我设法获得轮廓的线条(或边缘),所以我有每条线的限制:
我想要的是将第二个图中外部蓝线内轮廓的z轴中的所有级别相加,以将其与蓝线外部的值之和进行比较。有没有办法做到这一点?我到目前为止的代码是:
C = contourc(f, t, abs(tfr));
%extracts info from contour
sz = size(C,2); % Size of the contour matrix c
ii = 1; % Index to keep track of current location
jj = 1; % Counter to keep track of # of contour lines
while ii < sz % While we haven't exhausted the array
n = C(2,ii); % How many points in this contour?
s(jj).v = C(1,ii); % Value of the contour
s(jj).x = C(1,ii+1:ii+n); % X coordinates
s(jj).y = C(2,ii+1:ii+n); % Y coordinates
ii = ii + n + 1; % Skip ahead to next contour line
jj = jj + 1; % Increment number of contours
end
答案 0 :(得分:2)
因此,在运行问题中的代码后,您将获得数组S
中每个轮廓的坐标。假设您有f
和t
或类似形式的变量f = 94:0.1:101
和t = 0:1000
,并且您希望求和的值为abs(tfr)
,那么您应该能够使用
[fg, tg] = meshgrid(f,t)
is_inside = inpolygon(fg,tg, S(1).x, S(1).y)
integral = sum(abs(tfr(is_inside))
同样适用于S
中的其他条目。有关更多示例,请参阅inpolygon的帮助。您可以将~is_inside
用于曲线外的点