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
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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