How to find nth minimum of a numpy array

Question

I want to find indices in ascending order of elements in last row in a 2D numpy array. An example is as below.

ti = np.array([[255, 2, 255, 9], [4, 255, 6, 255], [23, 11, 5, 18]])

# output = function(ti[-1])
# output : 2, 1, 3, 0

How do I achieve this? np.argmin only gives one minimum, and I could not find any other way including using partition (which returns sorted elements instead). Kindly help.

I believe this is a duplicate of: stackoverflow.com/questions/22546180/… — Kevinj22
– Kevinj22, Commented Feb 20, 2019 at 17:58
What is "smaller" between [255, 2, 255, 9] and [4, 255, 6, 255]? How would you determine the "minimum"? — r.ook
– r.ook, Commented Feb 20, 2019 at 18:01
It's not at all clear what you're trying to do. You've got 4 numbers in the output, so you're not argsorting the rows. The numbers aren't right for argsorting the columns, or for taking the argmin of each column. I have no idea what that 2, 1, 3, 0 is supposed to be. — user2357112
– user2357112, Commented Feb 20, 2019 at 18:03
Are you trying to find the "nth minimum of a numpy array" as the title suggests or the " indices in ascending order" of all the elements as you wrote in the question? — Noam Peled
– Noam Peled, Commented Feb 20, 2019 at 18:09

Parthiban Rajendran · Accepted Answer · 2019-02-20 18:24:07Z

0

I tried np.argsort which sort of gives this output. Others please englighten if any more insights to be added.

np.argsort(ti[-1])

output:

array([2, 1, 3, 0], dtype=int64)

edited Feb 20, 2019 at 18:24

answered Feb 20, 2019 at 17:56

Parthiban Rajendran

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