从矩阵中的每一列获取最小值的最快方法? [英] fastest way to get Min from every column in a matrix?

问题描述

从矩阵的每一列中提取最小值的最快方法是什么?

What is the fastest way to extract the min from each column in a matrix?

---

将所有基准测试移至以下答案.

Moved all the benchmarks to the answer below.

  ##  TEST DATA
  set.seed(1)
  matrix.inputs <- list(
        "Square Matrix"     = matrix(sample(seq(1e6), 4^2*1e4, T), ncol=400),   #  400 x  400
        "Tall Matrix"       = matrix(sample(seq(1e6), 4^2*1e4, T), nrow=4000),  # 4000 x   40
        "Wide-short Matrix" = matrix(sample(seq(1e6), 4^2*1e4, T), ncol=4000),  #   40 x 4000
        "Wide-tall Matrix"  = matrix(sample(seq(1e6), 4^2*1e5, T), ncol=4000),   #  400 x 4000
        "Tiny Sq Matrix"    = matrix(sample(seq(1e6), 4^2*1e2, T), ncol=40)     #   40 x   40
  )

推荐答案

这是在平方矩阵和宽矩阵上速度更快的一种.它在矩阵的行上使用pmin. (如果您知道将矩阵分成几行的更快方法，请随时进行编辑)

Here is one that is faster on square and wide matrices. It uses pmin on the rows of the matrix. (If you know a faster way of splitting the matrix into its rows, please feel free to edit)

do.call(pmin, lapply(1:nrow(mat), function(i)mat[i,]))

使用与@RicardoSaporta相同的基准:

Using the same benchmark as @RicardoSaporta:

$`Square Matrix`
          test elapsed relative
3 pmin.on.rows   1.370    1.000
1          apl   1.455    1.062
2         cmin   2.075    1.515

$`Wide Matrix`
      test elapsed relative
3 pmin.on.rows   0.926    1.000
2         cmin   2.302    2.486
1          apl   5.058    5.462

$`Tall Matrix`
          test elapsed relative
1          apl   1.175    1.000
2         cmin   2.126    1.809
3 pmin.on.rows   5.813    4.947

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