I need assistance with matching spaces and subsequent matches in regex.
the example is as follows:
I want to match all of the following scenarios:
60 ml ( 1)
60ML (2 )
60ml(2) (a)
the regex I have used is:
(60\s?(?:ml)\s?(?:\w|\(.{0,3}\)){0,5})
link to the example: link to regex
the regex matches the first 2 examples, but not the instances where there is a space between (2) and (a).
any guidance would be appreciated.
Your regex doesn't allow for spaces between the parenthesised groups (2) and (a) in your last example. You can add <space>* to it to allow it to do so. Note you cannot use \s* unless you are only matching a single value at a time, otherwise the fact that \s will match newline can cause the first match to go too far.
(60\s?ml\s?(?:\w|\(.{0,3}\) *){0,5})
Note that without anchors counting repetitions doesn't really make sense. For example, this regex will match both 60ML (2 )(a)(a)(a)(a) and 60ML (2 )(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a)(a), returning 60ML (2 )(a)(a)(a)(a) in both cases. If that is not what you want, you will need to add an anchor to the end of the regex ($ perhaps) to prevent it matching the longer string.
Demo on regex101
Related
What I am trying to do is match values from one file to another, but I only need to match the first portion of the string and the last portion.
I am reading each file into a list, and manipulating these based on different Regex patterns I have created. Everything works, except when it comes to these type of values:
V-1\ZDS\R\EMBO-20-1:24
V-1\ZDS\R\EMBO-20-6:24
In this example, I only want to match 'V-1\ZDS\R\EMBO-20' and then compare the '24' value at the end of the string. The number x in '20-x:', can vary and doesn't matter in terms of comparisons, as long as the first and last parts of this string match.
This is the Regex I am using:
re.compile(r"(?:.*V-1\\ZDS\\R\\EMBO-20-\d.*)(:\d*\w.*)")
Once I filter down the list, I use the following function to return the difference between the two sets:
funcDiff = lambda x, y: list((set(x)- set(y))) + list((set(y)- set(x)))
Is there a way to take the list of differences and filter out the ones that have matching values after the
:
as mentioned above?
I apologize is this is an obvious answer, I'm new to Python and Regex!
The output I get is the differences between the entire strings, so even if the first and last part of the string match, if the number following the 'EMBO-20-x' doesn't also match, it returns it as being different.
Before discussing your question, regex101 is an incredibly useful tool for this type of thing.
Your issue stems from two issues:
1.) The way you used .*
2.) Greedy vs. Nongreedy matches
.* kinda sucks
.* is a regex expression that is very rarely what you actually want.
As a quick aside, a useful regex expression is [^c]* or [^c]+. These expressions match any character except the letter c, with the first expression matching 0 or more, and the second matched 1 or more.
.* will match all characters as many times as it can. Instead, try to start your regex patterns with more concrete starting points. Two good ways to do this are lookbehind expressions and anchors.
Another quick aside, it's likely that you are misusing regex.match and regex.find. match will only return a match that begins at the start of the string, while find will return matches anywhere in the input string. This could be the reason you included the .* in the first place, to allow a .match call to return a match deeper in the string.
Lookbehind Expressions
There are more complete explanations online, but in short, regex patterns like:
(?<=test)foo
will match the text foo, but only if test is right in front of it. To be more clear, the following strings will not match that regex:
foo
test-foo
test foo
but the following string will match:
testfoo
This will only match the text foo, though.
Anchors
Another option is anchors. ^ and $ are special characters, matching the start and end of a line of text. If you know your regex pattern will match exactly one line of text, start it with ^ and end it with $.
Leading patterns with .* and ending with .* are likely the source of your issue. Although you did not include full examples of your input or your code, you likely used match as opposed to find.
In regex, . matches any character, and * means 0 or more times. This means that for any input, your pattern will match the entire string.
