blocks|key|1069542|text|将lambda传递给sort时，需要返回整数，而不是布尔值。因此，您的代码应该改为如下所示：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1069543|xs.sort(lambda+x,y:+cmp(len(x),+len(y)))|code-block|syntax|javascript|1069544|请注意，cmp是一个内置函数，如果x小于y，则cmp(x,+y)返回-1；如果x等于y，则返回0；如果x大于y，则返回1。|1069545|当然，您也可以使用key参数：|1069546|xs.sort(key=lambda+s:+len(s))|1069547|这告诉sort方法根据键函数返回的内容进行排序。|1069548|编辑:感谢下面的balpha和Ruslan指出，您可以直接将len作为关键参数传递给函数，从而消除了对lambda的需要|1069549|xs.sort(key=len)|1069550|正如Ruslan在下面指出的，您还可以使用内置的sorted函数而不是list.sort方法，后者会创建一个新的列表，而不是对现有列表进行就地排序：|1069551|print(sorted(xs,+key=len))|1069552|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/library/functions.html#cmp|1|http://docs.python.org/library/functions.html#sorted^0|1|6|A|4|0|0|H|1|K|1|N|9|13|1|16|1|1F|1|1I|1|4|3|0|0|9|3|0|0|3|4|0|U|3|1F|6|0|0|Z|9|O|6|1|0|0^^$0|@$1|2|3|4|5|6|7|1A|8|@$9|1B|A|1C|B|C]|$9|1D|A|1E|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|1F|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|1G|8|@$9|1H|A|1I|B|C]|$9|1J|A|1K|B|C]|$9|1L|A|1M|B|C]|$9|1N|A|1O|B|C]|$9|1P|A|1Q|B|C]|$9|1R|A|1S|B|C]|$9|1T|A|1U|B|C]]|D|@$9|1V|A|1W|1|1X]]|E|$]]|$1|M|3|N|5|6|7|1Y|8|@$9|1Z|A|20|B|C]]|D|@]|E|$]]|$1|O|3|P|5|H|7|21|8|@]|D|@]|E|$I|J]]|$1|Q|3|R|5|6|7|22|8|@$9|23|A|24|B|C]]|D|@]|E|$]]|$1|S|3|T|5|6|7|25|8|@$9|26|A|27|B|C]|$9|28|A|29|B|C]]|D|@]|E|$]]|$1|U|3|V|5|H|7|2A|8|@]|D|@]|E|$I|J]]|$1|W|3|X|5|6|7|2B|8|@$9|2C|A|2D|B|C]]|D|@$9|2E|A|2F|1|2G]]|E|$]]|$1|Y|3|Z|5|H|7|2H|8|@]|D|@]|E|$I|J]]|$1|10|3|-4|5|6|7|2I|8|@]|D|@]|E|$]]]|11|$12|$5|13|14|15|E|$16|17]]|18|$5|13|14|15|E|$16|19]]]]

When you pass a <code>lambda</code> to <code>sort</code>, you need to return an integer, not a boolean. So your code should instead read as follows:
<pre><code>xs.sort(lambda x,y: cmp(len(x), len(y)))
</code></pre>
Note that <a href="http://docs.python.org/library/functions.html#cmp" rel="noreferrer">cmp</a> is a builtin function such that <code>cmp(x, y)</code> returns -1 if <code>x</code> is less than <code>y</code>, 0 if <code>x</code> is equal to <code>y</code>, and 1 if <code>x</code> is greater than <code>y</code>.
Of course, you can instead use the <code>key</code> parameter:
<pre><code>xs.sort(key=lambda s: len(s))
</code></pre>
This tells the <code>sort</code> method to order based on whatever the key function returns.
EDIT: Thanks to balpha and Ruslan below for pointing out that you can just pass <code>len</code> directly as the key parameter to the function, thus eliminating the need for a <code>lambda</code>:
<pre><code>xs.sort(key=len)
</code></pre>
And as Ruslan points out below, you can also use the built-in <a href="http://docs.python.org/library/functions.html#sorted" rel="noreferrer">sorted</a> function rather than the <code>list.sort</code> method, which creates a new list rather than sorting the existing one in-place:
<pre><code>print(sorted(xs, key=len))
</code></pre>

blocks|key|1069620|text|与Eli的答案相同-只是使用较短的表单，因为您可以跳过此处的lambda部分。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1069621|创建新列表：|1069622|>>>+xs+=+['dddd','a','bb','ccc']
>>>+sorted(xs,+key=len)
['a',+'bb',+'ccc',+'dddd']|code-block|syntax|javascript|1069623|就地排序：|1069624|>>>+xs.sort(key=len)
>>>+xs
['a',+'bb',+'ccc',+'dddd']|1069625|entityMap^0|U|6|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@$9|T|A|U|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|V|8|@]|D|@]|E|$]]|$1|H|3|I|5|J|7|W|8|@]|D|@]|E|$K|L]]|$1|M|3|N|5|6|7|X|8|@]|D|@]|E|$]]|$1|O|3|P|5|J|7|Y|8|@]|D|@]|E|$K|L]]|$1|Q|3|-4|5|6|7|Z|8|@]|D|@]|E|$]]]|R|$]]

The same as in Eli's answer - just using a shorter form, because you can skip a <code>lambda</code> part here. 

