曲线拟合:在 Matlab 中将具有不同 length/number 点的多条曲线合并为一条 3D 曲线
curve fitting: a number of curves with different length/number of points into one curve in 3D in Matlab
假设我有许多不同长度的曲线(每条曲线中的点数和点距离都不同)。我能否在 3D space 中找到最适合这组线条的曲线?
Matlab 中的代码示例将不胜感激。
示例数据集:
第一条曲线有10个点。
18.5860 18.4683 18.3576 18.2491 18.0844 17.9016 17.7709 17.6401 17.4617 17.2726
91.6178 91.5711 91.5580 91.5580 91.5701 91.6130 91.5746 91.5050 91.3993 91.2977
90.6253 91.1090 91.5964 92.0845 92.5565 93.0199 93.5010 93.9785 94.4335 94.8851
第二条曲线有8个点
15.2091 15.0894 14.9765 14.8567 14.7360 14.6144 14.4695 14.3017
90.1138 89.9824 89.8683 89.7716 89.6889 89.6040 89.4928 89.3624
99.4393 99.9066 100.3802 100.8559 101.3340 101.8115 102.2770 102.7296
想要的曲线是可以代表这两条存在曲线的曲线。
我想把这些曲线做成点散布并从中拟合出一条线。但是我只能从网上的许多代码片段中得到一条直线。
我是不是遗漏了什么或者有人可以提供一些提示。谢谢。
如果没有更多细节,很难想出万无一失的解决方案,但这里有一种适用于所提供示例数据的方法。我找到了所有点的最佳拟合线,然后参数化了沿着该最佳拟合线的所有点。然后我分别对每个维度进行了最小二乘多项式拟合。这产生了一个三维参数曲线,似乎很好地拟合了数据。
请注意,多项式最小二乘法以外的曲线拟合方法可能更适合某些情况——只需将首选拟合函数替换为 polyfit 和 polyval。
希望对您有所帮助!
clear;
close all;
pts1=[18.5860 18.4683 18.3576 18.2491 18.0844 17.9016 17.7709 17.6401 17.4617 17.2726;
91.6178 91.5711 91.5580 91.5580 91.5701 91.6130 91.5746 91.5050 91.3993 91.2977;
90.6253 91.1090 91.5964 92.0845 92.5565 93.0199 93.5010 93.9785 94.4335 94.8851]';
pts2=[ 15.2091 15.0894 14.9765 14.8567 14.7360 14.6144 14.4695 14.3017;
90.1138 89.9824 89.8683 89.7716 89.6889 89.6040 89.4928 89.3624;
99.4393 99.9066 100.3802 100.8559 101.3340 101.8115 102.2770 102.7296]';
%Combine all of our curves into a single point cloud
X = [pts1;pts2];
%=======================================================
%We want to first find the line of best fit
%This line will provide a parameterization of the points
%See accepted answer to
% calculate centroid
x0 = mean(X)';
% form matrix A of translated points
A = [(X(:, 1) - x0(1)) (X(:, 2) - x0(2)) (X(:, 3) - x0(3))];
% calculate the SVD of A
[~, S, V] = svd(A, 0);
% find the largest singular value in S and extract from V the
% corresponding right singular vector
[s, i] = max(diag(S));
a = V(:, i);
%=======================================================
a=a / norm(a);
%OK now 'a' is a unit vector pointing along the line of best fit.
%Now we need to compute a new variable, 't', for each point in the cloud
%This 't' value will parameterize the curve of best fit.
%Essentially what we're doing here is taking the dot product of each
%shifted point (contained in A) with the normal vector 'a'
t = A * a;
tMin = min(t);
tMax = max(t);
%This variable represents the order of our polynomial fit
%Use the smallest number that produces a satisfactory result
polyOrder = 8;
%Polynomial fit all three dimensions separately against t
pX = polyfit(t,X(:,1),polyOrder);
pY = polyfit(t,X(:,2),polyOrder);
pZ = polyfit(t,X(:,3),polyOrder);
%And that's our curve fit: (pX(t),pY(t),pZ(t))
%Now let's plot it.
tFine = tMin:.01:tMax;
fitXFine = polyval(pX,tFine);
fitYFine = polyval(pY,tFine);
fitZFine = polyval(pZ,tFine);
figure;
scatter3(X(:,1),X(:,2),X(:,3));
hold on;
plot3(fitXFine,fitYFine,fitZFine);
hold off;
假设我有许多不同长度的曲线(每条曲线中的点数和点距离都不同)。我能否在 3D space 中找到最适合这组线条的曲线?
