本文介绍了如何有效生成对称矩阵的下三角索引的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

我需要生成较低的三角形矩阵索引(行和列对).当前的实现方式效率低下(在内存方面),特别是当对称矩阵变大(超过50K行)时.有更好的方法吗?

I need to generate lower triangle matrix indices (row and columns pairs). The current implementation is inefficient (memory wise) specially when symmetric matrix gets big (more than 50K rows). Is there a better way?

rows <- 2e+01
id <- which(lower.tri(matrix(, rows, rows)) == TRUE, arr.ind=T)
head(id)

#      row col
# [1,]   2   1
# [2,]   3   1
# [3,]   4   1
# [4,]   5   1
# [5,]   6   1
# [6,]   7   1

推荐答案

这是另一种方法:

z <- sequence(rows)
cbind(
  row = unlist(lapply(2:rows, function(x) x:rows), use.names = FALSE),
  col = rep(z[-length(z)], times = rev(tail(z, -1))-1))


具有更大数据的基准:


Benchmarks with larger data:

library(microbenchmark)

rows <- 1000
m <- matrix(, rows, rows)

## Your current approach
fun1 <- function() which(lower.tri(m) == TRUE, arr.ind=TRUE)

## An improvement of your current approach
fun2 <- function() which(lower.tri(m), arr.ind = TRUE)

## The approach shared in this answer
fun3 <- function() {
  z <- sequence(rows)
  cbind(
    row = unlist(lapply(2:rows, function(x) x:rows), use.names = FALSE),
    col = rep(z[-length(z)], times = rev(tail(z, -1))-1))
}

## Sven's answer
fun4 <- function() {
  row <- rev(abs(sequence(seq.int(rows - 1)) - rows) + 1)
  col <- rep.int(seq.int(rows - 1), rev(seq.int(rows - 1)))
  cbind(row, col)
}

microbenchmark(fun1(), fun2(), fun3(), fun4())
# Unit: milliseconds
#    expr       min        lq   median       uq       max neval
#  fun1() 77.813577 85.343356 90.60689 95.71648 130.40059   100
#  fun2() 73.812204 82.103600 85.87555 90.59235 138.66547   100
#  fun3()  9.016237  9.382506 10.63291 13.20085  55.42137   100
#  fun4() 20.591863 24.999702 28.82232 31.90663  65.05169   100

这篇关于如何有效生成对称矩阵的下三角索引的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持!

08-11 23:09