Question

I have two 1D arrays, x & y, one smaller than the other. I’m trying to find the index of every element of y in x.

I’ve found two naive ways to do this, the first is slow, and the second memory-intensive.

The slow way

indices= []
for iy in y:
    indices += np.where(x==iy)[0][0]

The memory hog

xe = np.outer([1,]*len(x), y)
ye = np.outer(x, [1,]*len(y))
junk, indices = np.where(np.equal(xe, ye))

Is there a faster way or less memory intensive approach? Ideally the search would take advantage of the fact that we are searching for not one thing in a list, but many things, and thus is slightly more amenable to parallelization.
Bonus points if you don’t assume that every element of y is actually in x.

Answer 1

As Joe Kington said, searchsorted() can search element very quickly. To deal with elements that are not in x, you can check the searched result with original y, and create a masked array:

import numpy as np
x = np.array([3,5,7,1,9,8,6,6])
y = np.array([2,1,5,10,100,6])

index = np.argsort(x)
sorted_x = x[index]
sorted_index = np.searchsorted(sorted_x, y)

yindex = np.take(index, sorted_index, mode="clip")
mask = x[yindex] != y

result = np.ma.array(yindex, mask=mask)
print result

the result is:

[-- 3 1 -- -- 6]