如何在python中按最大两个轴对列表进行排序？

Question

I want to sort a list by the maximum of the 0th and 1st index of each element of the list.

For example, if I have a list like this:

[[2,1,2],
 [0,3,9],
 [1,2,9],
 [1,1,2],
 [3,2,7]]

After sorting, it should look like this:

[[0,3,9],
 [3,2,7],
 [2,1,2],
 [1,2,9],
 [1,1,2]]

So what's an efficient way to do this?

I prefer to use numpy but other solutions are also welcome.

Answer 1

Take the row-wise maxima of the first two columns, argsort, then reverse to get descending order.

You haven't specified any tie-breaker, so this doesn't match your desired output. In other words, there's no unique solution to your problem.

res = A[A[:, :2].max(1).argsort()[::-1]]

array([[3, 2, 7],
       [0, 3, 9],
       [1, 2, 9],
       [2, 1, 2],
       [1, 1, 2]])

Answer 2

Just use sort with a custom key function:

data = [[2,1,2],[0,3,9],[1,2,9],[1,1,2],[3,2,7]]


data.sort(key=lambda x: max(x[0:2]),reverse=True)