我使用以下代码生成一个随机矩阵，其中一些元素 = 1 在对角线附近，其余元素 = 0。(这基本上是沿着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
下一步，我想将 1 元素右侧/上侧的 0 元素替换为 1。对于上述矩阵，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
我已经尝试过
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吗？
干杯，
MCE
PS:如果第一行全为零(上面的代码可能会发生这种情况)，则应将其更改为全零。
最佳答案
为什么不只是:
t(apply(field,1,cummax))
一个实例:
dput(field)
structure(c(0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0,
0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0,
0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0,
0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0,
0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0), .Dim = c(10L,
10L))
> field
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 0 0 0 0 0 0 0 0 0 0
[2,] 1 1 1 1 1 1 0 0 0 0
[3,] 0 0 0 0 0 1 0 0 0 0
[4,] 0 0 0 0 0 1 0 0 0 0
[5,] 0 0 0 0 0 1 1 1 1 1
[6,] 0 0 0 0 0 0 0 0 0 0
[7,] 0 0 0 0 0 0 0 0 0 0
[8,] 0 0 0 0 0 0 0 0 0 0
[9,] 0 0 0 0 0 0 0 0 0 0
[10,] 0 0 0 0 0 0 0 0 0 0
输出:
> t(apply(field,1,cummax))
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] 0 0 0 0 0 0 0 0 0 0
[2,] 1 1 1 1 1 1 1 1 1 1
[3,] 0 0 0 0 0 1 1 1 1 1
[4,] 0 0 0 0 0 1 1 1 1 1
[5,] 0 0 0 0 0 1 1 1 1 1
[6,] 0 0 0 0 0 0 0 0 0 0
[7,] 0 0 0 0 0 0 0 0 0 0
[8,] 0 0 0 0 0 0 0 0 0 0
[9,] 0 0 0 0 0 0 0 0 0 0
[10,] 0 0 0 0 0 0 0 0 0 0
关于R:替换随机矩阵的 "off-diagonal"元素，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/26326259/