[Numpy-discussion] puzzle: generate index with many ranges
Rick White
rlw@stsci....
Fri Jan 30 16:20:01 CST 2009
Here's a technique that works:
Python 2.4.2 (#5, Nov 21 2005, 23:08:11)
[GCC 4.0.0 20041026 (Apple Computer, Inc. build 4061)] on darwin
Type "help", "copyright", "credits" or "license" for more information.
>>> import numpy as np
>>> a = np.array([0,4,0,11])
>>> b = np.array([-1,11,4,15])
>>> rangelen = b-a+1
>>> cumlen = rangelen.cumsum()
>>> c = np.arange(cumlen[-1],dtype=np.int32)
>>> c += np.repeat(a[1:]-c[cumlen[0:-1]], rangelen[1:])
>>> print c
[ 4 5 6 7 8 9 10 11 0 1 2 3 4 11 12 13 14 15]
The basic idea is that the difference of your desired output from a
simple range is an array with a bunch of constant values appended
together, and that is what repeat() does. I'm assuming that you'll
never have b < a. Notice the slight ugliness of prepending the
elements at the beginning so that the cumsum starts with zero.
(Maybe there is a cleaner way to do that.)
This does create a second array (via the repeat) that is the same
length as the result. If that uses too much memory, you could break
up the repeat and update of c into segments using a loop. (You
wouldn't need a loop for every a,b element -- do a bunch in each
iteration.)
-- Rick
Raik Gruenberg wrote:
> Hi there,
>
> perhaps someone has a bright idea for this one:
>
> I want to concatenate ranges of numbers into a single array (for
> indexing). So I
> have generated an array "a" with starting positions, for example:
>
> a = [4, 0, 11]
>
> I have an array b with stop positions:
>
> b = [11, 4, 15]
>
> and I would like to generate an index array that takes 4..11, then
> 0..4, then
> 11..15.
>
> In reality, a and b have 10000+ elements and the arrays to be
> "sliced" are very
> large so I want to avoid any for loops etc. Any idea how this could
> be done? I
> thought some combination of *repeat* and adding of *arange* should
> do the trick
> but just cannot nail it down.
>
> Thanks in advance for any hints!
>
> Greetings,
> Raik
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