Separating A String Into Characters

user_768619 0
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I have some ordered test results encoded in a character string. The string can be of arbitrary length. Each digit in the string represents a test result. In the following, for example, there are four test results represented:

2069

I want to tidy these up in R by splitting the string into individual observations. No problem with strsplit or string::str_split, which returns four values that will become my observations.

strsplit("2069" %>% as.character(), split = "") %>% unlist()
[1] "2" "0" "6" "9"

Now, however, I have realized that some results are values greater than 9. These two-digit values have been encoded with parentheses to make clear they are not individual results.

For example, in the following case I still have four values, but some have been enclosed in parentheses to group the values larger than 9.

2(10)1(12)

I'm struggling with a way to break these up so that I get

[1] "2" "10" "1" "12"

Appreciate any guidance. Thanks.

r regex string
3 Answers

Updated - pattern match based on the OP's new pattern showed in the comments. Here, we use str_extract to extract one or more digits that follow an open parentheses (regex lookaround ) or (|) any character that is not a parentheses ([^()])

library(stringr)
str_extract_all(str1, "(?<=[(])\\d+|[^()]")
[[1]]
[1] "2"  "10" "1"  "12"

[[2]]
[1] "2" "0" "6" "9"

[[3]]
[1] "2"  "15"

[[4]]
[1] "2" "1" "3" "1"

-testing on the OP's extra pattern

str_extract_all(str2, "(?<=[(])\\d+|[^()]")
[[1]]
[1] "2"  "10" "1"  "12"

[[2]]
[1] "2" "0" "6" "9"

[[3]]
[1] "2"  "15"

[[4]]
[1] "2" "1" "3" "1"

[[5]]
[1] "10" "0"  "2"  "0"  "1"

-Earlier solutions (Based on the assumption that all the numbers that are greater than 9 will be wrapped inside the parentheses)

We may split on the parentheses in base R

unlist(strsplit(str1[1], "\\(|\\)"))
[1] "2"  "10" "1"  "12"

Assuming if there are both cases, then an option is to get the index of those elements have the parentheses and do this separately

i1 <- grepl("\\(|\\)", str1)
lst1 <- vector('list', length(str1))
lst1[i1] <- strsplit(str1[i1], "\\(|\\)")
lst1[!i1] <- strsplit(str1[!i1], "")
unlist(lst1)
[1] "2"  "10" "1"  "12" "2"  "0"  "6"  "9"  "2"  "15" "2"  "1"  "3"  "1"

or another option is ifelse with grepl to create a single delimiter and then use strsplit

lst1 <- strsplit(trimws(ifelse(grepl("\\(|\\)", str1), 
    gsub("\\(|\\)", ",", str1), gsub("(?<=.)(?=.)", "\\1,\\2", 
       str1, perl = TRUE)), whitespace = ","), ",")
lst1
[[1]]
[1] "2"  "10" "1"  "12"

[[2]]
[1] "2" "0" "6" "9"

[[3]]
[1] "2"  "15"

[[4]]
[1] "2" "1" "3" "1"

data

str1 <- c("2(10)1(12)", "2069", "2(15)", "2131")
str2 <- c(str1, "(10)0201")
user_37322710

Maybe we can do like below (borrow str1 from @akrun)

> mapply(strsplit, str1, ifelse(grepl("[()]", str1), "\\(|\\)", ""))
$`2(10)1(12)`
[1] "2"  "10" "1"  "12"

$`2069`
[1] "2" "0" "6" "9"

$`2(15)`
[1] "2"  "15"

$`2131`
[1] "2" "1" "3" "1"
user_121587570

Use

(?<=\()\d+(?=\))|\d

See regex proof.

EXPLANATION

--------------------------------------------------------------------------------
  (?<=                     look behind to see if there is:
--------------------------------------------------------------------------------
    \(                       '('
--------------------------------------------------------------------------------
  )                        end of look-behind
--------------------------------------------------------------------------------
  \d+                      digits (0-9) (1 or more times (matching
                           the most amount possible))
--------------------------------------------------------------------------------
  (?=                      look ahead to see if there is:
--------------------------------------------------------------------------------
    \)                       ')'
--------------------------------------------------------------------------------
  )                        end of look-ahead
--------------------------------------------------------------------------------
 |                        OR
--------------------------------------------------------------------------------
  \d                       digits (0-9)

R code:

library(stringr)
str1 <- c("2(10)1(12)", "2069", "2(15)", "2131")
str_extract_all(str1, "(?<=\\()\\d+(?=\\))|\\d")

Results:

[1] "2"  "10" "1"  "12"

[[2]]
[1] "2" "0" "6" "9"

[[3]]
[1] "2"  "15"

[[4]]
[1] "2" "1" "3" "1"
user_113298900
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