Given an array:
big_array = np.array(['dog', 'cat', 'dog', 'dog', 'dog', 'cat', 'cat', 'dog'])
I want to get the position of each of these elements, in a smaller array:
small_array = np.array(['dog', 'cat'])
It should return:
[0, 1, 0, 0, 0, 1, 1, 0]
It would be the equivalent of:
[np.where(i == small_array)[0][0] for i in big_array]
Can it be done without list comprehension, preferably with a Numpy function?
If you're willing to sort small_array, this will do it:
small_array.sort() # in-place, or `x = np.sort(small_array)` for a sorted copy
np.searchsorted(small_array, big_array)
searchsorted() requires that the first argument is sorted, or you have to provide a third argument to tell it how to sort (as in Austin's answer). If you do neither of those things, it will give you wrong results (including for your example input).There's is no numpy function as far as I know, but you could also do a combination of argsort and searchsorted something like:
import numpy as np
big_array = np.array(['dog', 'cat', 'dog', 'dog', 'dog', 'cat', 'cat', 'dog'])
small_array = np.array(['dog', 'cat'])
sortd = np.argsort(small_array)
res = sortd[np.searchsorted(small_array[sortd], big_array)]
print(res)
# [0 1 0 0 0 1 1 0]
Compare with broadcasting and find index of True in the result along the last axis.
>>> a = np.array(['dog', 'cat', 'dog', 'dog', 'dog', 'cat', 'cat', 'dog'])
>>> b = np.array(['dog','cat'])
>>> c = a[:,None] == b
>>> c.argmax(axis=-1)
array([0, 1, 0, 0, 0, 1, 1, 0], dtype=int64)
>>> a[:,None] == b
array([[ True, False],
[False, True],
[ True, False],
[ True, False],
[ True, False],
[False, True],
[False, True],
[ True, False]])
