blocks|key|36141|text|试试这个：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|36142|sorted(l,+key=lambda+x:+float('inf')+if+x+is+None+else+x)|code-block|syntax|javascript|36143|因为无穷大比所有整数都大，所以None总是放在最后。|offset|length|style|CODE|36144|entityMap^0|0|0|F|4|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|P|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|Q|8|@$I|R|J|S|K|L]]|9|@]|A|$]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

Try this:

<pre><code>sorted(l, key=lambda x: float('inf') if x is None else x)
</code></pre>

Since infinity is larger than all integers, <code>None</code> will always be placed last.

blocks|key|252680|text|>>>+l+=+[1,+3,+2,+5,+4,+None,+7]
>>>+sorted(l,+key=lambda+x:+(x+is+None,+x))
[1,+2,+3,+4,+5,+7,+None]|type|code-block|depth|inlineStyleRanges|entityRanges|data|syntax|javascript|252681|这将为列表中的每个元素构造一个元组，如果值为None，则元组为(True,+None)，如果值为其他值，则为(False,+x)+(其中x为值)。由于元组是逐项排序的，这意味着所有非None元素将首先出现(从False+<+True开始)，然后按值排序。|unstyled|offset|length|style|CODE|252682|entityMap^0|0|M|4|V|C|1I|A|1W|1|2J|4|2W|C|0^^$0|@$1|2|3|4|5|6|7|M|8|@]|9|@]|A|$B|C]]|$1|D|3|E|5|F|7|N|8|@$G|O|H|P|I|J]|$G|Q|H|R|I|J]|$G|S|H|T|I|J]|$G|U|H|V|I|J]|$G|W|H|X|I|J]|$G|Y|H|Z|I|J]]|9|@]|A|$]]|$1|K|3|-4|5|F|7|10|8|@]|9|@]|A|$]]]|L|$]]

<pre><code>&gt;&gt;&gt; l = [1, 3, 2, 5, 4, None, 7]
&gt;&gt;&gt; sorted(l, key=lambda x: (x is None, x))
[1, 2, 3, 4, 5, 7, None]
</code></pre>

This constructs a tuple for each element in the list, if the value is <code>None</code> the tuple with be <code>(True, None)</code>, if the value is anything else it will be <code>(False, x)</code> (where <code>x</code> is the value). Since tuples are sorted item by item, this means that all non-<code>None</code> elements will come first (since <code>False &lt; True</code>), and then be sorted by value.

blocks|key|252705|text|我创建了一个函数，扩展了Andrew+Clark的答案和tutuDajuju的评论。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|252706|def+sort(myList,+reverse+=+False,+sortNone+=+False):
++++"""Sorts+a+list+that+may+or+may+not+contain+None.
++++Special+thanks+to+Andrew+Clark+and+tutuDajuju+for+how+to+sort+None+on+https://stackoverflow.com/questions/18411560/python-sort-list-with-none-at-the-end

++++reverse+(bool)+-+Determines+if+the+list+is+sorted+in+ascending+or+descending+order

++++sortNone+(bool)+-+Determines+how+None+is+sorted
++++++++-+If+True:+Will+place+None+at+the+beginning+of+the+list
++++++++-+If+False:+Will+place+None+at+the+end+of+the+list
++++++++-+If+None:+Will+remove+all+instances+of+None+from+the+list

++++Example+Input:+sort([1,+3,+2,+5,+4,+None,+7])
++++Example+Input:+sort([1,+3,+2,+5,+4,+None,+7],+reverse+=+True)
++++Example+Input:+sort([1,+3,+2,+5,+4,+None,+7],+reverse+=+True,+sortNone+=+True)
++++Example+Input:+sort([1,+3,+2,+5,+4,+None,+7],+sortNone+=+None)
++++"""

++++return+sorted(filter(lambda+item:+True+if+(sortNone+!=+None)+else+(item+!=+None),+myList),+
++++++++key+=+lambda+item:+(((item+is+None)+++++if+(reverse)+else+(item+is+not+None))+if+(sortNone)+else
++++++++++++++++++++++++++++((item+is+not+None)+if+(reverse)+else+(item+is+None)),+item),+
++++++++reverse+=+reverse)|code-block|syntax|javascript|252707|下面是一个如何运行它的示例：|252708|myList+=+[1,+3,+2,+5,+4,+None,+7]
print(sort(myList))
print(sort(myList,+reverse+=+True))
print(sort(myList,+sortNone+=+True))
print(sort(myList,+reverse+=+True,+sortNone+=+True))
print(sort(myList,+sortNone+=+None))
print(sort(myList,+reverse+=+True,+sortNone+=+None))|252709|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 created a function that expands on the answer by Andrew Clark and the comment by tutuDajuju.

<pre><code>def sort(myList, reverse = False, sortNone = False):
 """Sorts a list that may or may not contain None.
 Special thanks to Andrew Clark and tutuDajuju for how to sort None on https://stackoverflow.com/questions/18411560/python-sort-list-with-none-at-the-end

 reverse (bool) - Determines if the list is sorted in ascending or descending order

 sortNone (bool) - Determines how None is sorted
 - If True: Will place None at the beginning of the list
 - If False: Will place None at the end of the list
 - If None: Will remove all instances of None from the list

 Example Input: sort([1, 3, 2, 5, 4, None, 7])
 Example Input: sort([1, 3, 2, 5, 4, None, 7], reverse = True)
 Example Input: sort([1, 3, 2, 5, 4, None, 7], reverse = True, sortNone = True)
 Example Input: sort([1, 3, 2, 5, 4, None, 7], sortNone = None)
 """

 return sorted(filter(lambda item: True if (sortNone != None) else (item != None), myList), 
 key = lambda item: (((item is None) if (reverse) else (item is not None)) if (sortNone) else
 ((item is not None) if (reverse) else (item is None)), item), 
 reverse = reverse)
</code></pre>

Here is an example of how you can run it:

<pre><code>myList = [1, 3, 2, 5, 4, None, 7]
print(sort(myList))
print(sort(myList, reverse = True))
print(sort(myList, sortNone = True))
print(sort(myList, reverse = True, sortNone = True))
print(sort(myList, sortNone = None))
print(sort(myList, reverse = True, sortNone = None))
</code></pre>

I have a homogeneous list of objects with None, but it can contain any type of values. 
Example:

<pre><code>&gt;&gt;&gt; l = [1, 3, 2, 5, 4, None, 7]
&gt;&gt;&gt; sorted(l)
[None, 1, 2, 3, 4, 5, 7]
&gt;&gt;&gt; sorted(l, reverse=True)
[7, 5, 4, 3, 2, 1, None]
</code></pre>

Is there a way without reinventing the wheel to get the list sorted the usual python way, but with None values at the end of the list, like that:

<pre><code>[1, 2, 3, 4, 5, 7, None]
</code></pre>

I feel like here can be some trick with "key" parameter

Sort list while pushing None values to the end

我有一个不带None的同构对象列表，但它可以包含任何类型的值。示例：>>> l = [1, 3, 2, 5, 4, None, 7]>>> sorted(l)[None, 1, 2, 3, 4, 5, 7]>>> sorted(l, reverse=True)[7, 5, 4, 3, 2, 1, None]有没有一种方...

问将无值推送到末尾时对列表进行排序
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