How do the regular expression capture infinite groups?

How do the regular expression capture infinite groups?

By : Young
Date : November 18 2020, 01:01 AM
Hope this helps There is no way to have an unbounded number of capturing groups in Python regexes. However, you can use one regex to match the entire expression and then use a second regex to parse the options. For example:
code :
match = re.match(r'^([\w ]+)(?:\s-((?:\s\w+)+))?$', input)
if match:
    text = match.group(1)
    if match.group(2):
        options = [m.group(0) for m in re.finditer(r'\w+', match.group(2))]
        options = []

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Regular expression capture groups which is in a group

Regular expression capture groups which is in a group

By : user2576132
Date : March 29 2020, 07:55 AM
Hope this helps Regular expressions cannot handle repeating groups it can return any of the captured groups (in your case it returned "Ghi").
The example below will print:
code :
public static void main(String[] args) {

    String example = "AbcDefGhi12345";

    if (example.matches("(([A-Z][a-z]+)+)([0-9]+)")) {

        Scanner s = new Scanner(example);

        String m;
        while ((m = s.findWithinHorizon("[A-Z][a-z]+", 0)) != null)

        System.out.println(s.findWithinHorizon("[0-9]+", 0));
regular expression capture groups

regular expression capture groups

By : brucelin
Date : March 29 2020, 07:55 AM
With these it helps Yes, parentheses capture everything between them. Captured groups are numbered by their opening parenthesis. So if /(foo)((bar)baz)/ matches, your first captured group will contain foo, your second barbaz, and your third bar. In some dialects, only the first 9 capturing groups are numbered.
Captured groups can be used for backreferencing. If you want to match "foobarfoo", /(foo)bar\1/ will do that, where \1 means "the first group I captured".
code :
/                   # matching starts
  (                 # open 1st capturing group
    [a-zA-Z0-9]     # match 1 character that's in a-z, A-Z, or 0-9
    .*              # match as much of any character possible
    \s?             # optionally match a white space (this will generally never happen, since the .* before it will have gobbled it up)
    (               # open 2nd capturing group
      @classs?      # match '@class' or '@classs'
    )+              # close 2n group, matching it once or more
    \s+             # match one or more white space characters
    [a-zA-Z0-9]     # match 1 character that's in a-z, A-Z, or 0-9
    [^><]*          # match any number of characters that's not an angle bracket
  )                 # close 1st capturing group
/g                  # modifiers - g: match globally (repeatedly match throughout entire input)
Regular Expression and Capture Groups

Regular Expression and Capture Groups

By : user2575325
Date : March 29 2020, 07:55 AM
around this issue Just use a lookahead to capture the values of cs1 and cs2 fields, if you don't know the order of it's arrangement.
code :
Regular expression to capture groups delimited by an expression

Regular expression to capture groups delimited by an expression

By : Turbo
Date : March 29 2020, 07:55 AM
I wish did fix the issue. Assuming that you don't have sub-patterns like [[u'1', u'2'], [u'3',u'5']] (multiple nested sub-groups at the same level, in which case you need to use a stack and parse like pushdown automata) you could do this with regular expressions in two steps:
(1) split the expression with regex \]\s*,\s*\[ to get the groups first, you will get 3 groups for the example provided.
code :
str <- "[[[[u'1', u'2'], u'3'], u'4'], [[[u'1', u'2'], u'4'], [[u'1', u'5'], u'4']]]"
groups <- unlist(strsplit(str, split='\\]\\s*,\\s*\\['))
pattern <- "[^0-9u]*u'([0-9]+)'[^0-9u]*"
lapply(groups, function(str) gsub(pattern, "\\1", regmatches(str,gregexpr(pattern,str))[[1]]))

#[1] "1" "2" "3" "4"

#[1] "1" "2" "4"

#[1] "1" "5" "4"
import re
str = "[[[[u'1', u'2'], u'3'], u'4'], [[[u'1', u'2'], u'4'], [[u'1', u'5'], u'4']]]"
groups = re.split('\]\s*,\s*\[', str)
pattern = "[^0-9u]*u'([0-9]+)'[^0-9u]*"
print map(lambda x: re.findall(pattern, x), groups)
# [['1', '2', '3', '4'], ['1', '2', '4'], ['1', '5', '4']]
Capture groups with Regular Expression (Python)

Capture groups with Regular Expression (Python)

Date : March 29 2020, 07:55 AM
like below fixes the issue Kind of a noob here, apologies if I misstep. , You need the first captured group:
code :
In [8]: string_one = 'file_record_transcript.pdf'

In [9]: re.search(r'^(file.*)\.pdf$', string_one).group()
Out[9]: 'file_record_transcript.pdf'

In [10]: re.search(r'^(file.*)\.pdf$', string_one).group(1)
Out[10]: 'file_record_transcript'
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