Using a for loop to read in 16 data sheets and storing one value from each data sheet in a vector - full problem explained

I have currently got a for loop that reads in all the data values from 16 excel spreadsheets (4 columns, 300+ rows).

I then perform a series of calculations with these data points and I want to store just one value from each - creating a (16, 1) vector where the value from the first data sheet is stored in first row, the second value calculated from the second data sheet is stored in row 2 and so on.

The value I am trying to store is mSTRESS = min(y1) for each data sheet - the max stress experienced by each sample.

The code:

datafiles = dir('Synthetic_Pancreas_Sample_*.xlsx');
figure();
axis padded;
grid on;
xlabel('Strain', 'Interpreter', 'latex', 'fontsize', 16, 'Color', 'k');
ylabel('Stress / MPa', 'Interpreter', 'latex', 'fontsize', 16, 'Color', 'k');
set(gca, 'TickLabelInterpreter', 'latex', 'fontsize', 14);
lgd = legend('fontsize', 10, 'Interpreter', 'latex', 'location', 'Southeast');
title(lgd, 'Sample Number:');
legend('1', '2', '3', '4', '5', '6', '7', '8', '9', '10', ...
    '11','12','13','14','15','16');
hold on
A_A = zeros(16,1);
for i = 1:16
    A_A(i,:) = mSTRESS;
    for file = datafiles'
        
        currentfile = file.name; 
        RAW = readmatrix(currentfile);
        t = RAW(:,1);
        mm = RAW(:,2);
        N = RAW(:,3);
        d = RAW(:,4);
        D = max(d);
        r = (D/2);
        Area = pi*(r^2);
        h = RAW(:,5);
        H = max(h);
        c1 = RAW(:,6);
        c1 = max(c1);
        c2 = RAW(:,7);
        c2 = max(c2);
        c3 = RAW(:,8);
        c3 = max(c3);
        Stress = N./Area;
        Strain = mm./H;
        I = [0, 0];
        XY = [I; Stress Strain];
        X = any(XY(:,2)<-0.5,2);
        XY(X,:) = [];
        STRESS = XY(:,1);
        STRAIN = XY(:,2);
        STRESS(isnan(STRESS(:))) = [];
        STRAIN(isnan(STRAIN(:))) = [];
        %plot(STRAIN(1:50:end), STRESS(1:50:end),'color',[c1 c2 c3],'LineWidth',1.5);
        p = polyfit(STRAIN,STRESS,4);
        x1 = -0.5:0.025:0;
        y1 = polyval(p,x1);
        plot(x1,y1,'color',[c1 c2 c3],'LineWidth',1.5);
        %plot(D, H, '*','color',[c1 c2 c3],'LineWidth',1.5);
        mSTRESS = min(y1);
    end

Answer 1

[Mathieu NOE]

hello

I don't see why the code needs two for loops

seems to me one suffice and this code gives the answer in mSTRESS (vector of same dimension as number of files, whatever the quantity, 3 or 16 , no need to specify it )

datafiles = dir('Synthetic_Pancreas_Sample_*.xlsx');
figure();
axis padded;
grid on;
xlabel('Strain', 'Interpreter', 'latex', 'fontsize', 16, 'Color', 'k');
ylabel('Stress / MPa', 'Interpreter', 'latex', 'fontsize', 16, 'Color', 'k');
set(gca, 'TickLabelInterpreter', 'latex', 'fontsize', 14);
lgd = legend('fontsize', 10, 'Interpreter', 'latex', 'location', 'Southeast');
title(lgd, 'Sample Number:');
legend('1', '2', '3', '4', '5', '6', '7', '8', '9', '10', ...
    '11','12','13','14','15','16');
hold on

    for k = 1:length(datafiles)
        
        currentfile = datafiles(k).name; 
        RAW = readmatrix(currentfile);
        t = RAW(:,1);
        mm = RAW(:,2);
        N = RAW(:,3);
        d = RAW(:,4);
        D = max(d);
        r = (D/2);
        Area = pi*(r^2);
        h = RAW(:,5);
        H = max(h);
        c1 = RAW(:,6);
        c1 = max(c1);
        c2 = RAW(:,7);
        c2 = max(c2);
        c3 = RAW(:,8);
        c3 = max(c3);
        Stress = N./Area;
        Strain = mm./H;
        I = [0, 0];
        XY = [I; Stress Strain];
        X = any(XY(:,2)<-0.5,2);
        XY(X,:) = [];
        STRESS = XY(:,1);
        STRAIN = XY(:,2);
        STRESS(isnan(STRESS(:))) = [];
        STRAIN(isnan(STRAIN(:))) = [];
        %plot(STRAIN(1:50:end), STRESS(1:50:end),'color',[c1 c2 c3],'LineWidth',1.5);
        p = polyfit(STRAIN,STRESS,4);
        x1 = -0.5:0.025:0;
        y1 = polyval(p,x1);
        plot(x1,y1,'color',[c1 c2 c3],'LineWidth',1.5);
        %plot(D, H, '*','color',[c1 c2 c3],'LineWidth',1.5);
        mSTRESS(k) = min(y1); % <= here
    end

[Fabian Hickson]

I asked, you answered. Appreciate that - thanks.

[Mathieu NOE]

My pleasure !