r中矩阵累积标准差的高效计算 [英] Efficient calculation of matrix cumulative standard deviation in r

问题描述

我最近在 r-help 邮件列表上发布了这个问题，但没有得到答案，所以我想我也将它发布在这里，看看是否有任何建议.

I recently posted this question on the r-help mailing list but got no answers, so I thought I would post it here as well and see if there were any suggestions.

我正在尝试计算矩阵的累积标准偏差.我想要一个函数，它接受一个矩阵并返回一个相同大小的矩阵，其中输出单元格 (i,j) 设置为第 1 行和第 i 行之间输入列 j 的标准偏差.NA 应该被忽略，除非输入矩阵的单元格 (i,j) 本身就是 NA，在这种情况下，输出矩阵的单元格 (i,j) 也应该是 NA.

I am trying to calculate the cumulative standard deviation of a matrix. I want a function that accepts a matrix and returns a matrix of the same size where output cell (i,j) is set to the standard deviation of input column j between rows 1 and i. NAs should be ignored, unless cell (i,j) of the input matrix itself is NA, in which case cell (i,j) of the output matrix should also be NA.

我找不到内置函数，所以我实现了以下代码.不幸的是，这使用了一个循环，最终对于大型矩阵来说有点慢.有没有更快的内置函数，或者有人可以提出更好的方法吗?

I could not find a built-in function, so I implemented the following code. Unfortunately, this uses a loop that ends up being somewhat slow for large matrices. Is there a faster built-in function or can someone suggest a better approach?

cumsd <- function(mat)
{
    retval <- mat*NA
    for (i in 2:nrow(mat)) retval[i,] <- sd(mat[1:i,], na.rm=T)
    retval[is.na(mat)] <- NA
    retval
}

谢谢.

推荐答案

您可以使用 cumsum 来计算从方差/sd 的直接公式到矩阵矢量化运算所需的总和:

You could use cumsum to compute necessary sums from direct formulas for variance/sd to vectorized operations on matrix:

cumsd_mod <- function(mat) {
    cum_var <- function(x) {
        ind_na <- !is.na(x)
        nn <- cumsum(ind_na)
        x[!ind_na] <- 0
        cumsum(x^2) / (nn-1) - (cumsum(x))^2/(nn-1)/nn
    }
    v <- sqrt(apply(mat,2,cum_var))
    v[is.na(mat) | is.infinite(v)] <- NA
    v
}

仅供对比:

set.seed(2765374)
X <- matrix(rnorm(1000),100,10)
X[cbind(1:10,1:10)] <- NA # to have some NA's

all.equal(cumsd(X),cumsd_mod(X))
# [1] TRUE

关于时间:

X <- matrix(rnorm(100000),1000,100)
system.time(cumsd(X))
# user  system elapsed 
# 7.94    0.00    7.97 
system.time(cumsd_mod(X))
# user  system elapsed 
# 0.03    0.00    0.03 

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