Re: [Numpy-discussion] complex numbers

[David L G. <Dav...@no...>]

Thanks Tim,

DG

Tim Hochberg wrote:
> David L Goldsmith wrote:
>   
>> PS:  The Python built in function (i.e., you don't even need numpy for 
>> this) abs(x) is "vectorized" (i.e., accepts a (nested) sequence, incl. 
>> numpy array, argument) and overloaded to give the modulus (i.e., 
>> Pythagorean "length") of a complex number when such is its argument.
>>   
>>     
> This isn't quite right. The built in abs function looks for the special 
> method __abs__. So, abs(x) is equivalent to x.__abs__(). Arrays supply 
> an appropriate __abs__ method, lists do not. For example:
>
>  >>> l = [1,2,3,4]
>  >>> abs(l)
> Traceback (most recent call last):
>   File "<stdin>", line 1, in ?
> TypeError: bad operand type for abs()
>  >>> a = numpy.arange(5)
>  >>> abs(a)
> array([0, 1, 2, 3, 4])
>
>
> -tim
>   
>> DG
>>
>> Kenny Ortmann wrote:
>>   
>>     
>>> excuse my laziness for not looking this up, I googled it but could not find 
>>> a solution.
>>> matlab has a
>>> isreal(array)
>>> where if the array is full of real numbers the value returned is 1.
>>> I'm translating matlab code and ran across
>>>
>>> if ~isreal(array)
>>>     array = abs(array)
>>>
>>> Is there a way to check to see if a number is real or complex? and if so is 
>>> there a way to extract the(a + ib) because the absolute value of a complex 
>>> number is like the pythagorean therom on a and b?
>>>
>>> thanks for your help,
>>> Kenny
>>>
>>>
>>>
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>>   
>>     
>
>
>
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