blocks|key|1242025|text|将for+br+in+my_list+:替换为for+br+in+my_list[:]+:。因此，您将迭代源列表的副本。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1242026|entityMap^0|1|J|N|M|0^^$0|@$1|2|3|4|5|6|7|H|8|@$9|I|A|J|B|C]|$9|K|A|L|B|C]]|D|@]|E|$]]|$1|F|3|-4|5|6|7|M|8|@]|D|@]|E|$]]]|G|$]]

Replace <code>for br in my_list :</code> with <code>for br in my_list[:] :</code>. Thanks to that you are going to iterate over a copy of the source list.

blocks|key|227895|text|您不应该从正在迭代的列表中删除。如果你想要移除东西，就用这个。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|227896|while+list:
+++item+=+list.pop()
+++#Do+stuff|code-block|syntax|javascript|227897|编辑:如果您想了解更多关于pop()的信息，请查看python|offset|length|style|CODE|227898|如果订单很重要，请使用pop(0)。默认情况下，pop()将删除最后一个项，如果您想按顺序遍历列表，则应该使用pop(0)删除第一个(索引0)项并返回它。|227899|Edit2:感谢用户文森对while+list的建议。|227900|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/2/tutorial/datastructures.html#more-on-lists|1|https://stackoverflow.com/users/2846140/vincent^0|0|0|D|5|P|6|0|0|B|6|O|5|1J|6|0|D|A|A|2|1|0^^$0|@$1|2|3|4|5|6|7|10|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|11|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|12|8|@$I|13|J|14|K|L]]|9|@$I|15|J|16|1|17]]|A|$]]|$1|M|3|N|5|6|7|18|8|@$I|19|J|1A|K|L]|$I|1B|J|1C|K|L]|$I|1D|J|1E|K|L]]|9|@]|A|$]]|$1|O|3|P|5|6|7|1F|8|@$I|1G|J|1H|K|L]]|9|@$I|1I|J|1J|1|1K]]|A|$]]|$1|Q|3|-4|5|6|7|1L|8|@]|9|@]|A|$]]]|R|$S|$5|T|U|V|A|$W|X]]|Y|$5|T|U|V|A|$W|Z]]]]

You shouldn't delete from the list you are iterating. Use this if you want to remove stuff

<pre><code>while list:
 item = list.pop()
 #Do stuff
</code></pre>

Edit: if you want to know more about <code>pop()</code> look at the <a href="https://docs.python.org/2/tutorial/datastructures.html#more-on-lists" rel="nofollow noreferrer">python doc</a>

And if order is important, use <code>pop(0)</code>. <code>pop()</code> removes the last item by default and if you want to go through your list in order you should use <code>pop(0)</code> that removes the first (index 0) item and returns it.

Edit2: Thanks to user <a href="https://stackoverflow.com/users/2846140/vincent">Vincent</a> for the <code>while list</code> suggestion.

blocks|key|226659|text|tobias_k是正确的，从正在迭代的列表中删除项会导致各种问题。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|226660|在本例中，通过使用repl在每个循环上打印列表，可以相对容易地显示它导致迭代跳过：|226661|for+br+in+my_list:
++my_list.remove(br)
++print+my_list|code-block|syntax|javascript|226662|这就产生了：|226663|[1,+2,+3,+4,+5]
[1,+3,+4,+5]
[1,+3,+5]|226664|并最终离开含有:+1，3，5的my_list|226665|要做您想做的事情，mic4ael是正确的--最简单的(尽管可能不是最有效的)方法是在迭代列表之前获取它的副本，如下所示：|226666|my_list+=+[0,1,2,3,4,5]
for+br+in+my_list[:]:+++++#The+[:]+takes+a+copy+of+my_list+and+iterates+over+it
++my_list.remove(br)
++print+my_list|226667|226668|[1,+2,+3,+4,+5]
[2,+3,+4,+5]
[3,+4,+5]
[4,+5]
[5]
[]|226669|entityMap^0|0|0|0|0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|X|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|Y|8|@]|9|@]|A|$]]|$1|D|3|E|5|F|7|Z|8|@]|9|@]|A|$G|H]]|$1|I|3|J|5|6|7|10|8|@]|9|@]|A|$]]|$1|K|3|L|5|F|7|11|8|@]|9|@]|A|$G|H]]|$1|M|3|N|5|6|7|12|8|@]|9|@]|A|$]]|$1|O|3|P|5|6|7|13|8|@]|9|@]|A|$]]|$1|Q|3|R|5|F|7|14|8|@]|9|@]|A|$G|H]]|$1|S|3|J|5|6|7|15|8|@]|9|@]|A|$]]|$1|T|3|U|5|F|7|16|8|@]|9|@]|A|$G|H]]|$1|V|3|-4|5|6|7|17|8|@]|9|@]|A|$]]]|W|$]]

tobias_k is correct removing items from a list you are iterating over causes all sorts of problems. 

In this case it's relatively easy to show that it is causing the iteration to skip, by printing the list on each loop, using the repl:

<pre><code>for br in my_list:
 my_list.remove(br)
 print my_list
</code></pre>

This produces:

<pre><code>[1, 2, 3, 4, 5]
[1, 3, 4, 5]
[1, 3, 5]
</code></pre>

And finally leaves my_list containing: [1, 3, 5]

To do what you want, mic4ael is correct the simplest (although possibly not the most efficient) way is to take a copy of the list prior to iterating over it, as below:

<pre><code>my_list = [0,1,2,3,4,5]
for br in my_list[:]: #The [:] takes a copy of my_list and iterates over it
 my_list.remove(br)
 print my_list
</code></pre>

This produces:

<pre><code>[1, 2, 3, 4, 5]
[2, 3, 4, 5]
[3, 4, 5]
[4, 5]
[5]
[]
</code></pre>

blocks|key|227972|text|如果您真的需要尽快收回列表中的项目所使用的内存，您可以这样释放它们|type|unstyled|depth|inlineStyleRanges|entityRanges|data|227973|for+i,+br+in+enumerate(my_list):
++++#other+code+here
++++my_list[i]+=+None|code-block|syntax|javascript|227974|但在Jython中，Java会在它想要的时候释放它们。|227975|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

If you really need to reclaim the memory used by the items in the lists ASAP, you can free them like this

<pre><code>for i, br in enumerate(my_list):
 #other code here
 my_list[i] = None
</code></pre>

Except in Jython where Java will free them when it feels like it

Why does this for-loop not go through all items:

<pre><code> temp = 0

 for br in my_list :
 temp +=1
 #other code here
 #my list is not used at all, only br is used inside here
 my_list.remove(br)

 print temp
 assert len(my_list) == 0 , "list should be empty"
</code></pre>

So, assertion fires. Then I added the temp counter and I do see that despite my list having 202 elements, the for loop processes only 101 of them. Why is that?

For-loop does not go over all objects

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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为什么这个for-循环不贯穿所有项目： temp = 0 for br in my_list :    temp +=1    #other code here    #my list is not used at all, only br is used inside here    my_list.remove(b...

问For-循环并不会遍历所有对象。
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问For-循环并不会遍历所有对象。EN