如何在 R 中用交错字符替换另一个字符串

Question

我有以下字符串：

x  <- "??????????DRHRTRHLAK??????????"
x2 <- "????????????????????TRCYHIDPHH"
x3 <- "FKDHKHIDVK????????????????????TRCYHIDPHH"
x4 <- "FKDHKHIDVK????????????????????"

我想做的是用另一个字符串替换所有 ？ 字符

rep <- "ndqeegillkkkkfpssyvv"

结果是：

ndqeegillkDRHRTRHLAKkkkfpssyvv           # x
ndqeegillkkkkfpssyvvTRCYHIDPHH           # x2
FKDHKHIDVKndqeegillkkkkfpssyvvTRCYHIDPHH # x3
FKDHKHIDVKndqeegillkkkkfpssyvv           # x4

基本上，保持rep的顺序替换为x中的交错字符DRHRTRHLAK。

< code>rep的总长度与< code >的总长度相同？，20个字符。

请注意，我不想作为额外的步骤再次手动拆分< code>rep。

我尝试了这个，但失败了：

>gsub(pattern = "\\?+", replacement = rep, x = x)
[1] "ndqeegillkkkkfpssyvvDRHRTRHLAKndqeegillkkkkfpssyvv"

Answer 1

你可以数数 ？s，然后基于此削减代表：

x <- "??????????DRHRTRHLAK??????????"
rep <- "ndqeegillkkkkfpssyvv"

pattern <- "(\\?+)(DRHRTRHLAK)(\\?+)"
n <- nchar(gsub(pattern, "\\1", x))

gsub(pattern, paste0(substr(rep, 1, n), "\\2", substr(rep, n+1, nchar(rep))), x)
#[1] "ndqeegillk??????????kkkfpssyvv"

一个非常冗长的方法是执行if-else链，检查在哪里s是，并相应地替换rep。

if(grepl("^\\?.+\\?$", x)){ #?'s on both ends
  n <- gsub(pattern, "\\1", x) %>% nchar()
  gsub(pattern, paste0(substr(rep, 1, n), "\\2", substr(rep, n+1, nchar(rep))), x)
} else if(grepl("^\\?", x)){ #?'s only on start
  n <- gsub(pattern, "\\1", x) %>% nchar()
  gsub(pattern, paste0(substr(rep, 1, n), "\\2"), x)
} else if(grepl("\\?$", x)){ #?'s only on end
  n <- gsub(pattern, "\\2", x) %>% nchar()
  gsub(pattern, paste0("\\2", substr(rep, 1, n)), x)
} else if(grepl("^[A-Z]+\\?+[A-Z]+$", x)){ #?'s only on middle
  n <- gsub(pattern, "\\2", x) %>% nchar()
  gsub("([A-Z]+)\\?+([A-Z]+)", paste0("\\1", substr(rep, 1, n), "\\2"), x)
}

Answer 2

使用 substr（） 进行字符串拆分：

x <- "??????????DRHRTRHLAK??????????"
rep <- "ndqeegillkkkkfpssyvv"
x<-gsub(pattern = "^\\?+", replacement = substr(rep, 1, 10), x = x)
x<-gsub(pattern = "\\?+$", replacement = substr(rep, 11, 20), x = x)
x
#[1] "ndqeegillkDRHRTRHLAKkkkfpssyvv"

Regex^匹配开始，而$则匹配结束。

Answer 3

示例数据：

x <- c(
    "??????????DRHRTRHLAK??????????",
    "????????????????????TRCYHIDPHH",
    "FKDHKHIDVK????????????????????TRCYHIDPHH"
)
rep <- "ndqeegillkkkkfpssyvv"

使用正则表达式修复它

gr <- gregexpr("\\?+", x)
csml <- lapply(gr, \(x) cumsum(attr(x, "match.length")) )
regmatches(x, gr) <- lapply(csml, \(x) substring(rep, c(1,x[-length(x)]+1), x))
#[1] "ndqeegillkDRHRTRHLAKkkkfpssyvv"          
#[2] "ndqeegillkkkkfpssyvvTRCYHIDPHH"          
#[3] "FKDHKHIDVKndqeegillkkkkfpssyvvTRCYHIDPHH"