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?

Answer 1

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])