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?
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Answer
numpy.nonzero((a > 3) & (a < 8))
& does an element-wise boolean and.