| Re: All day FFT.... [message #29208 is a reply to message #29206] |
Wed, 06 February 2002 15:30   |
Robert Stockwell
Messages: 74
Registered: October 2001
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Member |
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Paul van Delst wrote:
> Hey there,
>
> I've had this process running for about, oh, 4-5 hours now where I'm reducing the resolution of
> an input spectrum of about 500K points (i.e. a lot, but not terribly so.). Earlier I did the
> same for a spectrum of about 700K points. Something has apparently gone haywire in the second
> one - is there anything I can do to interrupt the process, check out some numbers to see if
> everything is o.k. and if it is, start it up where it left off? I think that's what a ^C does
> but I thought I better consult the idl wizards out there...
>
> thanks for any info.
>
> paulv
I'm betting that ^C will interrupt the process as soon
as it finishes with the FFT. LOL! I'd just kill IDL.
My guess, regarding slowness, is "Prime Number" (number of points
in your time series). Try zeropadding up to, or truncating
down to, a nice factorable number.
I've attached my hackware factors.pro which will return the
factors of a number. (and its recursive, COOL!)`
FYI, 500K should take seconds: Here is a quicky example:
IDL> a = lindgen(1025L^2)
IDL> help,a
A LONG = Array[1050625]
IDL> tic & b = fft(a) & toc
% Compiled module: TIC.
% Compiled module: TOC.
Elapsed time: 3.9645100 Seconds.
IDL> print,factors(n_elements(a))
% Compiled module: FACTORS.
5.00000 5.00000 5.00000 5.00000 41.0000 41.0000
Cheers,
bob stockwell
; do factoring of a function
; development interrupted when i realized I didn't need it
; drops the last number
function factors, n,prevfactors=prevfactors
maxfactor = fix(sqrt(n))
if maxfactor le 1 then begin
if keyword_set(prevfactors) then begin
prevfactors = [prevfactors,n]
return,n
endif else begin
return,n
endelse
endif
fac = findgen(maxfactor-1)+2 ; 2 -- sqrt(n)
doloop = 1
factorflag = 0
counter = 0
while doloop do begin
if n mod fac(counter) eq 0 then begin
factorflag = 1
newfactor = fac(counter)
if keyword_set(prevfactors) then prevfactors = [prevfactors,newfactor] $
else prevfactors = newfactor
newnumber = n/newfactor
; to iterate is human, to recurse is divine
r = factors(newnumber,prevfactors=prevfactors)
doloop = 0
endif
counter = counter+1
if counter ge maxfactor-1 then doloop = 0
endwhile
if n_elements(prevfactors) eq 0 then prevfactors = n else begin
; only if n is prime do we add it here
if not(factorflag) then prevfactors = [prevfactors,n]
endelse
return,prevfactors
end
;;;;___________ test code here ___________________
n = 5001
r = factors(n)
print
print
print,'Finished calculating factors_______'
print,'Number: ',n
print,'Factors:'
print,r
end
-
Attachment: factors.pro
(Size: 1.21KB, Downloaded 82 times)
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