find index of values that exceed previous values in a numpy array

Question

I have a numpy array:

>>> n1 = np.array([[1, 4], [1, 5], [2, 4], [7, 2], [1, 3], [4, 7], [2, 9]])
>>> n1
array([[1, 4],
       [1, 5],
       [2, 4],
       [7, 2],
       [1, 3],
       [4, 7],
       [2, 9]])

I'm looking for a way to find the index of the occurrence of a value in the first column that follows the occurrence of a value greater than it. In this instance, I want:

array([4, 6])

because the value 1 is less than 7, and the value 2 is less than 4 (all in column 0)

Is there a nice Pythonic way of doing this?

Play with n1[1:,:]-n1[:-1,:], the difference between successive rows. — hpaulj
– hpaulj, Commented Mar 19, 2018 at 20:09
Hi @hpaulj, thanks. By assigning a new array n2 that contains only column 0 of n1, and using your suggestion, I built np.where(n2[1:]-n2[:-1] < 0) which gives me (array([3, 5]),), not array([4, 6]) — StatsSorceress
– StatsSorceress, Commented Mar 19, 2018 at 20:13
where gives a tuple, one element per dimension of the condition. Your case is 1d, so the result is a 1 element tuple. — hpaulj
– hpaulj, Commented Mar 19, 2018 at 20:35

busybear · Accepted Answer · 2018-03-19 20:08:16Z

1

You could use numpy.diff on the first column and find where values are negative.

np.where(np.diff(n1[:, 0]) < 0)

Add 1 to adjust the index if needed.

answered Mar 19, 2018 at 20:08

busybear

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3 Comments

StatsSorceress Over a year ago

Thanks @busybear, that gives me (array([3,5]),), which isn't quite what I'm looking for (of course I acknowledge you said to play with 1 to adjust the index)

busybear Over a year ago

Are you looking for a solution without needing to add 1? Seems like a trivial step to take.

StatsSorceress Over a year ago

True. It's a bit pedantic. Still, for anyone who's looking, do this: n2 = np.where(np.diff(n1[:, 0]) < 0), then n3 = np.sum(n2, 1) and it gets you array([4, 6]) as required.

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