CODEWITHSUNDEEP
pythonregex
issac
issac

Reputation: 175

Check the occurrence of zero or one substring

Could anyone please point me how to check a pattern in which a sub-pattern occurs zero or once?

For example,

Test 1-2 (many): blah blah
Test 1-2: blah blah

Both lines should be detected.

I tried with: 

sub = 'Test\s+(\d+\s*\-\s*\d+)\s*\((.*?)\)?(\:*)\s*(.*)'

But it doesn't work as expected.

Upvotes: 0

Views: 2144

Answers (2)

Emma
Emma

Reputation: 27743

Here, we can add an optional sub-expression, behind the :, collect our value after that in a capturing group and collect our numbers and dashes in another one, maybe similar to: 

Test\s+([0-9-]+)(.+)?:\s+(.+)

If we wish to add more boundaries, we can do that. The rest of our work can be programmed.

DEMO

Test

# coding=utf8
# the above tag defines encoding for this document and is for Python 2.x compatibility

import re

regex = r"Test\s+([0-9-]+)(.+?):\s+(.+)"

test_str = ("Test 1-2 (many): blah blah\n"
    "Test 1-2: blah blah")

matches = re.finditer(regex, test_str, re.MULTILINE)

for matchNum, match in enumerate(matches, start=1):

    print ("Match {matchNum} was found at {start}-{end}: {match}".format(matchNum = matchNum, start = match.start(), end = match.end(), match = match.group()))

    for groupNum in range(0, len(match.groups())):
        groupNum = groupNum + 1

        print ("Group {groupNum} found at {start}-{end}: {group}".format(groupNum = groupNum, start = match.start(groupNum), end = match.end(groupNum), group = match.group(groupNum)))

# Note: for Python 2.7 compatibility, use ur"" to prefix the regex and u"" to prefix the test string and substitution.

Demo

const regex = /Test\s+([0-9-]+)(.+)?:\s+(.+)/gm;
const str = `Test 1-2 (many): blah blah
Test 1-2: blah blah`;
let m;

while ((m = regex.exec(str)) !== null) {
    // This is necessary to avoid infinite loops with zero-width matches
    if (m.index === regex.lastIndex) {
        regex.lastIndex++;
    }
    
    // The result can be accessed through the `m`-variable.
    m.forEach((match, groupIndex) => {
        console.log(`Found match, group ${groupIndex}: ${match}`);
    });
}

Another Approach by Kenan Güler

RegExp: ^(Test\s+\d+-\d+)\b(?:.*?:\s*)(.*)$

Demo: https://repl.it/repls/LovableCaringBrowser

  import re

  base_sub_pattern = ["Test 1-2", "blah blah"]

  string = """\
  Test 1-2 (many): blah blah
  Test 1-2: blah blahGGG
  """

  pattern = re.compile(r"^(Test\s+\d+-\d+)\b(?:.*:\s*)(.*)$", re.MULTILINE)
  matches = pattern.findall(string)

  if matches:
    print("found matches:", matches, "\n")

    for match in matches:
      if set(base_sub_pattern).difference(match):
        print("sub-pattern not exist here", match)

Upvotes: 0

jxc
jxc

Reputation: 13998

You can use the following to match a sub-pattern which occurs zero or once:

(?:sub_pattern)?

where (?:...) is a non-capturing group. In your particular example, the question mark(to match zero or one sub-pattern) is set to \)?, this influences only the single preceding closing parenthesis ')'. you should put the whole optional sub-pattern into a non-capturing group, thus:

(?:\(.*?\))?

Note: Do NOT use the capturing groups (...) unless you want to extract their values separately.

Below is a testing code for a full regex pattern:

import re

# a list of testing strings
x = ['Test 1-2 (many): blah blah', 'Test 1-2: blah blah', 'Test 1: no match']

# regex pattern
sub = r'Test\s+\d+\s*-\s*\d+\s*(?:\(.*?\))?:.+'  

for i in x:
    m = re.match(sub, i)
    if m: print(m.group(0)) 
#Test 1-2 (many): blah blah
#Test 1-2: blah blah

Upvotes: 2

Related Questions