Python 工匠 第六章 循环和可迭代对象
基础知识
迭代器和可迭代对象
iter() 与 next()
>>> iter([1,2,3])
<list_iterator object at 0x0000022FEFEA73A0>
>>> iter('abc')
<str_iterator object at 0x0000022FEFEA7190>
>>> iter(1)
Traceback (most recent call last):
File "" , line 1, in <module>
TypeError: 'int' object is not iterable
迭代器：帮助我们迭代其他对象的对象
>>> my_list = ['foo', 'bar']
>>> iter_my_list = iter(my_list)
>>> iter_my_list
<list_iterator object at 0x0000022FEFEA73A0>
>>> next(iter_my_list)
'foo'
>>> next(iter_my_list)
'bar'
>>> next(iter_my_list)
Traceback (most recent call last):
File "" , line 1, in <module>
StopIteration
此外，对迭代器执行iter()函数，获取迭代器的迭代器对象时，返回的结果一定是迭代器本身。
>>> iter(iter_my_list) is iter_my_list
True
当用for循环遍历某个可迭代对象时：先调用iter()拿到了它的迭代器，然后不断地用next()从迭代器中获取值
### 1
names = ['a', 'b', 'c']
for name in names:
print(name)
### 2
iterator = iter(name)
while True:
try:
name = next(iterator)
print(name)
except StopInteration:
break
自定义迭代器
关键在实现：__iter__ and __next__ 两者分别在调用iter() 和 next() 时触发
需求：编写一个类似range()的迭代器对象Range7， 其可以返回某个范围内所有可被7整除或者包含7的整数：
class Range7:
"""generate the number in certain range which could be divided by 7 exactly or contains 7
:param start: start number
:param end: end number
"""
def __init__(self, start, end):
self.start = start
self.end = end
self.current = start
def __iter__(self):
return self
def __next__(self):
while True:
if self.current >= self.end:
raise StopIteration
if self.num_is_valid(self.current):
ret = self.current
self.current += 1
return ret
self.current += 1
def num_is_valid(self, num):
if num == 0:
return False
return num % 7 == 0 or '7' in str(num)
r = Range7(0, 20)
for num in r:
print(num)
r = Range7(0, 20)
for num in r:
print(num)
7
14
17
发现一个问题，每个Range7对象只能被完整遍历一次
r = Range7(0, 20)
print(tuple(r))
print(tuple(r))
(7, 14, 17)
()
因为每个Range7对象的current属性在init之后逐渐增长到end，不会再回来了
除非手动更改其值
r = Range7(0, 20)
print(tuple(r))
r.current = 0
print(tuple(r))
(7, 14, 17)
(7, 14, 17)
区分迭代器和可迭代对象
迭代器是可迭代对象的一种。只需要实现__iter__
class Range7:
def __init__(self, start, end):
self.start = start
self.end = end
def __iter__(self):
return Range7Iterator(self)
class Range7Iterator:
def __init__(self, range_obj):
self.range_obj = range_obj
self.current = range_obj.start
def __iter__(self):
return self
def __next__(self):
while True:
if self.current >= self.range_obj.end:
raise StopIteration
if self.num_is_valid(self.current):
ret = self.current
self.current += 1
return ret
self.current += 1
def num_is_valid(self, num):
if num == 0:
return False
return num % 7 == 0 or '7' in str(num)
r = Range7(0, 20)
print(tuple(r))
print(tuple(r))
(7, 14, 17)
(7, 14, 17)
新代码，每次遍历Range7对象时，都会创建出一个全新的迭代器对象Range7Iterator。
总结 迭代器和可迭代对象的比较
迭代器是可迭代对象（只需要实现__iter__）的子集
iter(可迭代对象)返回迭代器，iter(迭代器)返回自身
此外，如果一个类型没有定义__iter__, 但是有__getitem__, python也会认为其是可迭代的（遍历时0, 1, 2, 3…）
生成器是迭代器
def range_7_generator(start, end):
num = start
while num != 0 and (num % 7 == 0 or '7' in str(num)):
yield num
num += 1
修饰可迭代对象优化循环
使用生成器函数修饰可迭代对象
用生成器在循环外部包装原本的循环主体
def sum_even_only(numbers):
result = 0
for num in numbers:
if num % 2 == 0:
result += num
return result
to
def even_only(numbers):
for num in numbers:
if num % 2 == 0:
yield num
def sum_even_only_v2(numbers):
result = 0
for num in even_only(numbers):
result += num
return result
使用itertools模块优化循环
使用product() 扁平化多层嵌套循环
from itertools import product
list(product([1, 2], [3, 4]))
[(1, 3), (1, 4), (2, 3), (2, 4)]
def find_12(num_list1, num_list12, num_list13):
for num1 in num_list1:
for num2 in num_list2:
for num3 in num_list3:
if num1 + num2 + num3 == 12:
return num1, num2, num3
to
def find_12_v2(num_list1, num_list12, num_list13):
for num1, num2, num3 in product(num_list1, num_list12, num_list13):
if num1 + num2 + num3 == 12:
return num1, num2, num3
使用islice()实现循环内隔行处理
在循环内部实现隔行处理
from itertools import islice
# islice(seq, start, end, step)
for i in islice(range(10), 0, None, 2): # 实际用例中range(10)可能是个无法分片的可迭代对象
print(i)
使用takewhile()代替break
from itertools import takewhile
# takewhile(predicate, iterable) 会在迭代第二个参数iterable的过程中，不断使用当前值作为参数调用predicate()函数，并对返回结果进行真值测试，如果为True,则返回当前值并继续迭代，否则立即中断本地迭代
chain() zip_longest() etc.
循环语句的else关键字
案例故事
编程建议
中断嵌套循环的正确方式
一般是多个break
建议：把循环代码拆分为一个新函数，然后直接使用return
巧用 next() 函数
获取字典的第一个key
next(iter(d.keys()))