numpy - return first index of element in array [duplicate]

Question

What's the fastest way of returning the index of the FIRST match between a variable and an element within an ndarray? I see numpy.where used a lot, but that returns all indices.

match = 5000
zArray = np.array([[0,1200,200],[1320,24,5000],[5000,234,5230]])

>array([[   0, 1200,  200],
   [1320,   24, 5000],
   [5000,  234, 5230]])

numpy.where(zArray==match)
>(array([1, 2], dtype=int64), array([2, 0], dtype=int64))

I'd like the first index returned, i.e. just [1,2]. but numpy.where returns both [1,2] and [2,0]

How would you define the first match ? Row major or column major ? — ZdaR
– ZdaR, Commented Oct 25, 2017 at 3:57
Be aware that your example array happens to be set such, that the two arrays in the result appear to be the two x, y index pairs you're looking for. Instead, these are the [x1, x2] and [y1, y2] indices of the matches. Try e.g. [[0,5000,200],[1320,24,1200],[234,5000,5230]] instead to see. — user707650
– user707650, Commented Oct 25, 2017 at 4:11
Just to iterate on what @Evert has mentioned, "I'd like the first index returned, i.e. just [1,2]. but numpy.where returns both [1,2] and [2,0]" needs edits I believe. — Divakar
– Divakar, Commented Oct 25, 2017 at 4:29

Divakar · Accepted Answer · 2017-10-25 04:14:55Z

4

You can use np.argwhere to get the matching indices packed as a 2D array with each row holding indices for each match and then index into the first row, like so -

np.argwhere(zArray==match)[0]

Alternatively, faster one with argmax to get the index of the first match on a flattened version and np.unravel_index for per-dim indices tuple -

np.unravel_index((zArray==match).argmax(), zArray.shape)

Sample run -

In [100]: zArray
Out[100]: 
array([[   0, 1200, 5000], # different from sample for a generic one
       [1320,   24, 5000],
       [5000,  234, 5230]])

In [101]: match
Out[101]: 5000

In [102]: np.argwhere(zArray==match)[0]
Out[102]: array([0, 2])

In [103]: np.unravel_index((zArray==match).argmax(), zArray.shape)
Out[103]: (0, 2)

Runtime test -

In [104]: a = np.random.randint(0,100,(1000,1000))

In [105]: %timeit np.argwhere(a==50)[0]
100 loops, best of 3: 2.41 ms per loop

In [106]: %timeit np.unravel_index((a==50).argmax(), a.shape)
1000 loops, best of 3: 493 µs per loop

edited Oct 25, 2017 at 4:14

answered Oct 25, 2017 at 4:02

Divakar

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2 Comments

hpaulj Over a year ago

argwhere is just transpose(where(...)). It's handy for getting the first occurrence, but doesn't do any sort of short-circuiting. argmax might short circuit the boolean match (I known the nan test does).

kabammi Over a year ago

thanks Divakar. This works well, and is perfect for my needs.

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