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

因为如果缺少一个键就会引发一个键错误是合理的,我们甚至可以不检查它,让它像这样单一:

def get_dict(d, kl):
  cur = d[kl[0]]
  return get_dict(cur, kl[1:]) if len(kl) > 1 else cur

其他回答

递归解。它不是最有效的,但我发现它比其他示例更具可读性,而且它不依赖于functools。

def deep_get(d, keys):
    if not keys or d is None:
        return d
    return deep_get(d.get(keys[0]), keys[1:])

例子

d = {'meta': {'status': 'OK', 'status_code': 200}}
deep_get(d, ['meta', 'status_code'])     # => 200
deep_get(d, ['garbage', 'status_code'])  # => None

一个更精致的版本

def deep_get(d, keys, default=None):
    """
    Example:
        d = {'meta': {'status': 'OK', 'status_code': 200}}
        deep_get(d, ['meta', 'status_code'])          # => 200
        deep_get(d, ['garbage', 'status_code'])       # => None
        deep_get(d, ['meta', 'garbage'], default='-') # => '-'
    """
    assert type(keys) is list
    if d is None:
        return default
    if not keys:
        return d
    return deep_get(d.get(keys[0]), keys[1:], default)

你可以使用开源ndicts包中的NestedDict(我是作者),它有一个完全像字典一样的安全get方法。

>>> from ndicts import NestedDict
>>> nd = NestedDict({"key1": {"key2": 0}}
>>> nd.get(("key1", "key2))
0
>>> nd.get("asd")

通过把所有这些答案和我做的小改变结合起来,我认为这个函数会很有用。安全、快捷、易于维护。

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

例子:

from functools import reduce
def deep_get(dictionary, keys, default=None):
    return reduce(lambda d, key: d.get(key, default) if isinstance(d, dict) else default, keys.split("."), dictionary)

person = {'person':{'name':{'first':'John'}}}
print(deep_get(person, "person.name.first"))    # John

print(deep_get(person, "person.name.lastname")) # None

print(deep_get(person, "person.name.lastname", default="No lastname"))  # No lastname
def safeget(_dct, *_keys):
    if not isinstance(_dct, dict): raise TypeError("Is not instance of dict")
    def foo(dct, *keys):
        if len(keys) == 0: return dct
        elif not isinstance(_dct, dict): return None
        else: return foo(dct.get(keys[0], None), *keys[1:])
    return foo(_dct, *_keys)

assert safeget(dict()) == dict()
assert safeget(dict(), "test") == None
assert safeget(dict([["a", 1],["b", 2]]),"a", "d") == None
assert safeget(dict([["a", 1],["b", 2]]),"a") == 1
assert safeget({"a":{"b":{"c": 2}},"d":1}, "a", "b")["c"] == 2

因为如果缺少一个键就会引发一个键错误是合理的,我们甚至可以不检查它,让它像这样单一:

def get_dict(d, kl):
  cur = d[kl[0]]
  return get_dict(cur, kl[1:]) if len(kl) > 1 else cur