c - 使三角形(或通常为正方形)矩阵对称

标签 c r performance matrix

我正在尝试从 R 中的下三角矩阵创建对称矩阵。

在之前的问答中 ( Convert upper triangular part of a matrix to symmetric matrix ) 用户 李哲源说对于大型矩阵,这不应该在 R 中完成,并在 C 中提出了一个解决方案。但我不了解 C 并且之前从未使用过例如 Rccp 所以不知道如何解释答案。然而很明显,那里的 C 代码生成了我不想要的随机数 (rnorm)。 我想输入一个方阵,得到一个对称矩阵。

对于给定的方阵 A,其下三角中有数据,我如何在 C 中有效地创建对称矩阵并在 R 中使用它?

最佳答案

快速改编自 as.matrix on a distance object is extremely slow; how to make it faster? 中的 dist2mat 函数.

library(Rcpp)

cppFunction('NumericMatrix Mat2Sym(NumericMatrix A, bool up2lo, int bf) {

  IntegerVector dim = A.attr("dim");
  size_t n = (size_t)dim[0], m = (size_t)dim[1];
  if (n != m) stop("A is not a square matrix!");

  /* use pointers */
  size_t j, i, jj, ni, nj;
  double *A_jj, *A_ij, *A_ji, *col, *row, *end;

  /* cache blocking factor */
  size_t b = (size_t)bf;

  /* copy lower triangular to upper triangular; cache blocking applied */
  for (j = 0; j < n; j += b) {
    nj = n - j; if (nj > b) nj = b;
    /* diagonal block has size nj x nj */
    A_jj = &A(j, j);
    for (jj = nj - 1; jj > 0; jj--, A_jj += n + 1) {
      /* copy a column segment to a row segment (or vise versa) */
      col = A_jj + 1; row = A_jj + n;
      for (end = col + jj; col < end; col++, row += n) {
        if (up2lo) *col = *row; else *row = *col;
        }
      }
    /* off-diagonal blocks */
    for (i = j + nj; i < n; i += b) {
      ni = n - i; if (ni > b) ni = b;
      /* off-diagonal block has size ni x nj */
      A_ij = &A(i, j); A_ji = &A(j, i);
      for (jj = 0; jj < nj; jj++) {
        /* copy a column segment to a row segment (or vise versa) */
        col = A_ij + jj * n; row = A_ji + jj;
        for (end = col + ni; col < end; col++, row += n) {
          if (up2lo) *col = *row; else *row = *col;
          }
        }
      }
    }

  return A;
  }')

对于方阵 A,此函数 Mat2Sym 将其下三角部分(带转置)复制到其上三角部分以使其对称,如果 up2lo = FALSE,如果 up2lo = TRUE,反之亦然。

请注意,函数覆盖 A 而不使用额外的内存。要保留输入矩阵并创建新的输出矩阵,请将 A + 0 而不是 A 传递到函数中。

## an arbitrary asymmetric square matrix
set.seed(0)
A <- matrix(runif(25), 5)
#          [,1]      [,2]       [,3]      [,4]      [,5]
#[1,] 0.8966972 0.2016819 0.06178627 0.7698414 0.7774452
#[2,] 0.2655087 0.8983897 0.20597457 0.4976992 0.9347052
#[3,] 0.3721239 0.9446753 0.17655675 0.7176185 0.2121425
#[4,] 0.5728534 0.6607978 0.68702285 0.9919061 0.6516738
#[5,] 0.9082078 0.6291140 0.38410372 0.3800352 0.1255551

## lower triangular to upper triangular; don't overwrite
B <- Mat2Sym(A + 0, up2lo = FALSE, 128)
#          [,1]      [,2]      [,3]      [,4]      [,5]
#[1,] 0.8966972 0.2655087 0.3721239 0.5728534 0.9082078
#[2,] 0.2655087 0.8983897 0.9446753 0.6607978 0.6291140
#[3,] 0.3721239 0.9446753 0.1765568 0.6870228 0.3841037
#[4,] 0.5728534 0.6607978 0.6870228 0.9919061 0.3800352
#[5,] 0.9082078 0.6291140 0.3841037 0.3800352 0.1255551

## A is unchanged
A
#          [,1]      [,2]       [,3]      [,4]      [,5]
#[1,] 0.8966972 0.2016819 0.06178627 0.7698414 0.7774452
#[2,] 0.2655087 0.8983897 0.20597457 0.4976992 0.9347052
#[3,] 0.3721239 0.9446753 0.17655675 0.7176185 0.2121425
#[4,] 0.5728534 0.6607978 0.68702285 0.9919061 0.6516738
#[5,] 0.9082078 0.6291140 0.38410372 0.3800352 0.1255551

