Filling areas between curves

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Michael Thomas
Michael Thomas 2020 年 2 月 19 日
回答済み: darova 2020 年 2 月 19 日
I'm relatively inexperienced with MATLAB and I'm having a hard time understanding using patch or fill. I would like to shade each of the four areas generated by the intersection of these two curves.
Here's my code:
load 'twopar2gga.dat'
x = twopar2gga;
x1 = x(4001:4137,1);
y1 = x(4001:4137,2);
x2 = x(4138:4201,1);
y2 = x(4138:4201,2);
hold on
plot(x1,y1,'k')
plot(x2,y2,'r')
axis([0 1 0 1])
If someone could expain how to do this that'd be amazing. I've attached my data.
  4 件のコメント
darova
darova 2020 年 2 月 19 日
Do you have polyxpoly?
Michael Thomas
Michael Thomas 2020 年 2 月 19 日
I didn't, but I do now

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採用された回答

the cyclist
the cyclist 2020 年 2 月 19 日
If you put this line after your line plotting, you'll see the basics of using the patch command:
patch([x1; x1(1)],[y1; y1(1)],'g')
The first two arguments are the (x,y) coordinates of the polygon that you want to fill. Notice how I "closed the loop" by appending the initial x1 and y1 points again, at the end of the vector. Here is the result:
I realize that that is not the exact area you want filled. But it gives you the idea.
To do what you really want, you'll need to similarly trace the outline of each of the polygons you want filled in, based on your input vectors. You'll need to combine some pieces of both x1 and x2 as input to patch, for some regions.

その他の回答 (1 件)

darova
darova 2020 年 2 月 19 日
Example
clc,clear
x1 = linspace(0,1.5);
y1 = sin(x1);
x2 = linspace(0,2);
y2 = cos(x1);
cla
[xc,yc] = polyxpoly(x1,y1,x2,y2); % find intersection point
ii1 = x1 > xc; % indices for the first curve
ii2 = x2 > xc; % indices for the second curve
patch([x1(ii1) flip(x2(ii2))], [y1(ii1) flip(y2(ii2))], 'r'); % fill right side
hold on
patch([x1(ii1) x2(~ii2)], [y1(ii1) y2(~ii2)], 'g'); % fill up/down
patch([x2(ii2) x1(~ii1)], [y2(ii2) y1(~ii1)], 'b'); % fill up/down
patch([x1(~ii1) flip(x2(~ii2))], [y1(~ii1) flip(y2(~ii2))], 'm'); % fill left side
hold off
The result
THe hole inside

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