How to limit the output decimal places?

Hi everyone,

I required upto 10 decimal place answer for my output, but after few decimal place its shows zero. Earlier i attempted with longEng command but still the issue is same. May someone help put me here.

my script is as follows

clear all
clc

raw=readmatrix('Hourly_ev_05_04_2022');
data_a=raw';
data_a1=data_a(1:24,:);
rate=(mean(data_a1))';
data_a2=data_a(25:96,:);
ev_hr=data_a2;
bbb=rate;


position = 0; % position of plot in subplot
fig_num = 1;  % figure() number
for S=184:184
    uu=S;
    r=bbb(uu); % for first event
    data=ev_hr(:,uu);
    u=transpose(data);

    for n=1:length(u);  
    dt=n;
    for i=1:length(u)-(n-1);

        K=u(i:i+n-1);
        k=sum(K);

     b(i,n)=(k-(r)*(dt))/sqrt(r*dt);
    end
    
    end

    T=transpose(b);
    %format long
    
    for ii=1:72
       
        for jj=1:72-(ii-1)
            
           if(ii==1)
                b1(ii,jj)=T(ii,jj);  
           else
           b1(ii,jj+(ii-1))=T(ii,jj);         
           end
        end
        
    end
    ddd=max(b1);
    vvv=max(ddd);
    
    for ii=1:length(b)
        for jj=1:72
            if(b1(ii,jj)==0)
                b1(ii,jj)=nan;
            end
        end
    end
    
    yyy=max(b1);
    pp=max(yyy);
    beta_new=pp';
    c = NaN(72,24);
    c2=[c b1];
    x = [-23:72]' ;
    y = [1:72]' ;


    figure;
    ax1=subplot(2,1,1); 
    pcolor(x,y,c2);
    shading interp ;
    cb = colorbar('Location','west')
    cb.Label.String = '\beta';
    cb.Label.Rotation = 360;
    set(cb.Label,'FontSize',11);
    ylh=ylabel('Time interval (hr)', 'FontSize', 9);
    xline(0,'--r', 'LineWidth', 1.5);
    xline(-0.4,'-k', 'LineWidth', 1.5);
    set(gca,'XTickLabel',[]);
    ax1_pos = get(ax1,'Position');
    yl=ylim;
    ax = gca;
    ax.FontSize = 8;
    grid on
    patch([46 53 53 46], [yl(1) yl(1) yl(2) yl(2)], [0.01 0.01 0.01], 'FaceAlpha', 0.2 )
    %patch([27 35 35 27], [yl(1) yl(1) yl(2) yl(2)], [0.01 0.01 0.01], 'FaceAlpha', 0.2 )
    ylh.Position(1) = ylh.Position(1) - 0.2;
    text(50,yl(2)-5,sprintf('\\beta_{max} = %.1f',beta_new), ...
    'VerticalAlignment','top');
%     text(30,yl(2)-20,sprintf('\\beta_{max} = %.1f',16.1), ...
%     'VerticalAlignment','top');
    %middleCentered
    ax2=subplot(2,1,2); 
    set(ax2, 'Position', [.13 .3 .775 .25]);
    A=data_a(:,uu);
    ss= sum(A)+20;
    b=cumsum(A);
    h=[-23:72];
    plot(h,b, 'b', 'LineWidth', 1.5)
    ylabel('Cummulative events', 'FontSize', 9)
    xlabel('Time (hr)', 'FontSize', 9)
    xline(0,'--r', 'LineWidth', 1.5);
    xline(-0.4,'-k', 'LineWidth', 1.5);




    axis([-23 72 0 ss])
    yl=ylim;
%     text(25,yl(2),sprintf('\\beta_{max} = %.1f',beta_new), ...
%     'VerticalAlignment','top');
    grid on
    patch([46 53 53 46], [yl(1) yl(1) yl(2) yl(2)], [0.01 0.01 0.01], 'FaceAlpha', 0.2 )
    %patch([27 35 35 27], [yl(1) yl(1) yl(2) yl(2)], [0.01 0.01 0.01], 'FaceAlpha', 0.2 )
    ax = gca;
    ax.FontSize = 8;
    %f = gcf;
    %exportgraphics(gca,'Ev_01.png','Resolution',300)

end

Here is the output (3 and 5 coloumn and 12th rows showing zero).

