Regular expression to get a group of text

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I cant find the right regular expression:

print(re.compile(r'row_([0-9]+)(_[^_]+)*').split('row_0007_id_testa_testb'))
> ['', '0007', '_testb', '']

I tried with non-greedy regexp, didn't work too:

print(re.compile(r'row_([0-9]+)(_[^_]+)+?').split('row_0007_id_testa_testb'))
['', '0007', '_id', '_testa_testb']

I need to get this:

> ['', '0007', 'id', 'testa', 'testb']

You can use a simple regex _([^_]+) in findall with an inline if condition to assert that string starts with row_:

>>> reg = re.compile(r'_([^_]+)')

>>> s = 'row_0007_id_testa_testb'
>>> print re.findall(reg, s) if s.startswith('row_') else None
['0007', 'id', 'testa', 'testb']

>>> s = 'col_0007_id_testa_testb'
>>> print re.findall(reg, s) if s.startswith('row_') else None
None

Regex Tutorial, Regular expressions allow us to not just match text but also to extract information for further processing. This is done by defining groups of characters and  When you want to search and replace specific patterns of text, use regular expressions. They can help you in pattern matching, parsing, filtering of results, and so on. Once you learn the regex syntax, you can use it for almost any language. Press Ctrl+R to open the search and replace pane.

You did't indicate the host programming language, but I assume it is Python.

As I see, you want to split the source text on _ character, so just _ should be the content of the regex.

p = re.compile('_').split('row_0007_id_testa_testb')

gives sets ['row', '0007', 'id', 'testa', 'testb'] in p.

So the only thing to change is to set the starting element to an empty string. Then you can print the p array, getting the expected result.

Below you have the example script:

import re
p = re.compile('_').split('row_0007_id_testa_testb')
print p
p[0] = ''
print p

Capturing groups, How do Capture Groups Beyond \9 get Referenced? Normally, within a pattern, you create a back-reference to the content a capture group previously matched by  @MichaelMikowski that isn't meaningfully slower when you're not hitting the execution limit. When you are, it's clearly much slower. I'm not saying your code doesn't work, I'm saying that in practice I think it will cause more harm than good.

Assuming you want to match strings that contains letters and numbers only, but replace the first match with an empty string. If so, use

re.compile(r'^[^\W_]+_+|[\W_]+').split('row_0007_id_testa_testb')

Output:

['', '0007', 'id', 'testa', 'testb']

Test this code here.

Learn Regular Expressions - Lesson 11: Match groups, Gets a collection of groups matched by the regular expression. Text.​RegularExpressions.GroupCollection Groups { get; } member this.Groups : System. IgnoreCase); // Match the regular expression pattern against a text string. Match m = r. A regular expression pattern can include subexpressions, which are defined by enclosing a portion of the regular expression pattern in parentheses. Every such subexpression forms a group. The Groups property provides access to information about those subexpression matches.

Maybe try using a single re.split command:

re.split(r"^row_|_", "row_0007_id_testa_testb)

For a more generalized replace:

re.split(r"^[a-z]+_|_", "row_0007_id_testa_testb")

This would eliminate the first word characters preceded by the first _ and then split the rest.

It's unclear by the question, although if looking to prepend the split string list this should work:

r = re.split(r"_", s)[1:]
r.insert(0,'')

Regex Capture Groups and Back-References, Use ( ) in regexp and group(1) in python to retrieve the captured string title>' # text = '<title>hello</title>' if match := re.search(pattern, text, re. What is the regular expression to extract the words within the square brackets, ie. sample some another one Note: In my use case, brackets cannot be nested.

You could use an alternation | with findall, which would match row_ and capture not an underscore in a capturing group ([^_]+)

row_|([^_]+)

import re
print(re.findall(r"row_|([^_]+)", 'row_0007_id_testa_testb'))

That would give you:

['', '0007', 'id', 'testa', 'testb']

Demo

Match.Groups Property (System.Text.RegularExpressions , Matches a group after the main expression without including it in the result. not even the collective wisdom of the internet can make regular expressions funny. Regex to get the words after matching string. Regex Match text within a Capture Group. 3. Regular expression to match a line that doesn't contain a word.

Extract part of a regex match, For example, the regular expression (dog) creates a single group containing the To find out how many groups are present in the expression, call the non-​capturing groups that do not capture text and do not count towards the group total​. The result is an array of matches, but without details about each of them. But in practice we usually need contents of capturing groups in the result. To get them, we should search using the method str.matchAll (regexp). It was added to JavaScript language long after match, as its “new and improved version”.

Capture value and any unit, To enable the use of regular expressions in the Find what field during QuickFind, FindinFiles, Quick Replace, or Replace in Files operations, select the Use option under Find Options and choose Regular expressions. The triangular Reference List button next to the Find what field then becomes available.

2.4: Regular Expressions: Capturing Groups, RegExr is an online tool to learn, build, & test Regular Expressions (RegEx / RegExp). Supports JavaScript & PHP/PCRE RegEx. Results update in real-time as you type. Roll over a match or expression for details. Save & share expressions with others. Use Tools to explore your results. Full RegEx Reference with help & examples.

Comments
  • Where is __monitoring in your string?
  • @revo updated my question sorry
  • Try re.compile(r'row_|_').split('row_0007_id_testa_testb')
  • Do you need the first empty entry?
  • I believe he doesn't want to eliminate the first word,... he wants the first match to be an empty string.