Replace characters only if it is not repeating

Question

Is there a way to replace a character only if it is not repeating, or repeating a certain number of times?

str = c("ddaabb", "daabb", "aaddbb", "aadbb")
gsub("d{1}", "c", str)
[1] "ccaabb" "caabb"  "aaccbb" "aacbb" 

#Expected output
[1] "ddaabb" "caabb"  "aaddbb" "aacbb"

Tranbi · Accepted Answer · 2023-02-07T15:59:39.533

3

You can use negative lookarounds in your regex to exclude cases where d is preceeded or followed by another d:

gsub("(?<!d)d(?!d)", "c", str, perl=TRUE)

Edit: adding perl=TRUE as suggested by OP. For more info about regex engine in R see this question

edited Feb 07 '23 at 15:59

answered Feb 07 '23 at 15:53

Tranbi

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This throws an error on my console. Adding perl = TRUE fixed it – Maël Feb 07 '23 at 15:56
1

Thanks for the feedback! I'm not familiar with R so didn't know exactly what regex engine it uses... Added your comment to my answer! – Tranbi Feb 07 '23 at 15:59
If you're really into pain, consider `foo <- unlist(strsplit(str, ''); bar <- rle(foo)` and then look for instances of `bar$lengths ==1` :-) – Carl Witthoft Feb 07 '23 at 16:30
This regex works, but only specifically for single-instances of "d" . If @Maël wants to find **any** single char, then you'll need a rather more tricky regex whereas my hack in my previous comment will work. – Carl Witthoft Feb 07 '23 at 16:33

score 0 · Answer 2 · answered Feb 07 '23 at 18:38

Now that you've added "or repeating a specified number of times," the regex-based approaches may get messy. Thus I submit my wacky code from a previous comment.

foo <- unlist(strsplit(str, '')
bar <- rle(foo)

and then look for instances of bar$lengths == desired_length and use the returned indices to locate (by summing all bar$lengths[1:k] ) the position in the original sequence. If you only want to replace a specific character, check the corresponding value of bar$values[k] and selectively replace as desired.

Replace characters only if it is not repeating

2 Answers2