Question

我在我的一个项目中使用了kendoUI Grid。我使用他们的api检索了一段数据，发现它在我的json / dictionary中添加了一些“不需要的”数据。将json传递回我的Pyramid后端后，我需要删除这些键。问题是，字典可以是任何深度，我不提前知道深度。

示例：

product = {
    id: "PR_12"
    name: "Blue shirt",
    description: "Flowery shirt for boys above 2 years old",
    _event: {<some unwanted data here>},
    length: <some unwanted data>,
    items: [{_event: {<some rubbish data>}, length: <more rubbish>, price: 23.30, quantity: 34, color: "Red", size: "Large"}, {_event: {<some more rubbish data>}, length: <even more rubbish>, price: 34.50, quantity: 20, color: "Blue", size: "Large"} ....]
}

我想特别删除两个键：“_ event”＆amp; “长度”。我尝试编写一个递归函数来删除数据，但我似乎无法做到正确。有人可以帮忙吗？

这就是我所拥有的：

def remove_specific_key(the_dict, rubbish):
  for key in the_dict:
    if key == rubbish:
      the_dict.pop(key)
    else:
      # check for rubbish in sub dict
      if isinstance(the_dict[key], dict):
        remove_specific_key(the_dict[key], rubbish)

      # check for existence of rubbish in lists
      elif isinstance(the_dict[key], list):
        for item in the_dict[key]:
          if item == rubbish:
            the_dict[key].remove(item)
   return the_dict

Answer 1

如果允许remove_specific_key（重命名为remove_keys）接受任何对象作为其第一个参数，那么您可以简化代码：

def remove_keys(obj, rubbish):
    if isinstance(obj, dict):
        obj = {
            key: remove_keys(value, rubbish) 
            for key, value in obj.iteritems()
            if key not in rubbish}
    elif isinstance(obj, list):
        obj = [remove_keys(item, rubbish)
                  for item in obj
                  if item not in rubbish]
    return obj

由于您希望删除多个密钥，因此您可以将rubbish设置为一个而不是一个特定密钥。使用上面的代码，您可以使用

删除'_event'和'length'键

product = remove_keys(product, set(['_event', 'length']))

编辑：remove_key使用Python {2}中引入的dict comprehension。对于旧版本的Python，等效的是

    obj = dict((key, remove_keys(value, rubbish))
               for key, value in obj.iteritems()
               if key not in rubbish)

Answer 2

在迭代错误时修改dict，这是不必要的，因为你确切知道你要找的是什么键。此外，您的dicts列表未正确处理：

def remove_specific_key(the_dict, rubbish):
    if rubbish in the_dict:
        del the_dict[rubbish]
    for key, value in the_dict.items():
        # check for rubbish in sub dict
        if isinstance(value, dict):
            remove_specific_key(value, rubbish)

        # check for existence of rubbish in lists
        elif isinstance(value, list):
            for item in value:
                if isinstance(item, dict):
                    remove_specific_key(item, rubbish)

Answer 3

在迭代时不能删除dict或list，所以用测试函数替换迭代器。

def remove_specific_key(the_dict, rubbish):
    if the_dict.has_key(rubbish):
        the_dict.pop(rubbish)
    else:
        for key in the_dict:
            if isinstance(the_dict[key], dict):
                remove_specific_key(the_dict[key], rubbish)
            elif isinstance(the_dict[key], list):
                if the_dict[key].count(rubbish):
                    the_dict[key].remove(rubbish)
    return the_dict


d = {"a": {"aa": "foobar"}}
remove_specific_key(d, "aa")
print d

d = {"a": ["aa", "foobar"]}
remove_specific_key(d, "aa")
print d

如何递归删除多维（深度未知）python字典中的某些键？

3 个答案: