full width of a peak
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Mahsa Rm
le 7 Août 2021
Commenté : Martina Khunova
le 23 Avr 2023
Hi, I have an eog signal which I want to know the duration of its blinks. I want to obtain starting and ending point of a peak , I have used findpeaks function but it gives halfheight width. how can I find full durations?
any help would be appreciated.
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Image Analyst
le 7 Août 2021
Try this:
% Find both sides of a peak.
clc; % Clear the command window.
close all; % Close all figures (except those of imtool.)
clear; % Erase all existing variables. Or clearvars if you want.
workspace; % Make sure the workspace panel is showing.
format long g;
format compact;
fontSize = 15;
%==================================================================================
% % Mahsa forgot to upload data so I'm forced
% to try to recreate it from the image.
rgbImage = imread('image.png');
imshow(rgbImage)
axis on
grayImage = rgb2gray(rgbImage);
imshow(grayImage);
impixelinfo
mask = grayImage == 255;
props = regionprops(mask, 'BoundingBox');
% Shrink by 8 to get away from tick marks.
bb = props.BoundingBox
bb(1) = bb(1) + 8;
bb(2) = bb(2) + 8;
bb(3) = bb(3) - 16;
bb(4) = bb(4) - 16;
grayImage = imcrop(grayImage, bb);
imshow(grayImage);
[rows, columns] = size(grayImage);
y = zeros(columns, 1);
for col = 1 : columns
t = find(grayImage(:, col) < 255, 1, 'first');
if ~isempty(t)
y(col) = rows - t;
end
end
figure
x = linspace(2.54e5, 2.69e5, length(y));
% Scale y to -50 to +275
y = rescale(y, -50, 275);
plot(x, y, 'b-', 'LineWidth', 2);
grid on;
%==================================================================================
% Now we have our data and can begin.
[peakValues, indexesOfPeaks] = findpeaks(y, 'MinPeakHeight', 15);
numPeaks = length(peakValues);
peakLeft = ones(1, numPeaks);
peakRight = ones(1, numPeaks);
% Find valley to the left of the peak
hold on;
for k = 1 : numPeaks
plot(x(indexesOfPeaks(k)), y(indexesOfPeaks(k)), 'r+', 'LineWidth', 2, 'MarkerSize', 20);
for k2 = indexesOfPeaks(k) : -1 : 1
if y(k2) > y(k2+1)
break;
end
end
peakLeft(k) = k2;
end
% Find valley to the right of the peak
for k = 1 : numPeaks
for k2 = indexesOfPeaks(k) : length(y)-1
if y(k2+1) > y(k2)
break;
end
end
peakRight(k) = k2;
end
% If you want locations in terms of x instead of indexes:
xLeft = x(peakLeft);
xRight = x(peakRight);
% Plot red lines on the left and red lines on the right.
for k = 1 : length(xLeft)
xline(xLeft(k), 'Color', 'g', 'LineWidth', 3);
xline(xRight(k), 'Color', 'r', 'LineWidth', 3);
end
1 commentaire
Martina Khunova
le 23 Avr 2023
Can you please take a look at this?
Thank you :)
Plus de réponses (1)
Image Analyst
le 7 Août 2021
Have you tried just locating the peaks, and then "falling down" the slope on each side until the signal turns around? This would work for smooth peaks but not those that had lots of little tiny "shoulder peaks" on the main peak, but I don't see any of those in your signal.
Attaching your signal in a .mat file would have been helpful, but you forgot.
2 commentaires
Image Analyst
le 7 Août 2021
Modifié(e) : Image Analyst
le 7 Août 2021
Enclose your file in a .zip file and attach it with the paperclip icon. Something like (untested)
[peakValues, indexesOfPeaks] = findpeaks(y);
numPeaks = length(peakValues);
peakLeft = ones(1, numPeaks);
peakRight = ones(1, numPeaks);
% Find valley to the left of the peak
for k = 1 : numPeaks
for k2 = indexesOfPeaks(k) : -1 : 1
if y(k2) > y(k2+1)
break;
end
end
peakLeft(k) = k2;
end
% Find valley to the right of the peak
for k = 1 : numPeaks
for k2 = indexesOfPeaks(k) : length(y)
if y(k2+1) > y(k2)
break;
end
end
peakRight(k) = k2;
end
Now you have arrays giving the left and right indexes of the peaks.
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