How to sort Multi-Dimensional Array by Any Column Any Needed Order in Python

Question

I like to have a Python 2D array sorted by based on the first column ascending than the second column descending and the third column ascending. So, the sort function will first try to sort the items on the first column, if the value of the first column is same, then it will try to sort based on the second column value descending, even if the second column value is same, then it will try to sort based on the third value in ascending order.

So, if I have a Python array like this-

[[15, 18, 2], [1, 3, 3], [7, 10, 8], [1, 2, 1], [8, 11, 5], [8, 10, 5], [1, 6, 1], [1, 6, 2]]

Then for the given array, the output should be like this-

[[1, 6, 1], [1, 6, 2] [1, 3, 3], [1, 2, 1], [7, 10, 8], [8, 11, 5], [8, 10, 5], [15, 18, 2]]

I have to use python lambda function for this. What I have done is-

intervals = [[15, 18, 2], [1, 3, 3], [7, 10, 8], [1, 2, 1], [8, 11, 5], [8, 10, 5], [1, 6, 1], [1, 6, 2]]
................
................
intervals.sort(key=lambda x: x[2] or x[1] and x[0])

After this, what I am getting is like this-

[[1, 2, 1], [1, 6, 1], [15, 18, 2], [1, 6, 2], [1, 3, 3], [8, 11, 5], [8, 10, 5], [7, 10, 8]]

Which is not as expected. Is there any way of finding what I need to have?

Thanks in advance for helping.

Nick · Accepted Answer · 2020-09-24 13:27:53Z

2

Because your values are numeric, you can negate them to sort descending, so to sort according to x[0] ascending, x[1] descending and x[2] ascending you just need to sort on the tuple (x[0], -x[1], x[2]):

intervals = [[15, 18, 2], [1, 3, 3], [7, 10, 8], [1, 2, 1], [8, 11, 5], [8, 10, 5], [1, 6, 1], [1, 6, 2]]

intervals.sort(key=lambda x:(x[0], -x[1], x[2]))
print(intervals)

Output:

[[1, 6, 1], [1, 6, 2], [1, 3, 3], [1, 2, 1], [7, 10, 8], [8, 11, 5], [8, 10, 5], [15, 18, 2]]

answered Sep 24, 2020 at 13:27

Nick

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