This question already has answers here:
re.findall behaves weird
(3 answers)
Closed 4 years ago.
I want to match sth like 12.12a or 13.12b but the below regex match with 'a' and i have no clue why is like that
import re
pattern = re.compile('\d\d?\.\d\d?(a|b)')
txt = "12.12a"
pattern_list = re.findall(pattern,txt)
for item in pattern_list:
print(item) # result a
Put the expression into the brackets. When there are brackets, only staff in brackets (matching groups to be precize) are matched and returned
pattern = re.compile('(\d\d?\.\d\d?(a|b))')
The result is ('12.12b', 'a') because of internal brackets. To get rid of internal brackets matches, use item[0] or another appropriate operation. Or simply unfold the regex (might be a little bit slower)
pattern = re.compile('\d\d?\.\d\d?a|\d\d?\.\d\d?b')
Related
This question already has answers here:
Why do some regex engines match .* twice in a single input string?
(1 answer)
Reference - What does this regex mean?
(1 answer)
Closed 2 years ago.
I was doing some regex which simplifies to this code:
>>> import re
>>> re.sub(r'^.*$|', "xyz", "abc")
xyzxyz
I was expecting it to replace abc with xyz as the RE ^.*$ matches the whole string, the engine should just return that and exit. So I ran the same regex with re.findall().
>>> re.findall(r'^.*$|', 'abcd')
['abcd', '']
in the docs it says:
A|B, where A and B can be arbitrary REs. As the target string is scanned, REs separated by '|'
are tried from left to right. When one pattern completely matches,
that branch is accepted. This means that once A matches, B will not be
tested further, even if it would produce a longer overall match.
but than why is the regex matching an empty string?
This question already has answers here:
Order of regular expression operator (..|.. ... ..|..)
(1 answer)
Checking whole string with a regex
(5 answers)
Closed 2 years ago.
pattern = (1|2|3|4|5|6|7|8|9|10|11|12)
str = '11'
This only matches '1', not '11'. How to match the full '11'? I changed it to:
pattern = (?:1|2|3|4|5|6|7|8|9|10|11|12)
It is the same.
I am testing here first:
https://regex101.com/
It is matching 1 instead of 11 because you have 1 before 11 in your alternation. If you use re.findall then it will match 1 twice for input string 11.
However to match numbers from 1 to 12 you can avoid alternation and use:
\b[1-9]|1[0-2]?\b
It is safer to use word boundary to avoid matching within word digits.
RegEx Demo
Regex always matches left before right.
On an alternation you'd put the longest first.
However, factoring should take precedense.
(1|2|3|4|5|6|7|8|9|10|11|12)
then it turns into
1[012]?|[2-9]
https://regex101.com/r/qmlKr0/1
I purposely didn't add boundary parts as
everybody has their own preference.
do you mean this solution?
[\d]+
This question already has answers here:
Regular expression to return text between parenthesis
(11 answers)
Closed 2 years ago.
I have long string S, and I want to find value (numeric) in the following format "Value(**)", where ** is values I want to extract.
For example, S is "abcdef Value(34) Value(56) Value(13)", then I want to extract values 34, 56, 13 from S.
I tried to use regex as follows.
import re
regex = re.compile('\Value(.*'))
re.findall(regex, S)
But the code yields the result I did not expect.
Edit. I edited some mistakes.
You should escape the parentheses, correct the typo of Value (as opposed to Values), use a lazy repeater *? instead of *, add the missing right parenthesis, and capture what's enclosed in the escaped parentheses with a pair of parentheses:
regex = re.compile(r'Value\((.*?)\)')
Only one of your numbers follows the word 'Value', so you can extract anything inside parentheses. You also need to escape the parentheses which are special characters.
regex = re.compile('\(.*?\)')
re.findall(regex, S)
Output:
['(34)', '(56)', '(13)']
I think what you're looking for is a capturing group that can return multiple matches. This string is: (\(\d{2}\))?. \d matches an digit and {2} matches exactly 2 digits. {1,2} will match 1 or 2 digits ect. ? matches 0 to unlimited number of times. Each group can be indexed and combined into a list. This is a simple implementation and will return the numbers within parentheses.
eg. 'asdasd Value(102222), fgdf(20), he(77)' will match 20 and 77 but not 102222.
This question already has answers here:
My regex is matching too much. How do I make it stop? [duplicate]
(5 answers)
Closed 4 years ago.
I run the following script
a = r'[abc] [abc] [y78]'
paaa = re.compile(r'\[ab.*]')
paaa.findall(a)
I obtained
['[abc] [abc] [y78]']
Why the '[abc]' is missing? The '[abc]' clearly matches the pattern as well. Is there any bug in the python3 re.findall function?
Clarification:
Sorry the paaa should be paaa = re.compile(r'\[ab.*\]')
What I am looking for is something which will return
['[abc]', '[abc]', '[abc] [abc]', '[abc] [abc] [y78]']
Basically, any substring matches the pattern.
The repeated . in [ab.*] is greedy - it'll match as many characters as it can such that those characters are followed by a ]. So, everything in between the first [ and the last ] are matched.
Use lazy repetition instead, with .*?:
a = r'[abc] [abc] [y78]'
paaa = re.compile(r'\[ab.*?]')
print(paaa.findall(a))
['[abc]', '[abc]']
You should escape the right square bracket as well, and use non-greedy repeater *? in your regex:
import re
a = r'[abc] [abc] [y78]'
paaa = re.compile(r'\[ab.*?\]')
print(paaa.findall(a))
This outputs:
['[abc]', '[abc]']
This question already has an answer here:
match trailing slash with Python regex
(1 answer)
Closed 8 years ago.
I can't match the question mark character although I escaped it.
I tried escaping with multiple backslashes and also using re.escape().
What am I missing?
Code:
import re
text = 'test?'
result = ''
result = re.match(r'\?',text)
print ("input: "+text)
print ("found: "+str(result))
Output:
input: test?
found: None
re.match only matches a pattern at the begining of string; as in the docs:
If zero or more characters at the beginning of string match the regular expression pattern, return a corresponding match object.
so, either:
>>> re.match(r'.*\?', text).group(0)
'test?
or re.search
>>> re.search(r'\?', text).group(0)
'?'