## upper triangular to lower triangular; overwrite
D <- Mat2Sym(A, up2lo = TRUE, 128)
#           [,1]      [,2]       [,3]      [,4]      [,5]
#[1,] 0.89669720 0.2016819 0.06178627 0.7698414 0.7774452
#[2,] 0.20168193 0.8983897 0.20597457 0.4976992 0.9347052
#[3,] 0.06178627 0.2059746 0.17655675 0.7176185 0.2121425
#[4,] 0.76984142 0.4976992 0.71761851 0.9919061 0.6516738
#[5,] 0.77744522 0.9347052 0.21214252 0.6516738 0.1255551

## A has been changed; D and A are aliased in memory
A
#           [,1]      [,2]       [,3]      [,4]      [,5]
#[1,] 0.89669720 0.2016819 0.06178627 0.7698414 0.7774452
#[2,] 0.20168193 0.8983897 0.20597457 0.4976992 0.9347052
#[3,] 0.06178627 0.2059746 0.17655675 0.7176185 0.2121425
#[4,] 0.76984142 0.4976992 0.71761851 0.9919061 0.6516738
#[5,] 0.77744522 0.9347052 0.21214252 0.6516738 0.1255551

关于Matrix包的使用

Matrix 对于稀疏矩阵特别有用。为了兼容性,它还为密集矩阵定义了一些类,如“dgeMatrix”、“dtrMatrix”、“dtpMatrix”、“dsyMatrix”、“dspMatrix”。

给定一个方阵A,使其对称的Matrix方法如下。

set.seed(0)
A <- matrix(runif(25), 5)
#          [,1]      [,2]       [,3]      [,4]      [,5]
#[1,] 0.8966972 0.2016819 0.06178627 0.7698414 0.7774452
#[2,] 0.2655087 0.8983897 0.20597457 0.4976992 0.9347052
#[3,] 0.3721239 0.9446753 0.17655675 0.7176185 0.2121425
#[4,] 0.5728534 0.6607978 0.68702285 0.9919061 0.6516738
#[5,] 0.9082078 0.6291140 0.38410372 0.3800352 0.1255551

## equivalent to: Mat2Sym(A + 0, TRUE, 128)
new("dsyMatrix", x = base::c(A), Dim = dim(A), uplo = "U")
#5 x 5 Matrix of class "dsyMatrix"
#           [,1]      [,2]       [,3]      [,4]      [,5]
#[1,] 0.89669720 0.2016819 0.06178627 0.7698414 0.7774452
#[2,] 0.20168193 0.8983897 0.20597457 0.4976992 0.9347052
#[3,] 0.06178627 0.2059746 0.17655675 0.7176185 0.2121425
#[4,] 0.76984142 0.4976992 0.71761851 0.9919061 0.6516738
#[5,] 0.77744522 0.9347052 0.21214252 0.6516738 0.1255551

## equivalent to: Mat2Sym(A + 0, FALSE, 128)
new("dsyMatrix", x = base::c(A), Dim = dim(A), uplo = "L")
#5 x 5 Matrix of class "dsyMatrix"
#          [,1]      [,2]      [,3]      [,4]      [,5]
#[1,] 0.8966972 0.2655087 0.3721239 0.5728534 0.9082078
#[2,] 0.2655087 0.8983897 0.9446753 0.6607978 0.6291140
#[3,] 0.3721239 0.9446753 0.1765568 0.6870228 0.3841037
#[4,] 0.5728534 0.6607978 0.6870228 0.9919061 0.3800352
#[5,] 0.9082078 0.6291140 0.3841037 0.3800352 0.1255551

Matrix 方法不是最优的,原因有以下三个:

  1. 它需要将 x 槽作为数字 vector 传递,因此我们必须执行 base::c(A),这实际上是在 RAM 中创建矩阵的副本;
  2. 不能原地修改,所以创建一个新的矩阵副本作为输出矩阵;
  3. 转置复制时不阻塞缓存。

