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开发者_JAVA技巧[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?
numpy.nonzero((a > 3) & (a < 8))
& does an element-wise boolean and.
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)
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])
Side note, this format works for any bitwise operation.
numpy.nonzero((a > 3) | (a < 8)) #[or]
numpy.nonzero((a > 3) & (a < 8)) #[and]
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