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
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