Matlab 中的代码示例将不胜感激。
示例数据集:
第一条曲线有10个点。
18.5860 18.4683 18.3576 18.2491 18.0844 17.9016 17.7709 17.6401 17.4617 17.2726
91.6178 91.5711 91.5580 91.5580 91.5701 91.6130 91.5746 91.5050 91.3993 91.2977
90.6253 91.1090 91.5964 92.0845 92.5565 93.0199 93.5010 93.9785 94.4335 94.8851
第二条曲线有8个点
15.2091 15.0894 14.9765 14.8567 14.7360 14.6144 14.4695 14.3017
90.1138 89.9824 89.8683 89.7716 89.6889 89.6040 89.4928 89.3624
99.4393 99.9066 100.3802 100.8559 101.3340 101.8115 102.2770 102.7296
想要的曲线是可以代表这两条存在曲线的曲线。
我想把这些曲线做成点散布并从中拟合出一条线。但是我只能从网上的许多代码片段中得到一条直线。
我是不是遗漏了什么或者有人可以提供一些提示。谢谢。
如果没有更多细节,很难想出万无一失的解决方案,但这里有一种适用于所提供示例数据的方法。我找到了所有点的最佳拟合线,然后参数化了沿着该最佳拟合线的所有点。然后我分别对每个维度进行了最小二乘多项式拟合。这产生了一个三维参数曲线,似乎很好地拟合了数据。 请注意,多项式最小二乘法以外的曲线拟合方法可能更适合某些情况——只需将首选拟合函数替换为 polyfit 和 polyval。
希望对您有所帮助!
clear;
close all;
pts1=[18.5860 18.4683 18.3576 18.2491 18.0844 17.9016 17.7709 17.6401 17.4617 17.2726;
91.6178 91.5711 91.5580 91.5580 91.5701 91.6130 91.5746 91.5050 91.3993 91.2977;
90.6253 91.1090 91.5964 92.0845 92.5565 93.0199 93.5010 93.9785 94.4335 94.8851]';
pts2=[ 15.2091 15.0894 14.9765 14.8567 14.7360 14.6144 14.4695 14.3017;
90.1138 89.9824 89.8683 89.7716 89.6889 89.6040 89.4928 89.3624;
99.4393 99.9066 100.3802 100.8559 101.3340 101.8115 102.2770 102.7296]';
%Combine all of our curves into a single point cloud
X = [pts1;pts2];
%=======================================================
%We want to first find the line of best fit
%This line will provide a parameterization of the points
%See accepted answer to
% calculate centroid
x0 = mean(X)';
% form matrix A of translated points
A = [(X(:, 1) - x0(1)) (X(:, 2) - x0(2)) (X(:, 3) - x0(3))];
% calculate the SVD of A
[~, S, V] = svd(A, 0);
% find the largest singular value in S and extract from V the
% corresponding right singular vector
[s, i] = max(diag(S));
a = V(:, i);
%=======================================================
a=a / norm(a);
%OK now 'a' is a unit vector pointing along the line of best fit.
%Now we need to compute a new variable, 't', for each point in the cloud
%This 't' value will parameterize the curve of best fit.
%Essentially what we're doing here is taking the dot product of each
%shifted point (contained in A) with the normal vector 'a'
t = A * a;
tMin = min(t);
tMax = max(t);
%This variable represents the order of our polynomial fit
%Use the smallest number that produces a satisfactory result
polyOrder = 8;
%Polynomial fit all three dimensions separately against t
pX = polyfit(t,X(:,1),polyOrder);
pY = polyfit(t,X(:,2),polyOrder);
pZ = polyfit(t,X(:,3),polyOrder);
%And that's our curve fit: (pX(t),pY(t),pZ(t))
%Now let's plot it.
tFine = tMin:.01:tMax;
fitXFine = polyval(pX,tFine);
fitYFine = polyval(pY,tFine);
fitZFine = polyval(pZ,tFine);
figure;
scatter3(X(:,1),X(:,2),X(:,3));
hold on;
plot3(fitXFine,fitYFine,fitZFine);
hold off;