CODEWITHSUNDEEP
rregexstringrstrsplittidyverse
Brian Becker
Brian Becker

Reputation: 31

String Splitting and Truncating using Regular Expressions in R

I am looking for help implementing a function in R to truncate a level_stream string vector of my dataframe in R and haven't had much luck yet. Essentially when a row in the pre_quiz_score column is not NA, I want to truncate the beginning part of the string up until (and including) the first | character, and I want to truncate everything past the last | character if a post_quiz_score is not NA for that row.

df <- data.frame(ls = c('123 L0=38/42|425 L0=40/42', NA, '482 L7=7/12|789 L8=5/6|523 L9=2/6'), 
                 pre_quiz_score = c(88, NA, 12), 
                 post_quiz_score = c(NA, NA, 90))

I want to implement this in a "tidyverse" way and vectorized to get something like

----------------------------------------------------------------------------
|                 ls                  | pre_quiz_score | post_quiz_score   |
| 425 L0=40/42                        | 88             | NA                |
| NA                                  | NA             | NA                |
| 789 L8=5/6                          | 12             | 90                |

So far, I haven't gotten stringr::str_split, gsub, or sub to work correctly, mostly because I end up removing just the |'s or all the string but the last | and after.

I hope that makes sense, thanks!

Upvotes: 1

Views: 402

Answers (4)

Jonathan Carroll
Jonathan Carroll

Reputation: 3947

tidyr::separate() allows you to split up a column into sub-columns. With the extra = "drop" argument it will keep only up to length(into) columns.

library(tidyr)
separate(df, ls, c("remove", "keep"), sep="\\|", extra = "drop")

#>         remove         keep pre_quiz_score post_quiz_score
#> 1 123 L0=38/42 425 L0=40/42             88              NA
#> 2         <NA>         <NA>             NA              NA
#> 3  482 L7=7/12   789 L8=5/6             12              90

I've kept the remaining part after the first | but you can remove that too if you don't need it.

Upvotes: 0

joel.wilson
joel.wilson

Reputation: 8413

library(dplyr)
df %>% mutate(ls = sapply(strsplit(df$ls, "\\|"), function(x) x[2]))

#            ls pre_quiz_score post_quiz_score
#1 425 L0=40/42             88              NA
#2         <NA>             NA              NA
#3   789 L8=5/6             12              90

Upvotes: 2

akrun
akrun

Reputation: 886978

We can use sub from base R

df$ls <- sub("^[^|]+\\|([^|]+).*", "\\1", df$ls)
df
#            ls pre_quiz_score post_quiz_score
#1 425 L0=40/42             88              NA
#2         <NA>             NA              NA
#3   789 L8=5/6             12              90

Explanation

We match one or more characters that are not a | ([^|]+) from the start (^) of the string, followed by a | (escape it -\\| as a it is a metacharacter), then capture one or more characters that are not a | as a group (i.e. inside the parentheses ([^|]+)) followed by characters until the end of the string (.*) and replace it with the backreference of the captured group (\\1 - as there is only a single capture group and it is the first one, we denote it by 1)

Upvotes: 4

hrbrmstr
hrbrmstr

Reputation: 78792

Just implement the logic as you stated it:

library(stringi)
library(dplyr)

df <- data.frame(ls = c('123 L0=38/42|425 L0=40/42', NA, '482 L7=7/12|789 L8=5/6|523 L9=2/6'),
                 pre_quiz_score = c(88, NA, 12),
                 post_quiz_score = c(NA, NA, 90),
                 stringsAsFactors=FALSE)


df %>%
  mutate(ls=ifelse(!is.na(pre_quiz_score),
                   stri_replace_first_regex(ls, "^[[:alnum:][:blank:]=/]+\\|", ""), ls),
         ls=ifelse(!is.na(post_quiz_score),
                   stri_replace_last_regex(ls, "\\|[[:alnum:][:blank:]=/]+$", ""), ls))
##             ls pre_quiz_score post_quiz_score
## 1 425 L0=40/42             88              NA
## 2         <NA>             NA              NA
## 3   789 L8=5/6             12              90

Upvotes: 3

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