blocks|key|2064394|text|__iter__返回一个iterator+object。如果您需要在每次迭代中执行某些操作，则必须实现您自己的方法(它必须实现链接文档中描述的两个方法)。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2064395|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/library/stdtypes.html#typeiter^0|0|8|C|F|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@$9|O|A|P|B|C]]|D|@$9|Q|A|R|1|S]]|E|$]]|$1|F|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|G|$H|$5|I|J|K|E|$L|M]]]]

<code>__iter__</code> returns an <a href="http://docs.python.org/library/stdtypes.html#typeiter" rel="nofollow">iterator object</a>. If you need to do something in every iteration, you have to implement your own (it has to implement two methods described in the linked docs).

blocks|key|2883622|text|python类__iter__()方法实际上返回一个迭代器对象。请参阅以下内容作为参考：http://docs.python.org/library/stdtypes.html#iterator-types|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2883623|在迭代器对象上，将在循环的每个步骤上调用next()方法。您可以编写一个自定义迭代器，它将由您的自定义列表返回。|2883624|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/library/stdtypes.html#iterator-types^0|7|A|18|1N|0|0|K|6|0^^$0|@$1|2|3|4|5|6|7|P|8|@$9|Q|A|R|B|C]]|D|@$9|S|A|T|1|U]]|E|$]]|$1|F|3|G|5|6|7|V|8|@$9|W|A|X|B|C]]|D|@]|E|$]]|$1|H|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]]|I|$J|$5|K|L|M|E|$N|O]]]]

The python class <code>__iter__()</code> method actually returns an iterator object.
See the following for reference: <a href="http://docs.python.org/library/stdtypes.html#iterator-types" rel="nofollow">http://docs.python.org/library/stdtypes.html#iterator-types</a>

On the iterator object the <code>next()</code> method will be called on each step of the loop. You could write a custom iterator which would be returned by your custom list.

blocks|key|4998858|text|我不知道你到底想做什么，但你已经被告知的是真的。next()是在每次迭代中调用的方法。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4998859|class+MyList(list):
++++def+next(self):
++++++++print+"Do+something"
++++++++return+list.next(self)|code-block|syntax|javascript|4998860|(我没有测试它)|4998861|entityMap^0|O|6|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|S|8|@]|D|@]|E|$]]|$1|M|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|N|$]]

I don't understand exactly what you are trying to do, but what you've already been told is true. <code>next()</code> is the method called in every iteration.

<pre><code>class MyList(list):
 def next(self):
 print "Do something"
 return list.next(self)
</code></pre>

(I didn't tested it)

blocks|key|150861|text|type|unstyled|depth|inlineStyleRanges|entityRanges|data|150862|我希望列表的元素在列表的每个循环中都被初始化/完成|blockquote|150863|150864|150865|列表对此没有直接控制。多个循环可以并行访问列表(例如，在嵌套循环中)。每个循环都由一个迭代器控制，__iter__()应该在每次调用时返回迭代器。您不能总是返回相同的迭代器实例，因为这些并行循环会相互干扰。|offset|length|style|CODE|150866|但是在迭代器中，您可以根据需要自由地实例化项目。从迭代器内部完成它们几乎不是明智之举，因为一旦返回了一个项，就失去了对它的控制。如果您稍后完成它，仍然依赖于该项的外部代码可能会失败。如果你需要完成你的项目，让它成为项目消费者的责任。|150867|entityMap^0|0|0|0|0|1D|A|0|0^^$0|@$1|2|3|-4|4|5|6|P|7|@]|8|@]|9|$]]|$1|A|3|B|4|C|6|Q|7|@]|8|@]|9|$]]|$1|D|3|-4|4|5|6|R|7|@]|8|@]|9|$]]|$1|E|3|-4|4|5|6|S|7|@]|8|@]|9|$]]|$1|F|3|G|4|5|6|T|7|@$H|U|I|V|J|K]]|8|@]|9|$]]|$1|L|3|M|4|5|6|W|7|@]|8|@]|9|$]]|$1|N|3|-4|4|5|6|X|7|@]|8|@]|9|$]]]|O|$]]

<blockquote>
 I want the elements of the list to be initialized/finalized with every loop of the list
</blockquote>

A list has no direct control over this. Several loops can access a list in parallel (e.g. in nested loops). Each loop is controlled by an iterator, and <code>__iter__()</code> should return iterators, every time it's called. You can't return the same iterator instance all the time, because said parallel loops will then interfere with each other.

