1

我想用一组变量（例如1到10）在R中创建一个矩阵。这些变量应该在行和列上随机分配，但不应该在任何一个中重复（所以数字 1 应该在第 1 行中一次，在第 1 列中一次）！

例如：

1,2,3,4,5,6,7,8,9,10

2,3,4,5,6,7,8,9,10,1

3,4,5,6,7,8,9,10,1,2

4,5,6,7,8,9,10,1,2,3

5,6,7,8,9,10,1,2,3,4

6,7,8,9,10,1,2,3,4,5

7,8,9,10,1,2,3,4,5,6

8,9,10,1,2,3,4,5,6,7

9,10,1,2,3,4,5,6,7,8

10,1,2,3,4,5,6,7,8,9

但当然，在那个例子中，数字是递增的，我希望它们随机化。我尝试了简单的矩阵要求，但我无法弄清楚如何做到这一点。任何人都可以帮忙吗？提前致谢！

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2 回答 2

1

除非我误解了这个问题，否则有一种更简单的方法来创建这个混洗矩阵，没有任何循环或复杂的条件语句。

# number of rows and columns
n <- 10

# create ordered rows and columns
ordered.by.row <- matrix(1:n, n, n)
ordered.by.col <- matrix(1:n, n, n, byrow = T)

# offset the rows and columns relative to each other.
# no row or column has a repeated value, but the values are still ordered
offset <- (ordered.by.row + ordered.by.col) %% n + 1

# shuffle the columns, then shuffle the rows, this produces a randomized matrix
# 'shuffle.row' is the final, randomized matrix
set.seed(1222) # change this to change randomization
shuffle.col <- offset[,sample(1:n, n, replace = F)]
shuffle.row <- shuffle.col[sample(1:n, n, replace = F), ]

# verify solution
any(apply(shuffle.row, 1, function(r)any(duplicated(r)))) # FALSE
any(apply(shuffle.row, 2, function(r)any(duplicated(r)))) # FALSE

      [,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
 [1,]    1   10    6    9    2    8    3    5    7     4
 [2,]    3    2    8    1    4   10    5    7    9     6
 [3,]    7    6    2    5    8    4    9    1    3    10
 [4,]    9    8    4    7   10    6    1    3    5     2
 [5,]   10    9    5    8    1    7    2    4    6     3
 [6,]    2    1    7   10    3    9    4    6    8     5
 [7,]    8    7    3    6    9    5   10    2    4     1
 [8,]    6    5    1    4    7    3    8   10    2     9
 [9,]    5    4   10    3    6    2    7    9    1     8
[10,]    4    3    9    2    5    1    6    8   10     7
于 2017-03-20T21:53:31.390 回答
0

这似乎几乎就像生成一个Sudoku网格。下面的代码运行得非常快，但R可以做一些小的优化：

backtrack = function(n = 10){
  x = matrix(NA, ncol = n, nrow = n)
  cells = list()
  k = 1
  for (i in 1:n){
    for (j in 1:n){
      cells[[k]] = sample(1:n)
      k = k + 1
    }
  }

  i = 0
  while (i < n*n){
    candidates = cells[[i + 1]]
    idx = sample(1:length(candidates), 1)
    val = candidates[idx]

    if (length(candidates) == 0){
      cells[[i + 1]] = sample(1:n)
      i = i - 1
      x[as.integer(i/n) + 1,  i %% n + 1] = NA
    }
    else {
      rr = as.integer(i/n) + 1
      cc = i %% n + 1
      if ((val %in% x[rr, ]) || (val %in% x[, cc])){
        candidates = candidates[-idx]
        cells[[i + 1]] = candidates
      }
      else{
        x[as.integer(i/n) + 1, i %% n + 1] = val
        candidates = candidates[-idx]
        cells[[i + 1]] = candidates
        i = i + 1
      }
    }
  }
  x
}

测试：

set.seed(1) # Please change this
x = backtrack(10)
print(x)

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


any(apply(x, 1, function(r)any(duplicated(r)))) # FALSE
any(apply(x, 2, function(r)any(duplicated(r)))) # FALSE
于 2017-03-20T21:14:03.837 回答