# Specifically find the first substantial peak in a graph.

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Anthony 2023 年 6 月 5 日
コメント済み: Image Analyst 2023 年 6 月 6 日
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Anthony 2023 年 6 月 5 日
I want to find the value of the first occuring peak in this graph. How can I write a code that can specifcally do that for this graph or any graph I input into the code.

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### 回答 (2 件)

Diwakar Diwakar 2023 年 6 月 5 日
% Example usage
signal = [0, 0, 0, 1, 2, 3, 2, 1, 0, 0, 0, 0, 0, 0, 1, 2, 3, 2, 1, 0, 0, 0];
threshold = 0.5;
% Find the first peak in the signal above the threshold
localMaxima = islocalmax(signal);
peakIndex = find(localMaxima & signal > threshold, 1);
% Plot the signal
figure;
plot(signal, 'b-', 'LineWidth', 2);
hold on;
% Plot the threshold line
plot([1, numel(signal)], [threshold, threshold], 'r--', 'LineWidth', 1);
% Plot the first peak, if found
if ~isempty(peakIndex)
plot(peakIndex, signal(peakIndex), 'go', 'MarkerSize', 10, 'MarkerFaceColor', 'g');
legend('Signal', 'Threshold', 'First Peak');
else
legend('Signal', 'Threshold');
end
xlabel('Index');
ylabel('Signal Value');
title('Signal with First Substantial Peak');
hold off;
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Star Strider 2023 年 6 月 5 日
Perhaps something like this —
x = linspace(1.3, 3.3)*1E+4;
yfcn = @(x) x.*exp(-0.00025*x);
y = yfcn(x-x(1))/6 + [zeros(1,13) randn(1,87)*10];
[pks,locs] = findpeaks(y, 'NPeaks',1)
pks = 252.8592
locs = 18
figure
plot(x, y)
hold on
plot(x(locs), pks, 'rs')
hold off
legend('Data','First Peak','Location','best')
That should be robust to all your data sets.
.
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Star Strider 2023 年 6 月 5 日
It’s not stupid at all. The ‘yfcn’ anonymous function is my attempt to approximate the data you displayed. The relation essentially approximates, it, however I wanted to be as accurate as I could, so I wrote the function, allowing me to easily shift it to accommodate your independent variable values.
Using ‘NPeaks’,1 is the key here since it returns the first peak.
I noticed that you mentioned ‘substantial’ in the title. I would let 'Prominence' define ‘substantial’. You can return the prominence values for all the identified peaks (see the findpeaks documentation for that informaiton), and then choose the peak with the highest prominence, using the max functon with two outputs, the second output returning the index that you could then use to return the corresponding ‘locs’ value.
That would go something like this —
x = linspace(1.3, 3.3)*1E+4;
yfcn = @(x) x.*exp(-0.00025*x);
y = yfcn(x-x(1))/6 + [zeros(1,13) randn(1,87)*10];
[pks,locs,~,prm] = findpeaks(y);
[maxprom,idx] = max(prm)
maxprom = 262.3909
idx = 5
figure
plot(x, y)
hold on
plot(x(locs(idx)), pks(idx), 'rs')
hold off
legend('Data','Most Prominant Peak','Location','best')
.
To return only the first peak, us my original code.
Image Analyst 2023 年 6 月 6 日
@Anthony what is your definition of the first peak, out of all those peaks in the image. Is it the very first one, like islocalmax would return in the first element it returns? Or is it the peak of the noisy signal, as it you had denoised the signal? Or is it the overall max of the entire signal, like the red square in Star's last plot? Do you even want to denoise your signal, or do you want to use the original, noisy signal?
If you have any more questions, then attach your data and code to read it in with the paperclip icon after you read this:
and please indicate which (x,y) point on the curve you want to find as the "first peak".

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