Index exceeds array bounds at position 3

Question

Owen 2024-11-12，21:19

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/2165744-index-exceeds-array-bounds-at-position-3

评论： Voss 2024-11-14，21:56

The variable i on line 86 plot(vsine(begin(FREQ):begin(FREQ)+cycle(FREQ)-1,k),nsd(n,k,i,j)) is said to exceed array bounds once the variable indexes to 5. I am assuming this may be related to an issue at line 43 vsine(n,counter)=ampl(5-i)*sin(2*pi*freq(FREQ)*0.001*ptime(n))+vmean(j); since any value larger than 5 here would return a 0 or negative position variable for ampl. However, if I initialize i at line 79 as for i=1:4:16 then i should be able to index to 16. Is there a logic error I'm missing? Thank you!

freq=[100 250 500 1000];
cycle=[200 80 40 20];
begin=[753 785 795 739];
cnum=[2 4 8 16];
%tth must be accounted for in ZHAD
vhold=-90;
vr=-75;
%all the above are just initial values, trial stands for which model trial
%data was used
%gives us time(ptime)
step=0.05;           
ptime=0:step:1199*step;
%This gets sintrace value which is not needed. Left in just in case  
for FREQ=1:4
    SNUM=16;
    for i=1:SNUM
        %sweepnamestr=sprintf('%s%s%s', 'Trace_',int2str(9),'_',int2str(i),'_1_1'); %9 used to be 10-FREQ
        sweepnamestr=sprintf('%s%s%s', 'Trace_',int2str(10-FREQ),'_',int2str(i),'_1_1');
        tracetemp=eval(sweepnamestr);
        trace(:,i)=tracetemp(:,2)*10^9; %store sweeps in nA
        sintrace(:,i)=trace(301:1500,i);
    end
    init=14.175;
    inipt=285;
    finpt=1501;
    
    
    
    
    % to get the voltage amounts (vsine)
    
    ampl=[15 30 45 60];
    vmean=[0 -15 -30 -45];
    counter=1;                  
    for j=1:length(vmean)
        for i=1:length(ampl)
            for n=1:1200
                vsine(n,counter)=ampl(5-i)*sin(2*pi*freq(FREQ)*0.001*ptime(n))+vmean(j);
            end
            counter=counter+1;
        end
        
    end
    
    
    
    %generate current from parameters
    
    cycle=[200 80 40 20];
    ta=AM(trial,:);
    tb=BM(trial,:);
    tg=GM(trial,:);
    
    init=[1;0;0;0;0];
    
    for t=1:length(ptime);
        vv=vsine(t);
        coeff=[-0.000000000001456	0.000000000012566	0.000000013391643	-0.000000133240032	-0.000033555887315	0.000473197916512	0.038026092731197];                               
        m=aldamtrx(ta,tb,vv);
        p0=expm(m*50)*init;
        %ppd=p0;
        pgval=polyval(coeff,vv);    
        %ppd = expm(m*step)*ppd;
        ppd = expm(m*step)*p0;
        nsd(n,k,i,j) = (ppd(5))*(vv-vr)*pgval;      
    end
    
    
    
    %plot the voltage vs current
    %see sinavgplot for the use of the values in the first 2 loops
    
    for i=1:4:16
        %for i=1:4
        ii=(i-1)/4+1;
        %ii = (i-1)+1;
        for   j=1:4
            k=i+j;
            
            plot(vsine(begin(FREQ):begin(FREQ)+cycle(FREQ)-1,k),nsd(n,k,i,j))
            hold on
            
        end
        %pause
        hold off
        title([num2str(freq(FREQ)),'Hz, ','Mean V=',num2str(vmean(ii)),'V'])
        axis([-100 100 -.5 1,])
    end
    pause
end
Error using eval
Unrecognized function or variable 'Trace_9_1_1_1'.

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Owen 2024-11-13，16:17

编辑：Owen 2024-11-13，16:20

在 MATLAB Online 中打开

No. Variable n should be indexing through as well at 1:1200. Adding another for loop accounting for n after i and j where vsine is being plotted never fully runs to completion and I have to terminate the script.

for i=1:4:16
     %for i=1:4
         ii=(i-1)/4+1;
         %ii=(i-1)+1;
           for j=1:4
               k=i+j;
               for n=1:1200;
                plot(vsine(begin(FREQ):begin(FREQ)+cycle(FREQ)-1,k),nsd(n,k,i,j))
               hold on
               end
           end

Voss 2024-11-14，21:56

在 MATLAB Online 中打开

A couple of pointers:

1. This

sweepnamestr=sprintf('%s%s%s', 'Trace_',int2str(10-FREQ),'_',int2str(i),'_1_1');

would be better written like

sweepnamestr=sprintf('Trace_%d_%d_1_1',10-FREQ,i);

Example to show they are the same:

FREQ = 3;
i = 5;
sweepnamestr=sprintf('%s%s%s', 'Trace_',int2str(10-FREQ),'_',int2str(i),'_1_1')
sweepnamestr = 'Trace_7_5_1_1'
sweepnamestr=sprintf('Trace_%d_%d_1_1',10-FREQ,i)
sweepnamestr = 'Trace_7_5_1_1'

2. You can avoid using eval() by loading the Traces mat file into a structure (rather than loading all its variables directly into the workspace), and then using a dynamic field name to reference the relevant field of the structure. Example:

S = load('hyster_20140721_001sin100.mat');
% ...
tracetemp = S.(sweepnamestr);

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Answer 1

Voss 2024-11-12，22:02

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https://ww2.mathworks.cn/matlabcentral/answers/2165744-index-exceeds-array-bounds-at-position-3#answer_1544254

在 MATLAB Online 中打开

plot(vsine(begin(FREQ):begin(FREQ)+cycle(FREQ)-1,k),nsd(n,k,i,j))

Indexing nsd is the error, not indexing vsine. (nsd is the only variable indexed on that line using at least 3 subscripts.)

