[SciPy-user] once again about zeros() and ones()

dmitrey openopt@ukr....
Sun Jul 1 08:10:43 CDT 2007


I can't understand, which example of calling current numpy.zeros(...) is 
incompatible with mine, i.e. which one will yield other results?
D.

Matthieu Brucher wrote:
>
>
> 2007/7/1, dmitrey <openopt@ukr.net <mailto:openopt@ukr.net>>:
>
>     Hi all,
>     There already was some discussion about zeros() and ones() usage,
>     but I
>     still can't understand some things.
>     1) currently ones(4,5, **kwargs) just yields error. Why it can't be
>     translated to ones((4,5), **kwargs) ?
>     See my example of Ones() below.
>     ##############################
>     from numpy import ones
>     def Ones(*args, **kwargs):
>         if type(args[0]) in (type(()), type([])):
>             return ones(*args, **kwargs)
>         else: return ones(tuple(args), **kwargs)
>     #############################
>     if __name__ == '__main__':
>         print Ones((2,2))
>         print 2*Ones([2,2])
>         print 3*Ones(2,2)
>         print 4*Ones(2,2,2, dtype=float)
>         print 5*Ones((2,2,2), dtype='float')
>         print 6*Ones([2,2,2], dtype='int')
>         print 7*Ones(3,3, dtype=int, order = 'C')
>         print 8*Ones((3,3), dtype='int')
>         print 9*Ones(3, dtype='int')
>         print 10*Ones((3,), dtype='int')
>     #############################
>
>
> The first argument is a shape, the second is the type of the data. 
> Python has no way of knowing that the second argument is part of the 
> shape and not the type. This way, the arguments are defined correctly, 
> although you could make a wrapper *args, **kwargs and forcing people 
> to use explicitely dtype=float, which they won't do, thus leading to a 
> lot of trouble and errors.
>
> Matthieu
>
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