numpy where returns an array. I need just index

Question

I have a numpy array: k = np.array([100,20,25,10,1,2]) and I'm trying to use a np where as index=np.where(k<10) which gives me index (array([4, 5]),). I'm insterested in something to give me just the index so here I'd like to have index[0]=4 not index[0]=[4 5]

I couldn't find anything here on the numpy docs.

I believe you can change index = np.where(k<10) to index = np.where(k<10)[0] to get what you want. — Izaak Weiss
– Izaak Weiss, Commented Jul 12, 2017 at 22:41
It sounds like you may be confused by the fact that where returns a tuple of arrays of indices, instead of just an array of indices. — user2357112
– user2357112, Commented Jul 12, 2017 at 22:45

Miriam Farber · Accepted Answer · 2017-07-12 22:50:50Z

4

You can take the first element in the result that you get, as follows:

index=np.where(k<10)[0]

Then index is array([4, 5], dtype=int64), and you can access index[0] and index[1] as you wanted.

answered Jul 12, 2017 at 22:50

Miriam Farber

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akuiper · Accepted Answer · 2017-07-12 22:45:36Z

1

You can use numpy.flatnonzero, which returns an array of index instead of tuples of array(s):

k = np.array([100,20,25,10,1,2])

np.flatnonzero(k < 10)
# array([4, 5])

edited Jul 12, 2017 at 22:45

answered Jul 12, 2017 at 22:42

akuiper

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4 Comments

Alex Huszagh Over a year ago

This does the exact same thing as what the OP did not want, and just uses a (relatively) obscure method to do so.

akuiper Over a year ago

@AlexanderHuszagh I am not so sure. This does give the same output as numpy.where but for 1d array, it has the convenience of returning an array of non zero elements index, instead of a tuple.

roganjosh Over a year ago

@AlexanderHuszagh works for me as I understand the OP's desire, but also completely non-obvious as a method to get those results.

Alex Huszagh Over a year ago

@Psidom, the user clearly wants an element from the where tuple. This could either be flatnonzero(cond)[0], or where(cond)[0][0], or some other index.

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