Regex to extract numbers and trailing letter or white space

Question

I'm currently trying to extract data from strings that are always in the same format (scraped from social sites with no API support)

example of strings

53.2k Followers, 11 Following, 1,396 Posts
5m Followers, 83 Following, 1.1m Posts

I'm currently using the following regex expression: "[0-9]{1,5}([,.][0-9]{1,4})?" to get the numeric sections, preserving the comma and dot separators.

It yields results like

53.2, 11, 1,396 
5, 83, 1.1

I really need a regular expression that will also grab the character after the numeric sections, even if it's a white-space. i.e.

53.2k, 11 , 1,396
5m, 83 , 1.1m

Any help is greatly appreciated

R code for reproduction

  library(stringr)

  string1 <- ("536.2k Followers, 83 Following, 1,396 Posts")
  string2 <- ("5m Followers, 83 Following, 1.1m Posts")

  info <- str_extract_all(string1,"[0-9]{1,5}([,.][0-9]{1,4})?")
  info2 <- str_extract_all(string2,"[0-9]{1,5}([,.][0-9]{1,4})?")

  info 
  info2 

Answer 1

I would suggest the following regex pattern:

[0-9]{1,3}(?:,[0-9]{3})*(?:\\.[0-9]+)?[A-Za-z]*

This pattern generates the outputs you expect. Here is an explanation:

[0-9]{1,3}      match 1 to 3 initial digits
(?:,[0-9]{3})*  followed by zero or more optional thousands groups
(?:\\.[0-9]+)?  followed by an optional decimal component
[A-Za-z]*       followed by an optional text unit

I tend to lean towards base R solutions whenever possible, and here is one using gregexpr and regmatches:

txt <- "53.2k Followers, 11 Following, 1,396 Posts"
m <- gregexpr("[0-9]{1,3}(?:,[0-9]{3})*(?:\\.[0-9]+)?[A-Za-z]*", txt)
regmatches(txt, m)

[[1]]
[1] "53.2k"   "11"   "1,396"