[Numpy-discussion] sum and prod

[nicky van foreest]

Thanks for your hints.

NIcky


On 9 September 2012 00:30, eat <e.antero.tammi at gmail.com> wrote:
> Hi,
>
> On Sun, Sep 9, 2012 at 12:56 AM, nicky van foreest <vanforeest at gmail.com>
> wrote:
>>
>> Hi,
>>
>> I ran the following code:
>>
>>         args = np.array([4,8])
>>         print np.sum( (arg > 0) for arg in args)
>>         print np.sum([(arg > 0) for arg in args])
>>         print np.prod( (arg > 0) for arg in args)
>>         print np.prod([(arg > 0) for arg in args])
>
> Can't see why someone would write code like above, but anyway:
> In []: args = np.array([4,8])
> In []: print np.sum( (arg > 0) for arg in args)
> 2
> In []: print np.sum([(arg > 0) for arg in args])
> 2
> In []: print np.prod( (arg > 0) for arg in args)
> <generator object <genexpr> at 0x062BDA08>
> In []: print np.prod([(arg > 0) for arg in args])
> 1
> In []: print np.prod( (arg > 0) for arg in args).next()
> True
> In []: sys.version
> Out[]: '2.7.2 (default, Jun 12 2011, 15:08:59) [MSC v.1500 32 bit (Intel)]'
> In []: np.version.version
> Out[]: '1.6.0'
>
> My 2 cents,
> -eat
>>
>>
>> with this result:
>>
>> 2
>> 1
>> <generator object <genexpr> at 0x1c70410>
>> 1
>>
>> Is the difference between prod and sum intentional? I would expect
>> that  numpy.prod would also work on a generator, just like numpy.sum.
>>
>> BTW: the last line does what I need: the product over the truth values
>> of all elements of args. Is there perhaps a nicer (conciser) way to
>> achieve this?  Thanks.
>>
>> Nicky
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>
>
>
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