如何使列表列表中的每个列表按升序排序？

Question

I have the following lists:

[[1, 5], [3, 7], [4, 2], [7, 8], [6, 3], [2, 5], [4, 1]]

And I am trying to sort them by the first value, after making the list go in ascending order:

Desired output:

[[1, 4], [1, 5], [2, 4], [2, 5], [3, 6], [3, 7], [7, 8]]

However, list.sort() only gives the following:

>>> mylist = [[1, 5], [3, 7], [4, 2], [7, 8], [6, 3], [2, 5], [4, 1]]
>>> mylist.sort()
>>> mylist
[[1, 5], [2, 5], [3, 7], [4, 1], [4, 2], [6, 3], [7, 8]]
>>> 

Of course, I could always loop each list in the list of lists and sort it:

>>> mylist
[[1, 5], [2, 5], [3, 7], [4, 1], [4, 2], [6, 3], [7, 8]]
>>> for k in range(len(mylist)):
...     mylist[k] = sorted(mylist[k])
... 
>>> mylist
[[1, 5], [2, 5], [3, 7], [1, 4], [2, 4], [3, 6], [7, 8]]
>>> sorted(mylist)
[[1, 4], [1, 5], [2, 4], [2, 5], [3, 6], [3, 7], [7, 8]]

But is there a one liner to solve this?

Answer 1

你可以做：

sorted(sorted(sublist) for sublist in mylist)

Answer 2

This is a little better than your loop:

for sublist in mylist:
    sublist.sort()

mylist.sort()

Of course, this changes each sublist in-place . Judging by your examples, it looks like that is what you want, but I thought I should mention it just in case.

Answer 3

Here is a one liner that does the sort in-place

>>> mylist = [[1, 5], [3, 7], [4, 2], [7, 8], [6, 3], [2, 5], [4, 1]]
>>> mylist.sort(key=lambda x:x.sort() or x)
>>> mylist
[[1, 4], [1, 5], [2, 4], [2, 5], [3, 6], [3, 7], [7, 8]]