Retrieve indexes of multiple values with Numpy in a vectorization way

Question

In order to get the index corresponding to the "99" value in a numpy array, we do :

mynumpy=([5,6,9,2,99,3,88,4,7))
np.where(my_numpy==99)

What if, I want to get the index corresponding to the following values 99,55,6,3,7? Obviously, it's possible to do it with a simple loop but I'm looking for a more vectorization solution. I know Numpy is very powerful so I think it might exist something like that.

desired output :

searched_values=np.array([99,55,6,3,7])
np.where(searched_values in mynumpy)
[(4),(),(1),(5),(8)]

@Kasramvd Re-opening because this question additionally needs the masking for the elements that are not present in searched_values. Hope that looks fair. — Divakar
– Divakar, Commented Mar 2, 2018 at 10:32
Yeah Divakar's solution takes into account the special case where values are not present — hans glick
– hans glick, Commented Mar 2, 2018 at 10:45
A similar question involving multiple values, How to get a list of indexes selected by a specific value efficiently with numpy arrays?. The accepted answer used a default dictionary. — hpaulj
– hpaulj, Commented Mar 2, 2018 at 18:43

Divakar · Accepted Answer · 2018-03-02 17:58:35Z

2

Here's one approach with np.searchsorted -

def find_indexes(ar, searched_values, invalid_val=-1):
    sidx = ar.argsort()
    pidx = np.searchsorted(ar, searched_values, sorter=sidx)
    pidx[pidx==len(ar)] = 0
    idx = sidx[pidx]
    idx[ar[idx] != searched_values] = invalid_val
    return idx

Sample run -

In [29]: find_indexes(mynumpy, searched_values, invalid_val=-1)
Out[29]: array([ 4, -1,  1,  5,  8])

For a generic invalid value specifier, we could use np.where -

def find_indexes_v2(ar, searched_values, invalid_val=-1):
    sidx = ar.argsort()
    pidx = np.searchsorted(ar, searched_values, sorter=sidx)
    pidx[pidx==len(ar)] = 0
    idx = sidx[pidx]
    return np.where(ar[idx] == searched_values, idx, invalid_val)

Sample run -

In [35]: find_indexes_v2(mynumpy, searched_values, invalid_val=None)
Out[35]: array([4, None, 1, 5, 8], dtype=object)

# For list output
In [36]: find_indexes_v2(mynumpy, searched_values, invalid_val=None).tolist()
Out[36]: [4, None, 1, 5, 8]

edited Mar 2, 2018 at 17:58

answered Mar 2, 2018 at 10:28

Divakar

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5 Comments

hans glick Over a year ago

Waou, thks a lot, indeed it looks like to another question, but you take into account special cases so I do not know what to do.

Divakar Over a year ago

@hansglick Not sure what you meant by - "do not know what to do" thing :)

hans glick Over a year ago

Just see your stackoverflow profile, how can I send you a private message?

hans glick Over a year ago

What if it the element is present several times. I would like to be able to retrieve all the indexes. find_indexes(np.array([5,11,5]),np.array([5])) . Desired output [(0,2)]

hpaulj Over a year ago

@hansglick, you should have included some duplicates in the original question.

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