您可以使用apply()对矩阵进行逐行计算。

df <- matrix(c(9185, 3661, 9539, 4266, 9650, 3897, 9160, 4451), nrow = 2, ncol = 4)
t(apply(df, 1, function(x)(x-mean(x))/sd(x)))
#>            [,1]      [,2]       [,3]       [,4]
#> [1,] -0.8003056 0.6269397  1.0744658 -0.9010998
#> [2,] -1.1445034 0.5536562 -0.4820809  1.0729281

另外，scale()是一个执行所需计算的函数

t(apply(df, 1, scale))
#>            [,1]      [,2]       [,3]       [,4]
#> [1,] -0.8003056 0.6269397  1.0744658 -0.9010998
#> [2,] -1.1445034 0.5536562 -0.4820809  1.0729281

创建于2023-10-02使用reprex v2.0.2

如果你追求速度，你可以尝试定义一个自定义的“scale“函数，如下所示

f2 <- function(df) {
    (df - rowMeans(df)) / sqrt(1 / (1 - 1 / ncol(df)) * (rowMeans(df^2) - rowMeans(df)^2))
}

你会发现

> f2(df)
           [,1]      [,2]       [,3]       [,4]
[1,] -0.8003056 0.6269397  1.0744658 -0.9010998
[2,] -1.1445034 0.5536562 -0.4820809  1.0729281

基准测试

f1 <- function(df) {
    t(apply(df, 1, scale))
}
f2 <- function(df) {
    (df - rowMeans(df)) / sqrt(1 / (1 - 1 / ncol(df)) * (rowMeans(df^2) - rowMeans(df)^2))
}
microbenchmark(
    f_DaveArmstong = f1(df),
    f_TIC = f2(df),
    check = "equivalent",
    unit = "relative"
)

给

Unit: relative
           expr      min       lq     mean   median       uq       max neval
 f_DaveArmstong 17.09756 16.78235 2.380846 15.04688 14.59606 0.2508871   100
          f_TIC  1.00000  1.00000 1.000000  1.00000  1.00000 1.0000000   100