CODEWITHSUNDEEP
regexstringrstudiotext-extraction
SAJ
SAJ

Reputation: 320

Get the characters after a certain pattern in R - regex

I have a data frame with one column:

df <- data.frame(cat = c("c(\\\"BPT\\\", \"BP\")", "c(\"BP2\", \"BP\")", "c(\"BPT\", \"BP\")", "c(\"CN\", \"NC\")"))
df$cat <- as.character(df$cat)
df$cat

How can I extract the characters that appear after c(\", sometimes there is only one backslash and sometimes there's 2. Similarly with the characters, sometimes the characters are 2 and sometimes they are 3. e.g. BP2, BP etc.

So far I have tried:

substr(x = df$cat, start = 4, stop = 6)

But this results in:

 "\"BP" "BP2"  "BPT"  "CN\""

And I only want the output to show 

"BPT" "BP2"  "BPT"  "CN"

Upvotes: 1

Views: 544

Answers (1)

Wiktor Stribiżew
Wiktor Stribiżew

Reputation: 626738

You may use

df <- data.frame(cat = c("c(\\\"BPT\\\", \"BP\")", "c(\"BP2\", \"BP\")", "c(\"BPT\", \"BP\")", "c(\"CN\", \"NC\")"))
df$cat <- as.character(df$cat)
unlist(lapply(gsub('\\', '', df$cat, fixed=TRUE), function(x) eval(parse(text=x))[[1]]))
## => [1] "BPT" "BP2" "BPT" "CN"

See the R demo online.

Notes

  • gsub('\\', '', df$cat, fixed=TRUE) removes all backslashes. You may use gsub('\\\"', '"', df$cat, fixed=TRUE) if you only plan to remove backslashes before ".
  • eval(parse(text=x))[[1]] parses the vector and returns the first item
  • lapply helps traverse the whole data you have. See Using sapply and lapply.

Upvotes: 1

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