R：替换随机矩阵的“非对角线”元素

2023年5月7日 314次阅读

我正在使用以下代码生成一个随机矩阵,其中一些元素= 1靠近对角线,其余= 0.(这基本上是沿主对角线的随机游走.)

n <- 20
rw <- matrix(0, ncol = 2, nrow = n)
indx <- cbind(seq(n), sample(c(1, 2), n, TRUE))
rw[indx] <- 1
rw[,1] <- cumsum(rw[, 1])+1
rw[,2] <- cumsum(rw[, 2])+1
rw2 <- subset(rw, (rw[,1] <= 10 & rw[,2] <= 10))
field <- matrix(0, ncol = 10, nrow = 10)
field[rw2] <- 1
field

     [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
 [1,]    0    1    1    1    0    0    0    0    0     0
 [2,]    0    0    0    1    0    0    0    0    0     0
 [3,]    0    0    0    1    0    0    0    0    0     0
 [4,]    0    0    0    1    1    1    1    0    0     0
 [5,]    0    0    0    0    0    0    1    1    0     0
 [6,]    0    0    0    0    0    0    0    1    0     0
 [7,]    0    0    0    0    0    0    0    1    0     0
 [8,]    0    0    0    0    0    0    0    1    1     1
 [9,]    0    0    0    0    0    0    0    0    0     0
[10,]    0    0    0    0    0    0    0    0    0     0

接下来,我想将0元素替换为1元素的右侧/上侧1.对于上面的矩阵,所需的输出将是：

     [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
 [1,]    0    1    1    1    1    1    1    1    1     1
 [2,]    0    0    0    1    1    1    1    1    1     1
 [3,]    0    0    0    1    1    1    1    1    1     1
 [4,]    0    0    0    1    1    1    1    1    1     1
 [5,]    0    0    0    0    0    0    1    1    1     1
 [6,]    0    0    0    0    0    0    0    1    1     1
 [7,]    0    0    0    0    0    0    0    1    1     1
 [8,]    0    0    0    0    0    0    0    1    1     1
 [9,]    0    0    0    0    0    0    0    0    0     0
[10,]    0    0    0    0    0    0    0    0    0     0

我试过了

fill <- function(row) {first = match(1, row); if (is.na(first)) {row = rep(1, 10)} else {row[first:10] = 1}; return(row)}  
field2 <- apply(field, 1, fill)
field2

但这反过来给了我：

      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
 [1,]    0    0    0    0    0    0    0    0    1     1
 [2,]    1    0    0    0    0    0    0    0    1     1
 [3,]    1    0    0    0    0    0    0    0    1     1
 [4,]    1    1    1    1    0    0    0    0    1     1
 [5,]    1    1    1    1    0    0    0    0    1     1
 [6,]    1    1    1    1    0    0    0    0    1     1
 [7,]    1    1    1    1    1    0    0    0    1     1
 [8,]    1    1    1    1    1    1    1    1    1     1
 [9,]    1    1    1    1    1    1    1    1    1     1
[10,]    1    1    1    1    1    1    1    1    1     1

谁能帮我解决这个问题？

干杯,

MCE

PS：如果第一行全为零(因为它可能发生在上面的代码中),它应该改为全部.

最佳答案 这应该工作：

MaxFull <- which.max((apply(field,1,sum) > 0) * (1:10))
rbind(t(apply(field[1:MaxFull,], 1, fill)),matrix(0,ncol=10,nrow=10-MaxFull))

注意它使用你定义的填充.

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