Extract Part of a Regex Match

Extract part of a regex match

Use ( ) in regexp and group(1) in python to retrieve the captured string (re.search will return None if it doesn't find the result, so don't use group() directly):

title_search = re.search('<title>(.*)</title>', html, re.IGNORECASE)

if title_search:
    title = title_search.group(1)

Extract a regular expression match

Use the new stringr package which wraps all the existing regular expression operates in a consistent syntax and adds a few that are missing:

library(stringr)
str_locate("aaa12xxx", "[0-9]+")
#      start end
# [1,]     4   5
str_extract("aaa12xxx", "[0-9]+")
# [1] "12"

how can i extract a specific part of a string in c# with regex.match?

The part you need should be in the capture group (the part you put between ()). Try accessing it with

Match1.Groups[1].Value

How do I extract only part of a matched string with regex?

Yes, a capture group does this. You don't see it in your array because you've used String#match and the g flag. Either remove the g flag (and take the second entry from the array, which is the first capture group):

console.log('https://twitter.com/pixelhenk/status/891303699204714496'.match(/\/status\/(\d+)/)[1]);

Extracting part of string using regular expressions

You can use the following pattern with gsub:

> gsub("^(?:WER.*([a-zA-Z]\\d*)|.*)$", "\\1", test)
[1] ""        "H987654" "G789456" "F12"     ""

See the regex demo

This pattern matches:

^ - start of a string
(?: - start of an alternation group with 2 alternatives:
- WER.*([a-zA-Z]\\d*) - WER char sequence followed with 0+ any characters (.*) as many as possible up to the last letter ([a-zA-Z]) followed by 0+ digits (\\d*) (replace with \\d+ to match 1+ digits, to require at least 1 digit)
- | - or
- `.* - any 0+ characters
)$ - closing the alternation group and match the end of string with $.

With str_match from stringr, it is even tidier:

> library(stringr)
> res <- str_match(test, "^WER.*([a-zA-Z]\\d*)$")
> res[,2]
[1] NA        "H987654" "G789456" "F12"     NA       
>

See another regex demo

If there are newlines in the input, add (?s) at the beginning of the pattern: res <- str_match(test, "(?s)^WER.*([a-zA-Z]\\d*)$").

How to extract part of string using regex

Use regex to extract part of string from raw string, the following is the whole code

package main

import (
    "fmt"
    "regexp"
)

func main() {

    // extract part of string using regex
    str := "Upgrade starting on Mon Aug 9 06:46:00 UTC 2021 with ..."

    // extract string "Mon Aug 9 06:46:00 UTC 2021" using regex
    re := regexp.MustCompile(`(Mon|Tue|Wed|Thu|Fri|Sat|Sun) (Jan|Feb|Mar|Apr|May|Jun|Jul|Aug|Sep|Oct|Nov|Dec) \d{1,2} \d{2}:\d{2}:\d{2} (\S{3}) \d{4}`)
    t := re.FindString(str)
    fmt.Println(t)

}

Extract just the part of string that matches a regex pattern in R

I added a capture group and a back-reference "\\1":

sub("^([A-Z]{1}[a-z]{2}\\s\\d+).*", "\\1", vector)
[1] "May 20" "Dez 1"  "Oct 12"

The replacement argument accepts back-references like '\\1', but not typical regex patterns as you used. The back-reference refers back to the pattern you created and the capture group you defined. In this case our capture group was the abbreviated month and day which we outlined with parantheticals (..). Any text captured within those brackets are returned when "\\1" is placed in the replacement argument.

This quick-start guide may help

Python extract pattern matches

You need to capture from regex. search for the pattern, if found, retrieve the string using group(index). Assuming valid checks are performed:

>>> p = re.compile("name (.*) is valid")
>>> result = p.search(s)
>>> result
<_sre.SRE_Match object at 0x10555e738>
>>> result.group(1)     # group(1) will return the 1st capture (stuff within the brackets).
                        # group(0) will returned the entire matched text.
'my_user_name'

How to extract a substring using regex

Assuming you want the part between single quotes, use this regular expression with a Matcher:

"'(.*?)'"

Example:

String mydata = "some string with 'the data i want' inside";
Pattern pattern = Pattern.compile("'(.*?)'");
Matcher matcher = pattern.matcher(mydata);
if (matcher.find())
{
    System.out.println(matcher.group(1));
}

Result:


the data i want

Extract Part of a Regex Match