我有一本嵌套的字典。是否只有一种方法可以安全地传递价值观?

try:
    example_dict['key1']['key2']
except KeyError:
    pass

或者python有一个类似get()的方法用于嵌套字典?


当前回答

我建议你试试蟒蛇本尼迪克特。

它是一个dict子类,提供小键盘支持等功能。

安装:pip install python-benedict

from benedict import benedict

example_dict = benedict(example_dict, keypath_separator='.')

现在你可以使用keypath访问嵌套值:

val = example_dict['key1.key2']

# using 'get' method to avoid a possible KeyError:
val = example_dict.get('key1.key2')

或者使用键列表访问嵌套值:

val = example_dict['key1', 'key2']

# using get to avoid a possible KeyError:
val = example_dict.get(['key1', 'key2'])

它在GitHub上经过了很好的测试和开源:

https://github.com/fabiocaccamo/python-benedict

注:我是这个项目的作者

其他回答

在第一阶段,你可以得到一个空字典。

example_dict.get('key1',{}).get('key2')

如果您想使用另一个库来解决问题,这是最好的方法

https://github.com/maztohir/dict-path

from dict-path import DictPath

data_dict = {
  "foo1": "bar1",
  "foo2": "bar2",
  "foo3": {
     "foo4": "bar4",
     "foo5": {
        "foo6": "bar6",
        "foo7": "bar7",
     },
  }
}

data_dict_path = DictPath(data_dict)
data_dict_path.get('key1/key2/key3')

虽然reduce方法简洁而简短,但我认为简单的循环更容易理解。我还包含了一个默认参数。

def deep_get(_dict, keys, default=None):
    for key in keys:
        if isinstance(_dict, dict):
            _dict = _dict.get(key, default)
        else:
            return default
    return _dict

作为理解reduce一行程序如何工作的练习,我执行了以下操作。但最终循环方法对我来说似乎更直观。

def deep_get(_dict, keys, default=None):

    def _reducer(d, key):
        if isinstance(d, dict):
            return d.get(key, default)
        return default

    return reduce(_reducer, keys, _dict)

使用

nested = {'a': {'b': {'c': 42}}}

print deep_get(nested, ['a', 'b'])
print deep_get(nested, ['a', 'b', 'z', 'z'], default='missing')

已经有很多很好的答案,但我已经提出了一个类似于JavaScript领域的lodash get的函数,它也支持通过索引进入列表:

def get(value, keys, default_value = None):
'''
    Useful for reaching into nested JSON like data
    Inspired by JavaScript lodash get and Clojure get-in etc.
'''
  if value is None or keys is None:
      return None
  path = keys.split('.') if isinstance(keys, str) else keys
  result = value
  def valid_index(key):
      return re.match('^([1-9][0-9]*|[0-9])$', key) and int(key) >= 0
  def is_dict_like(v):
      return hasattr(v, '__getitem__') and hasattr(v, '__contains__')
  for key in path:
      if isinstance(result, list) and valid_index(key) and int(key) < len(result):
          result = result[int(key)] if int(key) < len(result) else None
      elif is_dict_like(result) and key in result:
          result = result[key]
      else:
          result = default_value
          break
  return result

def test_get():
  assert get(None, ['foo']) == None
  assert get({'foo': 1}, None) == None
  assert get(None, None) == None
  assert get({'foo': 1}, []) == {'foo': 1}
  assert get({'foo': 1}, ['foo']) == 1
  assert get({'foo': 1}, ['bar']) == None
  assert get({'foo': 1}, ['bar'], 'the default') == 'the default'
  assert get({'foo': {'bar': 'hello'}}, ['foo', 'bar']) == 'hello'
  assert get({'foo': {'bar': 'hello'}}, 'foo.bar') == 'hello'
  assert get({'foo': [{'bar': 'hello'}]}, 'foo.0.bar') == 'hello'
  assert get({'foo': [{'bar': 'hello'}]}, 'foo.1') == None
  assert get({'foo': [{'bar': 'hello'}]}, 'foo.1.bar') == None
  assert get(['foo', 'bar'], '1') == 'bar'
  assert get(['foo', 'bar'], '2') == None

根据Yoav的回答,一个更安全的方法是:

def deep_get(dictionary, *keys):
    return reduce(lambda d, key: d.get(key, None) if isinstance(d, dict) else None, keys, dictionary)