R 以4块为单位的导线矩阵（点网格）

此问题针对的是一个项目，与家庭作业/ACAD无关。我是一名在职统计员。 所以我的问题是，如果一个矩阵有400行和两列，每20行从第一行开始，形成一个点网格中的第一行坐标，你会如何编写一个R函数，我希望函数返回网格中每个正方形/矩形的四个角：

因此，输出将有四列，每行表示一个矩形。我只查看与中相同大小的相邻矩形，例如，如果下面的数字表示示例矩阵（有两列）的行索引：

行索引示例：

1  2  3
4  5  6
7  8  9

必须按照以下顺序进行遍历：

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

从示例输入数据集中返回相应的二维点 这将有9行和2个点。但是，这里的网格被指定为3乘3，而在我的示例中，网格是20乘20，输入数据集是400行乘2列。如果查看遍历结果，则会发现一种模式，其中每个4点块中的行索引递增1。我只是想把它推广到一个400乘2的或任何设置，其中有一个2列的点矩阵，并且提到了网格维度

---

如果您想创建一个可向下滚动和/或在400x400空间中滚动的20 x 20“可移动窗口”，请使用：

 mcorners <- function(xidx, yidx) mat[xidx:(xidx+19),
                                       yidx:(yidx+19])

 mcorners(1,1) # should return mat[1:20, mat1:20]

---

如果我理解正确，这里有一个解决办法。老实说，这是一个非常有趣的问题D

其思想是制作一个具有给定边长度的长方体，然后围绕网格移动该长方体并记录其顶点。请参见以下内容：

# Assuming the grid is always a square grid.

grid.size <- 20

# The matrix of row indices.

rindex.grid <- matrix(1:(grid.size * grid.size),
                  nrow=grid.size, ncol=grid.size, byrow=TRUE)

# We can traverse the grid by moving any given square either right or down in any 
# single  move. We choose to go right.

move.square.right <- function (this.square, steps=1) {
  new.square <- this.square + steps
}

# Going right, capture co-ordinates of all squares in this row.

collect.sq.of.edge.length.in.row.number <- function (grid.size, elength,
                                                    rownum=1) {
  first.square.in.row <- (rownum - 1) * grid.size + c(1, elength)
  first.square.in.row <- c(first.square.in.row,
                          first.square.in.row + grid.size * (elength - 1))
  squares.in.row <- t(sapply(X=seq_len(grid.size - (elength - 1)) - 1,
                            FUN=move.square.right,
                            this.square=first.square.in.row))
  squares.in.row
}

# Now we start going down the columns and using the function above to collect
# squares in each row. The we will rbind the list of squares in each row into a
# dataframe. So what we get is a (grid.size - (elength - 1) ^ 2) x 4 matrix where
# each row is the co-ordinates of a square of edge length elength.

collect.sq.of.edge.length.in.grid <- function (grid.size, elength) {
  all.squares=lapply(X=seq_len(grid.size - (elength - 1)),
          FUN=collect.sq.of.edge.length.in.row.number,
          grid.size=grid.size, elength=elength)
  all.squares <- do.call(rbind, all.squares)
  all.squares
}

---

这有点令人困惑。我认为一些人会从一个详细的描述中受益，其中包括要处理的样本数据，显示输入和期望输出的子集。我同意。我会把它放进去的。我现在好像已经有答案了。因此，我还将加入我编写的函数，看看是否可以提高效率。当我使用collect.sq.of.edge.length.in.grid（grid.size=20，elength=2）和grid.size 20时，它给出了正确的结果。好job@PraneethVepakomma：谢谢

# Assuming the grid is always a square grid.

grid.size <- 20

# The matrix of row indices.

rindex.grid <- matrix(1:(grid.size * grid.size),
                  nrow=grid.size, ncol=grid.size, byrow=TRUE)

# We can traverse the grid by moving any given square either right or down in any 
# single  move. We choose to go right.

move.square.right <- function (this.square, steps=1) {
  new.square <- this.square + steps
}

# Going right, capture co-ordinates of all squares in this row.

collect.sq.of.edge.length.in.row.number <- function (grid.size, elength,
                                                    rownum=1) {
  first.square.in.row <- (rownum - 1) * grid.size + c(1, elength)
  first.square.in.row <- c(first.square.in.row,
                          first.square.in.row + grid.size * (elength - 1))
  squares.in.row <- t(sapply(X=seq_len(grid.size - (elength - 1)) - 1,
                            FUN=move.square.right,
                            this.square=first.square.in.row))
  squares.in.row
}

# Now we start going down the columns and using the function above to collect
# squares in each row. The we will rbind the list of squares in each row into a
# dataframe. So what we get is a (grid.size - (elength - 1) ^ 2) x 4 matrix where
# each row is the co-ordinates of a square of edge length elength.

collect.sq.of.edge.length.in.grid <- function (grid.size, elength) {
  all.squares=lapply(X=seq_len(grid.size - (elength - 1)),
          FUN=collect.sq.of.edge.length.in.row.number,
          grid.size=grid.size, elength=elength)
  all.squares <- do.call(rbind, all.squares)
  all.squares
}

tmp <- sapply(1:20, collect.sq.of.edge.length.in.grid, grid.size=grid.size)
sapply(tt, nrow)

[1] 400 361 324 289 256 225 196 169 144 121 100  81  64  49  36  25  16   9   4   1

collect.sq.of.edge.length.in.grid(grid.size=3, elength=2)

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