在Python中排序列表

c2=[]
row1=[1,22,53]
row2=[14,25,46]
row3=[7,8,9]

c2.append(row2)
c2.append(row1)
c2.append(row3)

c2 is now:

c2是现在:

[[14, 25, 46], [1, 22, 53], [7, 8, 9]]

how do i sort c2 in such a way that for example:

如何对c2排序，例如:

for row in c2:

sort on row[2]

the result would be:

结果将会是:

[[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，如何首先对[2]行排序然后在这个集合中对[1]行排序

3 个解决方案

#1

The key argument to sort specifies a function of one argument that is used to extract a comparison key from each list element. So we can create a simple lambda that returns the last element from each row to be used in the sort:

排序的键参数指定一个参数的函数，用于从每个列表元素中提取比较键。所以我们可以创建一个简单的lambda来返回排序中每一行的最后一个元素:

c2.sort(key = lambda row: row[2])

A lambda is 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 lambda would be:

lambda是一个简单的匿名函数。当您想要创建这样一个简单的单用途函数时，它是很方便的。不使用lambda的等效代码为:

def sort_key(row):
    return row[2]

c2.sort(key = sort_key)

If you want to sort on more entries, just make the key function return a tuple containing the values you wish to sort on in order of importance. For example:

如果您想对更多的条目进行排序，只需让键函数返回一个元组，该元组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]))

#2

>>> 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]]

#3

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，你想要的例子似乎表明你想按列表中的最后一个索引排序，这可以用以下方法完成:

sorted_c2 = sorted(c2, lambda l1, l2: l1[-1] - l2[-1])

#1