

Answer 1

def delete_keys_from_dict(dict_del, lst_keys): 
    for k in lst_keys: 
     try: 
      del dict_del[k] 
     except KeyError: 
      pass 
    for v in dict_del.values(): 
     if isinstance(v, dict): 
      delete_keys_from_dict(v, lst_keys) 

    return dict_del 

Answer 2

既然你已经通过在字典每个元素都需要循环，我会用一个循环坚持，只是确保使用一组用于查找键删除

def delete_keys_from_dict(dict_del, the_keys): 
    """ 
    Delete the keys present in the lst_keys from the dictionary. 
    Loops recursively over nested dictionaries. 
    """ 
    # make sure the_keys is a set to get O(1) lookups 
    if type(the_keys) is not set: 
     the_keys = set(the_keys) 
    for k,v in dict_del.items(): 
     if k in the_keys: 
      del dict_del[k] 
     if isinstance(v, dict): 
      delete_keys_from_dict(v, the_keys) 
    return dict_del 

Answer 3

由于这个问题需要一个优雅的方式，我将提交我的通用解决方案来讨论嵌套结构。首先，安装boltons utility package与pip install boltons，则：

from boltons.iterutils import remap 

data = {'one': 'remains', 'this': 'goes', 'of': 'course'} 
bad_keys = set(['this', 'is', 'a', 'list', 'of', 'keys']) 

drop_keys = lambda path, key, value: key not in bad_keys 
clean = remap(data, visit=drop_keys) 
print(clean) 

# Output: 
{'one': 'remains'} 

总之，the remap utility是一个全功能的，但简洁的方法来处理这往往是嵌套在现实世界中的数据结构，甚至可以包含周期，特殊容器。

This page有更多的例子，包括从Github的API中处理更大的对象的例子。

它是纯Python，因此它可以在任何地方工作，并且已经在Python 2.7和3.3+中进行了全面测试。最重要的是，我为这样的案例编写了它，所以如果你发现它无法处理的情况，你可以修复我的错误right here。

Answer 4

从this后使用真棒代码并添加一个小声明：

def remove_fields(self, d, list_of_keys_to_remove): 
     if not isinstance(d, (dict, list)): 
      return d 
     if isinstance(d, list): 
      return [v for v in (self.remove_fields(v, list_of_keys_to_remove) for v in d) if v] 
     return {k: v for k, v in ((k, self.remove_fields(v, list_of_keys_to_remove)) for k, v in d.items()) if k not in list_of_keys_to_remove} 

Answer 5

我认为下面是更优雅：

def delete_keys_from_dict(dict_del, lst_keys): 
    if not isinstance(dict_del, dict): 
     return dict_del 
    return {key:value for key,value in ((key, delete_keys_from_dict(value)) for key,value in dict_del.items()) if key not in lst_keys}