如何在R中的每个位置生成具有给定字母集的所有字符串排列答案

【问题标题】：How to generate all string permutations with given sets of letters at each position in R如何在R中的每个位置生成具有给定字母集的所有字符串排列
【发布时间】：2020-06-18 16:01:18
【问题描述】：

我确信以前有人问过并解决了这个问题，但可能我正在寻找错误的术语。我找不到相关的线程。

在R，我想生成所有可能的单词/字符串，其中每个位置只能取一组值，比如

pos1 可以是 ABC
pos2 可以是 ABCD
pos3 可以是 ABC
pos4 可以是 BCD

等等。例如：BABC 是一个解决方案，但 DABC 不是。

如果您能指出我的解决方案，我将不胜感激！

感谢您的宝贵时间！

...timb！，timc！，timd！ ...

谢谢，芽

【问题讨论】：

你说 BABA 可以是一个解决方案，但在第 4 位没有可用的 A。这是一个错字还是我误解了你？
哈哈，确实如此，很好发现 :-D 感谢您的解决方案，现在检查一下！

标签： r string permutation

【解决方案1】：

在 Base R 中，我们可以执行以下操作

pos1 <- c('A','B','C')
pos2 <- c('A','B','C','D')
pos4 <- c('B','C','D')

AllPos <- list(pos1,pos2,pos3,pos4)

result <- AllPos[1]

for(i in AllPos[-1] ){
    result <-  apply(merge(result ,i),1,paste0,collapse="")
}

> result 
  [1] "AAAB" "BAAB" "CAAB" "ABAB" "BBAB" "CBAB" "ACAB" "BCAB" "CCAB" "ADAB"
 [11] "BDAB" "CDAB" "AABB" "BABB" "CABB" "ABBB" "BBBB" "CBBB" "ACBB" "BCBB"
 [21] "CCBB" "ADBB" "BDBB" "CDBB" "AACB" "BACB" "CACB" "ABCB" "BBCB" "CBCB"
 [31] "ACCB" "BCCB" "CCCB" "ADCB" "BDCB" "CDCB" "AAAC" "BAAC" "CAAC" "ABAC"
 [41] "BBAC" "CBAC" "ACAC" "BCAC" "CCAC" "ADAC" "BDAC" "CDAC" "AABC" "BABC"
 [51] "CABC" "ABBC" "BBBC" "CBBC" "ACBC" "BCBC" "CCBC" "ADBC" "BDBC" "CDBC"
 [61] "AACC" "BACC" "CACC" "ABCC" "BBCC" "CBCC" "ACCC" "BCCC" "CCCC" "ADCC"
 [71] "BDCC" "CDCC" "AAAD" "BAAD" "CAAD" "ABAD" "BBAD" "CBAD" "ACAD" "BCAD"
 [81] "CCAD" "ADAD" "BDAD" "CDAD" "AABD" "BABD" "CABD" "ABBD" "BBBD" "CBBD"
 [91] "ACBD" "BCBD" "CCBD" "ADBD" "BDBD" "CDBD" "AACD" "BACD" "CACD" "ABCD"
[101] "BBCD" "CBCD" "ACCD" "BCCD" "CCCD" "ADCD" "BDCD" "CDCD"

【讨论】：

我接受这个，因为这是长字符串最具可扩展性的解决方案！ require(stringr); AllPos <- lapply(AllPos, str_split_fixed, pattern = "", n=Inf); AllPos <- lapply(AllPos, as.character) 如果你从单词而不是字母向量开始会更容易

【解决方案2】：

快速而肮脏的基础 R 解决方案...

