Assigning multiple array indices at once in Python/Numpy

Question

I'm looking to quickly (hopefully without a for loop) generate a Numpy array of the form:

array([a,a,a,a,0,0,0,0,0,b,b,b,0,0,0, c,c,0,0....])

Where a, b, c and other values are repeated at different points for different ranges. I'm really thinking of something like this:

import numpy as np
a = np.zeros(100)
a[0:3,9:11,15:16] = np.array([a,b,c])

Which obviously doesn't work. Any suggestions?

Edit (jterrace answered the original question): The data is coming in the form of an N*M Numpy array. Each row is mostly zeros, occasionally interspersed by sequences of non-zero numbers. I want to replace all elements of each such sequence with the last value of the sequence. I'll take any fast method to do this! Using where and diff a few times, we can get the start and stop indices of each run.

raw_data = array([.....][....])
starts = array([0,0,0,1,1,1,1...][3, 9, 32, 7, 22, 45, 57,....])
stops = array([0,0,0,1,1,1,1...][5, 12, 50, 10, 30, 51, 65,....])
last_values = raw_data[stops]
length_to_repeat = stops[1]-starts[1]

Note that starts[0] and stops[0] are the same information (which row the run is occurring on). At this point, since the only route I know of is what jterrace suggest, we'll need to go through some contortions to get similar start/stop positions for the zeros, then interleave the zero start/stop with the values start/stops, and interleave the number 0 with the last_values array. Then we loop over each row, doing something like:

for i in range(N)
    values_in_this_row = where(starts[0]==i)[0]
    output[i] = numpy.repeat(last_values[values_in_this_row], length_to_repeat[values_in_this_row])

Does that make sense, or should I explain some more?

What's the form of the input? Tuples of the form (start, stop, value) ? — jterrace
– jterrace, Commented Aug 16, 2011 at 22:29

jterrace · Accepted Answer · 2011-08-16 22:32:23Z

4

If you have the values and repeat counts fully specified, you can do it this way:

>>> import numpy
>>> values = numpy.array([1,0,2,0,3,0])
>>> counts = numpy.array([4,5,3,3,2,2])
>>> numpy.repeat(values, counts)
array([1, 1, 1, 1, 0, 0, 0, 0, 0, 2, 2, 2, 0, 0, 0, 3, 3, 0, 0])

answered Aug 16, 2011 at 22:32

jterrace

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2 Comments

jeffalstott Over a year ago

This does exactly what I was looking for. Thanks! Now, would there be a way to push it one step further, to make: array([[a,a,a,a,0,0,0,0,0,b,b,b,...], [0,0,0, c,c,0,0, d d,d,....]]) As you imagined, the data is coming in with the form (start, stop, value). However, it's really in the form (start, stop, value, channel). Looping over each channel is probably acceptable, but if it can be pushed further, why not? :-) Numpy.repeat doesn't seem to be able to do that.

jterrace Over a year ago

Not really sure what you mean. If you could amend your question with the exact example input and output, that would be helpful.

sloth · Accepted Answer · 2012-08-03 06:45:18Z

0

you can use numpy.r_:

>>> np.r_[[a]*4,[b]*3,[c]*2]
array([1, 1, 1, 1, 2, 2, 2, 3, 3])

edited Aug 3, 2012 at 6:45

sloth

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answered Aug 16, 2011 at 22:15

HYRY

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