Question

我有一个像这样的矩阵：

http://i.imgur.com/3HiEBm4.png

您可以像这样加载：

matrix = structure(c("-", "-", "C", "G", "C", "A", "-", "0", "V", "V", 
"V", "V", "C", "H", "D", "V", "DV", "V", "A", "H", "H", "D", 
"DV", "D", "C", "H", "DH", "DH", "D", "V", "G", "H", "H", "D", 
"H", "D", "T", "H", "H", "H", "DH", "DH", "A", "H", "H", "H", 
"DH", "D", "T", "H", "H", "H", "DH", "H"), .Dim = c(6L, 9L))

从右下角开始，目标是遵循指示（D =对角线向0移动，H =向左移动，V =向上移动），以便所有路径都达到零。如您所见，有些细胞有多个方向（例如DH）。

我试图通过这样的矩阵找到所有可能的路径。我用递归做了。但我很难正确存储路径。看起来当函数返回到旧单元格以取向另一个方向时，它会将路径附加到错误的列表。

这是我的递归函数代码：

threading = function(matrix,i,j,list) { #Function wants the matrix to be threaded, number of rows and cols, and an empty list
  if (matrix[i,j] == 0) { #If the recursion has arrived at zero, stop and print out the path that arrived there
    print(list)
  }
  else { #If still elsewhere inside the matrix...
    for (move in strsplit(matrix[i,j],"")[[1]]) { #Iterate through each move in that cell
      if (move == "D") { #If a move is D...
        list = paste(list, "D", sep="") #Append that to the path
        threading(matrix,i-1,j-1,list) #Send the function to the diagonal cell
      }
      if (move == "V") { #If a move is V...
        list = paste(list, "V", sep="") #Append that to the path
        threading(matrix,i-1,j,list) #Send the function to the above cell
      }
      if (move == "H") { #If a move is H...
        list = paste(list, "H", sep="") #Append that to the path
        threading(matrix,i,j-1,list) #Send the function to the left cell
      }
    }
  }
}

所以当我使用上面的矩阵运行它时，它会将其作为输出：

> threading(matrix, 6,9, emptylist)
[1] "HDDDDHH"
[1] "HDDDDHHD"
[1] "HDDHHDDD"

踢球者是第二个两条路径的第二个字符是错误的，但其他一切都是正确的。我该如何避免？我无法弄清楚如何正确存储路径，而不会回到原来的路径。我认为它与附加的排序和将函数发送到下一个单元格有关，但如果我反转它们，那么附加的东西永远不会发生......

Answer 1

问题在于：

list = paste(list, "*", sep="")

当您点击有两个选项的单元格时，例如“VH”，for循环将经历两次迭代：第一次迭代修改list，然后将修改后的值传递给第二次迭代。相反，每次迭代必须使用原始值。所以你可以替换为：

l = paste(list, "*", sep="")

并将l而不是list传递给threading来电。

另外，最好避免命名变量matrix或list，因为它们也是函数的名称。

通过R中的方向矩阵追溯

1 个答案: