On Mi, 2015-12-30 at 20:21 +0100, Nicolas P. Rougier wrote: > In the end, I’ve only the list comprehension to work as expected > > A = [0,0,1,3] > B = np.arange(8) > np.random.shuffle(B) > I = [list(B).index(item) for item in A if item in B] > > > But Mark's and Sebastian's methods do not seem to work... >
Yeah, sorry had a mind slip with the sorter since it returns the sorted
version. I think this should do the correct thing (throws away invalid
ones as default, though I think it is a bad idea in general).
def index(A, B, fill_invalid=None):
B_sorter = np.argsort(B)
B_sorted = B[B_sorter]
B_sorted_index = np.searchsorted(B_sorted, A)
# Go back into the original index:
B_index = B_sorter[B_sorted_index]
if fill_invalid is None:
valid = B.take(B_index, mode='clip') == A
return B_index[valid]
else:
invalid = B.take(B_index, mode='clip') != A
B_index[invalid] = fill_invalid
return B_index
>
>
> > On 30 Dec 2015, at 19:51, Nicolas P. Rougier <
> > nicolas.roug...@inria.fr> wrote:
> >
> >
> > Unfortunately, this does not handle repeated entries in a.
> >
> > > On 30 Dec 2015, at 19:40, Mark Miller <markperrymil...@gmail.com>
> > > wrote:
> > >
> > > I was not familiar with the .in1d function. That's pretty handy.
> > >
> > > Yes...it looks like numpy.where(numpy.in1d(b, a)) does what you
> > > need.
> > >
> > > > > > numpy.where(numpy.in1d(b, a))
> > > (array([1, 2, 5, 7], dtype=int64),)
> > > It would be interesting to see the benchmarks.
> > >
> > >
> > > On Wed, Dec 30, 2015 at 10:17 AM, Nicolas P. Rougier <
> > > nicolas.roug...@inria.fr> wrote:
> > >
> > > Yes, it is the expected result. Thanks.
> > > Maybe the set(a) & set(b) can be replaced by
> > > np.where[np.in1d(a,b)], no ?
> > >
> > > > On 30 Dec 2015, at 18:42, Mark Miller <
> > > > markperrymil...@gmail.com> wrote:
> > > >
> > > > I'm not 100% sure that I get the question, but does this help
> > > > at all?
> > > >
> > > > > > > a = numpy.array([3,2,8,7])
> > > > > > > b = numpy.array([1,3,2,4,5,7,6,8,9])
> > > > > > > c = set(a) & set(b)
> > > > > > > c #contains elements of a that are in b (and vice versa)
> > > > set([8, 2, 3, 7])
> > > > > > > indices = numpy.where([x in c for x in b])[0]
> > > > > > > indices #indices of b where the elements of a in b occur
> > > > array([1, 2, 5, 7], dtype=int64)
> > > >
> > > > -Mark
> > > >
> > > >
> > > > On Wed, Dec 30, 2015 at 6:45 AM, Nicolas P. Rougier <
> > > > nicolas.roug...@inria.fr> wrote:
> > > >
> > > > I’m scratching my head around a small problem but I can’t find
> > > > a vectorized solution.
> > > > I have 2 arrays A and B and I would like to get the indices
> > > > (relative to B) of elements of A that are in B:
> > > >
> > > > > > > A = np.array([2,0,1,4])
> > > > > > > B = np.array([1,2,0])
> > > > > > > print (some_function(A,B))
> > > > [1,2,0]
> > > >
> > > > # A[0] == 2 is in B and 2 == B[1] -> 1
> > > > # A[1] == 0 is in B and 0 == B[2] -> 2
> > > > # A[2] == 1 is in B and 1 == B[0] -> 0
> > > >
> > > > Any idea ? I tried numpy.in1d with no luck.
> > > >
> > > >
> > > > Nicolas
> > > >
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