blocks|key|1491059|text|如果你想看整本字典，试试下面这一行：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1491060|newDic+=++{key:value[::-1]+for+key,value+in+dic.iteritems()}|offset|length|style|CODE|1491061|如果你只想做一个列表，试试这个：|1491062|list1+=++dic["foo"]
newDic+=+{}
newDic["foo"]+=+list1[::-1]|code-block|syntax|javascript|1491063|list.reverse()将更改列表本身。尝试list[::-1]，它将返回一个反转列表的新副本|1491064|entityMap^0|0|0|1O|0|0|0|0|E|O|A|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|T|8|@$D|U|E|V|F|G]]|9|@]|A|$]]|$1|H|3|I|5|6|7|W|8|@]|9|@]|A|$]]|$1|J|3|K|5|L|7|X|8|@]|9|@]|A|$M|N]]|$1|O|3|P|5|6|7|Y|8|@$D|Z|E|10|F|G]|$D|11|E|12|F|G]]|9|@]|A|$]]|$1|Q|3|-4|5|6|7|13|8|@]|9|@]|A|$]]]|R|$]]

If you want to do the whole dictionary try this one liner:

<code>newDic = {key:value[::-1] for key,value in dic.iteritems()}</code>

If you just want to do one list try this:

<pre><code>list1 = dic["foo"]
newDic = {}
newDic["foo"] = list1[::-1]
</code></pre>

<code>list.reverse()</code> will change the list itself. Try <code>list[::-1]</code> and it will return a new copy of a reversed list

blocks|key|1491078|text|list1.reverse()是一种就地逆转。请改用函数reversed(list1)，该函数不会修改调用该函数的列表。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1491079|具体地说：|1491080|dic+=+{"foo"+:+[1,3,5,6]+,+"bar"+:+[5,+3,+8,+4]}
list1+=+dic["foo"]
list2+=+list(reversed(list1))
newDic+=+{}
newDic["foo"]+=+list2|code-block|syntax|javascript|1491081|注意，reversed(list1)返回一个generator，而您想要完整的列表，这就是上面调用list的原因。|1491082|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/2/reference/expressions.html?highlight=yield#generator-expressions^0|0|F|S|F|0|0|0|3|F|1D|4|M|9|0|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|11|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|12|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|13|8|@$9|14|A|15|B|C]|$9|16|A|17|B|C]]|D|@$9|18|A|19|1|1A]]|E|$]]|$1|O|3|-4|5|6|7|1B|8|@]|D|@]|E|$]]]|P|$Q|$5|R|S|T|E|$U|V]]]]

<code>list1.reverse()</code> is an in-place reversal. Use the function <code>reversed(list1)</code> instead, which does not modify the list it is called on.

Specifically:

<pre><code>dic = {"foo" : [1,3,5,6] , "bar" : [5, 3, 8, 4]}
list1 = dic["foo"]
list2 = list(reversed(list1))
newDic = {}
newDic["foo"] = list2
</code></pre>

Note that <code>reversed(list1)</code> returns a <a href="https://docs.python.org/2/reference/expressions.html?highlight=yield#generator-expressions" rel="nofollow">generator</a> and you want the full list, so that's why <code>list</code> is called on it above.