Answer 1

[Walter Roberson]

The format command does not affect the Variable Browser.

The format for the variable browser can be changed for each variable. While you are browsing the variable, click on the View tab, and there is a drop-down there for Number Display Format.

You can change the default display format for the variable browser using Preferences -> Variables -> Format (Default array format)

[Andi]

The key problem is that i use this outputs as a input in the next step for plotting. And, the plots shows a missing point. I just need the output show be a number rather then zero.

[Walter Roberson]

It would help if you could attach your data and all code relevant to that section for us to test.

When you use the variable browser (which I can see you are using from the structure of the output window), then the variable browser never displays a value as plain 0 unless the value is exactly zero, or exactly negative zero. If the value was non-zero but too small to display to the digits of precision you asked for, then it would display as 0.0000

So those values that you point out are exactly zero, and we need to figure out why.

Your code is a bit difficult to follow, as it does some unnecessary transpose operations.

[Andi]

@Walter Roberson Please see the main question, I have uploaded the data along with fulll script.

[Walter Roberson]

When I run that code with that data, the lower right triangle of T is 0, but nothing else is. I do not get any holes in the output.

[Andi]

Yes, exactly, only few points shows empity hole, I want to fix them. Its only becsue that matlab consider those valuse as zero. This can be fix only if we be able to set a decimal place,

[Walter Roberson]

No, it is not a matter of decimal places being displayed. Put some debugging code in and you can see that with that data file, the values calculated at the locations you note are definitely not zero, and MATLAB doesn't show them as zero.

In the below, the "saved" values are there for comparison to be sure that we were examining the proper locations in the debugging phase.

Look at the spy() results: the zeros are strictly below the diagonal of T.

format long
filename = 'https://www.mathworks.com/matlabcentral/answers/uploaded_files/956880/Hourly_ev_05_04_2022.csv';
raw = readmatrix(filename);
data_a=raw';
data_a1=data_a(1:24,:);
rate=(mean(data_a1))';
data_a2=data_a(25:96,:);
ev_hr=data_a2;
bbb=rate;
position = 0; % position of plot in subplot
fig_num = 1;  % figure() number
for S=184:184
    uu=S;
    r=bbb(uu); % for first event
    data=ev_hr(:,uu);
    u=transpose(data);
    for n=1:length(u);  
    dt=n;
    for i=1:length(u)-(n-1);
        K=u(i:i+n-1);
        k=sum(K);
     b(i,n)=(k-(r)*(dt))/sqrt(r*dt);
     if n == 12 & ismember(i, [3 5])
         display(k); display(r); display(dt)
         fprintf('k-(r)*(dt) = %.15g\n', (k-(r)*(dt)));
         fprintf('b(i,n) = %.15g\n', b(i,n));
         save_this(i) = b(i,n);
     end
    end
    
    end
    T=transpose(b);
    fprintf('T(12,3) = %.15g\n', T(12,3));
    fprintf('T(12,5) = %.15g\n', T(12,5));
    fprintf('saved (12,3), (12,5) = %.15g, %.15g\n', save_this([3 5]));
end

k = 
    11

r = 
   1.458333333333333

dt = 
    12

k-(r)*(dt) = -6.5

b(i,n) = -1.55379719213471

k = 
     8

r = 
   1.458333333333333

dt = 
    12

k-(r)*(dt) = -9.5

b(i,n) = -2.27093435773535

T(12,3) = -1.55379719213471

T(12,5) = -2.27093435773535

saved (12,3), (12,5) = -1.55379719213471, -2.27093435773535

spy(T==0)

[Andi]

I see, let me try with this, I am not too much familir with matlab, so I just understand that some nonzero values are considered as zero by my script probably due to very small fraction.

[Walter Roberson]

Note that I did not change anything about the logic of the code: I just added display() and fprintf() and spy()

[Andi]

Yes, i see that but regarding my script more or less everthing is same then why i get some values as empty (NaN)

[Walter Roberson]

When you run the code with the debugging added, on your system, what output is shown by the debugging?

If you have a recent enough MATLAB, then you should be able to copy and run exactly the code I posted, without even modifying it to change the file name. If it works that way but fails when you change the file name to refer to your local system, then that would give us more information about what to look for.