Not sure it's trivial as this but "endif" is not a keyword in Julia.
> On Apr 6, 2015, at 9:05 AM, Jim Christoff <[email protected]> wrote: > > First, Julia is a great replacement for Octave and in most cases "C". I have > used Octave for decades and Julia for a year or so. > I still have problems making the conversions. Any assistance will be greatly > appreciated. > This is short example of the modified Octave code that I have tried to > convert. I realize there is a difference > in how Julia handles arrays, but I don't seem to be able to resolve this > difference. > The following code I use in Octave: > > function arburg1 (x, p) > > > a = zeros(p) > n = length(x) > v = sum(x.*x) > > ## f and b are the forward and backward error sequences > f = x[2:n] > b = x[1:n-1] > > ## remaining stages i=2 to p > for i=1:p > > ## get the i-th reflection coefficient > g = 2 * sum(f.*b)/(sum(f.*f)+sum(b.*b)); > > > ## generate next filter order > if i==1 > a = g > else > a = a-g*a[i-1:-1:1] **** this is my problem area***** > endif > > ## keep track of the error > v = v*(1-g^2); > > ## update the prediction error sequences > oldf = f > f = oldf[2:n-i]- g*b[2:n-i] > b = b[1:n-i-1] - g*oldf[1:n-i-1] > > end > a = -a[p:-1:1] > > end > > > Thanks for any suggestions
