Re: Parse error?

[Jaroslav Hajek]

On Thu, Aug 12, 2010 at 12:23 PM, dirac <address@hidden> wrote:
>
> Hi everyone,
>
> This is a bit of a long winded question: I am having problems with my code.
> I can't seem to get by it. Basically I am wanting to read in a file, do some
> analysis on it and then have a plot output.
>
> The piece of code below first reads in the file, splits the file into
> vectors and then the analysis. The function aver calculates the running mean
> of the data which, has many peaks. The idea is to find the points in the
> data set that just cross the running mean. I then want to fit a curve to the
> peak and find the maximum. This is essentially a peak finiding program.
>
>
> function gn = groupindcnt(filename);
>
>
> f = dlmread(filename);
> lambda = f(2:end,1);
> power = f(2:end,2:21);
> runmean = aver(lambda, power);
>
> %This for loop will loop over all the currents/columns in power
> for crnt = 1:1:20
>  data = power(:,crnt);
>  sign = 1;
>  cross = zeros(1000,1);
>  n = 1;
>
>
>  %This bit is finding the data points just above and below the running
> mean.
>  %and places a 1 in the vector cross.
>  for i = 1:1:9000
>    if (sign*(data(i+499)-runmean(i,crnt)) > 0)
>        sign = -1*sign;
>    cross(i) = 1;
>    n = n + 1;
>    end
>  end
>
>  %This bit is to remove the false crossing points, peaks are typically 30
> data
>  %points wide.
>  for j = 1:1:length(cross)
>    if (cross(j)==1)
>      cross(j+1) = 0;
>          cross(j+2) = 0;
>          cross(j+3) = 0;
>          cross(j+4) = 0;
>          cross(j+5) = 0;
>          cross(j+6) = 0;
>          cross(j) = j;
>    end
>  end
>
>  %removing zeros from cross
>  cross(cross==0)=[];
>  data_cp = data(cross+499,:);
>  lambda_cp = lambda(cross+499);
>
>
>
>  %This is the bit I am having problems with, it runs through all of the
> crossing points
>  %and calculates if between two crossing points there is a maximum (the if
> statement).
>  for k = 2:1:length(cross)-1
>    cp = (499+ cross(k) + round((cross(k+1) - cross(k))/2));
>        ll = data((499 + cross(k)));
>        hl = data((499 + cross(k+1)));
>        ml = data(cp);
>
>        if (ml > ((hl+ll)/2)
>


Unbalanced parentheses?



>               %ANYTHING ADDED IN HERE RETURNS PARSE ERROR
>               %This is where I want to do the curve fitting
>
>            end
>
>
>
>  end
> end
> end
>
>
>
> This may be a bit hard to understand; it is dificult reading other peoples
> code I know!
>
> Basically anything I put in the IF statement causes a 'parse' error.
>
> Any suggestions?
>
> Cheers
> Martin
>
> -----
> Pretty much convinced Octave>Excel for scientific data analysis!
> --
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-- 
RNDr. Jaroslav Hajek, PhD
computing expert & GNU Octave developer
Aeronautical Research and Test Institute (VZLU)
Prague, Czech Republic
url: www.highegg.matfyz.cz