Looking at how nsd is defined:

for t=1:length(ptime)
    % ...
    nsd(n,k,i,j) = (ppd(5))*(vv-vr)*pgval;
end

At that point in the code, j is length(vmean), i is length(ampl), and n is 1200, because the previous j/i/n nested for loops have all completed (and the loop iteration variables maintain their values after the loops). And k is undefined. (There's no telling what value of k you had in your workspace when you ran the script.) But k is not the problem (not right now, anyway), i is.

So i is length(ampl), which is 4 (and j is also 4). Which means size(nsd) is 1200-by-k-by-4-by-4 (at least - it could be larger because nsd could've been defined by some previously run code before this script was run). Thus, attempting to index nsd in the third dimension with anything greater than 4 will produce the error you got (assuming nsd was not previously defined to be larger than we see here). Indeed you see the error when i is 5.

What's the solution?

Define variables before you use them, in particular k.
Preallocate your variables to the correct size before defining them using indexing in loops.
Consider using clearvars at the top of the script to clear old variables left over from previously run code (i.e., start with a clean workspace).
Also, the definition of nsd is likely incorrect; assigning to nsd(n,k,i,j) suggests that that assignment should take place inside 4 nested loops (over i, j, k, and n) not one loop over t where the variable k is not previously defined in this script and i, j, and n happen to have values left over from previous loops. Maybe that's really what you want, but it doesn't seem likely.

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Voss 2024-11-13，20:28

在 MATLAB Online 中打开

untitled_modified.m

On your plotting line

plot(vsine(begin(FREQ):begin(FREQ)+cycle(FREQ)-1,k),nsd(n,k,i,j))

you're plotting a scalar (nsd(n,k,i,j)) vs a vector (vsine(begin(FREQ):begin(FREQ)+cycle(FREQ)-1,k)).

When you plot a scalar vs a vector, MATLAB creates as many lines as there are elements in the vector, and each of those lines contains a single point. Lines containing only a single point are not rendered unless they have a data marker, so that's why the plots are blank. Here's an example to illustrate:

x = 1:10; % vector

y = 4; % scalar

subplot(2,1,1)

plot(x,y) % plot without data marker -> nothing shows up

subplot(2,1,2)

plot(x,y,'o') % plot with a data marker -> 10 points show up

You can include a data marker in your plot call, in order to see what's plotted, but I don't think the result will be what you're looking for.

I guess that the calculation of nsd

nsd(n,k,i,j) = (ppd(5))*(vv-vr)*pgval;

should actually involve not vv (which is equivalent to vsine(end,1)) but all elements of vsine

nsd = ppd(5)*(vsine-vr)*pgval;

which can be done before the loop since now it doesn't depend on i,j (or k or n). Then nsd is the same size as vsine and the same indexing done to vsine should be done to nsd in order to plot one vs the other.

for i=1:4:16

ii=(i-1)/4+1;

for j=0:3

k=i+j;

idx = begin(FREQ):begin(FREQ)+cycle(FREQ)-1;

plot(vsine(idx,k),nsd(idx,k))

hold on

end

% ...

end

Plotting nsd vs vsine is not very interesting since they are linearly related, but maybe that's what you're supposed to get. I've attached a modified version of untitled.m, so you can run it and see if it produces something closer to what you're after.

Owen 2024-11-14，20:38

While this does present figures that are closer to what I am looking for, nsd must be dependent on voltage (vv and thus, vsine) directly because the current values (nsd) should be generated in response to the changing voltage values at each time interval. The figures to be produced should exhibit hysteretic behavior, not linear behavior. Could it could potentially be linear since the for loop at:

for n=1:1200

vsine(n,counter)=ampl(5-i)*sin(2*pi*freq(FREQ)*0.001*ptime(n))+vmean(j);

would be set at 1200 instead of indexing through the array? Thus, only the vsine value taken at ptime = 1200 would be retained.

Voss 2024-11-14，21:48

untitled_modified_modified.m

I think this is probably what you were going for with the nsd calculation:

nsd = zeros(size(vsine)); % pre-allocate nsd

init=[1;0;0;0;0];

coeff = [-0.000000000001456 0.000000000012566 0.000000013391643 -0.000000133240032 -0.000033555887315 0.000473197916512 0.038026092731197];

for t=1:length(ptime)

vv=vsine(t); % same as vsine(t,1)

m=aldamtrx(ta,tb,vv);

p0=expm(m*50)*init;

ppd=p0;

pgval=polyval(coeff,vv);

ppd = expm(m*step)*ppd;

nsd(t,:) = ppd(5)*(vsine(t,:)-vr)*pgval; % calculate one row of nsd from one row of vsine

end

Making that change (see attached untitled_modified_modified.m) gives plots that may be close to being correct. E.g.:

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Index exceeds array bounds at position 3

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Index exceeds array bounds at position 3

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