在字符串中创建所有字母替换组合

Question

我有一个字符串“ECET”，我想创建所有可能的字符串，其中我用“X”替换一个或多个字母（除第一个之外的所有字母）。

所以在这种情况下，我的结果是：

> result
[1] "EXET" "ECXT" "ECEX" "EXXT" "EXEX" "ECXX" "EXXX"

关于如何解决这个问题的任何想法？

这不仅仅是创建“X”的可能组合/排列，还包括如何将它们与现有字符串组合。

Answer 1

使用combn 的FUN 参数：

a <- "ECET"

fun <- function(n, string) {
  combn(nchar(string), n, function(x) {
    s <- strsplit(string, '')[[1]]
    s[x] <- 'X'
    paste(s, collapse = '')
  } )
}
lapply(seq_len(nchar(a)), fun, string = a)

[[1]]
[1] "XCET" "EXET" "ECXT" "ECEX"

[[2]]
[1] "XXET" "XCXT" "XCEX" "EXXT" "EXEX" "ECXX"

[[3]]
[1] "XXXT" "XXEX" "XCXX" "EXXX"

[[4]]
[1] "XXXX"

unlist 获取单个向量。可能有更快的解决方案。

保持第一个字符不变：

paste0(
  substring(a, 1, 1),
  unlist(lapply(seq_len(nchar(a) - 1), fun, string = substring(a, 2)))
)

[1] "EXET" "ECXT" "ECEX" "EXXT" "EXEX" "ECXX" "EXXX"

Answer 2

这是一个递归解决方案：

f <- function(x,pos=2){
  if(pos <= nchar(x))
    c(f(x,pos+1), f(`substr<-`(x, pos, pos, "X"),pos+1))
  else x
}
f(x)[-1]
# [1] "ECEX" "ECXT" "ECXX" "EXET" "EXEX" "EXXT" "EXXX"

或者使用expand.grid：

do.call(paste0, expand.grid(c(substr(x,1,1),lapply(strsplit(x,"")[[1]][-1], c, "X"))))[-1]
# [1] "EXET" "ECXT" "EXXT" "ECEX" "EXEX" "ECXX" "EXXX"

或者使用combn/Reduce/substr<-:

combs <- unlist(lapply(seq(nchar(x)-1),combn, x =seq(nchar(x))[-1],simplify = F),F)
sapply(combs, Reduce, f= function(x,y) `substr<-`(x,y,y,"X"), init = x)
# [1] "EXET" "ECXT" "ECEX" "EXXT" "EXEX" "ECXX" "EXXX"

---

解释第二个解决方案

pairs0 <- lapply(strsplit(x,"")[[1]][-1], c, "X") # pairs of original letter + "X"
pairs1 <- c(substr(x,1,1), pairs0)                # including 1st letter (without "X")
do.call(paste0, expand.grid(pairs1))[-1]          # expand into data.frame and paste

Answer 3

有点为了使用二进制逻辑添加另一个选项：

假设您的字符串总是 4 个字符长：

input<-"ECET"
invec <- strsplit(input,'')[[1]]
sapply(1:7, function(x) {
  z <- invec
  z[rev(as.logical(intToBits(x))[1:4])] <- "X"
  paste0(z,collapse = '')
})

[1] "ECEX" "ECXT" "ECXX" "EXET" "EXEX" "EXXT" "EXXX"

如果字符串必须更长，您可以用 2 的幂计算值，应该这样做：

input<-"ECETC"
pow <- nchar(input)
invec <- strsplit(input,'')[[1]]
sapply(1:(2^(pow-1) - 1), function(x) {
  z <- invec
  z[rev(as.logical(intToBits(x))[1:(pow)])] <- "X"
  paste0(z,collapse = '')
})

[1] "ECETX" "ECEXC" "ECEXX" "ECXTC" "ECXTX" "ECXXC" "ECXXX" "EXETC" "EXETX" "EXEXC" "EXEXX" "EXXTC" "EXXTX" "EXXXC"
[15] "EXXXX"

