Numpy -- maximum of each row of a 2D array after a certain index

Question

Consider this 2D array:

import numpy as np
a = np.array([[3,1,5,3],[1,4,8,2],[4,2,1,2],[9,2,4,4]])

I know how to find the maximum entry of each row: np.amax(a, axis=1). This returns array([5,8,4,9])

What I would like to do, however, is to get the maximum entry of each row after a certain index n, i.e. ignoring the first n terms and just looking for the maximum amongst the remaining entries in that row, for each row.

One complication: the point at which I "slice" each row is different for each row.

Example: I want to calculate the maximum of each row, but only amongst the last 2 entries in the first row, the last 3 entries in the second row, the last 2 in the third row, and the last 1 in the fourth row. This should return array([5,8,2,4])

If this can be done without for or while loops, that would be great -- I really cannot use them due to computational time limits.

@Prune that just prints [5,8,4,9] for n=0, [8,4,9] for n=1, [4,9] for n=2, etc — martin
– martin, Commented Sep 12, 2019 at 0:28

JacoSolari · Accepted Answer · 2019-09-12 00:23:22Z

1

You can use list comprehension.

   import numpy as np
   a = np.array([[3,1,5,3],[1,4,8,2],[4,2,1,2],[9,2,4,4]])
   inds = [-2,-3,-2,-1] # the indices in your example

   result = [max(line[i:]) for line,i in zip(a,inds)]
   print(result)
   #output
   [5, 8, 2, 4]

answered Sep 12, 2019 at 0:23

JacoSolari

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1 Comment

martin Over a year ago

Thanks! The only thing stopping me from approving this question is the for loop, I really cannot afford using many in my code and I'm already using a couple elsewhere. I am dealing with very large arrays and a lot of iterations...

Nils Werner · Accepted Answer · 2019-09-12 13:33:52Z

1

You can set the undesired entries to a value smaller than the minimum in your array and take the rowwise maximum value afterwards:

starts = np.array([2, 1, 2, 3])
mask = np.arange(a.shape[1]) < starts[:, None]
a[mask] = np.min(a) - 1
# array([[0, 0, 5, 3],
#        [0, 4, 8, 2],
#        [0, 0, 1, 2],
#        [0, 0, 0, 4]])
np.amax(a, axis=1)
# array([5, 8, 2, 4])

edited Sep 12, 2019 at 13:33

answered Sep 12, 2019 at 13:28

Nils Werner

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Comments

Brenlla · Accepted Answer · 2019-09-12 09:23:29Z

0

You could create a mask like in this answer:

a = np.array([[3,1,5,3],[1,4,8,2],[4,2,1,2],[9,2,4,4]])
n = np.array([2,3,2,1])
idx = a.shape[1]-n  #starting indices
mask=np.arange(a.shape[1]) >= idx[:,None]

# array([[False, False,  True,  True],
#       [False,  True,  True,  True],
#       [False, False,  True,  True],
#       [False, False, False,  True]])

Then apply reduceat with maximum over masked array:

n = np.roll(n,1) #indices for slices
n[0] = 0 #starts at zero

np.maximum.reduceat(a[mask],n.cumsum())
array([5, 8, 2, 4], dtype=int32)

answered Sep 12, 2019 at 9:23

Brenlla

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Comments

Derek Eden · Accepted Answer · 2019-09-12 15:11:02Z

0

can also mask / filter the array then find row-wise maximums like this:

row_limits = a.shape[1]-np.array([[2],[3],[2],[1]]) 
idx = np.repeat(np.arange(0,a.shape[1]), a.shape[0], 0).reshape(a.shape).T 
mask = idx >= row_limits
result = np.amax(np.multiply(a,mask),axis=1)

output:

array([5, 8, 2, 4])

edited Sep 12, 2019 at 15:11

answered Sep 12, 2019 at 13:16

Derek Eden

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