R从另一个矩阵中的信息创建一个二进制矩阵

Question

我有一个矩阵，指示哪一列的值应为 1。所有其他列的值应为 0。例如，

m = matrix(c(3,3,3,5,12,13),nrow = 3)

我应该如何有效地创建以下矩阵？

     [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14]
[,1]    0    0    1    0    1    0    0    0    0     0     0     0     0     0
[,2]    0    0    1    0    0    0    0    0    0     0     0     1     0     0
[,3]    0    0    1    0    0    0    0    0    0     0     0     0     1     0

谢谢

Answer 1

根据原始矩阵创建行 ID 和元素的矩阵 m1 创建一个用零填充的矩阵，其行数与原始矩阵相同，列数比原始矩阵中的最大元素多一个。



m1 <- matrix(c(rep(1:nrow(m), ncol(m)), as.vector(m)), ncol = 2)

mx <- matrix(0, nrow = 3, ncol = max(m) + 1)
mx[m1] <- 1
mx
#>      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14]
#> [1,]    0    0    1    0    1    0    0    0    0     0     0     0     0     0
#> [2,]    0    0    1    0    0    0    0    0    0     0     0     1     0     0
#> [3,]    0    0    1    0    0    0    0    0    0     0     0     0     1     0

^{由 reprex package (v2.0.1)}

于 2022-03-23 创建

Answer 2

您可以创建具有所需维度的 0 矩阵

out <- matrix(0, nrow = 3, ncol = 14)

然后 cbind 具有 c(m) 的行索引向量，即列位置

tmp <- cbind(seq_len(3), c(m))

tmp 是一个矩阵，您现在可以使用它来用 1

替换所需的位置

out[tmp] <- 1
out
#         [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10] [,11] [,12] [,13] [,14]
#    [1,]    0    0    1    0    1    0    0    0    0     0     0     0     0     0
#    [2,]    0    0    1    0    0    0    0    0    0     0     0     1     0     0
#    [3,]    0    0    1    0    0    0    0    0    0     0     0     0     1     0

另一种可能效率较低的方法可能是这样

t(apply(m, 1, tabulate, nbins = 14))

R从另一个矩阵中的信息创建一个二进制矩阵

R creating a binary matrix from information in another matrix

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