Cycle for the matrix

Hello! I have a matrix with dimensions of 600 per 1000, it is necessary to select a maximum of each column and count the first 50 to the maximum, without taking into account zero values ​​and the first 50 after it, also without zero values

here is my code:

SS=AA(1:600,:);
Lum=1;
a1 = SS(:,Lum); 
[d,g] = max(a1);
 % d maximum values, g their location
mn=a1((g-50:g),:) ;  
% creating a matrix from a maximum of -50 to that maximum
v1=mn(mn ~= 0); 
% remove null values
[LL,PP] = size(v1); 
%how many were not zero elements
nmm=a1(g:g+LL,:)  ; 
% from the maximum + how many elements were in the first part
v2=nmm(nmm ~= 0); % удаляем нулевые значения 
% remove null values
E1=trapz(v1) ;
E2=trapz(v2) ;
% integrals 1 and 2 parts

"Lum" is one of my entries, I can count everything, but when setting up the "for" loop where Lum from 1 to 1000 gives me an error, help me figure it out.

[Bob Thompson]

What error did you receive? Please copy the entire error message.

Also, you can use max() to do the maximums of all columns at the same time.

SS=AA(1:600,:);
d = max(SS,[],1);

[Lev Mihailov]

Errors in the results shown.

I don’t understand, but where you are going, "g" is exactly what I need for each column.

When i do without cycle :

[M,I] = max(AA); 
mn=Wscat((I-100:I),:) ;
v1=mn(mn ~= 0);
[LL,PP] = size(v1);
nmm=Wscat(I:I+LL,:) 
v2=nmm(nmm ~= 0);
E1=trapz(v1) ;
E2=trapz(v2) ;

My last values ​​are zeros, although their maxima are in the middle.

Error testimony in "v1",

when looking for maxima and their location of the whole matrix

Answer 1

[Bob Thompson]

Try this instead. It calculates all of the maximums at the same time and then loops through the results. I wasn't able to confirm that it worked because I ran into an error where the maximum value was too close to the beginning or end of the column, but that was because I used a bunch of random numbers. You may need to make some other minor modifications to this.

SS=AA(1:600,:);
[d,g] = max(SS,[],1);
 % d maximum values, g their location
 %% Using max(SS,[],1) determines all maximums of each column at the same time
 
 %% Now we need a loop.
 for i = 1:length(d)
     mn=SS((g(i)-50:g(i)),i) ;  
     % creating a matrix from a maximum of -50 to that maximum
v1=mn(mn ~= 0); 
% remove null values
[LL,PP] = size(v1); 
%how many were not zero elements
nmm=SS(g(i):g(i)+LL,i)  ; 
% from the maximum + how many elements were in the first part
v2=nmm(nmm ~= 0); % удаляем нулевые значения 
% remove null values
E1=trapz(v1) ;
E2=trapz(v2) ;
% integrals 1 and 2 parts
 end

[Lev Mihailov]

I realized what the error is, the value on Lum 400 is on 598 and there are 5 real numbers, then it goes beyond the matrix, can I specify how it would be no more than the end of the matrix?

[Bob Thompson]

The simplest way to accomplish this is with an if statement.

if g(i)+LL>size(SS,1)
    nmm=SS(g(i):end,i)  ; 
    mn=SS((g(i)-50:g(i)),i) ; 
elseif g(i)-50<0
    mn=SS(1:g(i),i) ;
    nmm=SS(g(i):g(i)+LL,i)  ; 
else
    nmm=SS(g(i):g(i)+LL,i)  ; 
    mn=SS((g(i)-50:g(i)),i) ; 
end

This is not a direct insertion example, because it doesn't account for other things you are trying to accomplish, I am just trying to demonstrate the logic necessary.

[Lev Mihailov]

Error on the last "v1" , In an assignment A(:) = B, the number of elements in A and B must be the same.

