python:自定义类中迭代行为的实现
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2022-03-15 19:38:32
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#!/usr/bin/env python
# -*- coding: utf-8 -*-
# @author : cat
# @date : 2017/6/23.
class Squares:
"""
单(个活跃的)迭代器的实现
"""
def __init__(self, start, stop):
self.value = start - 1
self.stop = stop
print("s1 init ---- ", start, stop)
def __iter__(self):
print('s1: _iter_')
return self
def __next__(self):
print("s1: _next_")
if self.value == self.stop:
raise StopIteration
self.value += 1
return self.value ** 2
class Squares2:
"""
多(个活跃的)迭代器的实现
"""
def __init__(self, start, stop):
self.start = start
self.value = start - 1
self.stop = stop
print("init: ", start, stop)
def __iter__(self):
"""
重写该方法,让自己成为:iterable
:return:
"""
print("__iter__ -- ")
return Squares2(self.value + 1, self.stop)
def __next__(self):
"""
重写该方法,让自己成为:iterator
:return:
"""
print("__next__")
if self.value == self.stop:
raise StopIteration
self.value += 1
return self.value ** 2
if __name__ == '__main__':
s1 = Squares(1, 4)
# 一次遍历仅调用一次 __iter__ 方法
print([x for x in s1])
# 一次遍历中,每返回一个值,都会调用一次__next__方法
print("\n-------------\n")
s2 = Squares2(1, 4)
print([x for x in s2])
print([x for x in s2])
输出如下:
iterttt.py
s1 init ---- 1 4
s1: _iter_
s1: _next_
s1: _next_
s1: _next_
s1: _next_
s1: _next_
[1, 4, 9, 16]
s1: _iter_
s1: _next_
[]
-------------
init: 1 4
__iter__ --
init: 1 4
__next__
__next__
__next__
__next__
__next__
[1, 4, 9, 16]
__iter__ --
init: 1 4
__next__
__next__
__next__
__next__
__next__
[1, 4, 9, 16]
Process finished with exit code 0
感谢:http://blog.csdn.net/raylee2007/article/details/48325653
感谢:http://blog.csdn.net/passionkk/article/details/49929887
感谢:《Learning Python》