blocks|key|1901314|text|因为for+i+in+li的工作方式是这样的：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1901315|for+idx+in+range(len(li)):
++++i+=+li[idx]
++++i+=+'foo'|code-block|syntax|javascript|1901316|所以如果你给i赋值，它不会影响li[idx]。|1901317|解决方案要么是您提出的解决方案，要么是遍历索引：|1901318|for+idx+in+range(len(li)):
++++li[idx]+=+'foo'|1901319|或者使用enumerate|1901320|for+idx,+item+in+enumerate(li):
++++li[idx]+=+'foo'|1901321|entityMap^0|2|B|0|0|6|1|F|7|0|0|0|4|9|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|Z|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|10|8|@$9|11|A|12|B|C]|$9|13|A|14|B|C]]|D|@]|E|$]]|$1|M|3|N|5|6|7|15|8|@]|D|@]|E|$]]|$1|O|3|P|5|H|7|16|8|@]|D|@]|E|$I|J]]|$1|Q|3|R|5|6|7|17|8|@$9|18|A|19|B|C]]|D|@]|E|$]]|$1|S|3|T|5|H|7|1A|8|@]|D|@]|E|$I|J]]|$1|U|3|-4|5|6|7|1B|8|@]|D|@]|E|$]]]|V|$]]

Because the way <code>for i in li</code> works is something like this:

<pre><code>for idx in range(len(li)):
 i = li[idx]
 i = 'foo'
</code></pre>

So if you assign anything to <code>i</code>, it won't affect <code>li[idx]</code>.

The solution is either what you have proposed, or looping through the indices:

<pre><code>for idx in range(len(li)):
 li[idx] = 'foo'
</code></pre>

or use <code>enumerate</code>:

<pre><code>for idx, item in enumerate(li):
 li[idx] = 'foo'
</code></pre>

blocks|key|1687806|text|实际上，使用列表理解并不是在修改列表，而是创建一个新的列表，然后将其赋给包含前一个列表的变量。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1687807|无论如何，当您执行for+i+in+li时，您将获得变量i中li的每个值的副本，您不会获得对li中某个位置的引用，因此您不会修改li中的任何值。|offset|length|style|CODE|1687808|如果你想修改你的列表，你可以使用enumerate|1687809|>>>+li+=+["spam",+"eggs"]
>>>+for+i,_+in+enumerate(li):
++++++++li[i]+=+"foo"
>>>+li
+['foo',+'foo']|code-block|syntax|javascript|1687810|或者使用xrange+(在Python2.7中，在Python3中使用range+)：|1687811|>>>+for+i+in+xrange(len(li)):
++++++++li[i]+=+"foo"
>>>+li+
+['foo',+'foo']|1687812|或者使用您在问题中显示的列表理解。|1687813|entityMap^0|0|9|B|S|1|U|2|1A|2|1S|2|0|G|9|0|0|4|6|0|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|X|8|@$D|Y|E|Z|F|G]|$D|10|E|11|F|G]|$D|12|E|13|F|G]|$D|14|E|15|F|G]|$D|16|E|17|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|18|8|@$D|19|E|1A|F|G]]|9|@]|A|$]]|$1|J|3|K|5|L|7|1B|8|@]|9|@]|A|$M|N]]|$1|O|3|P|5|6|7|1C|8|@$D|1D|E|1E|F|G]]|9|@]|A|$]]|$1|Q|3|R|5|L|7|1F|8|@]|9|@]|A|$M|N]]|$1|S|3|T|5|6|7|1G|8|@]|9|@]|A|$]]|$1|U|3|-4|5|6|7|1H|8|@]|9|@]|A|$]]]|V|$]]

In fact with list comprehension you are not modifying the list, you are creating a new list and then assigning it to the variable that contained the previous one.

Anyway, when you do <code>for i in li</code> you are getting a copy of each value of <code>li</code> in variable <code>i</code>, you don't get the reference to a position in <code>li</code>, so you are not modifying any value in <code>li</code>.

If you want to modify your list you can do it with <code>enumerate</code>:

<pre><code>&gt;&gt;&gt; li = ["spam", "eggs"]
&gt;&gt;&gt; for i,_ in enumerate(li):
 li[i] = "foo"
&gt;&gt;&gt; li
 ['foo', 'foo']
</code></pre>

or with <code>xrange</code> (in Python 2.7, use range in python 3):

<pre><code>&gt;&gt;&gt; for i in xrange(len(li)):
 li[i] = "foo"
&gt;&gt;&gt; li 
 ['foo', 'foo']
</code></pre>

or with the list comprehension you showed in your question.

blocks|key|1901368|text|我可以在循环时修改列表：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1901369|lst+=+range(10)++//+[0,+1,+2,+3,+4,+5,+6,+7,+8,+9]

for+i,+elem+in+enumerate(lst):
++++lst[i]+=+0+++//+[0,+0,+0,+0,+0,+0,+0,+0,+0,+0]|code-block|syntax|javascript|1901370|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I'm able to modify a list while looping:

<pre><code>lst = range(10) // [0, 1, 2, 3, 4, 5, 6, 7, 8, 9]

for i, elem in enumerate(lst):
 lst[i] = 0 // [0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
</code></pre>

blocks|key|1687829|text|for+element+in+li+-返回元素的副本，而不是元素本身。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1687830|针对您的案例的解决方案是：|1687831|for+i+in+range(len(li)):
++++li[i]+=+'foo'|code-block|syntax|javascript|1687832|entityMap^0|0|H|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|R|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|S|8|@]|D|@]|E|$K|L]]|$1|M|3|-4|5|6|7|T|8|@]|D|@]|E|$]]]|N|$]]

<code>for element in li</code> - returns you a copy of element, not the element itself.

Solution for your case would be:

<pre><code>for i in range(len(li)):
 li[i] = 'foo'
</code></pre>

blocks|key|1901392|text|也许使用字典可能会有所帮助。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1901393|>>>+li+=+{0:+"spam",+1:"eggs"}
++++for+k,+v+in+li.iteritems():
++++++++li[k]+=+"foo"

>>>+li
{0:+'foo',+1:+'foo'}|code-block|syntax|javascript|1901394|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

Maybe using dictionaries might be helpful. 

<pre><code>&gt;&gt;&gt; li = {0: "spam", 1:"eggs"}
 for k, v in li.iteritems():
 li[k] = "foo"

&gt;&gt;&gt; li
{0: 'foo', 1: 'foo'}
</code></pre>

While looping over a list in Python, I was unable to modify the elements without a list comprehension.
For reference:
<pre><code>li = [&quot;spam&quot;, &quot;eggs&quot;]
for i in li:
 i = &quot;foo&quot;

li
[&quot;spam&quot;, &quot;eggs&quot;]

li = [&quot;foo&quot; for i in li]
li 
[&quot;foo&quot;, &quot;foo&quot;]
</code></pre>
So, why can't I modify elements through a loop in Python? There's definitely something I'm missing, but I don't know what. I'm sure this is a duplicate, but I couldn't find a question about this, and if there is a link, that would be more than enough.

Can't modify list elements in a loop

Python

在Python中遍历列表时，如果不理解列表，我就无法修改元素。供参考：li = ["spam", "eggs"]for i in li:    i = "foo"li["spam", "eggs"]li = ["foo" for i in li]li ["foo", "foo"]那么，为什么我不能在Python中通过循...

问无法修改循环中的列表元素
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