Create binary array from list of indexes in Python

Question

I have a list of indexes, say [3,4,7,9]. What is the fastest way to create a binary array (in this case [0,0,0,1,1,0,0,1,0,1]). Thanks!

Each decimal digit corresponds to an index of 1 in a binary array. I can do this using "for" loop but there must be a faster approach. — stani
– stani, Commented Jul 20, 2021 at 19:03

Corralien · Accepted Answer · 2021-07-20 19:04:10Z

3

With numpy:

l = [3,4,7,9]
m = np.zeros(max(l)+1)
m[l] = 1

>>> m
array([0., 0., 0., 1., 1., 0., 0., 1., 0., 1.])

answered Jul 20, 2021 at 19:04

Corralien

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Comments

Reinderien · Accepted Answer · 2021-07-20 19:04:52Z

2

Simple list comprehension will do it:

indices = {3, 4, 7, 9}
[i in indices for i in range(10)]

answered Jul 20, 2021 at 19:04

Reinderien

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Jab · Accepted Answer · 2021-07-20 19:07:45Z

2

Using a comprehension:

indices = {3,4,7,9}
output = [int(i in indices) for i in range(max(indices)+1)]

answered Jul 20, 2021 at 19:06

Jab

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This has poor time complexity. indices should be a set.

very nice. Thanks!