blocks|key|762032|text|如下所示：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|762033|import+operator
list1+=+sorted(csv1,+key=operator.itemgetter(1,+2))|code-block|syntax|javascript|762034|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
list1 = sorted(csv1, key=operator.itemgetter(1, 2))
</code></pre>

blocks|key|977686|text|使用lambda函数时不需要导入任何内容。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|977687|下面的代码先按第一个元素对list排序，然后再按第二个元素排序。您还可以按一个字段升序和另一个字段降序进行排序，例如：|offset|length|style|CODE|977688|sorted_list+=+sorted(list,+key=lambda+x:+(x[0],+-x[1]))|code-block|syntax|javascript|977689|entityMap^0|0|D|4|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@$D|Q|E|R|F|G]]|9|@]|A|$]]|$1|H|3|I|5|J|7|S|8|@]|9|@]|A|$K|L]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

No need to import anything when using lambda functions. 
The following sorts <code>list</code> by the first element, then by the second element. You can also sort by one field ascending and another descending for example:
<pre class="lang-py prettyprint-override"><code>sorted_list = sorted(list, key=lambda x: (x[0], -x[1]))
</code></pre>

blocks|key|977616|text|list1+=+sorted(csv1,+key=lambda+x:+(x[1],+x[2])+)|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|977617|unstyled|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|G|8|@]|9|@]|A|$B|C]]|$1|D|3|-4|5|E|7|H|8|@]|9|@]|A|$]]]|F|$]]

<pre><code>list1 = sorted(csv1, key=lambda x: (x[1], x[2]) )
</code></pre>

blocks|key|977753|text|employees.sort(key+=+lambda+x:x[1])
employees.sort(key+=+lambda+x:x[0])|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|977754|我们还可以将.sort与lambda一起使用2次，因为python排序已经就位并且稳定。这将首先根据第二个元素x1对列表进行排序。然后，它将对第一个元素x0进行排序。|unstyled|offset|length|977755|employees[0]+=+Employee's+Name
employees[1]+=+Employee's+Salary|977756|这相当于执行以下操作:+employees.sort(key+=+lambda+x:(x，x1))|977757|entityMap|0|LINK|mutability|MUTABLE|url|highest+priority^0|0|25|1|0|0|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|V|8|@]|9|@$G|W|H|X|1|Y]]|A|$]]|$1|I|3|J|5|6|7|Z|8|@]|9|@]|A|$B|C]]|$1|K|3|L|5|F|7|10|8|@]|9|@]|A|$]]|$1|M|3|-4|5|F|7|11|8|@]|9|@]|A|$]]]|N|$O|$5|P|Q|R|A|$S|T]]]]

<pre><code>employees.sort(key = lambda x:x[1])
employees.sort(key = lambda x:x[0])
</code></pre>

We can also use .sort with lambda 2 times because python sort is in place and stable. This will first sort the list according to the second element, x[1]. Then, it will sort the first element, x[0] (highest priority).

<pre><code>employees[0] = Employee's Name
employees[1] = Employee's Salary
</code></pre>

This is equivalent to doing the following:
employees.sort(key = lambda x:(x[0], x[1]))

blocks|key|977826|text|使用下面的字典排序列表将按降序对列表进行排序，第一列为薪水，第二列为年龄|type|unstyled|depth|inlineStyleRanges|entityRanges|data|977827|d=[{'salary':123,'age':23},{'salary':123,'age':25}]
d=sorted(d,+key=lambda+i:+(i['salary'],+i['age']),reverse=True)|code-block|syntax|javascript|977828|输出：{‘工资’：123，‘年龄’：25}，{‘工资’：123，‘年龄’：23}|977829|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|$]]

Sorting list of dicts using below will sort list in descending order on first column as salary and second column as age

<pre><code>d=[{'salary':123,'age':23},{'salary':123,'age':25}]
d=sorted(d, key=lambda i: (i['salary'], i['age']),reverse=True)
</code></pre>

Output: [{'salary': 123, 'age': 25}, {'salary': 123, 'age': 23}]

blocks|key|977884|text|在升序中，您可以使用：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|977885|sorted_data=+sorted(non_sorted_data,+key=lambda+k:+(k[1],k[0]))|code-block|syntax|javascript|977886|或按降序排列，您可以使用：|977887|sorted_data=+sorted(non_sorted_data,+key=lambda+k:+(k[1],k[0]),reverse=True)|977888|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 ascending order you can use:

<pre><code>sorted_data= sorted(non_sorted_data, key=lambda k: (k[1],k[0]))
</code></pre>

or in descending order you can use:

<pre><code>sorted_data= sorted(non_sorted_data, key=lambda k: (k[1],k[0]),reverse=True)
</code></pre>

blocks|key|762488|text|在阅读了这篇文章中的答案后，我写了一个通用的解决方案，可以适用于任意数量的列：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|762489|def+sort_array(array,+*columns):
++++for+col+in+columns:
++++++++array.sort(key+=+lambda+x:x[col])|code-block|syntax|javascript|762490|操作员会这样称呼它：|762491|sort_array(list1,+2,+1)|762492|它首先按第2列排序，然后按第1列排序。|762493|(最重要的一列放在最后)|762494|entityMap^0|0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|Q|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|R|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|S|8|@]|9|@]|A|$]]|$1|I|3|J|5|D|7|T|8|@]|9|@]|A|$E|F]]|$1|K|3|L|5|6|7|U|8|@]|9|@]|A|$]]|$1|M|3|N|5|6|7|V|8|@]|9|@]|A|$]]|$1|O|3|-4|5|6|7|W|8|@]|9|@]|A|$]]]|P|$]]

After reading the answers in this thread, I wrote a general solution that will work for an arbitrary number of columns:
<pre><code>def sort_array(array, *columns):
 for col in columns:
 array.sort(key = lambda x:x[col])
</code></pre>
The OP would call it like this:
<pre><code>sort_array(list1, 2, 1)
</code></pre>
Which sorts first by column 2, then by column 1. 
(Most important column goes last)

blocks|key|762417|text|python+3+https://docs.python.org/3.5/howto/sorting.html#the-old-way-using-the-cmp-parameter|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|762418|from+functools+import+cmp_to_key

def+custom_compare(x,+y):
++++#+custom+comparsion+of+x[0],+x[1]+with+y[0],+y[1]
++++return+0

sorted(entries,+key=lambda+e:+(cmp_to_key(custom_compare)(e[0]),+e[1]))|code-block|syntax|javascript|762419|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3.5/howto/sorting.html#the-old-way-using-the-cmp-parameter^0|9|2A|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]]]]

python 3
<a href="https://docs.python.org/3.5/howto/sorting.html#the-old-way-using-the-cmp-parameter" rel="nofollow noreferrer">https://docs.python.org/3.5/howto/sorting.html#the-old-way-using-the-cmp-parameter</a>
<pre><code>from functools import cmp_to_key

def custom_compare(x, y):
 # custom comparsion of x[0], x[1] with y[0], y[1]
 return 0

sorted(entries, key=lambda e: (cmp_to_key(custom_compare)(e[0]), e[1]))
</code></pre>

I have the following list created from a sorted csv

<pre><code>list1 = sorted(csv1, key=operator.itemgetter(1))
</code></pre>

I would actually like to sort the list by two criteria: first by the value in field 1 and then by the value in field 2. How do I do this?

Sorting a Python list by two fields

我从排序后的csv中创建了以下列表list1 = sorted(csv1, key=operator.itemgetter(1))我实际上想要根据两个标准对列表进行排序:首先是字段1中的值，然后是字段2中的值。我该怎么做？

问按两个字段对Python列表进行排序
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