blocks|key|5573865|text|问题出在list2+=+list1.reverse()|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|5573866|list.reverse()方法将颠倒相同的列表。|5573867|喜欢|5573868|>>>+a+=+[66.25,+333,+333,+1,+1234.5]
>>>+a.reverse()
>>>+a
[1234.5,+1,+333,+333,+66.25]|code-block|syntax|javascript|5573869|reverse的返回类型为None|5573870|>>>+a+=+[66.25,+333,+333,+1,+1234.5]
>>>+b+=+a.reverse()
>>>+b
>>>+print+b
None|5573871|所以在你的例子中，list2+=+None。|5573872|另一个问题是|5573873|您正在访问f00，密钥为foo|BOLD|5573874|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/2/tutorial/datastructures.html#more-on-lists^0|4|N|0|0|E|0|0|0|0|0|7|D|4|0|0|9|C|0|0|5|3|C|3|5|3|C|3|0^^$0|@$1|2|3|4|5|6|7|17|8|@$9|18|A|19|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|1A|8|@]|D|@$9|1B|A|1C|1|1D]]|E|$]]|$1|H|3|I|5|6|7|1E|8|@]|D|@]|E|$]]|$1|J|3|K|5|L|7|1F|8|@]|D|@]|E|$M|N]]|$1|O|3|P|5|6|7|1G|8|@$9|1H|A|1I|B|C]|$9|1J|A|1K|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|L|7|1L|8|@]|D|@]|E|$M|N]]|$1|S|3|T|5|6|7|1M|8|@$9|1N|A|1O|B|C]]|D|@]|E|$]]|$1|U|3|V|5|6|7|1P|8|@]|D|@]|E|$]]|$1|W|3|X|5|6|7|1Q|8|@$9|1R|A|1S|B|Y]|$9|1T|A|1U|B|Y]|$9|1V|A|1W|B|C]|$9|1X|A|1Y|B|C]]|D|@]|E|$]]|$1|Z|3|-4|5|6|7|1Z|8|@]|D|@]|E|$]]]|10|$11|$5|12|13|14|E|$15|16]]]]

The problem is <code>list2 = list1.reverse()</code>

<a href="https://docs.python.org/2/tutorial/datastructures.html#more-on-lists" rel="nofollow">list.reverse()</a> method will reverse same list. 

like 

<pre><code>&gt;&gt;&gt; a = [66.25, 333, 333, 1, 1234.5]
&gt;&gt;&gt; a.reverse()
&gt;&gt;&gt; a
[1234.5, 1, 333, 333, 66.25]
</code></pre>

Return type of <code>reverse</code> is <code>None</code>

<pre><code>&gt;&gt;&gt; a = [66.25, 333, 333, 1, 1234.5]
&gt;&gt;&gt; b = a.reverse()
&gt;&gt;&gt; b
&gt;&gt;&gt; print b
None
</code></pre>

So in your case <code>list2 = None</code>.

Another problem is 

your are accessing <code>f00</code> and key is <code>foo</code>

blocks|key|1491103|text|你真正的问题是创建列表的副本，所以你可以这样做|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1491104|list2+=+list(list1)
list2.reverse()|code-block|syntax|javascript|1491105|或|1491106|list2+=+list1[:]
list2.reverse()|1491107|entityMap^0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|N|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|O|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|K|3|-4|5|6|7|Q|8|@]|9|@]|A|$]]]|L|$]]

Your real problem is creating a copy of the list, so you can do

<pre><code>list2 = list(list1)
list2.reverse()
</code></pre>

or

<pre><code>list2 = list1[:]
list2.reverse()
</code></pre>

blocks|key|1496138|text|另一种方式：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1496139|list2+=+[]
for+e+in+reversed(list1):
+++list2.append(e)|code-block|syntax|javascript|1496140|List1.verse()反转原始列表。相反，您可以在for循环中执行此操作。|1496141|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|$]]

Another way:

<pre><code>list2 = []
for e in reversed(list1):
 list2.append(e)
</code></pre>

list1.reverse() reverses the original list. rather you could do in a for loop.

I have a dictionary of lists say:

<pre><code>dic = {"foo" : [1,3,5,6] , "bar" : [5, 3, 8, 4]}
</code></pre>

I'm trying to create a new dictionary of the reverse of these lists without corrupting the original

I have even tried things like

<pre><code>list1 = dic["foo"]
list2 = list1.reverse()
newDic = {}
newDic["foo"] = list2
</code></pre>

But it seems that <code>dic["A"]</code>, <code>list1</code> and <code>list2</code> are all pointing to the same memory location.
In other words, <code>dic["foo"]</code> also gets reversed in this process, which is not what I want!!!
is there any easy way around this?

Python: How to reverse a list in a new dictionary without reversing the original

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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我有一本列表字典，上面写着：dic = {"foo" : [1,3,5,6] , "bar" : [5, 3, 8, 4]}我正在尝试创建一个与这些列表相反的新字典，同时又不破坏原来的字典我甚至尝试过像这样的东西list1 =  dic["foo"]list2 = list1.reverse()newDic = {}n...

问Python:如何在不颠倒原始词典的情况下颠倒新词典中的列表
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问Python:如何在不颠倒原始词典的情况下颠倒新词典中的列表EN