[英]R: List of indices, including empty ones, to binary matrix
假設我有一個索引列表,其中包括空元素,例如:
l <- list(c(1,2,3), c(1), c(1,5), NULL, c(2, 3, 5))
在矩陣中指定非零元素,例如:
(m <- matrix(c(1,1,1,0,0, 1,0,0,0,0, 1,0,0,0,5, 0,0,0,0,0, 0,1,1,0,1), nrow=5, byrow=TRUE))
[,1] [,2] [,3] [,4] [,5]
[1,] 1 1 1 0 0
[2,] 1 0 0 0 0
[3,] 1 0 0 0 5
[4,] 0 0 0 0 0
[5,] 0 1 1 0 1
給定矩陣非常大(例如50.000行和2000列),使用R從l
m
的最快方法是什么?
您可以從'l' Filter
非NULL列表元素,並使用melt
將其重塑為帶有'key / value'列或'row / column'索引列的'data.frame'。
library(reshape2)
d2 <- melt(Filter(Negate(is.null), setNames(l, seq_along(l))))
Un1 <- unlist(l)
m1 <- matrix(0, nrow=length(l), ncol=max(Un1))
m1[cbind(as.numeric(d2[,2]), d2[,1])] <- 1
m1
# [,1] [,2] [,3] [,4] [,5]
#[1,] 1 1 1 0 0
#[2,] 1 0 0 0 0
#[3,] 1 0 0 0 1
#[4,] 0 0 0 0 0
#[5,] 0 1 1 0 1
要么
library(Matrix)
as.matrix(sparseMatrix(as.numeric(d2[,2]), d2[,1], x=1))
# [,1] [,2] [,3] [,4] [,5]
#[1,] 1 1 1 0 0
#[2,] 1 0 0 0 0
#[3,] 1 0 0 0 1
#[4,] 0 0 0 0 0
#[5,] 0 1 1 0 1
你可以做:
do.call(rbind, lapply(l, function(x) (1:max(unlist(l)) %in% x)+0L))
# [,1] [,2] [,3] [,4] [,5]
#[1,] 1 1 1 0 0
#[2,] 1 0 0 0 0
#[3,] 1 0 0 0 1
#[4,] 0 0 0 0 0
#[5,] 0 1 1 0 1
即使akrun解決方案也應該更快!
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