p1 <- "ABC"
p2 <- "ABCD"
p3 <- "ABC"
p4 <- "BCD"

apply(expand.grid(strsplit(p1, "")[[1]], strsplit(p2, "")[[1]],
            strsplit(p3, "")[[1]], strsplit(p4, "")[[1]]), 1, paste0, 
      collapse = "")
#>   [1] "AAAB" "BAAB" "CAAB" "ABAB" "BBAB" "CBAB" "ACAB" "BCAB" "CCAB" "ADAB"
#>  [11] "BDAB" "CDAB" "AABB" "BABB" "CABB" "ABBB" "BBBB" "CBBB" "ACBB" "BCBB"
#>  [21] "CCBB" "ADBB" "BDBB" "CDBB" "AACB" "BACB" "CACB" "ABCB" "BBCB" "CBCB"
#>  [31] "ACCB" "BCCB" "CCCB" "ADCB" "BDCB" "CDCB" "AAAC" "BAAC" "CAAC" "ABAC"
#>  [41] "BBAC" "CBAC" "ACAC" "BCAC" "CCAC" "ADAC" "BDAC" "CDAC" "AABC" "BABC"
#>  [51] "CABC" "ABBC" "BBBC" "CBBC" "ACBC" "BCBC" "CCBC" "ADBC" "BDBC" "CDBC"
#>  [61] "AACC" "BACC" "CACC" "ABCC" "BBCC" "CBCC" "ACCC" "BCCC" "CCCC" "ADCC"
#>  [71] "BDCC" "CDCC" "AAAD" "BAAD" "CAAD" "ABAD" "BBAD" "CBAD" "ACAD" "BCAD"
#>  [81] "CCAD" "ADAD" "BDAD" "CDAD" "AABD" "BABD" "CABD" "ABBD" "BBBD" "CBBD"
#>  [91] "ACBD" "BCBD" "CCBD" "ADBD" "BDBD" "CDBD" "AACD" "BACD" "CACD" "ABCD"
#> [101] "BBCD" "CBCD" "ACCD" "BCCD" "CCCD" "ADCD" "BDCD" "CDCD"

^{由reprex package (v0.3.0) 于 2020-06-18 创建}

【讨论】：

【解决方案3】：

expand.grid是你的朋友。

一个简单的解决方案：

apply(expand.grid(list(
    LETTERS[1:3],
    LETTERS[1:4],
    LETTERS[1:3],
    LETTERS[2:4])), 1, paste, collapse = "")
#>   [1] "AAAB" "BAAB" "CAAB" "ABAB" "BBAB" "CBAB" "ACAB" "BCAB" "CCAB" "ADAB"
#>  [11] "BDAB" "CDAB" "AABB" "BABB" "CABB" "ABBB" "BBBB" "CBBB" "ACBB" "BCBB"
#>  [21] "CCBB" "ADBB" "BDBB" "CDBB" "AACB" "BACB" "CACB" "ABCB" "BBCB" "CBCB"
#>  [31] "ACCB" "BCCB" "CCCB" "ADCB" "BDCB" "CDCB" "AAAC" "BAAC" "CAAC" "ABAC"
#>  [41] "BBAC" "CBAC" "ACAC" "BCAC" "CCAC" "ADAC" "BDAC" "CDAC" "AABC" "BABC"
#>  [51] "CABC" "ABBC" "BBBC" "CBBC" "ACBC" "BCBC" "CCBC" "ADBC" "BDBC" "CDBC"
#>  [61] "AACC" "BACC" "CACC" "ABCC" "BBCC" "CBCC" "ACCC" "BCCC" "CCCC" "ADCC"
#>  [71] "BDCC" "CDCC" "AAAD" "BAAD" "CAAD" "ABAD" "BBAD" "CBAD" "ACAD" "BCAD"
#>  [81] "CCAD" "ADAD" "BDAD" "CDAD" "AABD" "BABD" "CABD" "ABBD" "BBBD" "CBBD"
#>  [91] "ACBD" "BCBD" "CCBD" "ADBD" "BDBD" "CDBD" "AACD" "BACD" "CACD" "ABCD"
#> [101] "BBCD" "CBCD" "ACCD" "BCCD" "CCCD" "ADCD" "BDCD" "CDCD"

^{由reprex package (v0.3.0) 于 2020 年 6 月 18 日创建}

【讨论】：

这不是和我之前发布的解决方案一样吗？