如何有效地生成对称矩阵的下三角指数

Question

我需要生成下三角矩阵索引(行和列对).当对称矩阵变大(超过50K行)时,当前实现是低效的(内存方式).有没有更好的办法？

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

Answer 1

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))

数据量较大的基准:

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