在 R 中提取 substring

Question

Suppose I have list of string "S[+229]EC[+57]VDSTDNSSK[+229]PSSEPTSHVAR" and need to get a vector of string that contains only numbers with bracket like eg. [+229][+57] .

Is there a convenient way in R to do this?

Answer 1

Using base R , then try it with

> unlist(regmatches(s,gregexpr("\\[\\+\\d+\\]",s)))
[1] "[+229]" "[+57]"  "[+229]"

Or you can use

> gsub(".*?(\\[.*\\]).*","\\1",gsub("\\].*?\\[","] | [",s))
[1] "[+229] | [+57] | [+229]"

Answer 2

We can use str_extract_all from stringr

stringr::str_extract_all(x, "\\[\\+\\d+\\]")[[1]]
#[1] "[+229]" "[+57]"  "[+229]"

Wrap it in unique if you need only unique values.

---

Similarly, in base R using regmatches and gregexpr

regmatches(x, gregexpr("\\[\\+\\d+\\]", x))[[1]]

data

x <- "S[+229]EC[+57]VDSTDNSSK[+229]PSSEPTSHVAR"

Answer 3

Seems like you want to remove the alphabetical characters, so

gsub("[[:alpha:]]", "", x)

where [:alpha:] is the class of alphabetical (lower-case and upper-case) characters, [[:alpha:]] says 'match any single alphabetical character', and gsub() says substitute, globally, any alphabetical character with the empty string "" . This seems better than trying to match bracketed numbers, which requires figuring out which characters need to be escaped with a (double!) \\ .

If the intention is to return the unique bracketed numbers, then the approach is to extract the matches (rather than remove the unwanted characters). Instead of using gsub() to substitute matches to a regular expression with another value, I'll use gregexpr() to identify the matches, and regmatches() to extract the matches. Since numbers always occur in [] , I'll simplify the regular expression to match one or more ( + ) characters from the collection +[:digit:] .

> xx <- regmatches(x, gregexpr("[+[:digit:]]+", x))
> xx
[[1]]
[1] "+229" "+57"  "+229"

xx is a list of length equal to the length of x . I'll write a function that, for any element of this list, makes the values unique, surrounds the values with [ and ] , and concatenates them

fun <- function(x)
    paste0("[", unique(x), "]", collapse = "")

This needs to be applied to each element of the list, and simplified to a vector, a task for sapply() .

> sapply(xx, fun)
[1] "[+229][+57]"

A minor improvement is to use vapply() , so that the result is robust (always returning a character vector with length equal to x ) to zero-length inputs

> x = character()
> xx <- regmatches(x, gregexpr("[+[:digit:]]+", x))
> sapply(xx, fun)               # Hey, this returns a list :(
list()
> vapply(xx, fun, "character")  # vapply() deals with 0-length inputs
character(0)