Assign value to numpy array with overlaid slices of indexes

Question

I want to assign a value to some segments of an array. I've got indexes of the segments as tuples (start_idx, end_idx). Segments may overlay or be sub-segments of each other.

a = np.zeros(12)
segments = np.array([(0, 3), (1, 2), (6, 8), (8, 10)])
a[segments] = 1

The results is:

a
>> array([1, 1, 1, 1, 0, 0, 1, 0, 1, 0, 1, 0])

How can I mask all segments to get this output:

a
>> array([1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0])

Are you sure the desired output is correct? It seems like the [3] element is wrong?! Shouldn't it be 0? — MSeifert
– MSeifert, Commented Aug 21, 2017 at 21:20
@Divakar But how should 10 be handled then? Normally it would be an IndexError at least for a = np.zeros(10). — MSeifert
– MSeifert, Commented Aug 21, 2017 at 21:23
@MSeifert Guessing that's a typo : a = np.zeros(12) as the final output seems to have 12 elems. — Divakar
– Divakar, Commented Aug 21, 2017 at 21:24
@Divakar Ah, that's a possibility as well. But seems like the answers exclude the end point - at least the accepted one. So I thought the desired output may be wrong (but wrong input would make sense too). — MSeifert
– MSeifert, Commented Aug 21, 2017 at 21:28

Divakar · Accepted Answer · 2017-08-21 21:17:33Z

2

Here's one vectorized approach with the idea being borrowed from this post -

def segment_arr(segments, L): # L being length of output array
    s = np.zeros(L,dtype=int)
    stop = segments[:,1]+1
    np.add.at(s,segments[:,0],1)
    np.add.at(s,stop[stop<len(s)],-1)
    return (s.cumsum()>0).astype(int)

Sample run -

In [298]: segments = np.array([(0, 3), (1, 2), (6, 8), (8, 10)])

In [299]: segment_arr(segments, L=12)
Out[299]: array([1, 1, 1, 1, 0, 0, 1, 1, 1, 1, 1, 0])

edited Aug 21, 2017 at 21:17

answered Aug 21, 2017 at 20:56

Divakar

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1 Comment

MSeifert Over a year ago

At least one solution that actually produces the "given expected output". :)

akuiper · Accepted Answer · 2017-08-21 20:50:50Z

1

One option is to simply loop through the segments, and convert ranges to actual index:

a = np.zeros(10)
segments = np.array([(0, 3), (1, 2), (6, 8), (8, 10)])

a[[i for s in segments for i in range(*s)]] = 1    
a
# array([ 1.,  1.,  1.,  0.,  0.,  0.,  1.,  1.,  1.,  1.])

answered Aug 21, 2017 at 20:50

akuiper

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Comments

d_void · Accepted Answer · 2017-08-21 20:47:20Z

1

Try this:

a = np.zeros(10)
segments = np.array([(0, 3), (1, 2), (6, 8), (8, 10)])
a[range(3)+range(1,2)+range(6,8)+range(8,10)] = 1
print (a)

answered Aug 21, 2017 at 20:47

d_void

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3 Comments

Katerina Over a year ago

How can I do that without assigning them manually? I can create ranges in a list generator, but how to feed them?

d_void Over a year ago

Well in that case you will have to Loop through

MSeifert Over a year ago

Note that this throws a TypeError: unsupported operand type(s) for +: 'range' and 'range' on python-3.x

MSeifert · Accepted Answer · 2017-08-21 21:33:43Z

1

Just to mention the trivial solution: Use a for-loop over the segments and assign to the slices:

import numpy as np
a = np.zeros(12)
segments = np.array([(0, 3), (1, 2), (6, 8), (8, 10)])

for seg in segments.tolist():  # the "tolist" is just an optimization here, you *could* omit it.
    a[seg[0]: seg[1]+1] = 1    # or just "seq[1]" if you want to exclude the end point
print(a)
# array([ 1.,  1.,  1.,  1.,  0.,  0.,  1.,  1.,  1.,  1.,  1.,  0.])

answered Aug 21, 2017 at 21:33

MSeifert

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