On Thu, Feb 19, 2009 at 12:12, Nicolas Pinto <pi...@mit.edu> wrote: > Grissiom, > > Using the following doesn't require any loop: > > In [9]: sqrt((a**2.).sum(1)) > Out[9]: array([ 5., 10.]) > > Best, > > Got it~ Thanks really ;)
-- Cheers, Grissiom
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