[SciPy-user] FFT speed ?

[Samuel GARCIA]

yes I was thinking of doing something like that but
fft(a,pow(2,18)).shape is of course 262144 (2**18)
and when I use ifft after that The length of my signal have changed I
have an interpolated signal ... new problem for me ...



Nils Wagner wrote:
> Samuel GARCIA wrote:
>   
>> Yes it is pretty fast.
>> The difference of time is not a problem for the end user (between 0.2
>> and 1 s).
>> But specialy the value of 186889 is really a problem, the computation
>> is infinit.
>> Did you try tis value especialy ?
>>
>>     
> You may try
>
> a = rand((186889))
> t1=time.time()
> fft(a,pow(2,18))
> t2=time.time()
> print t2-t1
>  
> Nils
>
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Samuel Garcia
Universite Claude Bernard LYON 1
CNRS - UMR5020, Laboratoire des Neurosciences et Systemes Sensoriels
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FRANCE
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