slice 2D numpy array based on condition

Question

I have an numpy array

import numpy as np

a = np.array([
[999, 999, 999, 999, 999, 999, 999, 999, 999, 999],
[999, 999, 999, 1, 2, 3, 4, 999, 999, 999],
[999, 999, 999, 5, 6, 7, 8, 999, 999, 999],
[999, 999, 999, 9, 10, 11, 12, 999, 999, 999],
[999, 999, 999, 999, 999, 999, 999, 999, 999, 999]])

how to return the filtered values, containing only the different values than 999 using numpy slicing?

filtered = np.where(a != 999)
In [5]: filtered
Out[5]: 
(array([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 2, 2, 2,
    2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4,
    4, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5, 5, 5, 5, 6, 6, 6, 6, 6, 6, 6, 6, 6,
    6, 7, 7, 7, 7, 7, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 8, 8, 8, 8, 8, 9, 9,
    9, 9, 9, 9, 9, 9, 9, 9]),
 array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2,
    3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4, 5,
    6, 7, 8, 9, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4, 5, 6, 7, 8,
    9, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 0, 1,
    2, 3, 4, 5, 6, 7, 8, 9]))

Desired output:

output = np.array([
    [1, 2, 3, 4],
    [5, 6, 7, 8],
    [9, 10, 11, 12]])

Daniel · Accepted Answer · 2014-11-26 16:27:44Z

1

You can do the following:

>>> mask = (a!=999)
>>> dim1 = np.any(mask, axis=1).sum()
>>> a[mask].reshape(dim1, -1)
array([[ 1,  2,  3,  4],
       [ 5,  6,  7,  8],
       [ 9, 10, 11, 12]])

This of course assume that you only have a single contiguous box in the whole array.

answered Nov 26, 2014 at 16:27

Daniel

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1 Comment

wim Over a year ago

A good solution for the special case. May raise ValueError for "other" cases

wim · Accepted Answer · 2014-11-26 16:22:09Z

1

Yours is a special case, because the subarray is rectangular. You can get the flat values using fancy indexing:

>>> a[filtered]
array([ 1,  2,  3,  4,  5,  6,  7,  8,  9, 10, 11, 12])

And if you know the shape already, you can reshape that:

>>> a[filtered].reshape(3,4)
array([[ 1,  2,  3,  4],
       [ 5,  6,  7,  8],
       [ 9, 10, 11, 12]])

However, there can be no guarantee that the input data will leave you with a rectangular array after the filtering in the general case. Consider, for example, what output array should look like if the input array had a[0,0] == 13.

answered Nov 26, 2014 at 16:22

wim

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7 Comments

arnaldo Over a year ago

Sorry. I did not read your modifications, before my comments. The main problem is that I really don't know the shape of the valid numbers inside the matrix. This is only an raw example. In my case I'll not get an outside number as pointed out by @wim (i.e. a[0,0] == 13)

wim Over a year ago

Your question is not well defined by the example you have given.

wim Over a year ago

Give the desired output where the valid numbers are not neatly contained inside a rectangle, for example if a[0,1] != 999 with all other things being the same.

wim Over a year ago

And what if an element inside the box is equal to 999 ?

arnaldo Over a year ago

Do you prefer that I edit my question and insert a statement which that I don't know the index of those valid values? In the case you have suggested the .reshape(3,4) visually based?

|

Gautam Sreekumar · Accepted Answer · 2019-02-08 21:53:22Z

1

You can also do this. Create a 2D mask using the condition. Typecast the condition mask to int or float, depending on the array, and multiply it with the original array.

In [8]: arr
Out[8]: 
array([[ 1.,  2.,  3.,  4.,  5.],
       [ 6.,  7.,  8.,  9., 10.]])

In [9]: arr*(arr % 2 == 0).astype(np.int) 
Out[9]: 
array([[ 0.,  2.,  0.,  4.,  0.],
       [ 6.,  0.,  8.,  0., 10.]])

answered Feb 8, 2019 at 21:53

Gautam Sreekumar

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