Python String.Replace Regular Expression

Python string.replace regular expression

str.replace() ^v2|v3 does not recognize regular expressions.

To perform a substitution using a regular expression, use re.sub() ^v2|v3.

For example:

import re

line = re.sub(
           r"(?i)^.*interfaceOpDataFile.*$", 
           "interfaceOpDataFile %s" % fileIn, 
           line
       )

In a loop, it would be better to compile the regular expression first:

import re

regex = re.compile(r"^.*interfaceOpDataFile.*$", re.IGNORECASE)
for line in some_file:
    line = regex.sub("interfaceOpDataFile %s" % fileIn, line)
    # do something with the updated line

How to replace string using regular expression in Python

This works.

import re
s3 = ['March/21/2019' , 'Mar/23/2019']
s3 = [re.sub(r'Mar[a-z]*', '03', item) for item in s3]

# ['03/21/2019', '03/23/2019']

Of course, you can also use a for loop for better readability.

import re
s3 = ['March/21/2019' , 'Mar/23/2019']
for i in range(len(s3)):
    s3[i] = re.sub(r'Mar[a-z]*', '03', s3[i])

# ['03/21/2019', '03/23/2019']

How to input a regex in string.replace?

This tested snippet should do it:

import re
line = re.sub(r"</?\[\d+>", "", line)

Edit: Here's a commented version explaining how it works:

line = re.sub(r"""
  (?x) # Use free-spacing mode.
  <    # Match a literal '<'
  /?   # Optionally match a '/'
  \[   # Match a literal '['
  \d+  # Match one or more digits
  >    # Match a literal '>'
  """, "", line)

Regexes are fun! But I would strongly recommend spending an hour or two studying the basics. For starters, you need to learn which characters are special: "metacharacters" which need to be escaped (i.e. with a backslash placed in front - and the rules are different inside and outside character classes.) There is an excellent online tutorial at: www.regular-expressions.info. The time you spend there will pay for itself many times over. Happy regexing!

python .replace() regex

No. Regular expressions in Python are handled by the re module.

article = re.sub(r'(?is)</html>.+', '</html>', article)

In general:

text_after = re.sub(regex_search_term, regex_replacement, text_before)

How can I replace a string match with part of itself in Python?

Instead of directly using the re.sub() method, you can use the re.findall() method to find all substrings (in a non-greedy fashion) that begins and ends with the proper square brackets.

Then, iterate through the matches and use the str.replace() method to replace each match in the string with the second character in the match:

import re

s = "alEhos[cr@e]sjt"

for m in re.findall("\[.*?\]", s):
    s = s.replace(m, m[1])

print(s)

Output:

alEhoscsjt

replace before and after a string using re in python

Here is a regex one-liner:

inp = "approved:rakeshc@IAD.GOOGLE.COM"
output = re.sub(r'^.*:|@.*$', '', inp)
print(output)  # rakeshc

The above approach is to strip all text from the start up, and including, the :, as well as to strip all text from @ until the end. This leaves behind the email ID.

Regular expression replace string in Python 3

* is a meta character, you need to escape it if you want to match a literal * character. You are also missing the literal * character just before the closing /:

s2 = re.sub(r'/\*\s*\d+\s*\*/', "", s1)

Your code was matching zero or more / characters, and zero or more \s spaces, but not any of the literal * characters at the start and end of the comment.

Demo:

>>> import re
>>> s1 = "/*  123  */"
>>> re.sub(r'/\*\s*\d+\s*\*/', "", s1)
''

Regular expressions in Python: Replace starting and closing brackets with dynamic/incremental strings

One approach, it to take advantage of the fact that re.sub can take a function as the repl argument and create a function that will remember how many times it has matched before:

import re
from itertools import count

s = "We want to visit that place (HBD) and meet our friends (DTO) after a long time. We really love having such gatherings at this time of year (XYZ). Let's do it."

def repl(match, counter=count(1)):
    val = next(counter)
    return f"sb{val}-{match.group(1)}-eb{val}"

res = re.sub("\((.*?)\)", repl, s)
print(res)

Output

We want to visit that place sb1-HBD-eb1 and meet our friends sb2-DTO-eb2 after a long time. We really love having such gatherings at this time of year sb3-XYZ-eb3. Let's do it.

The pattern "\((.*?)\)" will match anything inside parenthesis, you can find a better explanation for a similar pattern in the Greedy versus Non-Greedy section of the Regular Expression HOWTO.

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