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Re: FFT in 1 dimension [message #28498 is a reply to message #28494]

Fri, 07 December 2001 12:10

hradilv.nospam
Messages: 19
Registered: November 2001

Junior Member

OOPS - I think this one works:

function ft1d, data, dim, inverse=inverse

ds = size(data)
ndim = ds[0]
nlines = n_elements(data)/ds[dim]

dims = lindgen(ndim)+1
dims[dim-1] = -1
porder = sort(dims)

fdata = transpose(data,porder)
fdata = reform(fdata,ds[dim],nlines)

for i=0, nlines-1 do fdata[*,i] = fft(fdata[*,i],inverse=inverse)

dims[0] = dim
porder = sort(dims)
fdata = reform(fdata,(ds[1:ndim])[porder])
fdata = transpose(fdata,porder)

return, fdata
end

On Fri, 07 Dec 2001 19:05:36 GMT, hradilv.nospam@yahoo.com wrote:

> Here's what I came up with. Any comments?
>
> function ft1d, data, dim, inverse=inverse
>
> ds = size(data)
> ndim = ds[0]
> nlines = n_elements(data)/ds[dim]
>
> dims = lindgen(ndim)+1
> dims[dim-1] = -1
> porder = sort(dims)
>
> fdata = transpose(data,porder)
> fdata = reform(fdata,ds[dim],nlines)
>
> for i=0, nlines-1 do fdata[*,i] = fft(fdata[*,i],inverse=inverse)
>
> porder = (ds[1:ndim])[sort(dims)]
> fdata = reform(fdata,porder)
> fdata = transpose(fdata,sort(porder))
>
> return, fdata
> end
>

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[Message index]

		Re: FFT in 1 dimension By: Paul Woodford on Fri, 07 December 2001 20:09
		Re: FFT in 1 dimension By: Wayne Landsman on Fri, 07 December 2001 13:41
		Re: FFT in 1 dimension By: hradilv.nospam on Fri, 07 December 2001 12:10
		Re: FFT in 1 dimension By: hradilv.nospam on Fri, 07 December 2001 11:05

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