How can i define this in matlab

I want to define this function {s(-1)} in matlab. it is giving me error message. array index must be positive integers or logical value. i have search through mathworks answers on how to rewrite this but i did not get an answer peculiar to this kind of situation.

[Stephan]

Please explain what you want to achieve.

Read also here:

https://de.mathworks.com/matlabcentral/answers/6200-tutorial-how-to-ask-a-question-on-answers-and-get-a-fast-answer.html

Best regards

Stephan

[Folakemi Omotoye]

how i can re-write the following code s(-1)=0

such that i wont get the error message ''array index must be positive integers or logical value''

[Stephan]

The code

s(-1)

Tries to access the value saved in s at index -1. This is not possible. Please give more context and explain what you want to achieve.

[Folakemi Omotoye]

This is the code:

{s(-1)=j(-1)==0
t=0
for t=0
    s(t)=sum(t-1)+sum(t)
    j(t)=g(t-1)+s(t)
    if j(t)>10
        []=t(j>10)
    end
    t=t+1
end

[Stephan]

You really make helping hard... Please read the tutorial at the link i posted.

[Jan]

@Folakemi Omotoye: This is not Matlab code. Therefore we cannot know, what it is intended to do. "s(-1)=j(-1)==0" is not meaningful in Matlab. "for t=0" is not a loop, but the same as "t=0". "s(t)=sum(t-1)+sum(t)" is not meaningful, because sum is not defined here and because t is 0, you cannot use it as index. "[]=t(j>10)" is not clear also.

Please explain, what you want to achieve. Posting pseudo-code in an unknown notation does not clarify anything.

[Folakemi Omotoye]

this is the whole algorithm

m=zeros(10,10,30);  %define 3-D 10 by 10 by 30 matrix
n=30;               %number of matrix

for k=1:n

      T=linspace(0,1.0,n);    %specify the value an element will take in each iteration
      m(3,3,k) = T(k);    %the 3rd row and 3rd column in each matrix will have the value of t
      %     disp(m(:,:,k));
      vector=sum(sum(m));    %sum all the colums of the vector into a row
  end

[v]=vector(:,:);    %convert the vector into 1D vector

t=logical(T);
%CUSUM Algorithm
h=4;              %threshold
s(-1) = 0;
g(-1)=0 ;         %initial conditions

for    t=0         %start time

      s(t)=sum(v(t-1) + sum(v(t)));           %cumulative sum up to current value

      g(t)=max(g(t-1) + s(t));    %current value minus minimum value in vector up to current value

      if g(t) > h               %if output is greater than the specified threshold
          detect_tim=t;                 %store the detection time
          disp(['The detection time: (', num2str(detect_tim),')']);        %display the detection time

          change_time=(t(min(s(t))) + 1);   %estimate the change time

          disp(['The change time estimate: (', num2str(change_time),')'])          %display value of change
          % stop and reset algorithm
          return                %to re-start algorithm
      end
      t=t+1;
      xL = get(gca,'XLim');           %plot threshold line on same axis

      figure(1), subplot(2,1,1)
      line(xL,[0.6 0.6],'Color','r');   %horizontal line through the  threshold value
      hold on
      plot(1:t,v(1:t),'*b')           %the time vector plot
      title('signal change plot')
      xlabel('time (ms)')
      ylabel('Amplitude (v)')

end

Answer 1

[Image Analyst]

See the FAQ: https://matlab.wikia.com/wiki/FAQ#How_do_I_fix_the_error_.22Subscript_indices_must_either_be_real_positive_integers_or_logicals..22.3F

[Folakemi Omotoye]

I did convert to a logical but i a still getting an error for the line {s(t-1)=0}. this is the new code

{t=logical(T);

t=0;

 h=4;              %threshold

 s(t-1) = 0;

g(t-1)=0 ;         %initial conditions

for t=0

    s(t)=sum(t-1)+sum(t)

    j(t)=g(t-1)+s(t)

    if j(t)>10

        []=t(j>10)

    end

    t=t+1

end }

p.s T is of class double, t is of class logical

[Image Analyst]

That's because that's not valid MATLAB syntax. First of all T is not defined. Next, you can't have a line start with a brace. And you can't have a chunk of code end with a brace, like you have after the end in the final line. Braces are for doing something with a cell array. See the FAQ on cell arrays.

Next, if you do for t=0, you're only doing one iteration with a loop iterator value of zero. Essentially you're not doing a loop at all.

Next, you can't say []=t(j>10) because that is setting null to all elements of t, where the corresponding value of the j array is more than 10. You can't do that. Null is null and you can't set something equal to it. It's like if you said 300 = g. No, 300 is 300 and you can't change it. Maybe you meant t(j>10) = [], which would delete all elements of t where the corresponding value of j was more than 10, basically shortening t.

And saying sum(t) is just zero, because t is zero, and summing one element with a value of zero is just simply zero.

You really need to read the getting started section of the help because the code is full of syntax errors.

