Place a dot before a letter

user_13647125 0
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I need to put a dot before a letter in this type of strings

name of data set: V2

6K102 62D102 627Z102

I would like to get this:

6.K102 62.D102 627.Z102

I am using this regex:

mutate(V2 = gsub("^[A-Z]",'\\.', V2))
subset substring r regex
4 Answers

If the string has to start with 1 or more digits followed by a char A-Z, you could use 2 capturing groups

^(\d+)([A-Z])

In the replacement use "\\1.\\2"

sub("^([0-9]+)([A-Z])", "\\1.\\2", V2)
user_54249880

you could use sub("([A-Z])",".\\1", V2)

user_104424230

This apply to the question before it was updated.

Usually \u2022 prints a bullet in text items. So if your question is regarding a label, you may just inserted it there "\u2022 ..."

Otherwise, for text items in datasets such as V2, you can work around by applying paste0 in combination with ifelse. In this case, your text items the you need a black dot infront if, is stored in V2$name

V2$name <- ifelse(V2$name==1,1,paste0("\u2022", V2$name))
user_88113990

Your regex lacks a capturing group around the letter pattern (so that you could keep it after replacement) and contains a redundant ^ anchor that matches the string start location. Also, you are using a gsub function while you just need a sub, since only one replacement is expected.

Use

sub("([[:upper:]])", ".\\1", V2)

With stringr (see demo):

stringr::str_replace(V2, "[[:upper:]]", "\\.\\0")

Details

  • sub - only the first match is replaced
  • ([[:upper:]]) - matches and captures any uppercase letter into Group 1 (later referenced to with \1 from the replacement pattern)
  • \1 - the value of Group 1 (the uppercase letter matched)

Note that stingr solution uses \0, the placeholder for the whole match value, so no need to capture the uppercase letter in the regex pattern.

See the R demo:

V2 <- c("6K102","62D102","627Z102")
sub("([[:upper:]])", ".\\1", V2)
# => [1] "6.K102"   "62.D102"  "627.Z102"
user_38329700
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