Greedy vs. Non-Greedy qualifiers
The second issue is related to greediness. When your regex patterns have a * in them, they can match 0 or more characters. This can hide problems, as entire * expressions can be skipped. Your regex is likely matched several lines of text as one match, and hiding multiple records in a single .*.
The Actual Answer
Taking all of this in to consideration, let's assume that your input data looks like this:
V-1\ZDS\R\EMBO-20-1:24
V-1\ZDS\R\EMBO-20-6:24
V-1\ZDS\R\EMBO-20-3:93
V-1\ZDS\R\EMBO-20-6:22309
V-1\ZDS\R\EMBO-20-8:2238
V-1\ZDS\R\EMBO-20-3:28
A better regular expression would be:
^V-1\\ZDS\\R\\EMBO-20-\d:(\d+)$
To visualize this regex in action, follow this link.
There are several differences I would like to highlight:
Starting the expression with ^ and ending with $. This forces the regex to match exactly one line. Even though the pattern works without these characters, it's good practice when working with regex to be as explicit as possible.
No useless non-capturing group. Your example had a (?:) group at the start. This denotes a group that does not capture it's match. It's useful if you want to match a subpattern multiple times ((?:ab){5} matches ababababab without capturing anything). However, in your example, it did nothing :)
Only capturing the number. This makes it easier to extract the value of the capture groups.
No use of *, one use of +. + works like *, but it matches 1 or more. This is often more correct, as it prevents 'skipping' entire characters.
I wanted to extract experience from some line of text. It may contain some variation of years and months. I tried to make two non-capturing blocks using regex but it ends up in giving me several capturing instance.
Work Experience: 15 years 2 months
regex is:
((?:\d{1,3}(?:\.)?(?:\d{1})?\s+year(?:s)?\s+)?(?:\d{1,3}\s+month(?:s)?)?)
Though it captured the string that I want to find and it returns spurious matches as well.
One way to simply join the all instance as rest of matches are '' but that will not be justice to coding practice.
I need a small help to figure out where did I go wrong?
Apologies,
I have missed one scenario which has led to putting everyone off track. there are strings which are like
2 Months
1 year 3 months
1.5 year
15 year 2 months
The pattern matches a position before and after each character as well because all the parts in the pattern are optional.
You can write these parts like (?:s)? just as s? and you can omit {1}
If you don't want to match empty strings, you could either match an optional year part followed by the months, or match the months part.
You could either use a case insensitive match or match either a lowercase chars and upper case char using a character class [yY]
As you want the match only, you can omit the capturing group.
\b(?:\d+(?:\.\d+)? years? )?\d{1,3} months?\b|\b\d+(?:\.\d+)? years?\b
Explanation
\b(?:\d+(?:\.\d+)? years? )? Match an optional year part with optional decimal part
\d{1,3} months?\b Match the month part with 1-3 digits
| Or
\b\d+(?:\.\d+)? years?\b Match the years part
Regex demo
Note that \s could also match a newline
You can reduce your regex as below to capture only those groups that required. here \1 will have required string.
This will also match string separated with tabs and newlines.
^\D*((?:[\d.]+\s*[yY]ears?)?\s*(?:[\d.]+\s*[mM]onths?)?)
Demo
I am trying to create a regex expression in Python for non-hyphenated words but I am unable to figure out the right syntax.
The requirements for the regex are:
It should not contain hyphens AND
It should contain atleast 1 number
The expressions that I tried are:=
^(?!.*-)
This matches all non-hyphenated words but I am not able to figure out how to additionally add the second condition.
^(?!.*-(?=/d{1,}))
I tried using double lookahead but I am not sure about the syntax to use for it. This matches ID101 but also matches STACKOVERFLOW
Sample Words Which Should Match:
1DRIVE , ID100 , W1RELESS
Sample Words Which Should Not Match:
Basically any non-numeric string (like STACK , OVERFLOW) or any hyphenated words (Test-11 , 24-hours)
Additional Info:
I am using library re and compiling the regex patterns and using re.search for matching.
