OOP重构

Question

以下脚本从不同的文件格式派生架构。能给我看一下这个吗？这些改进是什么样子的：

    def _null_allowed_in_cols(self):
        for field in self.fields:
            field["type"] = self._field_type_converter(field["type"])
            if type(field["type"]) == list:
                if "null" not in field["type"]:
                    field["type"].insert(0, "null")
            else:
                field["type"] = ["null", field["type"]]

    def _field_type_converter(self, field_type) -> Union[str, list]:
        if type(field_type) is dict or type(field_type) is avro.schema.ImmutableDict:
            return field_type["type"]
        elif type(field_type) is list:
            return list(map(lambda x: self._field_type_converter(x), field_type))
        else:
            return field_type

Answer 1

_col_type

您可以在这里使用生成器表达式返回第一个匹配项或None：

    # go from this
    def _col_type(self, column_name):
        for column in self.fields:
            if column["name"] == column_name:
                return column
        return None

   # to this
   def _col_type(self, column_name):
       return next((
        col['name'] for col in self.fields if col['name'] == column_name
       ), None)

检查类型

使用isinstance而不是type(val) == cls：

# instead of this
if type(field["type"]) == list:
    # do something
# or this
if type(val) is dict:
    # do something

# do this instead
if isinstance(field['type'], list):
    # do something

# or
if isinstance(field_type, (dict, avro.schema.ImmutableDict)):
    # do something

何时使用lambda与函数

在本声明中：

list(map(lambda x: self._field_type_converter(x), field_type))

你其实不需要兰布达：

list(map(self._field_type_converter, field_type))