你也可以像这样使用defaultdict:

from collections import defaultdict
a = defaultdict(lambda: "default", key="some_value")
a["blabla"] => "default"
a["key"] => "some_value"

你可以传递任何普通函数而不是lambda:

from collections import defaultdict
def a():
  return 4

b = defaultdict(a, key="some_value")
b['absent'] => 4
b['key'] => "some_value"

(这是一个迟来的回答)

另一种方法是继承dict类的子类并实现__missing__()方法，如下所示:

class ConnectionDetails(dict):
    def __missing__(self, key):
        if key == 'host':
            return "localhost"
        raise KeyError(key)

例子:

>>> connection_details = ConnectionDetails(port=80)

>>> connection_details['host']
'localhost'

>>> connection_details['port']
80

>>> connection_details['password']
Traceback (most recent call last):
  File "python", line 1, in <module>
  File "python", line 6, in __missing__
KeyError: 'password'

你也可以像这样使用defaultdict:

from collections import defaultdict
a = defaultdict(lambda: "default", key="some_value")
a["blabla"] => "default"
a["key"] => "some_value"

你可以传递任何普通函数而不是lambda:

from collections import defaultdict
def a():
  return 4

b = defaultdict(a, key="some_value")
b['absent'] => 4
b['key'] => "some_value"

对于多个不同的默认值，请尝试以下方法:

connectionDetails = { "host": "www.example.com" }
defaults = { "host": "127.0.0.1", "port": 8080 }

completeDetails = {}
completeDetails.update(defaults)
completeDetails.update(connectionDetails)
completeDetails["host"]  # ==> "www.example.com"
completeDetails["port"]  # ==> 8080

是这样的:

host = connectionDetails.get('host', someDefaultValue)

这并不是我们要问的问题，但是在python字典中有一个方法:dict.setdefault

    host = connectionDetails.setdefault('host',someDefaultValue)

然而，这个方法将connectionDetails['host']的值设置为someDefaultValue，如果key host还没有定义，这与问题不同。