快速创建R中每行已知数目为1的二进制矩阵的快速方法 [英] Fast way to create a binary matrix with known number of 1 each row in R
问题描述
我有一个向量,提供矩阵的每一行有多少个"1".现在,我必须从向量中创建该矩阵.
I have a vector that provides how many "1" each row of a matrix has. Now I have to create this matrix out of the vector.
例如,假设我要使用下面的矢量v <- c(2,6,3,9)
创建一个4 x 9的矩阵out
.结果应该看起来像
For example, let say I want to create a 4 x 9 matrix out
with following vector v <- c(2,6,3,9)
. The result should look like
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9]
[1,] 1 1 0 0 0 0 0 0 0
[2,] 1 1 1 1 1 1 0 0 0
[3,] 1 1 1 0 0 0 0 0 0
[4,] 1 1 1 1 1 1 1 1 1
我已经使用for
循环完成了此操作,但是对于大型矩阵(100,000 x 500),我的解决方案很慢:
I've done this with a for
loop but my solution is slow for a large matrix (100,000 x 500):
out <- NULL
for(i in 1:length(v)){
out <- rbind(out,c(rep(1, v[i]),rep(0,9-v[i])))
}
有人想出一种更快的方法来创建这样的矩阵吗?
Has anyone an idea for a faster way to create such a matrix?
推荐答案
这是我使用sapply
和do.call
的方法以及一些小样本上的时间安排.
Here is my approach using sapply
and do.call
and some timings on a small sample.
library(microbenchmark)
library(Matrix)
v <- c(2,6,3,9)
microbenchmark(
roman = {
xy <- sapply(v, FUN = function(x, ncols) {
c(rep(1, x), rep(0, ncols - x))
}, ncols = 9, simplify = FALSE)
xy <- do.call("rbind", xy)
},
fourtytwo = {
t(vapply(v, function(y) { x <- numeric( length=9); x[1:y] <- 1;x}, numeric(9) ) )
},
akrun = {
m1 <- sparseMatrix(i = rep(seq_along(v), v), j = sequence(v), x = 1)
m1 <- as.matrix(m1)
})
Unit: microseconds
expr min lq mean median uq
roman 26.436 30.0755 36.42011 36.2055 37.930
fourtytwo 43.676 47.1250 55.53421 54.7870 57.852
akrun 1261.634 1279.8330 1501.81596 1291.5180 1318.720
还有更大的样本
v <- sample(2:9, size = 10e3, replace = TRUE)
Unit: milliseconds
expr min lq mean median uq
roman 33.52430 35.80026 37.28917 36.46881 37.69137
fourtytwo 37.39502 40.10257 41.93843 40.52229 41.52205
akrun 10.00342 10.34306 10.66846 10.52773 10.72638
随着对象尺寸的增加,spareMatrix
的优势逐渐显现.
With a growing object size, the benefits of spareMatrix
come to light.
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