On 16 March 2011 09:24, Paul Anton Letnes <paul.anton.let...@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]) > > > _______________________________________________ > NumPy-Discussion mailing list > NumPy-Discussion@scipy.org > http://mail.scipy.org/mailman/listinfo/numpy-discussion > -- AJC McMorland Post-doctoral research fellow Neurobiology, University of Pittsburgh _______________________________________________ NumPy-Discussion mailing list NumPy-Discussion@scipy.org http://mail.scipy.org/mailman/listinfo/numpy-discussion
