【Python】【重载运算符】

标签：一个   ret   short   abr   加法   tee   byte   math   numbers   

#
#例子13-3  一元运算符+得到一个新Counter实例，但是没有零值和负值计数器
from collections import Counter

ct = Counter(‘abracadabra‘)
print(ct)        #Counter({‘a‘: 5, ‘r‘: 2, ‘b‘: 2, ‘c‘: 1, ‘d‘: 1})
ct[‘r‘] = -3
ct[‘d‘] = 0
print(ct)        #Counter({‘a‘: 5, ‘b‘: 2, ‘c‘: 1, ‘d‘: 0, ‘r‘: -3})
print(+ct)       #Counter({‘a‘: 5, ‘b‘: 2, ‘c‘: 1})


import itertools
a = (1,2,3)
b = [4,5]
d = itertools.zip_longest(a,b,fillvalue=0.0)

#以下是Vector




from array import array
import reprlib
import math
import numbers
import functools
import operator
import itertools


class Vector:
    typecode = ‘d‘

    def __init__(self, components):
        self._components = array(self.typecode, components)

    def __iter__(self):
        return iter(self._components)

    def __repr__(self):
        components = reprlib.repr(self._components)
        components = components[components.find(‘[‘):-1]
        return ‘Vector({})‘.format(components)

    def __str__(self):
        return str(tuple(self))

    def __bytes__(self):
        return (bytes([ord(self.typecode)]) +
                bytes(self._components))

    def __eq__(self, other):
        return (len(self) == len(other) and
                all(a == b for a, b in zip(self, other)))

    def __hash__(self):
        hashes = (hash(x) for x in self)
        return functools.reduce(operator.xor, hashes, 0)

    def __abs__(self):
        return math.sqrt(sum(x * x for x in self))

    def __bool__(self):
        return bool(abs(self))

    def __len__(self):
        return len(self._components)

    def __getitem__(self, index):
        cls = type(self)
        if isinstance(index, slice):
            return cls(self._components[index])
        elif isinstance(index, numbers.Integral):
            return self._components[index]
        else:
            msg = ‘{.__name__} indices must be integers‘
            raise TypeError(msg.format(cls))

    shortcut_names = ‘xyzt‘

    def __getattr__(self, name):
        cls = type(self)
        if len(name) == 1:
            pos = cls.shortcut_names.find(name)
            if 0 <= pos < len(self._components):
                return self._components[pos]
        msg = ‘{.__name__!r} object has no attribute {!r}‘
        raise AttributeError(msg.format(cls, name))

    def angle(self, n):  # <2>
        r = math.sqrt(sum(x * x for x in self[n:]))
        a = math.atan2(r, self[n-1])
        if (n == len(self) - 1) and (self[-1] < 0):
            return math.pi * 2 - a
        else:
            return a

    def angles(self):  # <3>
        return (self.angle(n) for n in range(1, len(self)))

    def __format__(self, fmt_spec=‘‘):
        if fmt_spec.endswith(‘h‘):  # hyperspherical coordinates
            fmt_spec = fmt_spec[:-1]
            coords = itertools.chain([abs(self)],
                                     self.angles())  # <4>
            outer_fmt = ‘<{}>‘  # <5>
        else:
            coords = self
            outer_fmt = ‘({})‘  # <6>
        components = (format(c, fmt_spec) for c in coords)  # <7>
        return outer_fmt.format(‘, ‘.join(components))  # <8>

    @classmethod
    def frombytes(cls, octets):
        typecode = chr(octets[0])
        memv = memoryview(octets[1:]).cast(typecode)
        return cls(memv)
    ‘‘‘
    def __add__(self, other):
        pairs = itertools.zip_longest(self,other,fillvalue=0.0)
        return Vector(a+b for a,b in pairs)
     #print((3,40) + v1)     #TypeError: can only concatenate tuple (not "Vector") to tuple 【解析】如果左操作数是Vector之外的对象，则无法处理
     #...为了支持涉及不同类型的运算，python为中缀运算符特殊方法提供了特殊的分派机制。对表达式a+b来说，解释器会执行以下几步操作
     #...(1)如果a有__add__方法，而且返回值不是NotImplemented,调用a.__add__(b),然后返回结果
     #...(2)如果a没有__add__方法，或者调用__add__方法返回NotImplemented,检查b有没有__radd__方法，如果有，而且没有返回NotImplemented,调用b.__radd__(a)，然后返回结果
     #...(3)如果前两者条件都不满足，抛出TypeError，并在错误消息中指明操作数类型不支持
     #[注意]别把NotImplemented NotImplementedError搞混了。前者是特殊的单例值，如果中缀运算符特殊方法不能处理给定的操作数，那么要把它返回给解释器。后者是一种异常，抽象类中的占位方法把它抛出（raise），提醒子类必须覆盖
     ‘‘‘
    def __add__(self, other):
         pairs = itertools.zip_longest(self,other,fillvalue=0.0)
         return Vector(a+b for a,b in pairs)
    def __radd__(self, other):
        return self + other   #__radd__通常就这么简单：直接调用适当的运算符，在这里就委托__add__。任何可交换的运算符都能这么左。处理数字和向量时，+可以交换，但是拼接序列时不行
#例子13-8 Vector.__add__方法的操作数钥要是可迭代对象
v2 = Vector([3,4,5])
#print(v2 + 1)    #TypeError: zip_longest argument #2 must support iteration
#例子13-9 这个加法的操作数应该是可迭代的数值对象
print(v2 + ‘ABC‘)  #TypeError: unsupported operand type(s) for +: ‘float‘ and ‘str‘
#例子13-10 针对13-8 13-9 需要再进行优化，即有错误要抛出

v1 = Vector([3,4,5])
print(v1 + (10,20))   #(13.0, 24.0, 5.0)

【Python】【重载运算符】

标签：一个   ret   short   abr   加法   tee   byte   math   numbers   

原文地址：http://www.cnblogs.com/suren2017/p/7679805.html