Any assistance would be very helpful as I am new to regex matching and am stuck on this for quite a few hours.
Maybe,
(?!.*-)(?=.*\d)^.+$
might simply work OK.
Test
import re
string = '''
abc
abc1-
abc1
abc-abc1
'''
expression = r'(?m)(?!.*-)(?=.*\d)^.+$'
print(re.findall(expression, string))
Output
['abc1']
If you wish to simplify/modify/explore the expression, it's been explained on the top right panel of regex101.com. If you'd like, you can also watch in this link, how it would match against some sample inputs.
RegEx Circuit
jex.im visualizes regular expressions:
RegEx 101 Explanation
/
(?!.*-)(?=.*\d)^.+$
/
gm
Negative Lookahead (?!.*-)
Assert that the Regex below does not match
.* matches any character (except for line terminators)
* Quantifier — Matches between zero and unlimited times, as many times as possible, giving back as needed (greedy)
- matches the character - literally (case sensitive)
Positive Lookahead (?=.*\d)
Assert that the Regex below matches
.* matches any character (except for line terminators)
* Quantifier — Matches between zero and unlimited times, as many times as possible, giving back as needed (greedy)
\d matches a digit (equal to [0-9])
^ asserts position at start of a line
.+ matches any character (except for line terminators)
+ Quantifier — Matches between one and unlimited times, as many times as possible, giving back as needed (greedy)
$ asserts position at the end of a line
Global pattern flags
g modifier: global. All matches (don't return after first match)
m modifier: multi line. Causes ^ and $ to match the begin/end of each line (not only begin/end of string)
I came up with -
^[^-]*\d[^-]*$
so we need at LEAST one digit (\d)
We need the rest of the string to contain anything BUT a - ([^-])
We can have unlimited number of those characters, so [^-]*
but putting them together like [^-]*\d would fail on aaa3- because the - comes after a valid match- lets make sure no dashes can sneak in before or after our match ^[-]*\d$
Unfortunately that means that aaa555D fails. So we actually need to add the first group again- ^[^-]*\d[^-]$ --- which says start - any number of chars that aren't dashes - a digit - any number of chars that aren't dashes - end
Depending on style, we could also do ^([^-]*\d)+$ since the order of the digits/numbers dont matter, we can have as many of those as we want.
However, finally... this is how I would ACTUALLY solve this particular problem, since regexes may be powerful, but they tend to make the code harder to understand...
if ("-" not in text) and re.search("\d", text):
I have regex code
https://regex101.com/r/o5gdDt/8
As you see this code
(?<!\S)(?<![\d,])(?:(?!(?:1[2-9]\d\d|20[01]\d|2020))\d{4,}[\u00BC-\u00BE\u2150-\u215E]?|\d{1,3}(?:,\d{3})+)(?![\d,])[\u00BC-\u00BE\u2150-\u215E]?(?!x)(?!/)
can capture all digits which sperated by 3 digits in text like
"here is 100,100"
"23,456"
"1,435"
all more than 4 digit number like without comma separated
2345
1234 " here is 123456"
also this kind of number
65,656½
65,656½,
23,123½
The only tiny issue here is if there is a comma(dot) after the first two types it can not capture those. for example, it can not capture
"here is 100,100,"
"23,456,"
"1,435,"
unfortunately, there is a few number intext which ends with comma...can someone gives me an idea of how to modify this to capture above also?
I have tried to do it and modified version is so:
(?<!\S)(?<![\d,])(?:(?!(?:1[2-9]\d\d|20[01]\d|2020))\d{4,}[\u00BC-\u00BE\u2150-\u215E]?|\d{1,3}(?:,\d{3})+)(?![\d])[\u00BC-\u00BE\u2150-\u215E]?(?!x)(?!/)
basically I delete comma in (?![\d,]) but it causes to another problem in my context
it captures part of a number that is part of equation like this :
4,310,747,475x2
57,349,565,416,398x.
see here:
https://regex101.com/r/o5gdDt/10
I know that is kind of special question I would be happy to know your ides
The main problem here is that (?![\d,]) fails any match followed with a digit or comma while you want to fail the match when it is followed with a digit or a comma plus a digit.