这个想法是知道可能改变的数量，它是 3 个位置的二进制，所以 2^3 减去 1，因为我们不想保留无替换字符串：7

intToBits 返回整数的二进制值，为 5：

> intToBits(5)
 [1] 01 00 01 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00

R 默认使用 32 位，但是我们只想要一个与我们的字符串长度对应的逻辑向量，所以我们只保留原始字符串的 nchar。 然后我们转换为逻辑并反转这 4 个布尔值，因为我们永远不会触发最后一位（4 个字符为 8），它永远不会为真：

> intToBits(5)
 [1] 01 00 01 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00 00
> tmp<-as.logical(intToBits(5)[1:4])
> tmp
[1]  TRUE FALSE  TRUE FALSE
> rev(tmp)
[1] FALSE  TRUE FALSE  TRUE

为了避免覆盖我们的原始向量，我们将它复制到 z 中，然后使用这个逻辑向量替换 z 中的位置。

为了得到一个不错的输出，我们返回 paste0 并折叠为空，以重新创建单个字符串并检索字符向量。

Answer 4

另一个带有combn的版本，使用purrr：

s <- "ECET"
f <- function(x,y) {substr(x,y,y) <- "X"; x}
g <- function(x) purrr::reduce(x,f,.init=s)
unlist(purrr::map(1:(nchar(s)-1), function(x) combn(2:nchar(s),x,g)))

#[1] "EXET" "ECXT" "ECEX" "EXXT" "EXEX" "ECXX" "EXXX"

或没有咕噜声：

s <- "ECET"
f <- function(x,y) {substr(x,y,y) <- "X"; x}
g <- function(x) Reduce(f,x,s)
unlist(lapply(1:(nchar(s)-1),function(x) combn(2:nchar(s),x,g)))

Answer 5

这是一个基本的 R 解决方案，但我觉得它很复杂，有 3 个嵌套循环。

replaceChar <- function(x, char = "X"){
  n <- nchar(x)
  res <- NULL
  for(i in seq_len(n)){
    cmb <- combn(n, i)
    r <- apply(cmb, 2, function(cc){
      y <- x
      for(k in cc)
        substr(y, k, k) <- char
      y
    })
    res <- c(res, r)
  }
  res
}

x <- "ECET"

replaceChar(x)
replaceChar(x, "Y")
replaceChar(paste0(x, x))

Answer 6

带有布尔索引的矢量化方法：

permX <- function(text, replChar='X') {
    library(gtools)
    library(stringr)  
    # get TRUE/FALSE permutations for nchar(text)
    idx <- permutations(2, nchar(text),c(T,F), repeats.allowed = T)

    # we don't want the first character to be replaced
    idx <- idx[1:(nrow(idx)/2),]

    # split string into single chars
    chars <- str_split(text,'')

    # build data.frame with nrows(df) == nrows(idx)
    df = t(data.frame(rep(chars, nrow(idx))))

    # do replacing
    df[idx] <- replChar

    row.names(df) <- c()
    return(df)
}
permX('ECET')

[,1] [,2] [,3] [,4]  
[1,] "E"  "C"  "E"  "T"   
[2,] "E"  "C"  "E"  "X"  
[3,] "E"  "C"  "X"  "T"  
[4,] "E"  "C"  "X"  "X"  
[5,] "E"  "X"  "E"  "T"  
[6,] "E"  "X"  "E"  "X"  
[7,] "E"  "X"  "X"  "T"  
[8,] "E"  "X"  "X"  "X"  

Answer 7

一个更简单的解决方案

# expand.grid to get all combinations of the input vectors, result in a matrix
m <- expand.grid( c('E'), 
                  c('C','X'), 
                  c('E','X'), 
                  c('T','X') )

# then, optionally, apply to paste the columns together
apply(m, 1, paste0, collapse='')[-1]

[1] "EXET" "ECXT" "EXXT" "ECEX" "EXEX" "ECXX" "EXXX"