[Numpy-discussion] Can numpy catch this error for me?

Partridge, Matthew BGI SYD Matthew.Partridge@barclaysglobal....
Tue Apr 7 16:49:50 CDT 2009


 
> On Tue, Apr 7, 2009 at 14:19, Nathaniel Peterson 
> <nathanielpeterson08@gmail.com> wrote:
> >
> > import numpy as np
> > import operator
> > np.seterr(all='raise')
> > a=np.arange(1)+1
> > print(a.dtype)
> > # int32
> > for num in range(1,17):
> >     a=np.arange(num)+1
> >     b=np.multiply.reduce(a)
> >     print('%s! = %s'%(num,b))
> > #     c=reduce(operator.mul,range(1,num+1))
> > #     assert(b==c)
> >
> > The code above outputs
> >
> > int32
> > 1! = 1
> > 2! = 2
> > 3! = 6
> > 4! = 24
> > 5! = 120
> > 6! = 720
> > 7! = 5040
> > 8! = 40320
> > 9! = 362880
> > 10! = 3628800
> > 11! = 39916800
> > 12! = 479001600
> > 13! = 1932053504
> > 14! = 1278945280
> > 15! = 2004310016
> > 16! = 2004189184
> >
> > The results for 14! and above are wrong due to overflow of 
> the int32 
> > data type.
> > Is there a way to setup numpy so it will raise an error 
> when this occurs?
> 
> No, sorry.

Of course one can get the int behaviour by specifying dtype=object.
(This way you get the longer int behaviour and the *lack of speed* to go with it.)

for num in range(1,17):
    a=np.arange(num, dtype=object)+1
    b=np.multiply.reduce(a)
    print('%s! = %s'%(num,b))

1! = 1
2! = 2
3! = 6
4! = 24
5! = 120
6! = 720
7! = 5040
8! = 40320
9! = 362880
10! = 3628800
11! = 39916800
12! = 479001600
13! = 6227020800
14! = 87178291200
15! = 1307674368000
16! = 20922789888000

matt

 
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