python - python 类中的类方法和辅助函数

Question

我的目标是减少我创建的类中的一些冗余。我已将问题简化为一个非常简单的示例

我当前的类(class)

class BigNumbers(object):
      def __init__(self, big_number):
            self.number = big_number

      @classmethod
      def make_with_sums(cls, sum_to):
            fin_num = 0

            for i in range(1, sum_to):
                  fin_num += i

            return cls( fin_num )

      @classmethod
      def make_with_product(cls, mul_to):
            fin_num = 1

            for i in range(1, mul_to):
                  fin_num *= i

            return cls( fin_num )

我想要的类(class) (DNRY)

class dnryNumbers(object):
      def __init__(self, big_number):
            self.number = big_number

      @classmethod
      def _make_with_helper(cls, make_meth, fin_num):
            method_dict = {'sum': lambda x, y: x + y,
                           'prod': lambda x, y: x * y
                           }
            var = 0
            for i in range(1, fin_num):
                  var = method_dict[make_meth](var, i)

            return cls( var )

      def make_with_sums(self, sum_to):

            return self._make_with_helper( 'sum', sum_to )

      def make_with_product(self, mul_to):

            return self._make_with_helper('prod', mul_to )

我的目标是使用与使用 BigNumbers 类时相同的函数调用，例如:

In [60]: bn = BigNumbers.make_with_product(10)

In [61]: bn.number
Out[61]: 362880

-- 或 --

In [63]: bn = BigNumbers.make_with_sums(10)

In [64]: bn.number
Out[64]: 45

但是当前的功能不起作用:

In [65]: bn = dnryNumbers.make_with_product(10)
TypeError: unbound method make_with_product() must be called with dnryNumbers instance as first argument (got int instance instead)

Answer 1

简单的答案:make_with_sums和make_with_products是类方法，而不是实例方法，因此需要这样声明。另请注意，_make_with_helper 还需要采用起始 值作为参数；将 var 初始化为 0 将使 make_with_product 返回 cls(0)，无论其参数是什么。

无论将什么输入传递给 _make_with_helper，

method_dict 都是相同的字典，因此它应该是一个类变量:

class dnryNumbers(object):
    def __init__(self, big_number):
        self.number = big_number

    method_dict = {'sum': lambda x, y: x + y,
                   'prod': lambda x, y: x * y
                  }

    @classmethod
    def _make_with_helper(cls, make_meth, fin_num, starting_value):          
        var = starting_value
        for i in range(1, fin_num):
              var = dnryNumbers.method_dict[make_meth](var, i)
        return cls( var )

    @classmethod
    def make_with_sums(cls, sum_to):
        return cls._make_with_helper('sum', sum_to, 0)

    @classmethod
    def make_with_product(cls, mul_to):
        return cls._make_with_helper('prod', mul_to, 1)

但是现在，method_dict 只是添加了一个您不需要的额外间接层。由于这些函数不打算在类外部使用，因此只需将它们定义为“私有(private)”方法，并直接使用对它们的引用。

class dnryNumbers(object):
    def __init__(self, big_number):
        self.number = big_number

    @staticmethod
    def _sum(x, y):
        return x + y

    @staticmethod
    def _prod(x, y):
        return x * y

    @classmethod
    def _make_with_helper(cls, make_meth, fin_num, starting_value):        
        var = starting_value
        for i in range(1, fin_num):
              var = make_meth(var, i)
        return cls(var)

        @classmethod def make_with_sums(cls, sum_to): 返回 cls._make_with_helper(dnryNumbers._sum, sum_to, 0)

    @classmethod
    def make_with_product(cls, mul_to):
        return cls._make_with_helper(dnryNumbers._prod, mul_to, 1)

最后，值得指出的是，除非您的实际代码比此处显示的示例更复杂，否则 _sum 已经可以用作 operator.add、_prod 只是 operator.mul，而 _make_with_helper 只是 reduce 内置函数(或 functools.reduce，如果是 Python 3)。

import operator
try:
    reduce
except NameError:
    from functools import reduce

class dnryNumbers(object):
    def __init__(self, big_number):
        self.number = big_number

    @classmethod
    def _make_with_helper(cls, make_meth, fin_num, starting_value):
        return cls(reduce(make_meth, range(1, fin_num), starting_value))

    @classmethod
    def make_with_sums(cls, sum_to):
        return cls._make_with_helper(operator.add, sum_to, 0)

    @classmethod
    def make_with_product(cls, mul_to):
        return cls._make_with_helper(operator.add, mul_to, 1)