How to use a variable inside a regular expression

Question:

I’d like to use a variable inside a regex, how can I do this in Python?

TEXTO = sys.argv[1]

if re.search(r"b(?=w)TEXTOb(?!w)", subject, re.IGNORECASE):
    # Successful match
else:
    # Match attempt failed
Asked By: Pedro Lobito

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Source

Answers:

rx = r'b(?<=w){0}b(?!w)'.format(TEXTO)
Answered By: Cat Plus Plus
if re.search(r"b(?<=w)%sb(?!w)" % TEXTO, subject, re.IGNORECASE):

This will insert what is in TEXTO into the regex as a string.

Answered By: Bo Buchanan

You have to build the regex as a string:

TEXTO = sys.argv[1]
my_regex = r"b(?=w)" + re.escape(TEXTO) + r"b(?!w)"

if re.search(my_regex, subject, re.IGNORECASE):
    etc.

Note the use of re.escape so that if your text has special characters, they won’t be interpreted as such.

Answered By: Ned Batchelder

I agree with all the above unless:

sys.argv[1] was something like Chickend{2}-d{2}Ans*importants*anchor

sys.argv[1] = "Chickend{2}-d{2}Ans*importants*anchor"

you would not want to use re.escape, because in that case you would like it to behave like a regex

TEXTO = sys.argv[1]

if re.search(r"b(?<=w)" + TEXTO + "b(?!w)", subject, re.IGNORECASE):
    # Successful match
else:
    # Match attempt failed
Answered By: Max Carroll

I needed to search for usernames that are similar to each other, and what Ned Batchelder said was incredibly helpful. However, I found I had cleaner output when I used re.compile to create my re search term:

pattern = re.compile(r"("+username+".*):(.*?):(.*?):(.*?):(.*)"
matches = re.findall(pattern, lines)

Output can be printed using the following:

print(matches[1]) # prints one whole matching line (in this case, the first line)
print(matches[1][3]) # prints the fourth character group (established with the parentheses in the regex statement) of the first line.
Answered By: jdelaporte

I find it very convenient to build a regular expression pattern by stringing together multiple smaller patterns.

import re

string = "begin:id1:tag:middl:id2:tag:id3:end"
re_str1 = r'(?<=(S{5})):'
re_str2 = r'(idd+):(?=tag:)'
re_pattern = re.compile(re_str1 + re_str2)
match = re_pattern.findall(string)
print(match)

Output:

[('begin', 'id1'), ('middl', 'id2')]
Answered By: Deepak Nagarajan

You can use format keyword as well for this.Format method will replace {} placeholder to the variable which you passed to the format method as an argument. 

if re.search(r"b(?=w)**{}**b(?!w)".**format(TEXTO)**, subject, re.IGNORECASE):
    # Successful match**strong text**
else:
    # Match attempt failed
Answered By: Haneef Mohammed

you can try another usage using format grammer suger:

re_genre = r'{}'.format(your_variable)
regex_pattern = re.compile(re_genre)
Answered By: Kevin Chou

From python 3.6 on you can also use Literal String Interpolation, "f-strings". In your particular case the solution would be:

if re.search(rf"b(?=w){TEXTO}b(?!w)", subject, re.IGNORECASE):
    ...do something

EDIT:

Since there have been some questions in the comment on how to deal with special characters I’d like to extend my answer:

raw strings (‘r’):

One of the main concepts you have to understand when dealing with special characters in regular expressions is to distinguish between string literals and the regular expression itself. It is very well explained here:

In short:

Let’s say instead of finding a word boundary b after TEXTO you want to match the string boundary. The you have to write:

TEXTO = "Var"
subject = r"Varboundary"

if re.search(rf"b(?=w){TEXTO}\boundary(?!w)", subject, re.IGNORECASE):
    print("match")

This only works because we are using a raw-string (the regex is preceded by ‘r’), otherwise we must write "\\boundary" in the regex (four backslashes). Additionally, without ‘r’, b’ would not converted to a word boundary anymore but to a backspace!

re.escape:

Basically puts a backslash in front of any special character. Hence, if you expect a special character in TEXTO, you need to write:

if re.search(rf"b(?=w){re.escape(TEXTO)}b(?!w)", subject, re.IGNORECASE):
    print("match")

NOTE: For any version >= python 3.7: !, ", %, ', ,, /, :, ;, <, =, >, @, and ` are not escaped. Only special characters with meaning in a regex are still escaped. _ is not escaped since Python 3.3.(s. here)

Curly braces:

If you want to use quantifiers within the regular expression using f-strings, you have to use double curly braces. Let’s say you want to match TEXTO followed by exactly 2 digits:

if re.search(rf"b(?=w){re.escape(TEXTO)}d{{2}}b(?!w)", subject, re.IGNORECASE):
    print("match")
Answered By: airborne

more example

I have configus.yml
with flows files

"pattern":
  - _(d{14})_
"datetime_string":
  - "%m%d%Y%H%M%f"

in python code I use

data_time_real_file=re.findall(r""+flows[flow]["pattern"][0]+"", latest_file)
Answered By: Nikolay Baranenko

here’s another format you can use (tested on python 3.7)

regex_str = r'b(?<=w)%sb(?!w)'%TEXTO

I find it’s useful when you can’t use {} for variable (here replaced with %s)

Answered By: Ardhi
from re import search, IGNORECASE

def is_string_match(word1, word2):
    #  Case insensitively function that checks if two words are the same
    # word1: string
    # word2: string | list

    # if the word1 is in a list of words
    if isinstance(word2, list):
        for word in word2:
            if search(rf'b{word1}b', word, IGNORECASE):
                return True
        return False

    # if the word1 is same as word2
    if search(rf'b{word1}b', word2, IGNORECASE):
        return True
    return False

is_match_word = is_string_match("Hello", "hELLO") 
True

is_match_word = is_string_match("Hello", ["Bye", "hELLO", "@vagavela"])
True

is_match_word = is_string_match("Hello", "Bye")
False
Answered By: Luciano Narimatsu de Faria