[Numpy-discussion] curious FFFT timing

[Eric Hagemann]

I have an application that needs to perform close 1/2 million FFT's (actually inverse FFTs)  I was stymied by the time it was taking so I created a simple benchmark program to measure raw fft speed - hoping to extrapolate and see if the FFT algorithm was the time consuming portion of the algorithm.

The result I obtained surprised me

Inverse FFT (      4096) Number of seconds     5.608000 Time per FFT     0.005608
Forward FFT (      4096) Number of seconds     3.164000 Time per FFT     0.003164
Inverse FFT (      4095) Number of seconds     8.222000 Time per FFT     0.008222
Forward FFT (      4095) Number of seconds     5.468000 Time per FFT     0.005468
Inverse FFT (      8192) Number of seconds    12.578000 Time per FFT     0.012578
Forward FFT (      8192) Number of seconds     7.551000 Time per FFT     0.007551

Inverse FFTs were taking almost twice as long as forward FFTs !

Bottom line I found that in the inverse code the multiplication of 1/N at the end was the culprit.

When I removed the /n from the line 

return _raw_fft(a, n, axis, fftpack.cffti, fftpack.cfftb, _fft_cache)/n  

the code timing was more along the line of what was expected.

Any one want to speculate on the timing difference for the 4095 vice 4096 long transforms ?
Since 4095 is not a power of two I would have expected a greater time difference (DFT vice FFT)

N*N / N*log2(N) = N/log2(N) = 4096/12 = 341.33 which is >> than the 1.72 seen above

Cheers
Eric

------------------------------------ Code Snippet ---------------------------------------------

import time
from Numeric import *
from FFT import *


def f_fft(number,length):

    a = arange(float(length))
    start_time = time.time()
    
    for i in xrange(number):
        b = fft(a)

    durat = time.time() - start_time
    durat_per = durat / number
    print "Forward FFT (%10d) Number of seconds %12.6f Time per FFT %12.6f" % (length,durat,durat_per)
    
def i_fft(number,length):

    a = arange(float(length))
    start_time = time.time()
    
    for i in xrange(number):
        b = inverse_fft(a)

    durat = time.time() - start_time
    durat_per = durat / number
    print "Inverse FFT (%10d) Number of seconds %12.6f Time per FFT %12.6f" % (length,durat,durat_per)



i_fft(1000,4096)
f_fft(1000,4096)
i_fft(1000,4095)
f_fft(1000,4095)
i_fft(1000,8192)
f_fft(1000,8192)


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