python regular expression with "OR" and $ end of line

python regular expression with "OR" and $ end of line

By : Denis Patel
Date : November 21 2020, 04:03 PM
To fix the issue you can do I have 3 different types of string in which I'm trying to match a pattern: Patterns:
code :
strings = [

import re
pattern = re.compile(r"(?<=_)\d+(?=_|$)")
for item in strings:
    print pattern.search(item).group()

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Why does regular expression not match without boundary matcher "Beginning of line"?

Why does regular expression not match without boundary matcher "Beginning of line"?

By : Keyur Patel
Date : March 29 2020, 07:55 AM
will be helpful for those in need In Java, matches() validates the entire string. Your input probably has line breaks in them (which don't get matched by .+).
Try this instead:
code :
Pattern p = Pattern.compile(".+to\\s+date\\s*?:\\s*?([0-9]{2}[\\./][0-9]{2}[\\./][0-9]{2,4}).+", Pattern.CASE_INSENSITIVE);
Matcher m = p.matcher("... \n From Date :01/11/2011 To Date :30/11/2011;;;;;;;;;;;;; \n ...");

System.out.println(m.matches()); // prints false

if(m.find()) {
  System.out.println(m.group(1)); // prints 30/11/2011
Pattern.compile("to\\s+date\\s*?:\\s*?([0-9]{2}[./][0-9]{2}[./][0-9]{2,4})", Pattern.CASE_INSENSITIVE);
Django/Python Getting "error_message" : "multiple repeats" on using regular expression

Django/Python Getting "error_message" : "multiple repeats" on using regular expression

By : deepu
Date : March 29 2020, 07:55 AM
may help you . I don't think the problem is with the regex, but with your word list. It's likely the list contains characters which are interpreted as regex special characters. This works for me:
code :
import re

def replacement(match):
    return "*" * len(match.group(0))

def censored_string(dirty):
    clean = str(dirty)
    wordList = ["fuck", "shit", "damn"]

    for swear_word in wordList:
        target_word = swear_word.strip()
        result = re.sub("(?i)\\b(("+target_word+"){1,})(s{0,1})\\b",replacement, clean)
        clean = result
    return clean

print censored_string("god damn i love bananas and fucking fuckfuck shits")
# god **** i love bananas and fucking ******** *****
python regular expression: re.findall(r"(do|re|mi)+","mimi rere midore")

python regular expression: re.findall(r"(do|re|mi)+","mimi rere midore")

By : arun gangawat Gangaw
Date : March 29 2020, 07:55 AM
wish help you to fix your issue The difference is in the capturing group. With a capturing froup, findall() returns only what was captured. Without a capturing group, the whole match is returned.
In your first example, the group only captures the two characters, repeated or not. In the second example, the whole match includes any repetitions.
code :
r'Some example text: ((?:do|re|me)+)'
python regular expression: script to change "frame_20" to "frame_0020"

python regular expression: script to change "frame_20" to "frame_0020"

By : Sam
Date : March 29 2020, 07:55 AM
wish help you to fix your issue So I have a list of filenames. They are called: , You could use a regular expression as follows:
code :
import re

file_names = ["fgmask_frame_20.jpg", "fgmask_frame_40.jpg", "fgmask_frame_2000.jpg"]
file_names = [re.sub(r'(.*?_)(\d+)(\.jpg)', lambda x: '{}{:04}{}'.format(x.group(1), int(x.group(2)), x.group(3)), file_name) for file_name in file_names]
print file_names
['fgmask_frame_0020.jpg', 'fgmask_frame_0040.jpg', 'fgmask_frame_2000.jpg']
How to write regular expression of head of line, "^", and end of line, "$" in []?

How to write regular expression of head of line, "^", and end of line, "$" in []?

By : PeterPeter
Date : March 29 2020, 07:55 AM
around this issue You can try alternation with () instead of [] (class of characters or character sets)
code :
grep -q -E "(:|^)\$TARGET_PATH($|:)
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