Regular Expression in Python strings - python

I want to validate a string that satisfies the below three conditions using regular expression
The special characters allowed are (. , _ , - ).
Should contain only lower-case characters.
Should not start or end with special character.
To satisfy the above conditions, I have created a format as below
^[^\W_][a-z\.,_-]+
This pattern works fine up to second character. However, this pattern is failing for the 3rd and subsequent characters if those contains any special character or upper cases characters.
Example:
Pattern Works for the string S#yanthan but not for Sa#yanthan. I am expecting that pattern to pass even if the third and subsequent characters contains any special characters or upper case characters. Can you suggest me where this pattern goes wrong please? Below is the snippet of the code.
import re
a = "Sayanthan"
exp = re.search("^[^\W_][a-z\.,_-]+",a)
if exp:
print(True)
else:
print(False)

Based on you initial rules I'd go with:
^[a-z](?:[.,_-]*[a-z])*$
See the online demo.
However, you mentioned in the comments:
"Also the third condition is "should not start with Special character" instead of "should not start or end with Special character""
In that case you could use:
^[a-z][-.,_a-z]*$
See the online demo

The pattern that you tried ^[^\W_][a-z.,_-]+ starts with [^\W_] which will match any word char except an underscore, so it could also be an uppercase char.
Then [a-z.,_-]+ will match 1+ times any of the listed, which means the string can also end with a comma for example.
Looking at the conditions listed, you could use:
^[a-z](?:[a-z.,_-]*[a-z])?\Z
^ Start of string
[a-z] Match a lower case char a-z
(?: Non capture group
[a-z.,_-]*[a-z] Match 0+ occurrences of the listed ending with a-z
)? Close group and make it optional
\Z End of string
Regex demo

Related

Combining positive and negative lookahead in python

I'm trying to extract tokens that satisfy many conditions out of which, I'm using lookahead to implement the following two conditions:
The tokens must be either numeric/alphanumeric (i.e, they must have at least one digit). They can contain few special characters like - '-','/','\','.','_' etc.,
I want to match strings like: 165271, agya678, yah#123, kj*12-
The tokens can't have consecutive special characters like: ajh12-&
I don't want to match strings like: ajh12-&, 671%&i^
I'm using a positive lookahead for the first condition: (?=\w*\d\w*) and a negative lookahead for the second condition: (?!=[\_\.\:\;\-\\\/\#\+]{2})
I'm not sure how to combine these two look-ahead conditions.
Any suggestions would be helpful. Thanks in advance.
Edit 1 :
I would like to extract complete tokens that are part of a larger string too (i.e., They may be present in middle of the string).
I would like to match all the tokens in the string:
165271 agya678 yah#123 kj*12-
and none of the tokens (not even a part of a token) in the string: ajh12-& 671%&i^
In order to force the regex to consider the whole string I've also used \b in the above regexs : (?=\b\w*\d\w*\b) and (?!=\b[\_\.\:\;\-\\\/\#\+]{2}\b)
You can use
^(?!=.*[_.:;\-\\\/#+*]{2})(?=[^\d\n]*\d)[\w.:;\-\\\/#+*]+$
Regex demo
The negative lookahead (?=[^\d\n]*\d) matches any char except a digit or a newline use a negated character class, and then match a digit.
Note that you also have to add * and that most characters don't have to be escaped in the character class.
Using contrast, you could also turn the first .* into a negated character class to prevent some backtracking
^(?!=[^_.:;\-\\\/#+*\n][_.:;\-\\\/#+*]{2})(?=[^\d\n]*\d)[\w.:;\-\\\/#+*]+$
Edit
Without the anchors, you can use whitespace boundaries to the left (?<!\S) and to the right (?!\S)
(?<!\S)(?!=\S*[_.:;\-\\\/#+*]{2})(?=[^\d\s]*\d)[\w.:;\-\\\/#+*]+(?!\S)
Regex demo
You can use multiple look ahead assertions to only capture strings that
(?!.*(?:\W|_){2,}.*) - doesn't have consecutive special characters and
(?=.*\d.*) - has at least 1 digit
^(?!.*(?:\W|_){2,}.*)(?=.*\d.*).*$

Python regular expressions: disallowing characters with exceptions

In my actual project I want to get some input from a user. It cannot contain whitespace at the beginning, at the end or twice or more in a row and it also cannot contain anything that is not a letter or a number.
However I have 2 problems:
\W disallows whitespaces. I want to allow one but not in the beginning, the end or more than once.
I want to allow dots but \W disallows it. I can't replace it with [^a-zA-Z0-9\s\s.] because it disallows characters like 'äöüß' which I want to allow. How can I change my code to disallow everything that is in the pattern below except for single whitespaces and dots?
input_text = input("input: ")
if re.search(r"(^\s|\s{2,}|\s$|\W)", input_text)
print("invalid input")
You may use either of the two:
^[^\W_]+(?:[\s.][^\W_]+)*\Z
^(?:[^\W_]|\.)+(?:\s(?:[^\W_]|\.)+)*\Z
The first one will match single dots and whitespace only in between letters. The second one allows dots to be anywhere in the string and in any consecutive quantities.
See regex #1 demo and regex #2 demo.
Details
^ - start of string
[^\W_]+ - one or more Unicode letters or digits
(?:[^\W_]|\.)+ - a non-capturing group matching one or more Unicode letters/digits or a dot
(?:[\s.][^\W_]+)* - zero or more repetitions of
[\s.] - a single whitespace or dot
[^\W_]+ - one or more Unicode letters or digits
\Z - the very end of string.
In Python, use something like
if re.search(r'^[^\W_]+(?:[\s.][^\W_]+)*\Z', input_text):
print('Valid')

