在 Python 中对列表列表进行排序
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Sorting a list of lists in Python
提问by l--''''''---------''''''''''''
c2=[]
row1=[1,22,53]
row2=[14,25,46]
row3=[7,8,9]
c2.append(row2)
c2.append(row1)
c2.append(row3)
c2is now:
c2就是现在：
[[14, 25, 46], [1, 22, 53], [7, 8, 9]]
how do i sort c2in such a way that for example:
我如何排序c2，例如：
for row in c2:
sort on row[2]
the result would be:
结果将是：
[[7,8,9],[14,25,46],[1,22,53]]
the other question is how do i first sort by row[2] and within that set by row[1]
另一个问题是我如何首先按行 [2] 排序，并在按行 [1] 设置的范围内排序
采纳答案by Dave Webb
The keyargument to sortspecifies a function of one argument that is used to extract a comparison key from each list element. So we can create a simple lambdathat returns the last element from each row to be used in the sort:
所述key参数sort指定用于提取从每个列表元素的比较关键一个参数的函数。所以我们可以创建一个简单的lambda返回每行的最后一个元素以用于排序：
c2.sort(key = lambda row: row[2])
A lambdais a simple anonymous function.It's handy when you want to create a simple single use function like this. The equivalent code not using a lambdawould be:
Alambda是一个简单的匿名函数。当您想创建这样一个简单的一次性功能时，这很方便。不使用 a 的等效代码lambda是：
def sort_key(row):
return row[2]
c2.sort(key = sort_key)
If you want to sort on more entries, just make the keyfunction return a tuple containing the values you wish to sort on in order of importance. For example:
如果您想对更多条目进行排序，只需让key函数返回一个元组，其中包含您希望按重要性排序的值。例如：
c2.sort(key = lambda row: (row[2],row[1]))
or:
或者：
c2.sort(key = lambda row: (row[2],row[1],row[0]))
回答by mipadi
Well, your desired example seems to indicate that you want to sort by the last index in the list, which could be done with this:
好吧，您想要的示例似乎表明您想按列表中的最后一个索引进行排序，这可以通过以下方式完成：
sorted_c2 = sorted(c2, lambda l1, l2: l1[-1] - l2[-1])
回答by John La Rooy
>>> import operator
>>> c2 = [[14, 25, 46], [1, 22, 53], [7, 8, 9]]
>>> c2.sort(key=itemgetter(2))
>>> c2
[[7, 8, 9], [14, 25, 46], [1, 22, 53]]