[Folakemi Omotoye]

the braces is not part of the original code. i added it to seprate my comment from the code. t is supposed to increase by 1, then start the iteration again. sum(t) is only zero for the first iteration.subsequent iterations will have a value. i assigned a variable name to t(j>10), yet still have the error message.

one thing i observed was that, the line of converting double to logical (which i placed before the rest of the code) did evaluate but the rest of my code was evaluating t as a double.

[Walter Roberson]

When you use a logical in a numeric context it is treated as a double in most cases.

[Folakemi Omotoye]

any idea on what i can do please?

[Image Analyst]

I would fix it, but I honestly can't figure out what you want to do. Can you put in some comments that describe what you're trying to do? What is T? Is it an array or a single value? How many times do you want to iterate the loop? And we have no idea what you're attempting with this: "[]=t(j>10)". When you posted it you removed all the comments that I'm sure you put in there before, like any good programmer would have. Please put them back in.

[Walter Roberson]

Are you trying to define a timesequence written in terms of the unit step function, where [-1] by convention refers to the value at the previous time step?

[Folakemi Omotoye]

yes

[Folakemi Omotoye]

h=4;              %threshold
s(-1) = 0;
g(-1)=0 ;         %initial conditions

for t=0    %start time  
    s(t)=sum(v(t-1) + sum(v(t)));           %cumulative sum up to current value      
    g(t)=max(g(t-1) + s(t));    %current value minus minimum value in vector up to current value      
    if g(t) > h               %if output is greater than the specified threshold
        detect_tim=t;                 %store the detection time 
         disp(['The detection time: (', num2str(detect_tim),')']);        %display the detection time

           change_time=(t(min(s(t))) + 1);   %estimate the change time 

          disp(['The change time estimate: (', num2str(vee),')'])          %display value of change 
          % stop and reset algorithm
          return                %to re-start algorithm
      end
     t=t+1;
     xL = get(gca,'XLim');           %plot threshold line on same axis    
  end

[Dennis]

Maybe you can explain what you want to achieve. This line is NOT the cumulative sum up to current value:

   s(t)=sum(v(t-1) + sum(v(t)));

As Jan already stated the sum() does not do anything here.

Indexing in Matlab starts at 1, thus you can not assign any value to s(-1).

Your code looks still similar to the one you posted last week, but obviously my solution did not do what you were hoping.

Can you provide an example of input data (v=[1 3 2 4 5 1 3]) and expected output (s=[1 4 6 10 15 16 19])) g=[? ? ? ?] change_time=[? ? ? ????]).

Maybe explain what

change_time=(t(min(s(t))) + 1)

is supposed to do - I am really sure it is not working.

[Folakemi Omotoye]

I earlier had my code for the sum to be

s(t)=sum(v(1:t)) + sum(v(t))

but my instructor said it should be in the form

s(t)= sum(v(t-1)) + sum(v(t))

as written in the new codes i uploaded today. it is a challenge getting it in that form because matlab wouldn't take a negative index.

v = 1x30 double, with values-

v = [ 0    0.0345    0.0690    0.1034    0.1379
    0.1724    0.2069    0.2414    0.2759    0.3103
    0.3448    0.3793    0.4138    0.4483    0.4828
    0.5172    0.5517    0.5862    0.6207    0.6552
    0.6897    0.7241    0.7586    0.7931    0.8276

      0.8621    0.8966    0.9310    0.9655    1.0000]

while ''t'' is unspecified but is supposed to start from zero till the final possible iteration for ''v'' .

change_time is the time following the current minimum of the cumulative sum.

I hope my explanation is clear for you to understand.

[Image Analyst]

Please read this link so I won't have to keep fixing your formatting.

You might look at cumsum():

s = cumsum(v)

[Folakemi Omotoye]

the use of the cumsum command will nullify the need for the loop, which is not what the algorithm is design to use.

[Dennis]

v(t) and v(t-1) are scalars, this means v(t) might be 3 v(t-1) might be 5. What is sum(3) supposed to be (spoiler: its 3)?

If you want to sum up the vector one by one you need to do it like this:

s=zeros(numel(v),1);
for t=1:numel(v)
if t==1
    s(t)=v(t);
else
    s(t)=s(t-1)+v(t);
end
end

However the resulting vector will be the same as if you used

s=zeros(numel(v),1);
for t=1:numel(v)
    s(t)=sum(v(1:t));
end

or even cleaner as suggested by Image Analyst:

s=cumsum(v)

[Folakemi Omotoye]

Thanks for the idea, i have re-written the code in this form below but the line containing

g(t)=max(g(t) + s(t));

is producing error message ''index exceeds array bounds''

m=zeros(10,10,30);  %define 3-D 10 by 10 by 30 matrix
n=30;               %number of matrix

for k=1:n

      T=linspace(0,1.0,n);    %specify the value an element will take in each iteration
      m(3,3,k) = T(k);    %the 3rd row and 3rd column in each matrix will have the value of t
  %     disp(m(:,:,k));
     vector=sum(sum(m));    %sum all the colums of the vector into a row
  end