Python regex specific word with singe quote at end

Searching a large syslog repo and need to get a specific word to match with a certain condition.
I'm using regex to compile a search for this word. I've read the python docs on regex characters and I understand how to specify each criteria separately but somehow missing how to concatenate all together for my specific search. This is what I have so far but not working...
p = re.compile("^'[A-Z]\w+'$")
match = re.search(p, syslogline, )
the word is a username that can be alphanum, always beginning with an uppercase character (preceded by blank space), can contain chars or nums, is 3-12 in length and ends with single quote.
an example would be: Epresley01' or J98473'
Brief
Based on your requirements (also stated below), your regex doesn't work because:
^' Asserts the position at the start of the line and ensures a ' is the first character of that line.
$ Asserts the position at the end of the line.
Having said that you specify that it's preceded by a space character (which isn't present in your pattern). You pattern also checks for ' which isn't the first character of the username. Given that you haven't actually given us a sample of your file I can't confirm nor deny that your string starts before the username and ends after it, but if that's not the case the anchors ^$ are also not helping you here.
Requirements
The requirements below are simply copied from the OP's question (rewritten) to outline the username format. The username:
Is preceded by a space character.
Starts with an uppercase letter.
Contains chars or nums. I'm assuming here that chars actually means letters and that all letters in the username (including the uppercase starting character) are ASCII.
Is 3-12 characters in length (excluding the preceding space and the end character stated below).
Ends with an apostrophe character '.
Code
See regex in use here
(?<= )[A-Z][^\W_]{2,11}'
Explanation
(?<= ) Positive lookbehind ensuring what precedes is a space character
[A-Z] Match any uppercase ASCII letter
[^\W_]{2,11} Match any word character except underscore _ (equivalent to a-zA-Z0-9)
This appears a little confusing because it's actually a double-negative. It's saying match anything that's not in the set. The \W matches any non-word character. Since it's a double-negative, it's like saying don't match non-word characters. Adding _ to the set negates it.
' Match the apostrophe character ' literally
I think you can do it like this:
(Updated after the comment from #ctwheels)
See regex in use here
[A-Z][a-zA-Z0-9]{1,10}'
Explanation
Match a whitespace
Match an uppercase character [A-Z]
Match [a-zA-Z0-9]+
Match an apostrophe '
Demo

Python regex: How to make a group of words/character optional?

I am trying to make regex that can match all of them:
word
word-hyphen
word-hyphen-again
that is -\w+could be many depends on words in a term. How can I make it optional
Thing I made so far is given here:- https://regex101.com/r/Atpwze/1
Try using
\w+(-\w+)* for matching 0 or more hyphenated words after first word
\w+(-\w+){0,} same as first case
based on your exact requirement.
In order to eliminate some extreme cases like a-+-+---, you could use \w+(-\w+)*[^\W]
\W matches all non-word characters and ^\W negates the matching of non-word characters
To catch all of your examples, I think you could use:
^\w+(?:\w+\-?|\-\w+)+$
Beginning of the string ^
Match a word character one or more times \w+
Start a non capturing group (?:
Match a word character one or more times with an optional hyphen \w+\-?
Or |
A hyphen with one or more word characters \-\w+
Close the non capturing group )
End of the string $

Negative lookahead not working after character range with plus quantifier

I am trying to implement a regex which includes all the strings which have any number of words but cannot be followed by a : and ignore the match if it does. I decided to use a negative look ahead for it.
/([a-zA-Z]+)(?!:)/gm
string: lame:joker
since i am using a character range it is matching one character at a time and only ignoring the last character before the : .
How do i ignore the entire match in this case?
Link to regex101: https://regex101.com/r/DlEmC9/1
The issue is related to backtracking: once your [a-zA-Z]+ comes to a :, the engine steps back from the failing position, re-checks the lookahead match and finds a match whenver there are at least two letters before a colon, returning the one that is not immediately followed by :. See your regex demo: c in c:real is not matched as there is no position to backtrack to, and rea in real:c is matched because a is not immediately followed with :.
Adding implicit requirement to the negative lookahead
Since you only need to match a sequence of letters not followed with a colon, you can explicitly add one more condition that is implied: and not followed with another letter:
[A-Za-z]+(?![A-Za-z]|:)
[A-Za-z]+(?![A-Za-z:])
See the regex demo. Since both [A-Za-z] and : match a single character, it makes sense to put them into a single character class, so, [A-Za-z]+(?![A-Za-z:]) is better.
Preventing backtracking into a word-like pattern by using a word boundary
As #scnerd suggests, word boundaries can also help in these situations, but there is always a catch: word boundary meaning is context dependent (see a number of ifs in the word boundary explanation).
[A-Za-z]+\b(?!:)
is a valid solution here, because the input implies the words end with non-word chars (i.e. end of string, or chars other than letter, digits and underscore). See the regex demo.
When does a word boundary fail?
\b will not be the right choice when the main consuming pattern is supposed to match even if glued to other word chars. The most common example is matching numbers:
\d+\b(?!:) matches 12 in 12,, but not in 12:, and also 12c and 12_
\d+(?![\d:]) matches 12 in 12, and 12c and 12_, not in 12: only.
Do a word boundary check \b after the + to require it to get to the end of the word.
([a-zA-Z]+\b)(?!:)
Here's an example run.

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