Creating new list:

<pre><code>&gt;&gt;&gt; xs = ['dddd','a','bb','ccc']
&gt;&gt;&gt; sorted(xs, key=len)
['a', 'bb', 'ccc', 'dddd']
</code></pre>

In-place sorting:

<pre><code>&gt;&gt;&gt; xs.sort(key=len)
&gt;&gt;&gt; xs
['a', 'bb', 'ccc', 'dddd']
</code></pre>

blocks|key|1284289|text|编写一个函数lensort，根据长度对字符串列表进行排序。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1284290|def+lensort(a):
++++n+=+len(a)
++++for+i+in+range(n):
++++++++for+j+in+range(i%2B1,n):
++++++++++++if+len(a[i])+>+len(a[j]):
++++++++++++++++temp+=+a[i]
++++++++++++++++a[i]+=+a[j]
++++++++++++++++a[j]+=+temp
++++return+a
print+lensort(["hello","bye","good"])|code-block|syntax|javascript|1284291|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|$]]

Write a function lensort to sort a list of strings based on length. 

<pre><code>def lensort(a):
 n = len(a)
 for i in range(n):
 for j in range(i+1,n):
 if len(a[i]) &gt; len(a[j]):
 temp = a[i]
 a[i] = a[j]
 a[j] = temp
 return a
print lensort(["hello","bye","good"])
</code></pre>

blocks|key|1069639|text|def+lensort(list_1):
++++list_2=[];list_3=[]
for+i+in+list_1:
++++list_2.append([i,len(i)])
list_2.sort(key+=+lambda+x+:+x[1])
for+i+in+list_2:
++++list_3.append(i[0])
return+list_3|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|1069640|这对我来说很有效！|unstyled|1069641|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|J|8|@]|9|@]|A|$]]|$1|G|3|-4|5|F|7|K|8|@]|9|@]|A|$]]]|H|$]]

<pre><code>def lensort(list_1):
 list_2=[];list_3=[]
for i in list_1:
 list_2.append([i,len(i)])
list_2.sort(key = lambda x : x[1])
for i in list_2:
 list_3.append(i[0])
return list_3
</code></pre>

This works for me!

blocks|key|1284310|text|我可以使用以下两种方法来完成这项工作:使用function|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1284311|def+lensort(x):
++++list1+=+[]
++++for+i+in+x:
++++++++list1.append([len(i),i])
++++return+sorted(list1)

lista+=+['a',+'bb',+'ccc',+'dddd']
a=lensort(lista)
print([l[1]+for+l+in+a])|code-block|syntax|javascript|1284312|在一个使用Lambda的Liner中，如下所示，上面已经回答了。|1284313|+lista+=+['a',+'bb',+'ccc',+'dddd']
+lista.sort(key+=+lambda+x:len(x))
+print(lista)|1284314|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|$]]

I can do it using below two methods, using function

<pre><code>def lensort(x):
 list1 = []
 for i in x:
 list1.append([len(i),i])
 return sorted(list1)

lista = ['a', 'bb', 'ccc', 'dddd']
a=lensort(lista)
print([l[1] for l in a])
</code></pre>

In one Liner using Lambda, as below, a already answered above.

<pre><code> lista = ['a', 'bb', 'ccc', 'dddd']
 lista.sort(key = lambda x:len(x))
 print(lista)
</code></pre>

I want to sort a list of strings based on the string length. I tried to use sort as follows, but it doesn't seem to give me correct result.

<pre><code>xs = ['dddd','a','bb','ccc']
print xs
xs.sort(lambda x,y: len(x) &lt; len(y))
print xs

['dddd', 'a', 'bb', 'ccc']
['dddd', 'a', 'bb', 'ccc']
</code></pre>

What might be wrong?

Sorting Python list based on the length of the string

我想根据字符串长度对字符串列表进行排序。我尝试使用排序，如下所示，但它似乎没有给我正确的结果。xs = ['dddd','a','bb','ccc']print xsxs.sort(lambda x,y: len(x) < len(y))print xs['dddd', 'a', 'bb', 'ccc']['dddd'...

问根据字符串长度对Python列表进行排序
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