blocks|key|1091570|text|这是itemgetter的工作|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1091571|>>>+from+operator+import+itemgetter
>>>+L=[[0,+1,+'f'],+[4,+2,+'t'],+[9,+4,+'afsd']]
>>>+sorted(L,+key=itemgetter(2))
[[9,+4,+'afsd'],+[0,+1,+'f'],+[4,+2,+'t']]|code-block|syntax|javascript|1091572|在这里也可以使用lambda函数，但是lambda函数在这种简单的情况下会慢一些|1091573|entityMap|0|LINK|mutability|MUTABLE|url|http://docs.python.org/library/operator.html#operator.itemgetter^0|2|A|0|0|0|0^^$0|@$1|2|3|4|5|6|7|S|8|@]|9|@$A|T|B|U|1|V]]|C|$]]|$1|D|3|E|5|F|7|W|8|@]|9|@]|C|$G|H]]|$1|I|3|J|5|6|7|X|8|@]|9|@]|C|$]]|$1|K|3|-4|5|6|7|Y|8|@]|9|@]|C|$]]]|L|$M|$5|N|O|P|C|$Q|R]]]]

This is a job for <a href="http://docs.python.org/library/operator.html#operator.itemgetter" rel="noreferrer">itemgetter</a>

<pre><code>&gt;&gt;&gt; from operator import itemgetter
&gt;&gt;&gt; L=[[0, 1, 'f'], [4, 2, 't'], [9, 4, 'afsd']]
&gt;&gt;&gt; sorted(L, key=itemgetter(2))
[[9, 4, 'afsd'], [0, 1, 'f'], [4, 2, 't']]
</code></pre>

It is also possible to use a lambda function here, however the lambda function is slower in this simple case

blocks|key|1091626|text|就位|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1091627|>>>+l+=+[[0,+1,+'f'],+[4,+2,+'t'],+[9,+4,+'afsd']]
>>>+l.sort(key=lambda+x:+x[2])|code-block|syntax|javascript|1091628|未使用排序的位置：|1091629|>>>+sorted(l,+key=lambda+x:+x[2])|1091630|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|$]]

in place

<pre><code>&gt;&gt;&gt; l = [[0, 1, 'f'], [4, 2, 't'], [9, 4, 'afsd']]
&gt;&gt;&gt; l.sort(key=lambda x: x[2])
</code></pre>

not in place using sorted:

<pre><code>&gt;&gt;&gt; sorted(l, key=lambda x: x[2])
</code></pre>

blocks|key|1091677|text|Itemgetter允许您按多个条件/列进行排序：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1091678|sorted_list+=+sorted(list_to_sort,+key=itemgetter(2,0,1))|code-block|syntax|javascript|1091679|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|$]]

Itemgetter lets you to sort by multiple criteria / columns:

<pre><code>sorted_list = sorted(list_to_sort, key=itemgetter(2,0,1))
</code></pre>

blocks|key|1091746|text|还可以通过lambda函数实现多个条件|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1091747|sorted_list+=+sorted(list_to_sort,+key=lambda+x:+(x[1],+x[0]))|code-block|syntax|javascript|1091748|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|$]]

multiple criteria can also be implemented through lambda function

<pre><code>sorted_list = sorted(list_to_sort, key=lambda x: (x[1], x[0]))
</code></pre>

blocks|key|877198|text|array.sort(key+=+lambda+x:x[1])|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|877199|您可以使用此代码片段轻松地进行排序，其中1是元素的索引。|unstyled|877200|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>array.sort(key = lambda x:x[1])
</code></pre>

You can easily sort using this snippet, where 1 is the index of the element.

blocks|key|1091635|text|如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1091636|import+operator
l+=+[...]
sorted_list+=+sorted(l,+key=operator.itemgetter(desired_item_index))|code-block|syntax|javascript|1091637|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|$]]

Like this:

<pre><code>import operator
l = [...]
sorted_list = sorted(l, key=operator.itemgetter(desired_item_index))
</code></pre>

blocks|key|1091798|text|我认为lambda函数可以解决你的问题。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1091799|old_list+=+[[0,1,'f'],+[4,2,'t'],[9,4,'afsd']]

#let's+assume+we+want+to+sort+lists+by+last+value+(+old_list[2]+)
new_list+=+sorted(old_list,+key=lambda+x:+x[2])

#Resulst+of+new_list+will+be:

[[9,+4,+'afsd'],+[0,+1,+'f'],+[4,+2,+'t']]|code-block|syntax|javascript|1091800|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 think lambda function can solve your problem.

