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我想生成两个向量的所有组合，给定两个约束：第一个向量中的字符永远不能超过 3 个，并且第二个向量中必须始终至少有一个字符。我还想改变组合中的最终字符数。

例如，这里有两个向量：

vec1=c("A","B","C","D")
vec2=c("W","X","Y","Z")

假设我想要组合中的 3 个字符。可能的可接受的排列是："A" "B" "X"或"A" "Y" "Z"。一个不可接受的排列是："A" "B" "C"因为 . 中没有至少一个字符vec2。

现在说我想要组合中的 5 个字符。可能的可接受的排列是："A" "C" "Z" "Y"或"A" "Y" "Z" "X"。一个不可接受的排列是："A" "C" "D" "B" "X"因为有 >3 个字符来自vec2.

我想我可以expand.grid用来生成所有组合，然后以某种方式生成子集，但必须有一种更简单的方法。提前致谢！

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1 回答 1

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我不确定这是否更容易，但是您可以使用此策略放弃不满足您的条件的排列：

  1. 从中生成所有vec1可接受的组合。

  2. 从中生成所有vec2可接受的组合。

  3. 生成所有组合，从 1 中获取一种解决方案。+ 从 2 中获取一种解决方案。在这里，我将在之后使用条件 3 进行过滤。

  4. （如果你正在寻找组合，你就完成了，否则:) 在每个结果中产生所有字母排列。

现在，让我们

vec1 <- LETTERS [1:4]
vec2 <- LETTERS [23:26]

## lists can eat up lots of memory, so use character vectors instead.
combine <- function (x, y) 
  combn (y, x, paste, collapse = "")

res1 <- unlist (lapply (0:3, combine, vec1))
res2 <- unlist (lapply (1:length (vec2), combine, vec2))

现在我们有：

> res1
 [1] ""    "A"   "B"   "C"   "D"   "AB"  "AC"  "AD"  "BC"  "BD"  "CD"  "ABC"
[13] "ABD" "ACD" "BCD"
> res2
 [1] "W"    "X"    "Y"    "Z"    "WX"   "WY"   "WZ"   "XY"   "XZ"   "YZ"  
[11] "WXY"  "WXZ"  "WYZ"  "XYZ"  "WXYZ"

res3 <- outer (res1, res2, paste0)
res3 <- res3 [nchar (res3) == 5]

所以你在这里：

> res3
 [1] "ABCWX" "ABDWX" "ACDWX" "BCDWX" "ABCWY" "ABDWY" "ACDWY" "BCDWY" "ABCWZ"
[10] "ABDWZ" "ACDWZ" "BCDWZ" "ABCXY" "ABDXY" "ACDXY" "BCDXY" "ABCXZ" "ABDXZ"
[19] "ACDXZ" "BCDXZ" "ABCYZ" "ABDYZ" "ACDYZ" "BCDYZ" "ABWXY" "ACWXY" "ADWXY"
[28] "BCWXY" "BDWXY" "CDWXY" "ABWXZ" "ACWXZ" "ADWXZ" "BCWXZ" "BDWXZ" "CDWXZ"
[37] "ABWYZ" "ACWYZ" "ADWYZ" "BCWYZ" "BDWYZ" "CDWYZ" "ABXYZ" "ACXYZ" "ADXYZ"
[46] "BCXYZ" "BDXYZ" "CDXYZ" "AWXYZ" "BWXYZ" "CWXYZ" "DWXYZ"

如果您更喜欢将结果拆分为单个字母：

res <- matrix (unlist (strsplit (res3, "")), nrow = length (res3), byrow = TRUE)
> res
      [,1] [,2] [,3] [,4] [,5]
 [1,] "A"  "B"  "C"  "W"  "X" 
 [2,] "A"  "B"  "D"  "W"  "X" 
 [3,] "A"  "C"  "D"  "W"  "X" 
 [4,] "B"  "C"  "D"  "W"  "X"

（剪断）

[51,] "C"  "W"  "X"  "Y"  "Z" 
[52,] "D"  "W"  "X"  "Y"  "Z"

哪些是你的组合。

于 2012-05-09T17:32:09.400 回答