Dave - 1 year ago 111

Python Question

I have two arrays, a1 and a2. Assume

`len(a2) >> len(a1)`

I would like a quick way to return the a2 indices of all elements in a1. The time-intensive way to do this is obviously:

`from operator import indexOf`

indices = []

for i in a1:

indices.append(indexOf(a2,i))

This of course takes a long time where a2 is large. I could also use numpy.where() instead (although each entry in a1 will appear just once in a2), but I'm not convinced it will be quicker. I could also traverse the large array just once:

`for i in xrange(len(a2)):`

if a2[i] in a1:

indices.append(i)

But I'm sure there is a faster, more 'numpy' way - I've looked through the numpy method list, but cannot find anything appropriate.

Many thanks in advance,

D

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Answer Source

How about

```
numpy.nonzero(numpy.in1d(a2, a1))[0]
```

This should be fast. From my basic testing, it's about 7 times faster than your second code snippet for `len(a2) == 100`

, `len(a1) == 10000`

, and only one common element at index 45. This assumes that both `a1`

and `a2`

have no repeating elements.