R 查找字符串中重叠的长度

你们知道有什么现成的方法来获取两个字符串的长度和重叠吗？但是，只有使用

R

，可能是来自

stringr

？我在找，不幸的是没有成功

str1 <- 'ABCDE'
str2 <- 'CDEFG'

str_overlap(str1, str2)
'CDE'

str_overlap_len(str1, str2)
3

str1希望这有帮助：

library(stringr)

larsub<-function(x) {
  a<-x[1]
  b<-x[2]
  # get all forward substrings of a
  for(n in seq(1,nchar(a)))
    {
    sb<-unique(combn(strsplit(a, "")[[1]],n, FUN=paste, collapse=""))
    if(length(unlist(str_extract_all(b,sb)))==0){ 
      r<-prior
      return(r)
      }
    prior<-unlist(str_extract_all(b,sb))
    }

}

c1<-larsub(c('ABCD','BCDE'))
c2<-larsub(c('ABDFD','BCDE'))
c3<-larsub(c('CDEWQ','DEQ'))
c4<-larsub(c('BNEOYJBELMGY','BELM'))
print(c1)
print(c2)
print(c3)
print(c4)

库（stringr）
larsub在我看来，您（OP）并不十分关心代码的性能，而是更感兴趣的是在没有现成函数的情况下解决它的潜在方法。下面是我计算最长公共子串的一个例子。我必须注意，这只返回找到的第一个最大的公共子字符串，即使可能有几个相同长度的子字符串。这是你可以修改以适应你的需要的东西。请不要期望这是超级快-它不会

foo <- function(str1, str2, ignore.case = FALSE, verbose = FALSE) {

  if(ignore.case) {
    str1 <- tolower(str1)
    str2 <- tolower(str2)
  }

  if(nchar(str1) < nchar(str2)) {
    x <- str2
    str2 <- str1
    str1 <- x
  }

  x <- strsplit(str2, "")[[1L]]
  n <- length(x)
  s <- sequence(seq_len(n))
  s <- split(s, cumsum(s == 1L))
  s <- rep(list(s), n)

  for(i in seq_along(s)) {
    s[[i]] <- lapply(s[[i]], function(x) {
      x <- x + (i-1L)
      x[x <= n]
    })
    s[[i]] <- unique(s[[i]])
  }

  s <- unlist(s, recursive = FALSE)
  s <- unique(s[order(-lengths(s))])

  i <- 1L
  len_s <- length(s)
  while(i < len_s) {
    lcs <- paste(x[s[[i]]], collapse = "")
    if(verbose) cat("now checking:", lcs, "\n")
    check <- grepl(lcs, str1, fixed = TRUE)
    if(check) {
      cat("the (first) longest common substring is:", lcs, "of length", nchar(lcs), "\n")
      break
    } else {
      i <- i + 1L 
    }
  }
}

str1 <- 'ABCDE'
str2 <- 'CDEFG'
foo(str1, str2)
# the (first) longest common substring is: CDE of length 3 

str1 <- 'ATTAGACCTG'
str2 <- 'CCTGCCGGAA'
foo(str1, str2)
# the (first) longest common substring is: CCTG of length 4

str1 <- 'foobarandfoo'
str2 <- 'barand'
foo(str1, str2)
# the (first) longest common substring is: barand of length 6 

str1 <- 'EFGABCDE'
str2 <- 'ABCDECDE'
foo(str1, str2)
# the (first) longest common substring is: ABCDE of length 5 


set.seed(2018)
str1 <- paste(sample(c(LETTERS, letters), 500, TRUE), collapse = "")
str2 <- paste(sample(c(LETTERS, letters), 250, TRUE), collapse = "")

foo(str1, str2, ignore.case = TRUE)
# the (first) longest common substring is: oba of length 3 

foo(str1, str2, ignore.case = FALSE)
# the (first) longest common substring is: Vh of length 2 

