在 R 的 nnls 中实现附加约束
同时最小化 x 的“能量”：
||A x - b||^2 + m*||x||^2最小化“x 导数中的能量”
||A x - b||^2 + m ||H x||^2，其中 H 是恒等矩阵与第一个非对角线上为 -1 的矩阵之和最一般地，最小化
||A x - b||^2 + m ||H x - f||^2。
有没有办法通过一些巧妙的方式重述问题 1.-3 来诱导 nnls 做到这一点？多于？我之所以对这样的事情抱有希望，是因为 Whitall 等人（对付费墙表示抱歉）声称“幸运的是，NNLS 可以采用上面的原始形式来解决有问题的问题3”。
I am using the R interface to the Lawson-Hanson NNLS implementation of an algorithm for non-negative linear least squares that solves ||A x - b||^2 with the constraint that all elements of vector x ≥ 0. This works fine but I would like to add further constrains. Of interest to me are:
Also minimize "energy" of x:
||A x - b||^2 + m*||x||^2Minimize "energy in the x derivative"
||A x - b||^2 + m ||H x||^2, where H is the sum of identity and a matrix with -1 on the first off-diagonalMost generally, minimize
||A x - b||^2 + m ||H x - f||^2.
Is there are a way to coax nnls to do this by some clever way of restating the problems 1.-3. Above? The reason I have hope for such a thing is that there is a little-throw away comment in a paper by Whitall et al (sorry for the paywall) that claims that "fortunately, NNLS can be adopted from the original form above to accommodate something in problem 3".
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我认为 m 是一个标量，对吗？考虑简单的情况 m=1；您可以通过令 H* = sqrt(m) H 和 f* = sqrt(m) f 并使用此处给出的求解方法来推广 m 的其他值。
所以现在你试图最小化 ||A x - b||^2 + ||H x - f||^2。
设 A* = [A' | H']' 并令 b* = [b' | H']' f']'（即将A堆叠在H之上，b堆叠在f之上）并解决原来的问题
||A* x - b*||^2 上的非负线性最小二乘，约束条件是向量 x ≥ 0 。
I take it m is a scalar, right? Consider the simple case m=1; you can generalize for other values of m by letting H* = sqrt(m) H and f* = sqrt(m) f and using the solution method given here.
So now you're trying to minimise ||A x - b||^2 + ||H x - f||^2.
Let A* = [A' | H']' and let b* = [b' | f']' (i.e. stack up A on top of H and b on top of f) and solve the original problem of
non-negative linear least squares on ||A* x - b*||^2 with the constraint that all elements of vector x ≥ 0 .