Beautiful! That should do the trick. Now let's see how this performs against the list comprehension... Thanks a lot! Raik Rick White wrote:
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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