Python懒惰字典评估

Python传播者会说Python没有switch语句的原因是因为它有字典.那么……我怎么能用字典来解决这个问题呢？
问题是所有值都在评估中,并根据输入提出异常.

这只是一个存储数字或数字列表并提供乘法的类的一个愚蠢的例子.

class MyClass(object):

    def __init__(self, value):
        self._value = value

    def __mul__(self, other):
        return {
            (False, False): self._value * other._value                        ,
            (False, True ): [self._value * o for o in other._value]           ,
            (True , False): [v * other._value for v in self._value]           ,
            (True , True ): [v * o for v, o in zip(self._value, other._value)],
        }[(isinstance(self._value, (tuple, list)), isinstance(other._value, (tuple, list)))]

    def __str__(self):
        return repr(self._value)
    __repr__ = __str__



>>> x = MyClass(2.0)
>>> y = MyClass([3.0, 4.0, 5.0])
>>> print x
2.0
>>> print y
[3.0, 4.0, 5.0]
>>> print x * y
Traceback (most recent call last):
  File "<stdin>", line 1, in <module>
  File "<stdin>", line 8, in __mul__
TypeError: can't multiply sequence by non-int of type 'float'

我可以解决它的一种方法是为每个值添加“lambda：”前缀,并在字典查找后调用lambda函数….“}(isinsta …)”

有没有更好的办法？

解决方法:

是的,为这些不同的选项定义小lambdas：

    def __mul__(self, other): 
        scalar_times_scalar = lambda x,y: x*y
        scalar_times_seq    = lambda x,y: [x*y_i for y_i in y]
        seq_times_scalar    = lambda x,y: scalar_times_seq(y,x)
        seq_times_seq       = lambda x,y: [x_i*y_i for x_i,y_i in zip(x,y)]
        self_is_seq, other_is_seq = (isinstance(ob._value,(tuple, list)) 
                                                    for ob in (self, other))
        fn = {
            (False, False): scalar_times_scalar, 
            (False, True ): scalar_times_seq, 
            (True , False): seq_times_scalar, 
            (True , True ): seq_times_seq, 
            }[(self_is_seq, other_is_seq)] 
        return fn(self._value, other._value)

当然,理想情况下,您只能在类或模块范围内定义这些lambda.为了便于参考,我在这里用__mul__方法展示了它们.

标签：python,dictionary,lazy-evaluation,switch-statement
来源： https://codeday.me/bug/20190827/1739667.html