一个简单的解决方案是:

item['key'] if 'key' in item else None

这在尝试将字典值添加到新列表并希望提供默认值时非常有用:

eg.

row = [item['key'] if 'key' in item else 'default_value']

如果你有一个更复杂的需求，等价于缓存，这个类可能会派上用场:

class Cache(dict):
    """ Provide a dictionary based cache

        Pass a function to the constructor that accepts a key and returns
        a value.  This function will be called exactly once for any key
        required of the cache.
    """

    def __init__(self, fn):
        super()
        self._fn = fn

    def __getitem__(self, key):
        try:
            return super().__getitem__(key)
        except KeyError:
            value = self[key] = self._fn(key)
            return value

构造函数接受一个用键调用的函数，该函数应返回字典的值。然后存储这个值，并在下次从字典中检索。像这样使用它……

def get_from_database(name):
    # Do expensive thing to retrieve the value from somewhere
    return value

answer = Cache(get_from_database)
x = answer(42)   # Gets the value from the database
x = answer(42)   # Gets the value directly from the dictionary

正如上面其他人所说，您可以使用get()。

但要检查密钥，你还可以这样做:

d = {}
if 'keyname' in d:

    # d['keyname'] exists
    pass

else:

    # d['keyname'] does not exist
    pass

一个简单的解决方案是:

item['key'] if 'key' in item else None

这在尝试将字典值添加到新列表并希望提供默认值时非常有用:

eg.

row = [item['key'] if 'key' in item else 'default_value']

如果你想要一个更透明的解决方案，你可以子类dict来获得这个行为:

class NoneDict(dict):
    def __getitem__(self, key):
        return dict.get(self, key)

>>> foo = NoneDict([(1,"asdf"), (2,"qwerty")])
>>> foo[1]
'asdf'
>>> foo[2]
'qwerty'
>>> foo[3] is None
True

d1 ={“一”:1、“两个”:2,“三”:3} d1.get(“四”)

如果你要运行这段代码，将不会有'Keyerror'，这意味着你可以使用'dict.get()'来避免错误并执行你的代码