[Numpy-discussion] *= operator not intuitive

Dag Sverre Seljebotn d.s.seljebotn@astro.uio...
Wed Mar 16 09:57:45 CDT 2011


On 03/16/2011 02:35 PM, Paul Anton Letnes wrote:
> Heisann!

Hei der,

> On 16. mars 2011, at 14.30, Dag Sverre Seljebotn wrote:
>
>> On 03/16/2011 02:24 PM, Paul Anton Letnes 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!
>>
>> In [1]: a = np.ones(5)
> Here, a is numpy.float64:
>>>> numpy.ones(5).dtype
> dtype('float64')
>
>> In [2]: b = a
>>
>> In [3]: c = a * 2
>>
>> In [4]: b
>> Out[4]: array([ 1.,  1.,  1.,  1.,  1.])
>>
>> In [5]: a *= 2
> So since a is already float, and b is the same object as a, the resulting a and b are of course floats.
>> In [6]: b
>> Out[6]: array([ 2.,  2.,  2.,  2.,  2.])
>>
> This is not the case I am describing, as in my case, a was of dtype integer.
> Or did I miss something?

I was just trying to demonstrate that it is NOT the case that "a = a * 2 
is exactly the same as a *= 2". If you assume that the two statements 
are the same, then it does not make sense that b = [1, 1, ...] the first 
time around, but b = [2, 2, 2...] the second time around. And in trying 
to figure out why that happened, perhaps you'd see how it all fits 
together...

OK, it perhaps wasn't a very good explanation...

Dag Sverre


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