Array of ranges with numpy

Question

I have the following array:

>>> x = numpy.array([2,4,2,3,1])
>>> x
array([2, 4, 2, 3, 1])

I would like an array of ranges of these values. I can create it like this:

>>> numpy.hstack( (numpy.arange(v) for v in x) )
array([0, 1, 0, 1, 2, 3, 0, 1, 0, 1, 2, 0])

Given x, is there a faster way to generate this with numpy without having to use a for loop?

Looks like you found your solution... It would be interesting to know why you needed this, and why in this particular format. — Benjamin
– Benjamin, Commented Mar 14, 2011 at 15:09
I was working on speeding up the pycollada library (github.com/pycollada/pycollada). One of the geometry types in the collada format is called a polylist and it stores the vertex count of each polygon in an array like the x above. I needed the output range set to triangulate the polygons quickly in a single array. — jterrace
– jterrace, Commented Mar 14, 2011 at 18:58
You can see how it works in the triangleset() function I created in this commit: github.com/pycollada/pycollada/commit/… — jterrace
– jterrace, Commented Mar 14, 2011 at 19:00

jterrace · Accepted Answer · 2011-03-13 01:30:44Z

1

I figured it out:

>>> x
array([2, 4, 2, 3, 1])
>>> ends = numpy.cumsum(x)
>>> ranges = numpy.arange(ends[-1])
>>> ranges = ranges - numpy.repeat(ends-x, x)
>>> ranges
array([0, 1, 0, 1, 2, 3, 0, 1, 0, 1, 2, 0])
>>>

answered Mar 13, 2011 at 1:30

jterrace

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Comments

Yann · Accepted Answer · 2018-01-11 18:16:32Z

0

Is this actually faster ? I have a similar need, and

concatenate([range(l, r) for l, r in array((left, right)).T])

is twice as fast as

range(end[-1]) + repeat(left + end, right-left)

(where end = cumsum(right - left) just like yours).

(in my very short experience, repeat is very slow -- at least in python 3.6)

answered Jan 11, 2018 at 18:16

Yann

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