foo当输入为str1时，预期的输出是什么？您正在寻找最长的公共子字符串吗？检查一下，我不明白你自己做和避免动态编程是什么意思这里是我的实现（改编自@gaurav taneja，如果字符串a比b长，效率更高）：
stru重叠=函数（a，b）{if（nchar（a）>nchar（b））{a0=a；a=b；b=a0}；对于（n在seq（1，nchar（a））{sb=unique（combn（strsplit（a），“”[1]]，n，FUN=paste，collapse=“”）；if（长度（unlist（str_extract_all（b，sb））=0{r=prior；return（r）}；prior=unlist（str_extract_all（b，sb））；prior}
感谢您的努力。目前我正在尝试创建相同的函数，我将比较结果。当然，在这种情况下，性能根本不重要。我只是想为自己的实践和训练解决一些信息问题。干杯
str1 <- 'EFGABCDE'
str2 <- 'ABCDECDE'

str_overlap(str1, str2)
'ABCDE'

str_overlap_len(str1, str2)
5

library(stringr)

larsub<-function(x) {
  a<-x[1]
  b<-x[2]
  # get all forward substrings of a
  for(n in seq(1,nchar(a)))
    {
    sb<-unique(combn(strsplit(a, "")[[1]],n, FUN=paste, collapse=""))
    if(length(unlist(str_extract_all(b,sb)))==0){ 
      r<-prior
      return(r)
      }
    prior<-unlist(str_extract_all(b,sb))
    }

}

c1<-larsub(c('ABCD','BCDE'))
c2<-larsub(c('ABDFD','BCDE'))
c3<-larsub(c('CDEWQ','DEQ'))
c4<-larsub(c('BNEOYJBELMGY','BELM'))
print(c1)
print(c2)
print(c3)
print(c4)

foo <- function(str1, str2, ignore.case = FALSE, verbose = FALSE) {

  if(ignore.case) {
    str1 <- tolower(str1)
    str2 <- tolower(str2)
  }

  if(nchar(str1) < nchar(str2)) {
    x <- str2
    str2 <- str1
    str1 <- x
  }

  x <- strsplit(str2, "")[[1L]]
  n <- length(x)
  s <- sequence(seq_len(n))
  s <- split(s, cumsum(s == 1L))
  s <- rep(list(s), n)

  for(i in seq_along(s)) {
    s[[i]] <- lapply(s[[i]], function(x) {
      x <- x + (i-1L)
      x[x <= n]
    })
    s[[i]] <- unique(s[[i]])
  }

  s <- unlist(s, recursive = FALSE)
  s <- unique(s[order(-lengths(s))])

  i <- 1L
  len_s <- length(s)
  while(i < len_s) {
    lcs <- paste(x[s[[i]]], collapse = "")
    if(verbose) cat("now checking:", lcs, "\n")
    check <- grepl(lcs, str1, fixed = TRUE)
    if(check) {
      cat("the (first) longest common substring is:", lcs, "of length", nchar(lcs), "\n")
      break
    } else {
      i <- i + 1L 
    }
  }
}

str1 <- 'ABCDE'
str2 <- 'CDEFG'
foo(str1, str2)
# the (first) longest common substring is: CDE of length 3 

str1 <- 'ATTAGACCTG'
str2 <- 'CCTGCCGGAA'
foo(str1, str2)
# the (first) longest common substring is: CCTG of length 4

str1 <- 'foobarandfoo'
str2 <- 'barand'
foo(str1, str2)
# the (first) longest common substring is: barand of length 6 

str1 <- 'EFGABCDE'
str2 <- 'ABCDECDE'
foo(str1, str2)
# the (first) longest common substring is: ABCDE of length 5 


set.seed(2018)
str1 <- paste(sample(c(LETTERS, letters), 500, TRUE), collapse = "")
str2 <- paste(sample(c(LETTERS, letters), 250, TRUE), collapse = "")

foo(str1, str2, ignore.case = TRUE)
# the (first) longest common substring is: oba of length 3 

foo(str1, str2, ignore.case = FALSE)
# the (first) longest common substring is: Vh of length 2