[v]=vector(:,:);    %convert the vector into 1D vector

t=logical(T);

%CUSUM Algorithm

   h=4;              %threshold

if t==0
    s=0;
    g=0;
else
    if t==1
        s(t)=v(t);
        g(t)=s(t);
    else
    end
end
for    t=2         %start time

      s(t)=sum(v(t-1) + sum(v(t)));           %cumulative sum up to current value

      g(t)=max(g(t) + s(t));            %maximum value 

      if g(t) > h               %if output is greater than the specified threshold
         detect_tim=t;                 %store the detection time 
          disp(['The detection time: (', num2str(detect_tim),')'])        
  %display the detection time

           change_time=(t(min(s(t))) + 1);   %estimate the change time 

         disp(['The change time estimate: (', num2str(change_time),')'])          %display value of change 

          % stop and reset algorithm
      return                %to re-start algorithm

      end

     t=t+1;

      xL = get(gca,'XLim');           %plot threshold line on same axis

     figure(1), subplot(2,1,1)
    line(xL,[0.6 0.6],'Color','r');   %horizontal line through the  threshold value
     hold on
      plot(1:t,v(1:t),'*b')           %the time vector plot
      title('signal change plot')
      xlabel('time (ms)')
      ylabel('Amplitude (v)')
      end

[Dennis]

Now its getting kinda weird. However lets go through your code (i added a comment to most lines):

m=zeros(10,10,30);  %10x10x30 zeros - nothing wrong here
n=30;               %probably size(m,3) ->we do not need n, but no problem
for k=1:n  %k=1:size(m,3) is safer
    T=linspace(0,1.0,n);    %this does not change better calculate outside loop
    m(3,3,k) = T(k);    %see alternative below
%     disp(m(:,:,k));
   vector=sum(sum(m));   %this is more or less the same as linspace(0,1,30), and should not calculate this in every iteration
end
[v]=vector(:,:); %now its exactly the same as linspace(0,1,30), you do not need the [] here

I guess there are usually more non-zero elements in m, however you can replace the entire loop with

m(3,3,:)=linspace(0,1,30);

This line creates a vector of 1 zero and 29 ones:

t=logical(T); %not sure what we want with this

Now those lines don't do anything at all:

if t==0   %t is a logical vector, this will only be true when all values in t are 0 - and this is not happening
    s=0; %never gets executed, is not indexed either
    g=0; %never gets executed, is not indexed either
else %better use elseif in this case
    if t==1   %one value in t is 0 so this is never true
        s(t)=v(t); %looks okayish
        g(t)=s(t); %looks okayish
    else %you can omit this
    end
end

This is not working:

for    t=2        %this is not a loop, for t=2:numel(v)
    s(t)=sum(v(t-1) + sum(v(t)));    %the sum does not do anything...really and this is no cumulative sum since you only add up the last 2 values!!!
    g(t)=max(g(t) + s(t));   %this  throws an error because g(2) does not exist, maybe g(t)=max(g(t-1)+s(t); 
    if g(t) > h              
       detect_tim=t;             %you overwrite detect_tim in every iteration 
        disp(['The detection time: (', num2str(detect_tim),')'])        
         change_time=(t(min(s(t))) + 1);   %i don't think that this can work: min(s(t)) is the same as s(t) and might be a floating point number, which can not be used for indexing (t(1.5) will not work). Additionally t (at this point) is not a vector anymore, so anything other than t(1) will throw out an error
       disp(['The change time estimate: (', num2str(change_time),')'])         
      return               % this probably does not do what you think it might do
    end
  t=t+1; %omit this the for loop is incrementing on its own

[Folakemi Omotoye]

t=logical(T) %to convert T of class double to logical

    s(t)=sum(v(t-1) + sum(v(t)));    %if (t-1) was execuatble, it would have been the current sum plus cumulative sum of last iteration

    g(t)=max(g(t) + s(t));   %same explanation as above.the max value of past iterations of g(t) till current iteration

    if g(t) > h              

       detect_tim=t;             %it will only write when g(t) is greater than h, else it restarts iteration from beginning

        disp(['The detection time: (', num2str(detect_tim),')'])

         change_time=(t(min(s(t))) + 1);   %it should be the time following the time at which g(t) was greater than h

       disp(['The change time estimate: (', num2str(change_time),')'])

      return               % it should re-start the algorithm from the beginning rather than continuing with the next iteration

    end

t=t+1; %noted

[Dennis]

s(t)=sum(v(t-1) + sum(v(t)));    %if (t-1) was execuatble, it would have been the current sum plus cumulative sum of last iteration

No.

If v=[1 2 3 4] was your vector s(4) would be sum(v(3)) + sum(v(4)) which is the same as v(3) + v(4) -> 7

The cummulative sum would be v(1)+v(2)+v(3)+v(4) =1+2+3+4=10, but thats no what your code is calculating.

This works if you write

s(t)=s(t)-1+v(t) %s(t-1) !!!!!

And again sum(v(t)) is the same as v(t).