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华夏千秋木虫 (小有名气)
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[求助]
算法求助，关于“极小范数最小二乘法” 已有1人参与
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| 遇到欠定线性方程组了，怎么用极小范数最小二乘法呢？各位师兄师姐学弟学妹们帮帮忙吧，能有代码最好了。 |
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mathstudy
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华夏千秋: 金币+5, ★有帮助 2014-03-31 22:16:47
华夏千秋: 金币+5, ★★★★★最佳答案 2014-07-13 17:51:10
感谢参与，应助指数 +1
华夏千秋: 金币+5, ★有帮助 2014-03-31 22:16:47
华夏千秋: 金币+5, ★★★★★最佳答案 2014-07-13 17:51:10
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在matlab 中可以直接调用 ”\ “矩阵左除运算 可以算最小二乘广义解 下面是关于 \ help 的内容 \ Backslash or left matrix divide. A\B is the matrix division of A into B, which is roughly the same as INV(A)*B , except it is computed in a different way. If A is an N-by-N matrix and B is a column vector with N components, or a matrix with several such columns, then X = A\B is the solution to the equation A*X = B. A warning message is printed if A is badly scaled or nearly singular. A\EYE(SIZE(A)) produces the inverse of A. If A is an M-by-N matrix with M < or > N and B is a column vector with M components, or a matrix with several such columns, then X = A\B is the solution in the least squares sense to the under- or overdetermined system of equations A*X = B. The effective rank, K, of A is determined from the QR decomposition with pivoting. A solution X is computed which has at most K nonzero components per column. If K < N this will usually not be the same solution as PINV(A)*B. A\EYE(SIZE(A)) produces a generalized inverse of A. |
2楼2014-03-08 15:55:56
