用特征值求解最小二乘方程的正确方法是什么？ [英] What is the correct way to solve a least square equation with eigen?

问题描述

我有以下简单的示例执行最小二乘，但出现以下断言错误。

I have the following simple example to perform a least square, but got the following assertion error.

断言失败：（i> = 0）&& （（（（BlockRows == 1）&&
（BlockCols == XprType :: ColsAtCompileTime）&& i

Assertion failed: (i>=0) && ( ((BlockRows==1) && (BlockCols==XprType::ColsAtCompileTime) && i

什么是正确的方法吗？

typedef Eigen::ArrayXd arr;
typedef Eigen::ArrayXXd arr2;

arr2 A(3, 3);
arr B(3);
A << 1, 0, 0, 0, 1, 0, 0, 0, 1;
B << 1, 2, 3;
auto x = A.matrix().colPivHouseholderQr().solve(B.matrix());

推荐答案

正如我的评论所述，问题在于 x 是一个抽象表达式，它存储对QR对象的引用，但是在最后一行之后，它将引用一个死对象，然后任何事情都可能发生！

As said in my comment, the problem is that x is an abstract expression storing a reference to the QR object, but right after the last line it will reference a dead object and then anything can happen!

更确切地说， A.matrix（）。colPivHouseholderQr（）创建一个临时对象，我们称其为 tmp_qr 。然后 tmp_qr.solve（B）创建另一个对象，该对象变为 Solve<类型的 x 。 ..> 。该对象实际上存储了两个r参考：一个到 tmp_qr ，另一个到 B 。在此行之后，临时对象 tmp_qr 被删除，因此 Solve< ...> 对象 x 有一个无效引用。就像引用已删除缓冲区的指针一样。最后，如果您稍后 x ，例如：

More precisely, A.matrix().colPivHouseholderQr() creates a temporary object, let's call it tmp_qr. Then tmp_qr.solve(B) creates another object which becomes x of type Solve<...>. This object essentially stores two references: one to tmp_qr, and one to B. Right after this line, the temporary object tmp_qr is deleted, so the Solve<...> object x has one dead reference. It is like a pointer referencing a deleted buffer. Finally, if you x later, like:

VectorXd y = x;

operator = 将触发使用 x 引用的QR分解和 x <引用的右侧 B / code>也是，但是，等等... QR分解对象已被删除，因此充其量您会得到一个段错误。

operator= will trigger the solve operation using the QR decomposition referenced by x and the right-hand-side B referenced by x too, but, wait... the QR decomposition object has been deleted, so at best you'll get a segfault.

因此，解决方案是这样写的：

So the solution is to write:

VectorXd x = A.matrix().colPivHouseholderQr().solve(B.matrix());

请参阅Eigen的文档详细了解自动如果您不知道得到什么，将很危险。

See Eigen's doc for more details on how auto is dangerous if you don't know what you get.

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