<pre><code>old_list = [[0,1,'f'], [4,2,'t'],[9,4,'afsd']]

#let's assume we want to sort lists by last value ( old_list[2] )
new_list = sorted(old_list, key=lambda x: x[2])

#Resulst of new_list will be:

[[9, 4, 'afsd'], [0, 1, 'f'], [4, 2, 't']]
</code></pre>

blocks|key|1091890|text|更容易理解(+Lambda实际在做什么)：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1091891|ls2=[[0,1,'f'],[4,2,'t'],[9,4,'afsd']]
def+thirdItem(ls):
++++#return+the+third+item+of+the+list
++++return+ls[2]
#Sort+according+to+what+the+thirdItem+function+return+
ls2.sort(key=thirdItem)|code-block|syntax|javascript|1091892|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|$]]

More easy to understand (What is Lambda actually doing):

<pre><code>ls2=[[0,1,'f'],[4,2,'t'],[9,4,'afsd']]
def thirdItem(ls):
 #return the third item of the list
 return ls[2]
#Sort according to what the thirdItem function return 
ls2.sort(key=thirdItem)
</code></pre>

blocks|key|877305|text|**old_list+=+[[0,1,'f'],+[4,2,'t'],[9,4,'afsd']]
++++#let's+assume+we+want+to+sort+lists+by+last+value+(+old_list[2]+)
++++new_list+=+sorted(old_list,+key=lambda+x:+x[2])**|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|877306|如果我错了，请纠正我，但是'x2‘不是在调用列表中的第三项，而不是嵌套列表中的第三项吗？应该是x2吗？|unstyled|877307|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>**old_list = [[0,1,'f'], [4,2,'t'],[9,4,'afsd']]
 #let's assume we want to sort lists by last value ( old_list[2] )
 new_list = sorted(old_list, key=lambda x: x[2])**
</code></pre>

correct me if i'm wrong but isnt the 'x[2]' calling the 3rd item in the list, not the 3rd item in the nested list? should it be x[2][2]?

blocks|key|1091944|text|对多维数组execute+here进行排序|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1091945|arr=[[2,1],[1,2],[3,5],[4,5],[3,1],[5,2],[3,8],[1,9],[1,3]]



arr.sort(key=lambda+x:x[0])
la=set([i[0]+for+i+in+Points])

for+i+in+la:
++++tempres=list()
++++for+j+in+arr:
++++++++if+j[0]==i:
++++++++++++tempres.append(j[1])

++++for+j+in+sorted(tempres,reverse=True):
++++++++print(i,j)|code-block|syntax|javascript|1091946|entityMap|0|LINK|mutability|MUTABLE|url|http://tpcg.io/ccf5kj^0|5|C|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@$A|R|B|S|1|T]]|C|$]]|$1|D|3|E|5|F|7|U|8|@]|9|@]|C|$G|H]]|$1|I|3|-4|5|6|7|V|8|@]|9|@]|C|$]]]|J|$K|$5|L|M|N|C|$O|P]]]]

Sorting a Multidimensional Array <a href="http://tpcg.io/ccf5kj" rel="nofollow noreferrer">execute here</a>

<pre><code>arr=[[2,1],[1,2],[3,5],[4,5],[3,1],[5,2],[3,8],[1,9],[1,3]]



arr.sort(key=lambda x:x[0])
la=set([i[0] for i in Points])

for i in la:
 tempres=list()
 for j in arr:
 if j[0]==i:
 tempres.append(j[1])

 for j in sorted(tempres,reverse=True):
 print(i,j)
</code></pre>

blocks|key|877372|text|确保要排序的列表中没有任何null或NaN值。如果有NaN值，那么您的排序将为off，从而影响非NULL值的排序。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|877373|查看Python:+sort+function+breaks+in+the+presence+of+nan|offset|length|877374|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/4240050/python-sort-function-breaks-in-the-presence-of-nan^0|0|2|1F|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|O|8|@]|9|@$D|P|E|Q|1|R]]|A|$]]|$1|F|3|-4|5|6|7|S|8|@]|9|@]|A|$]]]|G|$H|$5|I|J|K|A|$L|M]]]]

Make sure that you do not have any null or NaN values in the list you want to sort. If there are NaN values, then your sort will be off, impacting the sorting of the non-null values.
Check out <a href="https://stackoverflow.com/questions/4240050/python-sort-function-breaks-in-the-presence-of-nan">Python: sort function breaks in the presence of nan</a>

blocks|key|877380|text|使用自定义键函数，您可以根据需要轻松地对列表的任何列表进行排序：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|877381|L+=+[[0,1,'f'],+[4,2,'t'],+[9,4,'afsd']]

def+sorter(lst):
++++return+lst[2].casefold()

L.sort(key=sorter)

#+result:+[[9,+4,+'afsd'],+[0,+1,+'f'],+[4,+2,+'t']]|code-block|syntax|javascript|877382|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|$]]

Using a custom key function you can easily sort any list of lists as you want:
<pre><code>L = [[0,1,'f'], [4,2,'t'], [9,4,'afsd']]

def sorter(lst):
 return lst[2].casefold()

L.sort(key=sorter)

# result: [[9, 4, 'afsd'], [0, 1, 'f'], [4, 2, 't']]
</code></pre>

I have a list of lists. For example,

<pre><code>[
[0,1,'f'],
[4,2,'t'],
[9,4,'afsd']
]
</code></pre>

If I wanted to sort the outer list by the string field of the inner lists, how would you do that in python?

How to sort a list of lists by a specific index of the inner list?

我有一份清单。例如,[[0,1,'f'],[4,2,'t'],[9,4,'afsd']]如果我想按内部列表的字符串字段对外部列表进行排序，那么在python中该如何做呢？

问如何根据内部列表的特定索引对列表进行排序？
EN

回答 12

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问如何根据内部列表的特定索引对列表进行排序？EN

回答 12

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问如何根据内部列表的特定索引对列表进行排序？
EN