[d,g] = max(SS,[],1);
 % d maximum values, g their location
 %% Using max(Wscat,[],1) determines all maximums of each column at the same time
 %% Now we need a loop.
 for i = 1:length(d)
% creating a matrix from a maximum of -50 to that maximum
if g(i)+LL>size(SS,1)
    nmm=SS(g(i):end,i)  ; 
    mn=SS((g(i)-50:g(i)),i) 
    % creating a matrix from a maximum of -50 to that maximum;
    v1(i)=mn(mn ~= 0);
    v2(i)=nmm(nmm ~= 0);
elseif g(i)-50<0
    mn=SS(1:g(i),i) ;
    nmm=SS(g(i):g(i)+LL,i)  ;
    v1(i)=mn(mn ~= 0);
    v2(i)=nmm(nmm ~= 0);
else
    nmm=SS(g(i):g(i)+LL,i)  ; 
    mn=SS((g(i)-50:g(i)),i) ;
    v1(i)=mn(mn ~= 0); % Error ???
    v2(i)=nmm(nmm ~= 0);
    
end
E1=trapz(v1(i)) ;
E2=trapz(v2(i)) ;
 end
fclose(fileID);

[Bob Thompson]

Yes, you're going to get that error because you're trying to put an array into a single element. You either need to just do v1, v2, or to change the class of v1 and v2 and make them cells. If you want to do the entire array, then just get rid of the indexing (you won't capture the different values for use outside the loop though), and if you want to do the cells method then change all of the v1 indexing marks to use curly braces.

% Array method
v1 = mn(mn ~= 0);

% Cell method
v1{i} = mn(mn ~= 0);

Answer 2

[Lev Mihailov]

I replaced v1 and v2 with T and B, but why don't they give out all my columns, but only the last one, and how can I fix this?

[d,g] = max(SS,[],1);
 % d maximum values, g their location
 %% Using max(Wscat,[],1) determines all maximums of each column at the same time
 %% Now we need a loop.
 for i = 1:length(d)
% creating a matrix from a maximum of -50 to that maximum
% remove null values
if g(i)+LL>size(SS,1);
    nmm=SS(g(i):end,i)  ; 
    mn=SS((g(i)-50:g(i)),i) ;
    % creating a matrix from a maximum of -50 to that maximum;
    V=mn(mn ~= 0);
    T=nmm(nmm ~= 0);
elseif g(i)-50<0
    mn=SS(1:g(i),i) ;
    nmm=SS(g(i):g(i)+LL,i)  ;
    V=mn(mn ~= 0);
    T=nmm(nmm ~= 0);
else
    nmm=SS(g(i):g(i)+LL,i)  ; 
    mn=SS((g(i)-50:g(i)),i) ;
    V=mn(mn ~= 0);
    T=nmm(nmm ~= 0);
    
end
E1=trapz(V) ;
E2=trapz(T) ;
 end
fclose(fileID);

[Bob Thompson]

As I mentioned previously, though not as explicitly, removing the indexing from v1 and v2 allows you to overwrite the entire matrix with mn or nmm. Overwriting the results removes the previous results, so the only visible outcome is the final column, as you say. If you would like to capture all results, I would suggest indexing with cells, because you can't affirm that the results from each column will be the same size.

Answer 3

[Lev Mihailov]

[d,g] = max(Wscat,[],1);
 LL=size(cell(V));  %% LL - ????
 V=cell(1);
 T=cell(1);
 for i = 1:length(d)
if g(i)+LL>size(Wscat,1); %% LL ????
    nmm=Wscat(g(i):end,i)  ; 
    mn=Wscat((g(i)-50:g(i)),i) ;
    V{i}=mn(mn ~= 0);
    T{i}=nmm(nmm ~= 0);
elseif g(i)-50<0
    mn=Wscat(1:g(i),i) ;
    nmm=Wscat(g(i):g(i)+LL,i)  ;
    V{i}=mn(mn ~= 0);
    T{i}=nmm(nmm ~= 0);
else
    nmm=Wscat(g(i):g(i)+LL,i)  ; 
    mn=Wscat((g(i)-50:g(i)),i) ;
    V{i}=mn(mn ~= 0);
    T{i}=nmm(nmm ~= 0);
    
end
E1=trapz(V{i}) ;
E2=trapz(T{i}) ;
     
 end

I need to know V to determine the value of LL, but to check whether the values ​​of the matrix are not exceeded, we use the parameter LL, help get rid of this recruitment

  LL=size(cell(V));

I tried to turn the code, but it did not help