Develop Reference

Prevent last duplicate character from string [duplicate]

My regex pattern looks something like
<xxxx location="file path/level1/level2" xxxx some="xxx">
I am only interested in the part in quotes assigned to location. Shouldn't it be as easy as below without the greedy switch?
/.*location="(.*)".*/
Does not seem to work.

You need to make your regular expression lazy/non-greedy, because by default, "(.*)" will match all of "file path/level1/level2" xxx some="xxx".
Instead you can make your dot-star non-greedy, which will make it match as few characters as possible:
/location="(.*?)"/
Adding a ? on a quantifier (?, * or +) makes it non-greedy.
Note: this is only available in regex engines which implement the Perl 5 extensions (Java, Ruby, Python, etc) but not in "traditional" regex engines (including Awk, sed, grep without -P, etc.).

location="(.*)" will match from the " after location= until the " after some="xxx unless you make it non-greedy.
So you either need .*? (i.e. make it non-greedy by adding ?) or better replace .* with [^"]*.
[^"] Matches any character except for a " <quotation-mark>
More generic: [^abc] - Matches any character except for an a, b or c

How about
.*location="([^"]*)".*
This avoids the unlimited search with .* and will match exactly to the first quote.

Use non-greedy matching, if your engine supports it. Add the ? inside the capture.
/location="(.*?)"/

Use of Lazy quantifiers ? with no global flag is the answer.
Eg,
If you had global flag /g then, it would have matched all the lowest length matches as below.

Here's another way.
Here's the one you want. This is lazy [\s\S]*?
The first item:
[\s\S]*?(?:location="[^"]*")[\s\S]* Replace with: $1
Explaination: https://regex101.com/r/ZcqcUm/2
For completeness, this gets the last one. This is greedy [\s\S]*
The last item:[\s\S]*(?:location="([^"]*)")[\s\S]*
Replace with: $1
Explaination: https://regex101.com/r/LXSPDp/3
There's only 1 difference between these two regular expressions and that is the ?

The other answers here fail to spell out a full solution for regex versions which don't support non-greedy matching. The greedy quantifiers (.*?, .+? etc) are a Perl 5 extension which isn't supported in traditional regular expressions.
If your stopping condition is a single character, the solution is easy; instead of
a(.*?)b
you can match
a[^ab]*b
i.e specify a character class which excludes the starting and ending delimiiters.
In the more general case, you can painstakingly construct an expression like
start(|[^e]|e(|[^n]|n(|[^d])))end
to capture a match between start and the first occurrence of end. Notice how the subexpression with nested parentheses spells out a number of alternatives which between them allow e only if it isn't followed by nd and so forth, and also take care to cover the empty string as one alternative which doesn't match whatever is disallowed at that particular point.
Of course, the correct approach in most cases is to use a proper parser for the format you are trying to parse, but sometimes, maybe one isn't available, or maybe the specialized tool you are using is insisting on a regular expression and nothing else.

Because you are using quantified subpattern and as descried in Perl Doc,
By default, a quantified subpattern is "greedy", that is, it will
match as many times as possible (given a particular starting location)
while still allowing the rest of the pattern to match. If you want it
to match the minimum number of times possible, follow the quantifier
with a "?" . Note that the meanings don't change, just the
"greediness":
*? //Match 0 or more times, not greedily (minimum matches)
+? //Match 1 or more times, not greedily
Thus, to allow your quantified pattern to make minimum match, follow it by ? :
/location="(.*?)"/

import regex
text = 'ask her to call Mary back when she comes back'
p = r'(?i)(?s)call(.*?)back'
for match in regex.finditer(p, str(text)):
print (match.group(1))
Output:
Mary

Regular expression .*(\d+x\d+)$ must be greedy

https://regex101.com/r/BjO6H1/1/
Regular expression: .*(\d+x\d+)$
Test string: android-chrome-192x192
Could you tell me why I get 2x192 as Group 1? I expected 192x192. And how to get the expected result?

