How find values in an array that meet two conditions using Python

Question

I have an array

a=[1,2,3,4,5,6,7,8,9]

and I want to find the indices of the element s that meet two conditions i.e.

a>3 and a<8
ans=[3,4,5,6]
a[ans]=[4,5,6,7]

I can use numpy.nonzero(a>3) or numpy.nonzero(a<8) but not numpy.nonzero(a>3 and a<8) which gives the error:

ValueError: The truth value of an array with more than one element is
ambiguous. Use a.any() or a.all()

When I try to use any or all I get the same error. Is it possible to combine two conditional tests to get the ans?

why you needs numpy, you can't do it like this way ! filter(lambda a: 3 < a < 8, a) — shahjapan
– shahjapan, Commented Jul 14, 2010 at 17:41
@shahjapan - likely because they need the increased speed of a numpy array because they probably really have a much much larger dataset. — Wayne Werner
– Wayne Werner, Commented Jul 14, 2010 at 17:59

Matthew Flaschen · Accepted Answer · 2010-07-14 17:03:34Z

27

numpy.nonzero((a > 3) & (a < 8))

& does an element-wise boolean and.

answered Jul 14, 2010 at 17:03

Matthew Flaschen

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tssch · Accepted Answer · 2014-10-23 07:15:53Z

3

An alternative (which I ended up using) is numpy.logical_and:

choice = numpy.logical_and(np.greater(a, 3), np.less(a, 8))
numpy.extract(choice, a)

edited Oct 23, 2014 at 7:15

answered Oct 22, 2014 at 15:34

tssch

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Georgy · Accepted Answer · 2019-05-06 11:47:30Z

0

if you use numpy array, you can directly use '&' instead of 'and'.

a=array([1,2,3,4,5,6,7,8,9])
a[(a>3) & (a<8)]
ans=array([3,4,5,6])

edited May 6, 2019 at 11:47

Georgy

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answered Jun 14, 2017 at 14:59

jaydeepsb

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rpuri · Accepted Answer · 2022-03-05 01:25:28Z

0

Side note, this format works for any bitwise operation.

numpy.nonzero((a > 3) | (a < 8)) #[or]
numpy.nonzero((a > 3) & (a < 8)) #[and]

answered Mar 5, 2022 at 1:25

rpuri

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