R正則表達式 - 在括號之間分割

Question

假設我有一個字符串x ，我想像這樣拆分它：

x <- "(A|C|T)AG(C|T)(A|C|G|T)(A|C|G|T)(A|C|G|T)(A|C|G|T)(A|C|G|T)GCC(C|T)(A|C|G|T)(A|C|G|T)(A|C|G)"

# Desired output
[1]  "(A|C|T)"  "A"  "G"  "(C|T)"  "(A|C|G|T)"  "(A|C|G|T)"  "(A|C|G|T)"  
[8]  "(A|C|G|T)"  "(A|C|G|T)"  "G"  "C"  "C"  "(C|T)"  "(A|C|G|T)"  
[15] "(A|C|G|T)"  "(A|C|G)"  

我正在使用這個分裂函數，但我無法分割不在括號中的字符串。 處理這個正則表達式問題的最佳方法是什么？

splitme <- function(x) {
  x <- unlist(strsplit(x, "(?=\\()", perl=TRUE))
  x <- unlist(strsplit(x, "(?<=\\))", perl=TRUE))
  for (i in which(x=="(")) {
    x[i+1] <- paste(x[i], x[i+1], sep="")
  }
  x[-which(x=="(")]
}

splitme(x)
 [1] "(A|C|T)"   "AG"        "(C|T)"     "(A|C|G|T)" "(A|C|G|T)" "(A|C|G|T)" "(A|C|G|T)" "(A|C|G|T)" "GCC"      
[10] "(C|T)"     "(A|C|G|T)" "(A|C|G|T)" "(A|C|G)"  

Answer 1

這樣的事情應該有效：

> library(stringi)

> unlist(stri_extract_all_regex(x, "\\([ACGT\\|]*\\)|[ACGT]"))
 [1] "(A|C|T)"   "A"         "G"         "(C|T)"     "(A|C|G|T)" "(A|C|G|T)"
 [7] "(A|C|G|T)" "(A|C|G|T)" "(A|C|G|T)" "G"         "C"         "C"        
[13] "(C|T)"     "(A|C|G|T)" "(A|C|G|T)" "(A|C|G)"  

\\\\([ACGT\\\\|]*\\\\)將匹配括號中的所有內容和[ACGT]剩余的基數。

Answer 2

它看起來像你想拆分后每串) ，而這之后無論是另一封信或由每個字母后(如果這是你想要的行為，您可以使用此：

pat <- "(?<=\\))|(?<=[[:alpha:]])(?=[[:alpha:]\\(])"
strsplit(x, pat, perl=TRUE)[[1]]
#  [1] "(A|C|T)"   "A"         "G"         "(C|T)"     "(A|C|G|T)" "(A|C|G|T)"
#  [7] "(A|C|G|T)" "(A|C|G|T)" "(A|C|G|T)" "G"         "C"         "C"        
# [13] "(C|T)"     "(A|C|G|T)" "(A|C|G|T)" "(A|C|G)" 

Answer 3

要拆分單個字母，您只需運行strsplit(x, "") 。 您所要做的就是確保不要將其應用於“已完成”的字符串（即帶括號的字符串）。

y = splitme(x)
Indices = !which(grepl(y, "\\("))
y[Indices] = strsplit(y[Indices], "")
unlist(y)
 [1] "(A|C|T)"   "A"         "G"         "(C|T)"     "(A|C|G|T)" "(A|C|G|T)" "(A|C|G|T)" "(A|C|G|T)"
 [9] "(A|C|G|T)" "G"         "C"         "C"         "(C|T)"     "(A|C|G|T)" "(A|C|G|T)" "(A|C|G)"