numpy: return boolean with true if array elements are in a given set

Question

>>> s = {0, 4, 27}
>>> a = numpy.arange(10)
>>> t = some_func(a)
>>> t
[1, 0, 0, 0, 1, 0, 0, 0, 0, 0]

What's the canonical some_func needed to be to make this work?

What I've tried

I've tried vectorizing a lambda function, which works ... it just doesn't feel like the right way to do this.

>>> f = lambda i: i in s
>>> vf = numpy.vectorize(f)
>>> t = numpy.fromfunction(vf, a.shape)
>>> t.astype(int)
array([1, 0, 0, 0, 1, 0, 0, 0, 0, 0])

I think what you want is in1d from numpy. docs.scipy.org/doc/numpy/reference/generated/… — mjspier
– mjspier, Commented Apr 7, 2014 at 9:27
Dont' really do numpy, but would something like [i in s for i in a] work here? — anon582847382
– anon582847382, Commented Apr 7, 2014 at 9:29

YXD · Accepted Answer · 2014-04-07 09:24:29Z

3

Use in1d with s as a NumPy array

>>> import numpy as np
>>> s = np.array([0, 4, 27])
>>> a = np.arange(10)
>>> t = np.in1d(a, s)
>>> t
array([ True, False, False, False,  True, False, False, False, False, False], dtype=bool)
>>>

answered Apr 7, 2014 at 9:24

YXD

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Comments

askewchan · Accepted Answer · 2014-04-07 12:37:46Z

0

This would be faster:

t = np.zeros(10, int)
sa = np.array(list(s))
t[sa[sa<10]] = 1

Unless of course, a is not np.arange()

answered Apr 7, 2014 at 12:37

askewchan

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