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 > > _______________________________________________ > Numpy-discussion mailing list > Numpy-discussion@scipy.org > http://projects.scipy.org/mailman/listinfo/numpy-discussion > > -- ________________________________ Dr. Raik Gruenberg http://www.raiks.de/contact.html ________________________________ _______________________________________________ Numpy-discussion mailing list Numpy-discussion@scipy.org http://projects.scipy.org/mailman/listinfo/numpy-discussion
