如何在 R 中的对称矩阵(12k X 12k)中找到前 10,000 个元素的索引 [英] How to find the indices of the top 10,000 elements in a symmetric matrix(12k X 12k) in R

问题描述

我有一个 12000X12000 的非零对称矩阵matr".我需要在 R 中的 'matr' 中找到前 10000 个元素的索引.我编写的代码需要很长时间 - 我想知道是否有任何指针可以使它更快.

I have a nonzero symmetric matrix 'matr' that is 12000X12000. I need to find the indices of the top 10000 elements in 'matr' in R. The code I have written takes a long time - I was wondering if there was any pointers to make it faster.

listk <- numeric(0)
for( i in 1:10000) {
    idx <- which(matr == max(matr), arr.ind=T)
    if( length(idx) != 0) {
        listk <- rbind( listk, idx[1,])
        matr[idx[1,1], idx[1,2]] <- 0
        matr[idx[2,1], idx[2,2]] <- 0
    } 
}

推荐答案

参加聚会有点晚，但我想出了这个，避免了排序.

A bit late into the party, but I came up with this, which avoids the sort.

假设您想要 12k x 12k 矩阵中的前 10k 个元素.这个想法是将数据剪辑"到与该大小的分位数对应的元素.

Say you want the top 10k elements from you 12k x 12k matrix. The idea is to "clip" the data to the elements corresponding to a quantile of that size.

find_n_top_elements <- function( x, n ){

  #set the quantile to correspond to n top elements
  quant <- n / (dim(x)[1]*dim(x)[2])

  #select the cutpoint to get the quantile above quant
  lvl <- quantile(x, probs=1.0-quant)

  #select the elements above the cutpoint
  res <- x[x>lvl[[1]]]
}

#create a 12k x 12k matrix (1,1Gb!)
n <- 12000
x <- matrix( runif(n*n), ncol=n)

system.time( res <- find_n_top_elements( x, 10e3 ) )

导致

system.time( res <- find_n_top_elements( x, 10e3 ) )
 user  system elapsed
 3.47    0.42    3.89 

为了比较，在我的系统上只对 x 排序需要

For comparison, just sorting x on my system takes

system.time(sort(x))
   user  system elapsed 
  30.69    0.21   31.33 

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