Regex R正则表达式-括号之间的拆分
假设我有一个字符串Regex R正则表达式-括号之间的拆分,regex,r,Regex,R,假设我有一个字符串x,我想像这样拆分它: 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
x
,我想像这样拆分它:
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)"
x像这样的东西应该可以工作:
> 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\\\\\\]*\\)
将匹配括号中的所有内容,并将[ACGT]
剩余的基数进行匹配。要拆分单个字母,只需运行strsplit(x,”)
。您所要做的只是确保不要将其应用于“完成”字符串(即带括号的字符串)
看起来您希望在每个)
后拆分字符串,并在每个字母后拆分后面跟有另一个字母或(
)的字符串。如果这是您想要的行为,您可以使用以下方法:
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)"
pat我认为我们发布了相同的方法。:\Rock,paper,剪刀,为谁删除sample(c(“石头”,“布”,“剪刀”),1)
paper(需要更多字符),所以应该将其作为一项官方政策来实施;)可以从问题的日期中随机生成种子。这个范例可以扩展到包括。regmatches(x,gregexpr(“\\\([^\\)]+)\\\”[^\\(\\)\\\\\\\\\\\\\\\\\]],x))[[1]]
这确实是一种比绝对尝试使用拆分方式更好的方法。
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)"
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)"