(译)Python关键字yield的解释--下篇(stackov
2016-11-30 本文已影响34人
艾尔温
6.回到你的代码
(译者注:这是回答这对问题的具体解释)
生成器:
# Here you create the method of the node object that will return the generator
def node._get_child_candidates(self, distance, min_dist, max_dist):
# Here is the code that will be called each time you use the generator object :
# If there is still a child of the node object on its left
# AND if distance is ok, return the next child
if self._leftchild and distance - max_dist < self._median:
yield self._leftchild
# If there is still a child of the node object on its right
# AND if distance is ok, return the next child
if self._rightchild and distance + max_dist >= self._median:
yield self._rightchild
# If the function arrives here, the generator will be considered empty
# there is no more than two values : the left and the right children
调用者:
# Create an empty list and a list with the current object referenceresult, candidates = list(), [self]
# Loop on candidates (they contain only one element at the beginning)
while candidates:
# Get the last candidate and remove it from the list
node = candidates.pop()
# Get the distance between obj and the candidate
distance = node._get_dist(obj) # If distance is ok, then you can fill the result
if distance <= max_dist and distance >= min_dist:
result.extend(node._values)
# Add the children of the candidate in the candidates list
# so the loop will keep running until it will have looked
# at all the children of the children of the children, etc. of the candidate
candidates.extend(node._get_child_candidates(distance, min_dist, max_dist))
return result
这个代码包含了几个小部分:
- 我们对一个列表进行迭代,但是迭代中列表还在不断的扩展。它是一个迭代这些嵌套的数据的简洁方式,即使这样有点危险,因为可能导致无限迭代。candidates.extend(node._get_child_candidates(distance, min_dist, max_dist))
穷尽了生成器的所有值,但 while 不断地在产生新的生成器,它们会产生和上一次不一样的值,既然没有作用到同一个节点上. - extend() 是一个迭代器方法,作用于迭代器,并把参数追加到迭代器的后面。
通常我们传递它一个列表参数:
>>> a = [1, 2]
>>> b = [3, 4]
>>> a.extend(b)
>>> print(a)
[1, 2, 3, 4]
但是在你的代码中的是一个生成器,这是不错的,因为:
- 你不必读两次所有的值
- 你可以有很多子对象,但不必叫他们都存储在内存里面。
并且这很奏效,因为Python不关心一个方法的参数是不是个列表。Python只希望它是个可以迭代的,所以这个参数可以是列表,元组,字符串,生成器... 这叫做 ducktyping,这也是为何Python如此棒的原因之一,但这已经是另外一个问题了...
你可以在这里停下,来看看生成器的一些高级用法:
7.控制生成器穷尽
>>> class Bank(): # let's create a bank, building ATMs
... crisis = False... def create_atm(self) :
... while not self.crisis :
... yield "$100"
>>> hsbc = Bank() # when everything's ok the ATM gives you as much as you want
>>> corner_street_atm = hsbc.create_atm()
>>> print(corner_street_atm.next())
$100
>>> print(corner_street_atm.next())
$100
>>> print([corner_street_atm.next() for cash in range(5)])
['$100', '$100', '$100', '$100', '$100']
>>> hsbc.crisis = True # crisis is coming, no more money!
>>> print(corner_street_atm.next())
<type 'exceptions.StopIteration'>
>>> wall_street_atm = hsbc.create_atm() # it's even true for new ATMs
>>> print(wall_street_atm.next())
<type 'exceptions.StopIteration'>
>>> hsbc.crisis = False # trouble is, even post-crisis the ATM remains empty
>>> print(corner_street_atm.next())
<type 'exceptions.StopIteration'>
>>> brand_new_atm = hsbc.create_atm() # build a new one to get back in business
>>> for cash in brand_new_atm :
... print cash
$100
$100
$100
$100
$100
$100
$100
$100
$100
...
对于控制一些资源的访问来说这很有用。
8.Itertools,你最好的朋友
itertools包含了很多特殊的迭代方法。是不是曾想过复制一个迭代器?串联两个迭代器?把嵌套的列表分组?不用创造一个新的列表的 zip/map?
只要 import itertools
需要个例子?让我们看看比赛中4匹马可能到达终点的先后顺序的可能情况:
>>> horses = [1, 2, 3, 4]
>>> races = itertools.permutations(horses)
>>> print(races)
<itertools.permutations object at 0xb754f1dc>
>>> print(list(itertools.permutations(horses)))
[(1, 2, 3, 4),
(1, 2, 4, 3),
(1, 3, 2, 4),
(1, 3, 4, 2),
(1, 4, 2, 3),
(1, 4, 3, 2),
(2, 1, 3, 4),
(2, 1, 4, 3),
(2, 3, 1, 4),
(2, 3, 4, 1),
(2, 4, 1, 3),
(2, 4, 3, 1),
(3, 1, 2, 4),
(3, 1, 4, 2),
(3, 2, 1, 4),
(3, 2, 4, 1),
(3, 4, 1, 2),
(3, 4, 2, 1),
(4, 1, 2, 3),
(4, 1, 3, 2),
(4, 2, 1, 3),
(4, 2, 3, 1),
(4, 3, 1, 2),
(4, 3, 2, 1)]
9.了解迭代器的内部机制
迭代是一个实现可迭代对象(实现的是 iter() 方法)和迭代器(实现的是next() 方法)的过程。可迭代对象是你可以从其获取到一个迭代器的任一对象。迭代器是那些允许你迭代可迭代对象的对象。