blocks|key|1390706|text|是的，这是默认设置。事实上，这是Python中经典的"DSU“(装饰-排序-取消装饰)习惯用法的基础。参见Code+Like+a+Pythonista。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1390707|entityMap|0|LINK|mutability|MUTABLE|url|https://david.goodger.org/projects/pycon/2007/idiomatic/handout.html^0|1H|M|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@$A|M|B|N|1|O]]|C|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]]]

Yes, this is the default. In fact, this is the basis of the classic "DSU" (Decorate-Sort-Undecorate) idiom in Python. See <a href="https://david.goodger.org/projects/pycon/2007/idiomatic/handout.html" rel="nofollow noreferrer">Code Like a Pythonista</a>.

blocks|key|1390766|text|它自动按照元组中的第一个元素对元组列表进行排序，然后按第二个元素排序，依此类推，tuple(1,2,3)将排在tuple(1,2,4)之前。如果您想覆盖此行为，请将一个callable作为第二个参数传递给排序方法。此可调用函数应返回1，-1，0。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1390767|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

It automatically sorts a list of tuples by the first elements in the tuples, then by the second elements and so on tuple([1,2,3]) will go before tuple([1,2,4]). If you want to override this behaviour pass a callable as the second argument to the sort method. This callable should return 1, -1, 0.

blocks|key|1605729|text|不，元组就像字符串一样是sequence+types。通过依次比较每个元素，对它们进行相同的排序：|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1605730|>>>+import+random
>>>+sorted([(0,0,0,int(random.getrandbits(4)))+for+x+in+xrange(10)])
[(0,+0,+0,+0),+(0,+0,+0,+4),+(0,+0,+0,+5),+(0,+0,+0,+7),+(0,+0,+0,+8),
(0,+0,+0,+9),+(0,+0,+0,+12),+(0,+0,+0,+12),+(0,+0,+0,+12),+(0,+0,+0,+14)]|code-block|syntax|javascript|1605731|三个零的存在只是为了表明必须检查第一个元素以外的其他元素。|1605732|entityMap|0|LINK|mutability|MUTABLE|url|http://www.python.org/doc/2.5.2/lib/typesseq.html^0|C|E|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]]]]

No, tuples are <a href="http://www.python.org/doc/2.5.2/lib/typesseq.html" rel="noreferrer">sequence types</a> just like strings. They are sorted the same, by comparing each element in turn:

<pre><code>&gt;&gt;&gt; import random
&gt;&gt;&gt; sorted([(0,0,0,int(random.getrandbits(4))) for x in xrange(10)])
[(0, 0, 0, 0), (0, 0, 0, 4), (0, 0, 0, 5), (0, 0, 0, 7), (0, 0, 0, 8),
(0, 0, 0, 9), (0, 0, 0, 12), (0, 0, 0, 12), (0, 0, 0, 12), (0, 0, 0, 14)]
</code></pre>

The three zeroes are only there to show that something other than the first element must be getting inspected.

blocks|key|1390862|text|尝试使用内部列表排序方法并传递一个lambda。如果您的元组的第一个元素是一个整数，这应该是可行的。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1390863|#+l+is+the+list+of+tuples
l.sort(lambda+x,y:+x-y)|code-block|syntax|javascript|1390864|您可以对compare函数使用任何可调用的函数，而不一定是lambda。但是，它需要返回-1+(小于)、0(等于)或1(大于)。|1390865|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|$]]

Try using the internal list sort method and pass a lambda. If your tuples first element is a integer, this should work.

<pre><code># l is the list of tuples
l.sort(lambda x,y: x-y)
</code></pre>

You can use any callable for the compare function, not necessarily a lambda. However it needs to return -1 (less than), 0 (equal) or 1 (greater than).

blocks|key|1605846|text|查看"Devin+Jeanpierre“这个问题的答案sort-a-dictionary-in-python-by-the-value，他说要使用一个元组，并展示了如何按第二个值排序|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1605847|entityMap|0|LINK|mutability|MUTABLE|url|https://stackoverflow.com/questions/613183/sort-a-dictionary-in-python-by-the-value^0|R|14|0|0^^$0|@$1|2|3|4|5|6|7|L|8|@]|9|@$A|M|B|N|1|O]]|C|$]]|$1|D|3|-4|5|6|7|P|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]]]

Check out "Devin Jeanpierre" answer to this question <a href="https://stackoverflow.com/questions/613183/sort-a-dictionary-in-python-by-the-value">sort-a-dictionary-in-python-by-the-value</a> where he says to use a tuple and shows how to sort by the second value

Empirically, it seems that Python's default list sorter, when passed a list of tuples, will sort by the first element in each tuple. Is that correct? If not, what's the right way to sort a list of tuples by their first elements?

How does Python sort a list of tuples?

Python

从经验上看，当传递一个元组列表时，Python的默认列表排序器将按每个元组中的第一个元素排序。对吗？如果不是，那么按照元组列表的第一个元素排序的正确方法是什么？

问Python如何对元组列表进行排序？
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