这是一个快速比较:

library(bench)
A <- matrix(runif(5000 * 5000), 5000)
bench::mark("Mat2Sym" = Mat2Sym(A, FALSE, 128),
            "Matrix" = new("dsyMatrix", x = base::c(A), Dim = dim(A), uplo = "L"),
            check = FALSE)
#  expression      min     mean   median     max `itr/sec` mem_alloc  n_gc n_itr
#  <chr>      <bch:tm> <bch:tm> <bch:tm> <bch:t>     <dbl> <bch:byt> <dbl> <int>
#1 Mat2Sym      56.8ms   57.7ms   57.4ms  59.4ms     17.3     2.48KB     0     9
#2 Matrix      334.3ms  337.4ms  337.4ms 340.6ms      2.96  190.74MB     2     2

请注意 Mat2Sym 有多快。在“覆盖”模式下也没有内存分配。

As G. Grothendieck mentions ,我们也可以使用“dspMatrix”。

set.seed(0)
A <- matrix(runif(25), 5)
#          [,1]      [,2]       [,3]      [,4]      [,5]
#[1,] 0.8966972 0.2016819 0.06178627 0.7698414 0.7774452
#[2,] 0.2655087 0.8983897 0.20597457 0.4976992 0.9347052
#[3,] 0.3721239 0.9446753 0.17655675 0.7176185 0.2121425
#[4,] 0.5728534 0.6607978 0.68702285 0.9919061 0.6516738
#[5,] 0.9082078 0.6291140 0.38410372 0.3800352 0.1255551

## equivalent to: Mat2Sym(A + 0, TRUE, 128)
new("dspMatrix", x = A[upper.tri(A, TRUE)], Dim = dim(A), uplo = "U")
#5 x 5 Matrix of class "dspMatrix"
#           [,1]      [,2]       [,3]      [,4]      [,5]
#[1,] 0.89669720 0.2016819 0.06178627 0.7698414 0.7774452
#[2,] 0.20168193 0.8983897 0.20597457 0.4976992 0.9347052
#[3,] 0.06178627 0.2059746 0.17655675 0.7176185 0.2121425
#[4,] 0.76984142 0.4976992 0.71761851 0.9919061 0.6516738
#[5,] 0.77744522 0.9347052 0.21214252 0.6516738 0.1255551

## equivalent to: Mat2Sym(A + 0, FALSE, 128)
new("dspMatrix", x = A[lower.tri(A, TRUE)], Dim = dim(A), uplo = "L")
#5 x 5 Matrix of class "dspMatrix"
#          [,1]      [,2]      [,3]      [,4]      [,5]
#[1,] 0.8966972 0.2655087 0.3721239 0.5728534 0.9082078
#[2,] 0.2655087 0.8983897 0.9446753 0.6607978 0.6291140
#[3,] 0.3721239 0.9446753 0.1765568 0.6870228 0.3841037
#[4,] 0.5728534 0.6607978 0.6870228 0.9919061 0.3800352
#[5,] 0.9082078 0.6291140 0.3841037 0.3800352 0.1255551

同样,Matrix 是次优方法,因为使用了 upper.trilower.tri

library(bench)
A <- matrix(runif(5000 * 5000), 5000)
bench::mark("Mat2Sym" = Mat2Sym(A, FALSE, 128),
            "Matrix" = new("dspMatrix", x = A[lower.tri(A, TRUE)], Dim = dim(A),
                           uplo = "L"),
            check = FALSE)
#  expression      min     mean   median     max `itr/sec` mem_alloc  n_gc n_itr
#  <chr>      <bch:tm> <bch:tm> <bch:tm> <bch:t>     <dbl> <bch:byt> <dbl> <int>
#1 Mat2Sym      56.5ms   57.9ms     58ms  58.7ms     17.3     2.48KB     0     9
#2 Matrix      934.7ms  934.7ms    935ms 934.7ms      1.07  524.55MB     2     1

特别是,我们看到使用“dspMatrix”的效率甚至低于使用“dsyMatrix”的效率。

关于c - 使三角形(或通常为正方形)矩阵对称,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/52571101/

相关文章:

performance - 最有效的 R 余弦计算

R:训练数据集的 k 折交叉验证

R 对 data.frame 中两个变量的惰性求值

c# - 在无限循环性能中声明变量类型?

android - 可见性设置为 "gone"的 View 是否是测量和布局过程的一部分?

python - 将位串转换为字节(霍夫曼编码)

c - 调用采用非常量指针参数、const_cast、reinterpret_cast、launder 的 c 函数的正确方法

c - C 中的数组初始化问题

c - C中的文件创建权限

objective-c - 随机 C/Objective-C 问题