Getting index for numpy array

Question

Suppose I have a list of names

names = ['Alex','Brad', 'Camilla']

If I had a numpy array like

norder = array([0, 1, 2, 1, 0, 1, 0, 2, 1, 1])

Then I could use norder to access names by doing

names[norder]

>>> array(['Alex', 'Brad', 'Camilla', 'Brad', 'Alex', 'Brad', 'Alex',
       'Camilla', 'Brad', 'Brad'], dtype='<U7')

How do I go in the reverse? Given

order= array(['Alex', 'Brad', 'Camilla', 'Brad', 'Alex', 'Brad', 'Alex', 'Camilla', 'Brad', 'Brad'], dtype='<U7')

and names, how do I return something that looks like norder?

You can do a broadcasted comparison: (np.array(names) == order[:, None]).argmax(1) — cs95
– cs95, Commented Sep 30, 2018 at 0:28
Will all names be used and in the order presented? For the specific example you have provided pd.factorize(order)[0] is sufficient. — jpp
– jpp, Commented Sep 30, 2018 at 19:34

cs95 · Accepted Answer · 2018-09-30 00:33:08Z

2

One option with numpy is using broadcasted comparison.

>>> (np.array(names) == order[:, None]).argmax(1)
array([0, 1, 2, 1, 0, 1, 0, 2, 1, 1])

While being fast, the drawback here is that it is memory intensive.

An alternative would be using the pandas Index API:

>>> import pandas as pd
>>> idx = pd.Index(names)
>>> idx.get_indexer(order)
array([0, 1, 2, 1, 0, 1, 0, 2, 1, 1])

This should be much more memory efficient.

answered Sep 30, 2018 at 0:33

cs95

406k106 gold badges744 silver badges794 bronze badges

Sign up to request clarification or add additional context in comments.

1 Answer 1