def convert(dictionary我的類中有一個輔助函數 ( )__init__來協助配置設置。定義如下：class Configuration: def __init__(self, config_file=None, config=None): if config_file is not None: with open(config_file) as in_file: self._config = yaml.load(in_file, Loader=yaml.FullLoader) elif config is not None: self._config = config else: raise ValueError("Could not create configuration. Must pass either location of config file or valid " "config.") def convert(dictionary): return namedtuple('Config', dictionary.keys())(**dictionary)這使我可以按如下方式撥打電話__init__： self.input = convert(self._config["input"]) self.output = convert(self._config["output"]) self.build = convert(self._config["build_catalog"])由于我要設置多個配置，因此我想從他的類中繼承如下：class BuildConfiguration(Configuration): def __init__(self, config_file=None, config=None): super().__init__(config_file, config) self.input = convert(self._config["input"]) self.output = convert(self._config["output"]) self.build = convert(self._config["build_catalog"])convert但是，我無法從父類訪問。我也試過這個：self.input = super().__init__.convert(self._config["input"])這似乎也行不通。所以問題是如何訪問super().__init__子類中定義的函數？
1 回答
牧羊人nacy
TA貢獻1862條經驗 獲得超7個贊
你不能。每次調用都會創建一個新函數，__init__然后將其丟棄，它不存在于函數之外。請注意，這也適用于由創建的類namedtuple('Config', dictionary.keys())(**dictionary)。繼續創建所有這些不必要的類確實不好，這完全違背了namedtuple創建內存高效記錄類型的目的。在這里，每個實例都有自己的類！
以下是您應該如何定義它：
Config = namedtuple('Config', "foo bar baz")
def convert(dictionary): # is this really necessary?
return Config(**dictionary)
class Configuration:
def __init__(self, config_file=None, config=None):
if config_file is not None:
with open(config_file) as in_file:
self._config = yaml.load(in_file, Loader=yaml.FullLoader)
elif config is not None:
self._config = config
else:
raise ValueError("Could not create configuration. Must pass either location of config file or valid "
"config.")
self.input = convert(self._config["input"])
self.output = convert(self._config["output"])
self.build = convert(self._config["build_catalog"])
雖然在這一點上，使用它似乎更干凈
Config(**self._config["input"])
etc 而不是 helper convert。
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