创建一个二进制矩阵的快速方法，该矩阵在R中的每行数为1
我有一个向量，它提供了矩阵每行有多少个“1”。现在我要用向量创建这个矩阵创建一个二进制矩阵的快速方法，该矩阵在R中的每行数为1,r,matrix,R,Matrix,我有一个向量，它提供了矩阵每行有多少个“1”。现在我要用向量创建这个矩阵 例如，假设我想创建一个4x9矩阵out，其中一个选项是sparseMatrixfrommatrix library(Matrix) m1 <- sparseMatrix(i = rep(seq_along(v), v), j = sequence(v), x = 1) m1 #4 x 9 sparse Matrix of class "dgCMatrix" #[1,] 1 1 . . . . . . . #[2,]
例如，假设我想创建一个4x9矩阵
outsparseMatrixmatrixlibrary(Matrix)
m1 <- sparseMatrix(i = rep(seq_along(v), v), j = sequence(v), x = 1)
m1
#4 x 9 sparse Matrix of class "dgCMatrix"
#[1,] 1 1 . . . . . . .
#[2,] 1 1 1 1 1 1 . . .
#[3,] 1 1 1 . . . . . .
#[4,] 1 1 1 1 1 1 1 1 1
as.matrix(m1)
vapplysapply> t( vapply( c(2,6,3,9), function(y) { x <- numeric( length=9); x[1:y] <- 1;x}, numeric(9) ) )
[,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
>t（vapply（c（2,6,3,9），函数（y）{x下面是我使用sapply
和do.call
的方法，以及一些小样本的计时
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
库（微基准）
图书馆（矩阵）
v2016-11-24更新
今天回答时，我得到了另一个解决方案：
<>这与我的初始答案中的<代码> f>代码>函数具有相同的内存使用，它不会比<代码> f>代码>慢。请考虑我原来答案中的基准：
microbenchmark(my_old = f(v, n), my_new = outer(v, n, ">=") + 0L, unit = "ms")
#Unit: milliseconds
# expr min lq mean median uq max neval cld
# my_old 109.3422 111.0355 121.0382120 111.16752 112.44472 210.36808 100 b
# my_new 0.3094 0.3199 0.3691904 0.39816 0.40608 0.45556 100 a
注意这个新方法的速度有多快，但我的旧方法已经是现有解决方案中速度最快的（见下文）
2016-11-07的原始答案
以下是我的“尴尬”解决方案：
f <- function (v, n) {
# n <- 9 ## total number of column
# v <- c(2,6,3,9) ## number of 1 each row
u <- n - v ## number of 0 each row
m <- length(u) ## number of rows
d <- rep.int(c(1,0), m) ## discrete value for each row
asn <- rbind(v, u) ## assignment of `d`
fill <- rep.int(d, asn) ## matrix elements
matrix(fill, byrow = TRUE, ncol = n)
}
n <- 9 ## total number of column
v <- c(2,6,3,9) ## number of 1 each row
f(v, n)
# [,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
f我甚至不打算尝试这一点，因为akrun将在一秒钟后在data.table中提供速度快1000倍的解决方案。顺便说一句，您的解决方案很慢，因为您正在增长对象。请参阅，我告诉过您。：）@ZheyuanLi使用42-的数据，我发现1.69比系统的2.20更快。时间即v，我模拟了大小矩阵。
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
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
outer(v, 1:9, ">=") + 0L
# [,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
microbenchmark(my_old = f(v, n), my_new = outer(v, n, ">=") + 0L, unit = "ms")
#Unit: milliseconds
# expr min lq mean median uq max neval cld
# my_old 109.3422 111.0355 121.0382120 111.16752 112.44472 210.36808 100 b
# my_new 0.3094 0.3199 0.3691904 0.39816 0.40608 0.45556 100 a
f <- function (v, n) {
# n <- 9 ## total number of column
# v <- c(2,6,3,9) ## number of 1 each row
u <- n - v ## number of 0 each row
m <- length(u) ## number of rows
d <- rep.int(c(1,0), m) ## discrete value for each row
asn <- rbind(v, u) ## assignment of `d`
fill <- rep.int(d, asn) ## matrix elements
matrix(fill, byrow = TRUE, ncol = n)
}
n <- 9 ## total number of column
v <- c(2,6,3,9) ## number of 1 each row
f(v, n)
# [,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
n <- 500 ## 500 columns
v <- sample.int(n, 10000, replace = TRUE) ## 10000 rows
microbenchmark(
my_bad = f(v, n),
roman = {
xy <- sapply(v, FUN = function(x, ncols) {
c(rep(1, x), rep(0, ncols - x))
}, ncols = n, simplify = FALSE)
do.call("rbind", xy)
},
fourtytwo = {
t(vapply(v, function(y) { x <- numeric( length=n); x[1:y] <- 1;x}, numeric(n) ) )
},
akrun = {
sparseMatrix(i = rep(seq_along(v), v), j = sequence(v), x = 1)
},
unit = "ms")
#Unit: milliseconds
# expr min lq mean median uq max neval cld
# my_bad 105.7507 118.6946 160.6818 138.5855 186.3762 327.3808 100 a
# roman 176.9003 194.7467 245.0450 213.8680 305.9537 435.5974 100 b
# fourtytwo 235.0930 256.5129 307.3099 273.2280 358.8224 587.3256 100 c
# akrun 316.7131 351.6184 408.5509 389.9576 456.0704 604.2667 100 d