MATLAB: Problem to find inflection point

index exceeds array boundsinflection point

Hi, i previously follow the guidelines and tutorial provided by Stryker to solve the inflection points problem. At first, i used 19 values in an array and it come out fine. However, when i have 20 values inside my array, i got an error message: "Index exceeds array bounds". Can someone help me? i really appreciate any help.

My array with 20 values inside:

xNew = [-433.08032 -422.60339 -435.8176 -447.13495 -458.09866 -469.62628 -476.03198 -478.79868 -482.30139 -481.55859 -479.44727 -474.50156 -472.0987 -463.40927 -456.15408 -449.5589 -442.93631 -442.58139 -445.34409 -447.53372];

yNew = [1443.96497 1399.23254 1369.25476 1351.823 1329.43958 1299.13647 1268.26343 1234.25989 1201.34021 1174.68372 1145.37134 1118.427 1089.32715 1061.75366 1034.36926 1008.59052 982.44653 956.20862 933.88422 905.15338];

Tutorial and guidelines followed previously: https://ww2.mathworks.cn/matlabcentral/answers/383721-how-to-find-inflection-point-please-help#answer_306101

Best Answer

Note that with your data, in order to calculate the inflection points correctly, you need to reverse ‘x’ and ‘y’ in my code:

x = yNew;
y = xNew;

The index range worked for the original example (How to find inflection point, PLEASE HELP!), however it needs to be modified in the event that it over-writes the end of the vector.

Try this:

dydx = gradient(y) ./ gradient(x);                                      % Derivative Of Unevenly-Sampled Data
zci = @(v) find(v(:).*circshift(v(:), [-1 0]) <= 0);                    % Returns Approximate Zero-Crossing Indices Of Argument Vector
zxidx = zci(dydx);                                                      % Approximate Indices Where ‘dydx=0’
for k1 = 1:numel(zxidx)                                                 % Loop Finds ‘x’ & ‘y’ For ‘dydx=0’
    ixrng = max(zxidx(k1)-2,1):min(zxidx(k1)+2,numel(x));
    inflptx(k1) = interp1(dydx(ixrng), x(ixrng), 0, 'linear');
    inflpty(k1) = interp1(x(ixrng), y(ixrng), inflptx(k1), 'linear');
end
figure(1)
plot(x, y)
hold on
plot(x, dydx*10)
plot(inflptx, inflpty, 'pg', 'MarkerFaceColor','g')
hold off
grid
legend('Data', 'Derivative', 'Inflection Points', 'Location','best')
inflpts = sprintfc('(%5.3f,%5.3f)', [inflptx; inflpty].');
text(inflptx, inflpty, inflpts,'FontSize',8, 'HorizontalAlignment','center', 'VerticalAlignment','bottom')

I edited my original post to include this change (‘ixrng’).

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MATLAB: How to find inflection point, PLEASE HELP!

Inflection points are defined where the curve changes direction, and the derivative is equal to zero.

See if this does what you want:

x = [ 7.0   7.2   7.4    7.6    8.4    8.8   9.2    9.6  10.0];
y = [ 0.692 0.719  0.723   0.732  0.719   0.712    1.407  1.714  1.99];
dydx = gradient(y) ./ gradient(x);                                      % Derivative Of Unevenly-Sampled Data
zci = @(v) find(v(:).*circshift(v(:), [-1 0]) <= 0);                    % Returns Approximate Zero-Crossing Indices Of Argument Vector
zxidx = zci(dydx);                                                      % Approximate Indices Where ‘dydx=0’
for k1 = 1:numel(zxidx)                                                 % Loop Finds ‘x’ & ‘y’ For ‘dydx=0’

    inflptx(k1) = interp1(dydx(zxidx(k1):zxidx(k1)+1), x(zxidx(k1):zxidx(k1)+1), 0, 'linear');
    inflpty(k1) = interp1(x(zxidx(k1):zxidx(k1)+1), y(zxidx(k1):zxidx(k1)+1), inflptx(k1), 'linear');
end
figure(1)
plot(x, y)
hold on
plot(x, dydx)
plot(inflptx, inflpty, 'pg', 'MarkerFaceColor','g')
hold off
grid
legend('Data', 'Derivative', 'Inflection Points')
inflpts = regexp(sprintf('(%5.3f,%5.3f)\n', [inflptx; inflpty]), '\n', 'split');
text(inflptx, inflpty, inflpts(1:end-1),'FontSize',8, 'HorizontalAlignment','center', 'VerticalAlignment','bottom')

The ‘inflptx’ and ‘inflpty’ vectors (derived from a linear interpolation of the derivative) are the coordinates of the inflection points. They are approximate because your data are discrete. The code does not find an inflection point where what is apparently a spline interpolation might create one, because that is not in your original data.

EDIT — (8 Jun 2020 at 12:21)

While the original code works for the original data here, if the identified zero-crossings of the derivative are close to the ends of the vector, it will crash with indices that exceed the number of vector elements.

This version of the loop fixes that:

for k1 = 1:numel(zxidx)                                                 % Loop Finds ‘x’ & ‘y’ For ‘dydx=0’
    ixrng = max(zxidx(k1)-2,1):min(zxidx(k1)+2,numel(x));
    inflptx(k1) = interp1(dydx(ixrng), x(ixrng), 0, 'linear');
    inflpty(k1) = interp1(x(ixrng), y(ixrng), inflptx(k1), 'linear');
end

The ‘inflpts’ assignment also needs to be modified:

inflpts = sprintfc('(%5.3f, %5.3f)', [inflptx; inflpty].');
text(inflptx, inflpty, inflpts,'FontSize',8, 'HorizontalAlignment','center', 'VerticalAlignment','bottom')

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