Numpy equivalent of list.index

Question

In a low-level function that is called many times, I need to do the equivalent of python's list.index, but with a numpy array. The function needs to return when it finds the first value, and raise ValueError otherwise. Something like:

>>> a = np.array([1, 2, 3]) >>> np_index(a, 1) 0 >>> np_index(a, 10) Traceback (most recent call last):       File "<stdin>", line 1, in <module> ValueError: 10 not in array 

I want to avoid a Python loop if possible. np.where isn't an option as it always iterates through the entire array; I need something that stops once the first index is found.

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EDIT: Some more specific information related to the problem.

• About 90% of the time, the index I'm searching for is in the first 1/4 to 1/2 of the array. So there's potentially a factor of 2-4 speedup at stake here. The other 10% of the time the value is not in the array at all.

• I've profiled things already, and the call to np.where is the bottleneck, taking up at least 50% of the total runtime.

• It is not essential that it raise a ValueError; it just has to return something that obviously indicates that the value isn't in the array.

I will probably code up a solution in Cython, as suggested.

Answer 1

See my comment on the OP's question for caveats, but in general, I would do the following:

import numpy as np a = np.array([1, 2, 3]) np.min(np.nonzero(a == 2)[0]) 

if the value you are looking for is not in the array, you'll get a ValueError due to:

ValueError: zero-size array to ufunc.reduce without identity 

because you are trying to take the min value of an empty array.

I would profile this code and see if it is an actual bottleneck, because in general when numpy searches through an entire array using a built-in function rather than an explicit python loop, it is relatively fast. An insistence on halting the search when it finds the first value may be functionally irrelevant.