Since * is greedy, you should change to:
.*?(\d+x\d+)$
Check the documentation for more explanation:
When repeating a regular expression, as in a*, the resulting action is to consume as much of the pattern as possible. This fact often bites you when you’re trying to match a pair of balanced delimiters, such as the angle brackets surrounding an HTML tag. The naive pattern for matching a single HTML tag doesn’t work because of the greedy nature of .*

Add a ? to make the .* lazy:
.*?(\d+x\d+)$

Python 2.7 regular expression match issue

Suppose I am using the following regular expression to match, logically the regular expression means match anything with prefix foo: and ends with anything which is not a space. Match group will be the parts exclude prefix foo
My question is what exactly means anything in Python 2.7? Any ASCII or? If anyone could share some document, it will be great. Thanks.
a = re.compile('foo:([^ ]+)')
thanks in advance,
Lin

Try:
a = re.compile('foo:\S*')
\S means anything but whitespace.
I recommend you check out http://pythex.org.
It's really good for testing out regular expresions and has a decent cheat-sheet.
UPDATE:
Anything (.) matches anything, all unicode/UTF-8 characters.

The regular expression metacharacter which matches any character is . (dot).
a = re.compile('foo:(.+)')
The character class [^ ] matches any one character which isn't one of the characters between the square brackets (a literal space, in this example). The quantifier + specifies one or more repetitions of the preceding expression.

search for string embedded in {} after keyword

How can I get the string embedded in {} after a keyword, where the number of characters between the keyword and the braces {} is unknown. e.g.:
includegraphics[x=2]{image.pdf}
the keyword would be includegraphics and the string to be found is image.pdf, but the text in between [x=2] could have anything between the two [].
So I want to ignore all characters between the keyword and { or I want to ignore everything between []

Use re.findall
>>> sample = 'includegraphics[x=2]{image.pdf}'
>>> re.findall('includegraphics.*?{(.*?)}',sample)
['image.pdf']
Explanation:
The re module deals with regular expressions in Python. Its findall method is useful to find all occurences of a pattern in a string.
A regular expression for the pattern you are interested in is 'includegraphics.*?{(.*?)}'. Here . symbolizes "any character", while the * means 0 or more times. The question mark makes this a non-greedy operation. From the documentation:
The *, +, and ? qualifiers are all greedy; they match as much
text as possible. Sometimes this behaviour isn’t desired; if the RE
<.*> is matched against <H1\>title</H1>, it will match the entire
string, and not just <H1>. Adding ? after the qualifier makes it
perform the match in non-greedy or minimal fashion; as few characters
as possible will be matched. Using .*? in the previous expression will
match only <H1>.
Please note that while in your case using .*? should be fine, in general it's better to use more specialized character groups such as \w for alphanumerics and \d for digits, when you know what the content is going to consist of in advance.

Use re.search
re.search(r'includegraphics\[[^\[\]]*\]\{([^}]*)\}', s).group(1)

Python Regular Expressions to match option of strings

I am new to Python and Python RE.
I am trying to make a parser for ARM assembly code. I want to make regular expression for matching conditional branch instructions which could be:
beq, bne, blt, bgt
I tried a regular expression of the form
'b[eq|ne|lt|gt]'
But this does not match. Can someone please help me with this?

You should be using parentheses for options, not square brackets:
b(eq|ne|lt|gt)
And you'd usually want a non-capture group:
b(?:eq|ne|lt|gt)
And you can also make it a little more optimised too:
b(?:eq|ne|[lg]t)
Square brackets will be understood as being any of the characters or range of characters. So [eq|ne|lt|gt] effectively means either one of e, q, |, n, e (again, so it becomes redundant), etc.

Try the following pattern: b(?:eq|ne|lt|gt)
[] Character set: Will only match any one character inside the brackets. You can specify a range of characters by using the metacharacter -, eg: [a-e] or even negate the expression by using the metacharacter ^, eg: [^aeiou]
() Capturing parentesis: Used for grouping part & for creating number capturing group, you can disable this feature by using the following char-set ?: within the capturing parentesis, eg(?:)
As mentioned above, you should be using the capturing parentesis for more than one character to be matched, so, that is why your pattern using brackets did not match your string.
Please note that using the non capturing parentesis was meant to no save any data being matched, however you can remove the metacharacters ?: in order to capture the group.
As python performs perl compatible regular expression engine, you are able to use named captured groups & numbered backreferences, the main advantage of using it, is to keep your expression easy to maintain, read, edit, etc.
Eg:
(?P<opcode>b(?:eq|ne|lt|gt)) - Will capture the match of your pattern b(?:eq|ne|lt|gt) into the backreference name opcode