Get the indexes of a numpy array inside another numpy array

Question

I'm trying to get the indexes of a numpy array inside another numpy array, for example , I have

 y = array([1, 2, 3, 4, 5, 8])
 x = np.array([3, 1, 8])

x is included in y, so what I would want to get is the indexes idx of the x array on the y array, in the same order, so here it would be idx = array([2, 0, 5]) so that we have np.array_equal(y[idx] ,x) yields True, I tried to use np.argwhere(np.in1d(y,x)) , but obviously I don't get the same order , I know I can always use list comprehension idx = [ list(y).index(el) for el in x], but I prefer to use numpy. Any thoughts ?

score 0 · Answer 1 · answered Jun 20 '19 at 08:58

0

From Is there a NumPy function to return the first index of something in an array?

>>> t = array([1, 1, 1, 2, 2, 3, 8, 3, 8, 8])
>>> nonzero(t == 8)
(array([6, 8, 9]),)
>>> nonzero(t == 8)[0][0]
6

So adapting for your case:

lst=[]
for item in x:
    lst.append( list(nonzero(y == item)) )

answered Jun 20 '19 at 08:58

Jkind9

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yeah you're right, but I'd prefer to avoid for loops as it slows down the process – Ayoub ZAROU Jun 20 '19 at 08:59
You probably will have to use a loop somewhere, you could compress it into one line though through comprehension – Jkind9 Jun 20 '19 at 09:01

Get the indexes of a numpy array inside another numpy array

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