But inside an iterator you're free to instantiate items as you need. Finalizing them from inside the iterator is hardly wise, because once you returned an item, you lost control over it. If you finalize it at a later moment, outer code that still relies on the item may fail. If you need to finalize your items, make it a responsibility of items' consumer.

blocks|key|2883747|text|您可以简单地从__iter__()返回生成器表达式|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2883748|class+MyList(list):
++++def+__iter__(self):
++++++++return+(self.do_something(x)+for+x+in+list.__iter__(self))

++++def+do_something(self,+x):
++++++++print+'do+something',+x
++++++++return+x

my_list+=+MyList(range(10))
print+my_list

for+item+in+my_list:
++++print+item|code-block|syntax|javascript|2883749|ncoghlan建议使用生成器来代替生成器表达式，这使得调试更容易|2883750|class+MyList(list):

++++def+__iter__(self):
++++++++for+x+in+list.__iter__(self):
++++++++++++yield+self.do_something(x)

++++def+do_something(self,+x):
++++++++print+'do+something',+x
++++++++return+x

my_list+=+MyList(range(10))
print+my_list

for+item+in+my_list:
++++print+item|2883751|另外，您也可以在此处使用imap|2883752|from+itertools+import+imap

class+MyList(list):
++++def+__iter__(self):
++++++++return+imap(self.do_something,+list.__iter__(self))

++++def+do_something(self,+x):
++++++++print+'do+something',+x
++++++++return+x


my_list+=+MyList(range(10))
print+my_list

for+item+in+my_list:
++++print+item|2883753|entityMap^0|7|A|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|Y|8|@]|D|@]|E|$]]|$1|M|3|N|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|O|3|P|5|6|7|10|8|@]|D|@]|E|$]]|$1|Q|3|R|5|H|7|11|8|@]|D|@]|E|$I|J]]|$1|S|3|-4|5|6|7|12|8|@]|D|@]|E|$]]]|T|$]]

You can simply return a generator expression from <code>__iter__()</code>

<pre><code>class MyList(list):
 def __iter__(self):
 return (self.do_something(x) for x in list.__iter__(self))

 def do_something(self, x):
 print 'do something', x
 return x

my_list = MyList(range(10))
print my_list

for item in my_list:
 print item
</code></pre>

ncoghlan suggests using a generator in place of the generator expression which makes for easier debugging

<pre><code>class MyList(list):

 def __iter__(self):
 for x in list.__iter__(self):
 yield self.do_something(x)

 def do_something(self, x):
 print 'do something', x
 return x

my_list = MyList(range(10))
print my_list

for item in my_list:
 print item
</code></pre>

alternatively you could use imap here

<pre><code>from itertools import imap

class MyList(list):
 def __iter__(self):
 return imap(self.do_something, list.__iter__(self))

 def do_something(self, x):
 print 'do something', x
 return x


my_list = MyList(range(10))
print my_list

for item in my_list:
 print item
</code></pre>

I am trying to make a class which inherits from a python list. I want the elements of the list to be initialized/finalized with every loop of the list. I thought this could be done by overriding the <code>__iter__</code> method of the python list but I can't seem to get it to work. The <code>__iter__</code> method appears to called only once ? (see below)

<pre><code>class MyList(list):
 def __iter__(self):
 print 'do something'
 return list.__iter__(self)

my_list = MyList(range(10))
print my_list

for item in my_list:
 print item
</code></pre>

Output

<pre><code>[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
do something
0
1
2
3
4
5
6
7
8
9
</code></pre>

Any idea how I can achieve what I want to do?

python list __iter__ method called on every loop?

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我正在尝试创建一个继承自python列表的类。我希望列表的元素在列表的每个循环中都被初始化/完成。我认为这可以通过重写python列表的__iter__方法来完成，但是我似乎无法让它工作。__iter__方法似乎只调用了一次？(见下文)class MyList(list):    def __iter__(self):...

问python list iter方法会在每个循环中调用吗？
EN

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问python list __iter__方法会在每个循环中调用吗？EN