[Numpy-discussion] *= operator not intuitive

Angus McMorland amcmorl@gmail....
Wed Mar 16 08:29:05 CDT 2011


On 16 March 2011 09:24, Paul Anton Letnes <paul.anton.letnes@gmail.com> wrote:
> Hi!
>
> This little snippet of code tricked me (in a more convoluted form). The *= operator does not change the datatype of the left hand side array. Is this intentional? It did fool me and throw my results quite a bit off. I always assumed that 'a *= b' means exactly the same as 'a = a * b' but this is clearly not the case!

This is intentional: a *= b works inplace, i.e. it's the equivalent,
not of a = a * b, but of a[:] = a * b

Angus.

> Paul.
>
> ++++++++++++++++++
>>>> from numpy import *
>>>> a = arange(10)
>>>> b = linspace(0,1,10)
>>>> a
> array([0, 1, 2, 3, 4, 5, 6, 7, 8, 9])
>>>> b
> array([ 0.        ,  0.11111111,  0.22222222,  0.33333333,  0.44444444,
>       0.55555556,  0.66666667,  0.77777778,  0.88888889,  1.        ])
>>>> a * b
> array([ 0.        ,  0.11111111,  0.44444444,  1.        ,  1.77777778,
>       2.77777778,  4.        ,  5.44444444,  7.11111111,  9.        ])
>>>> a *= b
>>>> a
> array([0, 0, 0, 1, 1, 2, 4, 5, 7, 9])
>
>
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-- 
AJC McMorland
Post-doctoral research fellow
Neurobiology, University of Pittsburgh


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