r - 从一列制作两列，涵盖所有组合

Question

这看似简单的问题却让我很头疼(这不是作业，而是实际研究中的症结所在)

我有一个包含 2266 个级别的列表。该列表看起来有点像这样:

[1] ~/folder1/folder1/a.bin
[2] ~/folder1/folder1/b.bin
[3] ~/folder1/folder1/c.bin
[4] ~/folder1/folder2/a.bin
[5] ~/folder1/folder2/b.bin
[6] ~/folder1/folder2/c.bin

解释一下:该列表是我使用 readBin 函数读取的二进制文件的文件名。我想将每一行与其他每一行进行比较，所以我想要的是两列，其中包含从我的单列派生的所有唯一组合。

(choose 2266,2) 告诉我，我们的单列有 2566245 种组合成两列。

`expand.grid()似乎让我成功了一半。但组合数量是我需要的四倍:我得到两行，每行 5132490。这意味着存在重复:1 + 2 和 2 + 1 对于我的目的来说是相同的。

expand.grid.df 与 unique=TRUE 似乎也没有帮助。

我的最后一个想法是对 500 万行中的每一行进行 md5 散列，并尝试以这种方式检测重复项。

我正在寻找某种方法来制作两个列表，涵盖我的列表的 2566245 种组合。或者通过某种方式删除所有重复项。 我想我并不完全热衷于使用 R，并且已经研究过 awk 或 sed 来做同样的事情。但尚未成功。

Answer 1

我认为您正在寻找类似于 expand.grid 的 combn，使用 @Arun 数据，

v <- c("~/folder1/folder1/a.bin", 
       "~/folder1/folder1/b.bin", 
       "~/folder1/folder1/c.bin", 
       "~/folder1/folder2/a.bin", 
       "~/folder1/folder2/b.bin", 
       "~/folder1/folder2/c.bin")
do.call(rbind,combn(v,2,simplify=F))

    [,1]                      [,2]                     
 [1,] "~/folder1/folder1/a.bin" "~/folder1/folder1/b.bin"
 [2,] "~/folder1/folder1/a.bin" "~/folder1/folder1/c.bin"
 [3,] "~/folder1/folder1/a.bin" "~/folder1/folder2/a.bin"
 [4,] "~/folder1/folder1/a.bin" "~/folder1/folder2/b.bin"
 [5,] "~/folder1/folder1/a.bin" "~/folder1/folder2/c.bin"
 [6,] "~/folder1/folder1/b.bin" "~/folder1/folder1/c.bin"
 [7,] "~/folder1/folder1/b.bin" "~/folder1/folder2/a.bin"
 [8,] "~/folder1/folder1/b.bin" "~/folder1/folder2/b.bin"
 [9,] "~/folder1/folder1/b.bin" "~/folder1/folder2/c.bin"
[10,] "~/folder1/folder1/c.bin" "~/folder1/folder2/a.bin"
[11,] "~/folder1/folder1/c.bin" "~/folder1/folder2/b.bin"
[12,] "~/folder1/folder1/c.bin" "~/folder1/folder2/c.bin"
[13,] "~/folder1/folder2/a.bin" "~/folder1/folder2/b.bin"
[14,] "~/folder1/folder2/a.bin" "~/folder1/folder2/c.bin"
[15,] "~/folder1/folder2/b.bin" "~/folder1/folder2/c.bin"

编辑

我认为路径格式使问题变得过于复杂。如果我们使用例如字母代替文件名，我们会得到:

do.call(rbind,combn(letters[1:4],2,simplify=F))
     [,1] [,2]
[1,] "a"  "b" 
[2,] "a"  "c" 
[3,] "a"  "d" 
[4,] "b"  "c" 
[5,] "b"  "d" 
[6,] "c"  "d"  

所以如您所见，没有重复。