How to cut out specific segment from findpeaks function?
8 ビュー (過去 30 日間)
古いコメントを表示
Hi All,
I am trying to figure out how to cut out the specific segment of the signal where I draw the border using findpeaks function.
I am using example of findpeaks link below:
x = linspace(0,1,1000);
Pos = [1 2 3 5 7 8]/10;
Hgt = [4 4 2 2 2 3];
Wdt = [3 8 4 3 4 6]/100;
for n = 1:length(Pos)
Gauss(n,:) = Hgt(n)*exp(-((x - Pos(n))/Wdt(n)).^2);
end
PeakSig = sum(Gauss);
plot(x,Gauss,'--',x,PeakSig)
grid
Then, using findpeaks function,
findpeaks(PeakSig,x,'Annotate','extents','WidthReference','halfheight')
title('Signal Peak Widths')
I want to cut the segment of the border into 3 segment.
The method I am thinking is find the X value corresponding the Y value, which is not an effective way to do it.
0 件のコメント
採用された回答
Star Strider
2019 年 2 月 8 日
You can get the ‘Borders’ x-coordinates using:
(Previous Code)
findpeaks(PeakSig,x,'Annotate','extents','WidthReference','halfheight')
title('Signal Peak Widths')
Ax = gca;
Kids = Ax.Children;
Borders = Kids(1:2);
Line = Borders.XData;
You have to search the ‘Axes.Children’ result manually to find the ‘(Border)’ elements. There appears to be no way to get them automatically, or even to specifically search for a ‘Line (Border)’ or ‘Border’ entry in the ‘Kids’ variable. (There may be a way, but it would take more patience than I have tonight to find it. I’ve tried every cell and structure addressing approach I can think of or can find in the documentation, including using strcmp and related functions, and couldn’t get a good result.) It would be very nice if this was more straightforward.
16 件のコメント
Star Strider
2019 年 2 月 14 日
‘I am not sure can I plot only specific range and then find the pks in that specific range?’
You can. Starting with:
[pks,locs] = findpeaks(-avSpots,year)
(or something similar) then calculate the indices as as:
idx = find((year >= yr1) & (year <= yr2));
where ‘yr1’ is the beginning year and ‘yr2’ is the end year, then analyse your data as:
spots_sub = avSpots(idx);
year_sub = year(idx);
That should work (I did not test it).
その他の回答 (3 件)
Giacomo Echevers
2022 年 1 月 18 日
Well, I kind of needed exactly this feature so I ventured and fought with the findpeaks code, luckily I found what I needed. I attached the modified version of the findpeaks function (now called Findpeaks.m).
Note: It seems that because MATLAB recognized that this is a custom function, the plotting features are not working, but that's no issue, because you can simply use the orginial function and that'll do, this mod is just to extract the x position for the borders.
I'm going to include here a small example using the previous presented case.
x = linspace(0,1,1000);
Pos = [1 2 3 5 7 8]/10;
Hgt = [4 4 2 2 2 3];
Wdt = [3 8 4 3 4 6]/100;
for n = 1:length(Pos)
Gauss(n,:) = Hgt(n)*exp(-((x - Pos(n))/Wdt(n)).^2);
end
PeakSig = sum(Gauss);
findpeaks(PeakSig,x,'Annotate','extents','WidthReference','halfheight');
Now, let's use the modified version of the findpeaks function to get the borders x values. Please note the use of the pair parameters 'BorderExtraction' and logical 1. Basically, logical 1 to get the x values for borders in the loc output variable position and logical 0 to tell it to behave exactly the same as the original findpeaks function.
[~,X] = Findpeaks(PeakSig,x,'Annotate','extents','WidthReference','halfheight','BorderExtraction',1)
And now let's use this to get the second peak just for fun.
plot(PeakSig(x>=X(2) & x<=X(3)))
Hope this helps
A94 out.
2 件のコメント
Image Analyst
2022 年 1 月 18 日
Helps me anyway. I never knew findpeaks() had those nice annotation lines available for display.
Giacomo Echevers
2022 年 1 月 18 日
Glad to hear that! Just noticed those borders, literally yesterday and got quite triggered that it didn't have a native way to extract those x coordinates.
João
2019 年 2 月 7 日
Try this,
[pks,locs] = findpeaks(PeakSig,x,'Annotate','extents','WidthReference','halfheight')
The variable locs have the indexes of the peaks found. Use them like this to remove a specific peak (your case the third):
PeakSig(locs(3)) = [];
3 件のコメント
João
2019 年 2 月 8 日
I didn't notice your were using x in findpeaks. You need to find the index that corresponds to locs. Do it like this then:
[pks,locs] = findpeaks(PeakSig,x,'Annotate','extents','WidthReference','halfheight')
peakToFind = 1; % the peak you want to delete (in this case the first one)
% where you get the postion in x corresponding to locs
idx_x = find(x==locs(peakToFind));
PeakSig(idx_x) = [];
Image Analyst
2019 年 2 月 9 日
The borders appear to be available in findpeaks() at line 558 (in R2018b version):
% get the x-coordinates of the half-height width borders of each peak
[wxPk,iLBh,iRBh] = getPeakWidth(yFinite,x,iPk,bPk,iLB,iRB,refW);
See it with edit, then search for border
>> edit findpeaks.m
Make a copy of findpeaks.m and call it findpeaksandborders() in some utilities folder on your path with all your other m-files. BE SURE NOT TO ALTER THE ORIGINAL ONE!!!
Then have findpeaksandborders() return the borders (iLB and iRB) in the list of output arguments.
7 件のコメント
参考
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!