问题
I would like to know how can I extract the values of the first diagonal from a distance matrix.
For example:
> mymatrix
[,1] [,2]
[1,] 1 2
[2,] 3 4
[3,] 6 4
[4,] 8 6
> dist(mymatrix)
1 2 3
2 2.828427
3 5.385165 3.000000
4 8.062258 5.385165 2.828427
I want to get in a vector the values: 2.828427, 3.000000, 2.828427
Thanks!
回答1:
One work around is to convert the dist
object to matrix
and then extract elements where row index is one larger than the column index:
mat = as.matrix(dist(mymatrix))
mat[row(mat) == col(mat) + 1]
# [1] 2.828427 3.000000 2.828427
回答2:
A seemingly complicated but extremely efficient solution for extracting the d
th sub-diagonal of a "dist" matrix.
subdiag <- function (dist_obj, d) {
if (!inherits(dist_obj, "dist")) stop("please provide a 'dist' object!")
n <- attr(dist_obj, "Size")
if (d < 1 || d > (n - 1)) stop(sprintf("'d' needs be between 1 and %d", n - 1L))
j_1 <- c(0, seq.int(from = n - 1, by = -1, length = n - d - 1))
subdiag_ind <- d + cumsum(j_1)
dist_obj[subdiag_ind]
}
See R - How to get row & column subscripts of matched elements from a distance matrix for details of packed storage of the "dist" object. Inside this function, j_1
is the number of "X
" in the (j - 1)
th column. cumsum
gives the 1D index for main diagonals (whose values are all zeros). A further offset by d
gives 1D index of the d
th sub-diagonal.
set.seed(0)
x <- dist(matrix(runif(10), 5))
# 1 2 3 4
#2 0.9401067
#3 0.9095143 0.1162289
#4 0.5618382 0.3884722 0.3476762
#5 0.4275871 0.6968296 0.6220650 0.3368478
subdiag(x, 1)
#[1] 0.9401067 0.1162289 0.3476762 0.3368478
lapply(1:4, subdiag, dist_obj = x)
#[[1]]
#[1] 0.9401067 0.1162289 0.3476762 0.3368478
#
#[[2]]
#[1] 0.9095143 0.3884722 0.6220650
#
#[[3]]
#[1] 0.5618382 0.6968296
#
#[[4]]
#[1] 0.4275871
Good performance for big to large "dist" matrix.
## mimic a "dist" object without actually calling function `dist`
n <- 2000
x <- structure(numeric(n * (n - 1) / 2), class = "dist", Size = n)
library(bench)
bench::mark("Psidom" = {mat = as.matrix(x); mat[row(mat) == col(mat) + 1]},
"zheyuan" = subdiag(x, 1))
## A tibble: 2 x 14
# expression min mean median max `itr/sec` mem_alloc n_gc n_itr
# <chr> <bch:tm> <bch:tm> <bch:t> <bch:tm> <dbl> <bch:byt> <dbl> <int>
#1 Psidom 553ms 553ms 553ms 552.74ms 1.81 251.8MB 5 1
#2 zheyuan 106µs 111µs 108µs 3.85ms 9045. 62.7KB 2 4519
## ... with 5 more variables: total_time <bch:tm>, result <list>, memory <list>,
## time <list>, gc <list>
subdiag
is 5120 times faster (553ms / 108µs), and 4112 times more memory efficient (251.8MB / 62.7KB).
来源:https://stackoverflow.com/questions/39231961/extract-diagonals-from-a-distance-matrix-in-r