我有我的 3D 数据 X、Y、Z（大小为 NxM 的矩阵）
我想把它安装到我所做的最合适的平面上：
X = X(isfinite(X));% deleting the NaN because svd Doesn't accept them
Y = Y(isfinite(Y));
Z = Z(isfinite(Z));
G = [X,Y,Z,ones(size(X(:)))];
[u s v] = svd(G,0);
P = v(:,4);
scalar = 2*P./P(1);
P = P./scalar; % supposed to be my plane equation but there is something wrong
然后从 X 和 Y 重新计算 Z
Z = -(P(1)*X + P(2)*Y + P(4)) / P(3);
不知道是什么问题！！
好的，所以目标是对您的数据进行平面拟合。我们假设某些数组中可能有 NaN 元素。因此，让我们在适合数据的变量中做一个错误。
% First, remove NaNs. All the NaN (if any) will be in the same places,
% so we need do only one test. Testing each of X,Y,Z leads to confusion
% and leads to the expectation that the NaNs are NOT in the same places
% with some potential.
k = isfinite(X);
if all(k(:))
% in this case there were no nans, so reshape the arrays into vectors
% While X(k) would convert an array into a column vector anyway in
% this case, it seems far more sane to do the reshape explicitly,
% rather than let X(k) do it implicitly. Again, a source of ambiguity
% removed.
X = X(:);
Y = Y(:);
Z = Z(:);
else
X = X(k);
Y = Y(k);
Z = Z(k);
end
% Combine X,Y,Z into one array
XYZ = [X,Y,Z];
% column means. This will allow us to do the fit properly with no
% constant term needed. It is necessary to do it this way, as
% otherwise the fit would not be properly scale independent
cm = mean(XYZ,1);
% subtract off the column means
XYZ0 = bsxfun(@minus,XYZ,cm);
% The "regression" as a planar fit is now accomplished by SVD.
% This presumes errors in all three variables. In fact, it makes
% presumptions that the noise variance is the same for all the
% variables. Be very careful, as this fact is built into the model.
% If your goal is merely to fit z(x,y), where x and y were known
% and only z had errors in it, then this is the wrong way
% to do the fit.
[U,S,V] = svd(XYZ0,0);
% The singular values are ordered in decreasing order for svd.
% The vector corresponding to the zero singular value tells us
% the direction of the normal vector to the plane. Note that if
% your data actually fell on a straight line, this will be a problem
% as then there are two vectors normal to your data, so no plane fit.
% LOOK at the values on the diagonal of S. If the last one is NOT
% essentially small compared to the others, then you have a problem
% here. (If it is numerically zero, then the points fell exactly in
% a plane, with no noise.)
diag(S)
% Assuming that S(3,3) was small compared to S(1,1), AND that S(2,2)
% is significantly larger than S(3,3), then we are ok to proceed.
% See that if the second and third singular values are roughly equal,
% this would indicate a points on a line, not a plane.
% You do need to check these numbers, as they will be indicative of a
% potential problem.
% Finally, the magnitude of S(3,3) would be a measure of the noise
% in your regression. It is a measure of the deviations from your
% fitted plane.
% The normal vector is given by the third singular vector, so the
% third (well, last in general) column of V. I'll call the normal
% vector P to be consistent with the question notation.
P = V(:,3);
% The equation of the plane for ANY point on the plane [x,y,z]
% is given by
%
% dot(P,[x,y,z] - cm) == 0
%
% Essentially this means we subtract off the column mean from our
% point, and then take the dot product with the normal vector. That
% must yield zero for a point on the plane. We can also think of it
% in a different way, that if a point were to lie OFF the plane,
% then this dot product would see some projection along the normal
% vector.
%
% So if your goal is now to predict Z, as a function of X and Y,
% we simply expand that dot product.
%
% dot(P,[x,y,z]) - dot(P,cm) = 0
%
% P(1)*X + P(2)*Y + P(3)*Z - dot(P,cm) = 0
%
% or simply (assuming P(3), the coefficient of Z) is not zero...
Zhat = (dot(P,cm) - P(1)*X) - P(2)*Y)/P(3);
请注意，如果 Z 的系数很小，那么结果预测将对该系数中的小扰动非常敏感。从本质上讲，这意味着该平面是垂直平面，或者几乎是垂直平面。
同样，所有这些都假设您的目标确实是变量计算中的错误，其中我们在所有三个变量 X、Y、Z 中都有噪声。如果噪声仅在 Z 中，则正确的解决方案要简单得多。