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
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
但是这给了我：
[,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
任何人都可以帮我解决这个问题吗？
干杯，
MCE
PS：如果第一行全部为零（可能会出现上述代码），则应将其更改为全部。
答案 0 :(得分：2)
为什么不呢：
t(apply(field,1,cummax))
一个例子：
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
答案 1 :(得分：0)
这应该有效：
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))
注意它使用了你定义的填充。
答案 2 :(得分：0)
在apply的值的帮助中，＆＃34;如果每次调用FUN都返回一个长度为n的向量，则apply返回一个维数c（n，dim（X）[MARGIN]）＆＃34的数组;。所以，你想要这个转置。打印语句被添加到填充功能以确认操作。您可能想要检查您的函数是否隐藏了另一个函数，还有一个名为fill的函数，但在这种情况下它并不重要。
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
myfill <- function(row) {
print("Function start")
print(row)
first = match(1, row)
print(paste("Match", first))
if (is.na(first)) {
row = rep(1, 10)
} else {
row[first:10] = 1
};
print(row)
flush.console()
return(row)
}
field2 = t(apply(field, 1, myfill))
field2