[Numpy-discussion] Inverted indices
Nicolas P. Rougier
Nicolas.Rougier at inria.fr
Thu Aug 7 08:09:43 EDT 2014
Oh thanks, I would never have imagined a one-line solution...
Here are the benchmarks:
In : %timeit stefan(S)
100000 loops, best of 3: 10.8 µs per loop
In : %timeit gregor(S)
10000 loops, best of 3: 48.1 µs per loop
In : %timeit gregor2(S)
100000 loops, best of 3: 3.23 µs per loop
On 07 Aug 2014, at 14:04, Gregor Thalhammer <gregor.thalhammer at gmail.com> wrote:
> Am 07.08.2014 um 13:59 schrieb Gregor Thalhammer <gregor.thalhammer at gmail.com>:
>> Am 07.08.2014 um 13:16 schrieb Nicolas P. Rougier <Nicolas.Rougier at inria.fr>:
>>> I've a small problem for which I cannot find a solution and I'm quite sure there is an obvious one:
>>> I've an array Z (any dtype) with some data.
>>> I've a (sorted) array I (of integer, same size as Z) that tells me the index of Z[i] (if necessary, the index can be stored in Z).
>>> Now, I have an arbitrary sequence S of indices (in the sense of I), how do I build the corresponding data ?
>>> Here is a small example:
>>> Z = [(0,0), (1,1), (2,2), (3,3), (4,4))
>>> I = [0, 20, 23, 24, 37]
>>> S = [ 20,20,0,24]
>>> -> Result should be [(1,1), (1,1), (0,0),(3,3)]
>>> S = [15,15]
>>> -> Wrong (15 not in I) but ideally, I would like this to be converted to [(0,0), (0,0)]
>>> Any idea ?
>> If I is sorted, I would propose to use a bisection algorithm, faster than linear search:
>> Z = array([(0,0), (1,1), (2,2), (3,3), (4,4)])
>> I = array([0, 20, 23, 24, 37])
>> S = array([ 20,20,0,24,15,27])
>> a = zeros(S.shape,dtype=int)
>> b = a + S.shape-1
>> for i in range(int(log2(S.shape))+2):
>> c = (a+b)>>1
>> sel = I[c]<=S
>> a[sel] = c[sel]
>> b[~sel] = c[~sel]
> or even simpler:
> c = searchsorted(I, S)
>> If I[c] != S, then there is no corresponding index entry in I to match S.
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