正则表达式提取数字和尾随字母或空格

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. ie

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"

Answer 2

We can add an optional character argument in the regex

stringr::str_extract_all(string1,"[0-9]{1,5}([,.][0-9]{1,4})?[A-Za-z]?")[[1]]
#[1] "536.2k" "83"     "1,396" 
stringr::str_extract_all(string2,"[0-9]{1,5}([,.][0-9]{1,4})?[A-Za-z]?")[[1]]
#[1] "5m"   "83"   "1.1m"

Answer 3

( Updated my earlier post that selected extraneous commas/space)
This works to meet the OP's requirement to extract trailing letter or white space after the numeric sections (without the extraneous commas and white_spaces of my previous version):

(?:[\\d]+[.,]?(?=\\d*)[\\d]*[km ]?)

previous version: \\b(?:[\\d.,]+[km\\s]?)

Explanation:  
- (?:          indicates non-capturing group
- [\d]+        matches 1 or more digits
- [.,]?(?=\d*) matches 0 or 1 decimal_point or comma that is immediately followed ("Positive Lookahead") by 1 or more digits
- [\d]*        matches 0 or more digits
- [km\s]?      matches 0 or 1 of characters within []

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

note the spaces matched after 11 and 83, as intended by OP.

Answer 4

Another stringr option:

new_s<-str_remove_all(unlist(str_extract_all(string2,"\\d{1,}.*\\w")),"[A-Za-z]{2,}")
strsplit(new_s," , ")

    #[[1]]
    #[1] "5m"    "83"    "1.1m "

Original

str_remove_all(unlist(str_extract_all(string2,"\\d{1,}\\W\\w+")),"[A-Za-z]{2,}")
#[1] "83 "  "1.1m"
str_remove_all(unlist(str_extract_all(string1,"\\d{1,}\\W\\w+")),"[A-Za-z]{2,}")
#[1] "536.2k" "83 "    "1,396" 

Answer 5

If you also want to grap the character after the numeric section even if it is a space, you could use your pattern and an optional character class [mk ]? including the space:

[0-9]{1,5}(?:[,.][0-9]{1,4})?[mk ]?

Regex demo | R demo

You might expand the range of characters in the the character class to match [a-zA-Z ]? instead. If you want to use a quantifier to match either 1+ times a char OR a single space you could use an alternation:

[0-9]{1,5}(?:[,.][0-9]{1,4})?(?:[a-zA-Z]+| )?