integrate a partial area under a plotted curve

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amirouche oumaziz 2020-2-24

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https://ww2.mathworks.cn/matlabcentral/answers/507270-integrate-a-partial-area-under-a-plotted-curve

评论： Star Strider 2020-2-25

采纳的回答： Star Strider

Hello every one,

I'm trying to calculate a partial area under a curve.

I've seen some asked questions about the subsject, but still struggling with some points.

I plotted a cuvre using numerical column vectors (electrical current versus time), then using this proposed solution, i tried to calculate the area under the curve:

yi = interp1(x, y, [x1:x2]); % x1 and x2 are the boundaries

shaded_area = trapz([4:5],yi+5)

Where y represents my electrical current vector and x my time vector, which contains by the way negative values, but i don't think that causes any problem.

When i execute the first line, it returns 'Nan' values!! So, I tried other column vectors which are extracted from the former ones (x and y) but i took only the positive values and again it returns NaN.

I thinking about extrapolating problem, but i have non idea about how to start with this issu?

It'll be very helpful if some one could give me some advice. Thank you!

Please find the attached source data file, function allowing importing data and the commands file (in the commands file, i changed the name of the dataset).

Using Matlab_2018b

https://fr.mathworks.com/matlabcentral/answers/314577-how-to-calculate-partial-area-under-a-graph

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amirouche oumaziz 2020-2-25

they are the x axis boundaries, indexMax is the index of the cell corresponding to the max. value of my Y axis (ifuse2

here).

darova 2020-2-25

I think these are the same values?

endingIndex = find(x==x(indexMax));

I suggest you use this instruction

% startingIndex = find(x==-2*1e-6);
startingIndex = find(x>-2e-6,1,'first');

Why not '=='? Example

>> a = [1 2 1/3];
ix = find(a==0.333)
ix =
   Empty matrix: 1-by-0

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采纳的回答

Star Strider 2020-2-25

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此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/507270-integrate-a-partial-area-under-a-plotted-curve#answer_417286

To get the area, add these lines to your code:

[TF,Q0] = ischange(ifuse, 'linear', 'Threshold', 1.5E+6);
[vl,vli] = min(ifuse(TF));
[pk,pki] = max(ifuse(TF));
Idx = find(TF);
idxrng = Idx(vli):Idx(pki);
Int_ifuse_pk = trapz(ox_axis1(idxrng), ifuse(idxrng))
figure
plot(ox_axis1,ifuse,'r') 
hold on
% plot(ox_axis1(TF),ifuse(TF),'+g')
patch([ox_axis1(idxrng); flipud(ox_axis1(idxrng))], [ifuse(idxrng); ones(size(ifuse(idxrng)))*ifuse(Idx(vli))], 'k', 'FaceAlpha',0.25)
hold off
([min(ox_axis1) max(ox_axis1)])
text(max(ox_axis1(idxrng)), mean(ifuse(idxrng)), sprintf('\\leftarrow Area = %.3f', Int_ifuse_pk), 'HorizontalAlignment','left')

producing the desired area (as ‘Int_ifuse_pk’), and this plot:

Experiment with this to get different results.

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amirouche oumaziz 2020-2-25

Thank you, very simple!

Though, i do have a little information concerning the value 1.5E6.

When i try with another data file (still the same kind of shape), it doesn't calculates the area, so i lowered the threshould value to 1.5E4, and it worked.

It may be helpful for someone who wants to try this!

Amir

Star Strider 2020-2-25

The best 'Threshold' parameter value may vary with different data. A value of 1.5E4 could work for all of them.

The ischange call just has to identify the beginning and peak of the pulse, with the min and max detection steps finding the correct indices, regardless of how many changes ischange returns.

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