Replace (?![\d,]) with (?!,?\d).
Also, (?<!\S)(?<![\d,]) looks redundant, as (?<!\S) requires a whitespace or start of string and that is certainly not a digit or ,. Either use (?<!\S) or (?<!\d)(?<!\d,) depending on your requirements.
Join the negative lookaheads with OR: (?!x)(?!/) => (?!x|/) => (?![x/]).
You wnat to avoid matching years, but you just fail all numbers that start with them, so 2020222 won't get matched. Add (?!\d) to the lookahead, (?!(?:1[2-9]\d\d|20[01]\d|2020)(?!\d)).
So, the pattern might look like
(?<!\S)(?:(?!(?:1[2-9]\d\d|20[01]\d|2020)(?!\d))\d{4,}[\u00BC-\u00BE\u2150-\u215E]?|\d{1,3}(?:,\d{3})+)(?!,?\d)[\u00BC-\u00BE\u2150-\u215E]?(?![x/])
See the regex demo.
IMPORTANT: You have [\u00BC-\u00BE\u2150-\u215E]?(?![x/]) at the end, a negative lookahead after an optional pattern. Once the engine fails to find the match for x or /, it will backtrack and will most probably find a partial match. If you do not want to match 65,656 in 65,656½x, replace [\u00BC-\u00BE\u2150-\u215E]?(?![x/]) with (?![\u00BC-\u00BE\u2150-\u215E]?[x/])[\u00BC-\u00BE\u2150-\u215E]?.
See another regex demo.
Not sure if this is something that should be a bounty. II just want to understand regex better.
I checked the responses in the Regex to match pattern.one skip newlines and characters until pattern.two and Regex to match if given text is not found and match as little as possible threads and read about Tempered Greedy Token Solutions and Explicit Greedy Alternation Solutions on RexEgg, but admittedly the explanations baffled me.
I spent the last day fiddling mainly with re.sub (and with findall) because re.sub's behaviour is odd to me.
.
Problem 1:
Given Strings below with characters followed by / how would I produce a SINGLE regex (using only either re.sub or re.findall) that uses alternating capture groups which must use [\S]+/ to get the desired output
>>> string_1 = 'variety.com/2017/biz/news/tax-march-donald-trump-protest-1202031487/'
>>> string_2 = 'variety.com/2017/biz/the/life/of/madam/green/news/tax-march-donald-trump-protest-1202031487/'
>>> string_3 = 'variety.com/2017/biz/the/life/of/news/tax-march-donald-trump-protest-1202031487/the/days/of/our/lives'
Desired Output Given the Conditions(!!)
tax-march-donald-trump-protest-
CONDITIONS: Must use alternating capture groups which must capture ([\S]+) or ([\S]+?)/ to capture the other groups but ignore them if they don't contain -
I'M WELL AWARE that it would be better to use re.findall('([\-]*(?:[^/]+?\-)+)[\d]+', string) or something similar but I want to know if I can use [\S]+ or ([\S]+) or ([\S]+?)/ and tell regex that if those are captured, ignore the result if it contains / or doesn't contain - While also having used an alternating capture group
I KNOW I don't need to use [\S]+ or ([\S]+) but I want to see if there is an extra directive I can use to make the regex reject some characters those two would normally capture.
Posted per request:
(?:(?!/)[\S])*-(?:(?!/)[\S])*
https://regex101.com/r/azrwjO/1
Explained
(?: # Optional group
(?! / ) # Not a forward slash ahead
[\S] # Not whitespace class
)* # End group, do 0 to many times
- # A dash must exist
(?: # Optional group, same as above
(?! / )
[\S]
)*
You could use
/([-a-z]+)-\d+
and take the first capturing group, see a demo on regex101.com.