使用LAPACK访问子矩阵 [英] Accessing submatrices using LAPACK

问题描述

在LAPACK中有一个函数，它会给我一个特定子矩阵的元素吗？

Is there a function in LAPACK, which will give me the elements of a particular submatrix? If so how what is the syntax in C++?

或者我需要编写它吗？

推荐答案

没有访问子矩阵的功能。但是，由于矩阵数据存储在LAPACK例程中的方式，您不需要一个。这节省了大量的复制，并且由于这个原因（部分地）选择了数据布局：

There is no function for accessing a submatrix. However, because of the way matrix data is stored in LAPACK routines, you don't need one. This saves a lot of copying, and the data layout was (partially) chosen for this reason:

回想一下密集的（即不是带状，三角形， ）矩阵由四个值定义：

Recall that a dense (i.e., not banded, triangular, hermitian, etc) matrix in LAPACK is defined by four values:

• 指向矩阵左上角的指针


• 矩阵中的行数


• 矩阵中的列数


• 矩阵的通常这是内存中相邻元素之间的距离。

• a pointer to the top left corner of the matrix
• the number of rows in the matrix
• the number of columns in the matrix
• the "leading dimension" of the matrix; typically this is the distance in memory between adjacent elements of a row.

大多数时候，大多数人只使用领先维度等于行数; 3×3矩阵通常存储如下：

Most of the time, most people only ever use a leading dimension that is equal to the number of rows; a 3x3 matrix is typically stored like so:

a[0] a[3] a[6] 
a[1] a[4] a[7]
a[2] a[5] a[8]

$ b b

假设我们想要一个3x3的子矩阵，该矩阵的前导维 lda 。假设我们特别想要左上角位于（15,42）的3x3子矩阵：

Suppose instead that we wanted a 3x3 submatrix of a huge matrix with leading dimension lda. Suppose we specifically want the 3x3 submatrix whose top-left corner is located at a(15,42):

         .             .            .
         .             .            .
... a[15+42*lda] a[15+43*lda] a[15+44*lda] ...
... a[16+42*lda] a[16+43*lda] a[16+44*lda] ...
... a[17+42*lda] a[17+43*lda] a[17+44*lda] ...
         .             .            .
         .             .            .

我们可以将此3x3矩阵复制到连续的存储空间，将它作为输入（或输出）矩阵传递给LAPACK例程，我们不需要;我们只需要适当地定义参数。让我们把这个子矩阵称为 b ;我们然后定义：

We could copy this 3x3 matrix into contiguous storage, but if we want to pass it as an input (or output) matrix to an LAPACK routine, we don't need to; we only need to define the parameters appropriately. Let's call this submatrix b; we then define:

// pointer to the top-left corner of b:
float *b = &a[15 + 42*lda];
// number of rows in b:
const int nb = 3;
// number of columns in b:
const int mb = 3;
// leading dimension of b:
const int ldb = lda;

唯一可能令人惊讶的是 ldb ;通过使用大矩阵的值 lda ，我们可以在不复制的情况下寻址子矩阵，并在其上就地操作。

The only thing that might be surprising is the value of ldb; by using the value lda of the "big matrix", we can address the submatrix without copying, and operate on it in-place.

但
我骗了（排序）。有时你真的不能操作一个子矩阵到位，真正需要复制它。我不想谈论这个，因为它是罕见的，你应该尽可能使用就地操作，但我会觉得糟糕，不告诉你，这是可能的。例程：

However I lied (sort of). Sometimes you really can't operate on a submatrix in place, and genuinely need to copy it. I didn't want to talk about that, because it's rare, and you should use in-place operations whenever possible, but I would feel bad not telling you that it is possible. The routine:

SLACPY(UPLO,M,N,A,LDA,B,LDB)

复制 M x N 矩阵，其左上角为 A ，并以前导维 LDA 存储到<$ c $其左上角为 B 且具有前导维度的矩阵C> M x N LDB 。 UPLO 参数指示是否复制上三角形，下三角形或整个矩阵。

copies the MxN matrix whose top-left corner is A and is stored with leading dimension LDA to the MxN matrix whose top-left corner is B and has leading dimension LDB. The UPLO parameter indicates whether to copy the upper triangle, lower triangle, or the whole matrix.

我给上面的，你会使用像这样（假设clapack绑定）：

In the example I gave above, you would use it like this (assuming the clapack bindings):

...
const int m = 3;
const int n = 3;
float b[9];
const int ldb = 3;
slacpy("A",  // anything except "U" or "L" means "copy everything"
       &m,   // number of rows to copy
       &n,   // number of columns to copy
       &a[15 + 42*lda], // pointer to top-left element to copy
       lda,  // leading dimension of a (something huge)
       b,    // pointer to top-left element of destination
       ldb); // leading dimension of b (== m, so storage is dense)
...

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