Why has my function returned zero?

I have a functio which has returned zero sometimes when it must not. I saw the function and it is running normal, but when it pass the values to variables, they are zero.

A is a column vector with more than 1000 lines. This is the input

function [amf,pf,wf]=la(A)
        %pks vetor com os picos
        %locs vetor com as  localizações
        %w  vetor com as larguras
        [pks,locs,w] = findpeaks(A,'Annotate','extents','WidthReference','halfheight'); 
        %'findpeaks' função que indica os picos e comprimentos dos picos.
        tm = size(pks);
        m = 0;
        %Variáveis usadas para armazenar as amplitudes (am), picos (p) e larguras (we). 
        am1 =0;
        p1 = 0;
        we1 = 0;
        am2 =0;
        p2 = 0;
        we2 = 0;
        
       for x = 1:tm(1,1) %Analisando os picos maiores que 1000
            if 1000 < pks(x,1)
                m = m+1;   
                am1(m,1) = pks(x,1);
                p1(m,1) = locs(x,1);
                we1(m,1) = w(x,1);
            end
       end
       
%Agrupando os picos ao redor do maior valor
v=0;%indica a quantidade de elementos formando o pico
u=1; %indica a quantidade de picos
indice=0; %Armazena os indices dos picos 
tm = size(p1);
    for x=tm(1,1):-1:2 %For para analisar as difrenças de posições
       a= p1(x,1) - p1(x-1,1);
       if a<200 %Se a diferença for menor que 200 canais, agruparemmos os indices na variável indice
        v=v+1;
        indice(v,u)= x;
        indice(v+1,u) = x-1;       
       else
           u=u+1; v=0;
       end
    end
    tm=size(indice);
    if tm(1,1)>1       
    %For para analisar qual maior pico dentro de um grupo
     for u=1:tm(1,2)
        for v=1:tm(1,1)-1
             i = indice(v,u);
            if am1(i,1)<am1(i-1,1)
                 am2(u,1)=am1(i-1,1);
                 p2(u,1)=p1(i-1,1);
                 we2(u,1)=we1(i-1,1);
                            
             end
         end
     end
    
    end
    
    tm=size(am2);
    %analisando entre os picos, qual é o último
    if tm(1,1)>1
        for x=1:tm(1,1)-1
            if p2(x,1)<p2(x+1,1)
                amf=am2(x+1,1);
                pf=p2(x+1,1);
                wf=we2(x+1,1);
              
            end
        end
    else  %Caso tm(1,1) seja igual a 1
         amf=am2(1,1);
         pf=p2(1,1);
         wf=we2(1,1);
    end
    
end

[James Tursa]

How can we possibly answer this without seeing your variables and your actual function code?

[Rik]

Everything depends on your input. Please make sure we can independently run your code and reproduce the issue.

[madhan ravi]

upload A(input) as a text file to test

Answer 1

[Cris LaPierre]

Is your code returning 0 or empty? Since we don't know anything about your input, here are some suspicious things to look into.

You create am2, p2, we2 and set them equal to zero at the top. There are two obvious scenarios when 0 is returned. When tm(1,1) == 1. This happens when size(p1,1) < 3 or when size(am2,1) == 1.

I also see that your code does not assign any values when p2(x,1)>=p2(x+1,1).

if p2(x,1)<p2(x+1,1)

That might be another area to investigate.

[Daniel Fonsêca]

It returns 0.

Answer 2

[Manuel Infante Francés]

To me the same thing happens to me. I have a matrix A of (6x501) and I want to get a vector (3006x1) with all the columns of A next to each other. The last row of all the columns of A are negative numbers and the function returns zeros in the rows that are multiples of 6 of the obtained vector. I attach the code. I would appreciate any help in that regard.

function y = fcn(a,b,c,d,e,f)

y=[a b c d e f]';

[Walter Roberson]

What is class() of each of your inputs?

[Torsten]

Like this ?

A = [0 4 7; 11 15 19]

A = 2×3

0 4 7 11 15 19

A = A(:)

A = 6×1

0 11 4 15 7 19

[Manuel Infante Francés]

Thank you. Yes, but when the values ​​are negative the answer is zero. For example if the elements of the second row are negative [0 4 7;-11 -15 -19], the fourth, fifth and sixth row of the answer are zero [0;11;4;0;0;0]

[Torsten]

For example if the elements of the second row are negative [0 4 7;-11 -15 -19], the fourth, fifth and sixth row of the answer are zero [0;11;4;0;0;0]

???

A = [0 4 7;-11 -15 -19];
A = A(:)

A = 6×1

0 -11 4 -15 7 -19

Seems you use a different code. You might want to include it here.

[Manuel Infante Francés]

Thank you, but how can I know the class ()? The input vectors [a b c d e f] come from MATLAB Function blocks, which in turn come from seector blocks which in turn come from an extended kalmam filter block.

[Walter Roberson]

I wonder if those negative entries happen to be less than 5/10000 and you have "format short" in effect?

format short
A = [0 4 7;-.0011 -.00015 -.000019];
A

A = 2×3

0 4.0000 7.0000 -0.0011 -0.0001 -0.0000

format long g
A

A = 2×3

0 4 7 -0.0011 -0.00015 -1.9e-05

[Torsten]

Manuel Infante Francés comment moved here:

Like this ?

A = [0 4 7; 11 15 19]

A = A(:)

Yes, now:

when I load the data from workspace and execute the code in command window the result is as desired. The problem arises when applying it in Matlab Function of the model in Simulink.

[Manuel Infante Francés]

I wonder if those negative entries happen to be less than 5/10000 and you have "format short" in effect?

I indicate the first three columns:

12.2549615216114 12.2500232146314 12.2352847886914

58.6971528224936 58.7743679178343 58.8736987076398

1.11001172339046 1.11000659363486 1.10995585012337

5.88001232319817 5.88000221999864 5.87972551295995

0 0 0

-1.46785244195337 -1.46785804038749 -1.46787014889430

and the first 6 rows of the response:

12.2549615216114

58.6971528224936

1.11001172339046

5.88001232319817

0

0

The sixth row should be (-1.46785244195337) instead of zero

[Torsten]

We don't know how you use your function "fcn" to transform your matrix into the column vector.

What about

function Y = fcn(A)
  Y = A(:);
end

?

[Manuel Infante Francés]

We don't know how you use your function "fcn" to transform your matrix into the column vector.

What about

function y = fcn(a,b,c,d,e,f)
y=[a b c d e f]';
y=y(:);
end

[Manuel Infante Francés]

No matter how you put it, the MATLAB function returns negative values ​​with value equal to zero.

[Walter Roberson]

function y = fcn(a,b,c,d,e,f)
    assert(isfloat(a), 'a must be float')
    assert(isfloat(b), 'b must be float')
    assert(isfloat(c), 'c must be float')
    assert(isfloat(d), 'd must be float')
    assert(isfloat(e), 'e must be float')
    assert(isfloat(f), 'f must be float')
    y=[a b c d e f]';
    y=y(:);
end

[Manuel Infante Francés]

Thank you Mr. Walter. Indeed, the solution provided by you gives the desired value. But, I must say and also thank Mr. Torsten's solution:

function y = fcn(a,b,c,d,e,f)
y=[a b c d e f]';
y=y(:);
end

After reviewing the model many times, I have realized that the vector (f) that entered the function was not the one that should enter (supposedly with negative values) but one with values equal to zero. My apologies for the confusion and thanks again.