The truth value of an array with more than one element is ambiguous. Use a.any() or a.all() - Numpy Library

Question

I have seen this same topic some other place but no real answer to my question. I have a numpy array and I need to find the index of a number.

a=np.argsort(cosine_similarity(tfidf_matrix[11:12], tfidf_matrix)) #numbers are from 0 to 11

b=np.equal(a,10)

# b values are [[False False False False False False False False  True False False False]]

How do I get it to return index 8? (The index for the true value in the array)

Answer 1

You are looking for numpy.where

b = np.where(a==10)

Here b will be an array that contains the indices of the items that matched your condition. You can select the first element (b[0]) if you are interested only in the first occurrence of the item.

The documentation on numpy.equal says that:

Return (x1 == x2) element-wise.

And that is exactly what you received, an array that contains an element wise comparison of the array a and the value 10.