# [Numpy-discussion] Fast decrementation of indices

Jaime Fernández del Río jaime.frio at gmail.com
Sun Feb 2 17:57:51 EST 2014

```Cannot test right now, but np.unique(b, return_inverse=True)[1].reshape(2,
-1) should do what you are after, I think.
On Feb 2, 2014 11:58 AM, "Mads Ipsen" <mads.ipsen at gmail.com> wrote:

> Hi,
>
> I have run into a potential 'for loop' bottleneck. Let me outline:
>
> The following array describes bonds (connections) in a benzene molecule
>
>     b = [[0, 0, 0, 1, 1, 1, 2, 2, 2, 3, 3, 3,  4, 4, 4, 5,  5, 5, 6, 7,
> 8, 9, 10, 11],
>          [5, 6, 1, 0, 2, 7, 3, 8, 1, 4, 9, 2, 10, 5, 3, 4, 11, 0, 0, 1,
> 2, 3,  4,  5]]
>
> ie. bond 0 connects atoms 0 and 5, bond 1 connects atom 0 and 6, etc. In
> practical examples, the list can be much larger (N > 100.000 connections.
>
> Suppose atoms with indices a = [1,2,3,7,8] are deleted, then all bonds
> connecting those atoms must be deleted. I achieve this doing
>
> i_0 = numpy.in1d(b[0], a)
> i_1 = numpy.in1d(b[1], a)
> b_i = numpy.where(i_0 | i_1)[0]
> b = b[:,~(i_0 | i_1)]
>
> If you find this approach lacking, feel free to comment.
>
> This results in the following updated bond list
>
> b = [[0,  0,  4,  4,  5,  5,  5,  6, 10, 11]
>      [5,  6, 10,  5,  4, 11,  0,  0,  4,  5]]
>
> This list is however not correct: Since atoms [1,2,3,7,8] have been
> deleted, the remaining atoms with indices larger than the deleted atoms
> must be decremented. I do this as follows:
>
> for i in a:
>     b = numpy.where(b > i, bonds-1, bonds)  (*)
>
> yielding the correct result
>
> b = [[0, 0, 1, 1, 2, 2, 2, 3, 5, 6],
>      [2, 3, 5, 2, 1, 6, 0, 0, 1, 2]]
>
> The Python for loop in (*) may easily contain 50.000 iteration. Is there
> a smart way to utilize numpy functionality to avoid this?
>
> Thanks and best regards,
>
>
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