Develop Reference

Python Regex: Remove optional characters

I have a regex pattern with optional characters however at the output I want to remove those optional characters. Example:
string = 'a2017a12a'
pattern = re.compile("((20[0-9]{2})(.?)(0[1-9]|1[0-2]))")
result = pattern.search(string)
print(result)
I can have a match like this but what I want as an output is:
desired output = '201712'
Thank you.

You've already captured the intended data in groups and now you can use re.sub to replace the whole match with just contents of group1 and group2.
Try your modified Python code,
import re
string = 'a2017a12a'
pattern = re.compile(".*(20[0-9]{2}).?(0[1-9]|1[0-2]).*")
result = re.sub(pattern, r'\1\2', string)
print(result)
Notice, how I've added .* around the pattern, so any of the extra characters around your data is matched and gets removed. Also, removed extra parenthesis that were not needed. This will also work with strings where you may have other digits surrounding that text like this hello123 a2017a12a some other 99 numbers
Output,
201712
Regex Demo

You can just use re.sub with the pattern \D (=not a number):
>>> import re
>>> string = 'a2017a12a'
>>> re.sub(r'\D', '', string)
'201712'

Try this one:
import re
string = 'a2017a12a'
pattern = re.findall("(\d+)", string) # this regex will capture only digit
print("".join(p for p in pattern)) # combine all digits
Output:
201712

If you want to remove all character from string then you can do this
import re
string = 'a2017a12a'
re.sub('[A-Za-z]+','',string)
Output:
'201712'

You can use re module method to get required output, like:
import re
#method 1
string = 'a2017a12a'
print (re.sub(r'\D', '', string))
#method 2
pattern = re.findall("(\d+)", string)
print("".join(p for p in pattern))
You can also refer below doc for further knowledge.
https://docs.python.org/3/library/re.html

How can I "divide" words with regular expressions?

I have a sentence in which every token has a / in it. I want to just print what I have before the slash.
What I have now is basic:
text = less/RBR.....
return re.findall(r'\b(\S+)\b', text)
This obviously just prints the text, how do I cut off the words before the /?

Assuming you want all characters before the slash out of every word that contains a slash. This would mean e.g. for the input string match/this but nothing here but another/one you would want the results match and another.
With regex:
import re
result = re.findall(r"\b(\w*?)/\w*?\b", my_string)
print(result)
Without regex:
result = [word.split("/")[0] for word in my_string.split()]
print(result)

Simple and straight-forward:
rx = r'^[^/]+'
# anchor it to the beginning
# the class says: match everything not a forward slash as many times as possible
In Python this would be:
import re
text = "less/RBR....."
print re.match(r'[^/]+', text)
As this is an object, you'd probably like to print it out, like so:
print re.match(r'[^/]+', text).group(0)
# less

This should also work
\b([^\s/]+)(?=/)\b
Python Code
p = re.compile(r'\b([^\s/]+)(?=/)\b')
test_str = "less/RBR/...."
print(re.findall(p, test_str))
Ideone Demo

Combining two patterns with named capturing group in Python?

I have a regular expression which uses the before pattern like so:
>>> RE_SID = re.compile(r'(?P<sid>(?<=sid:)([A-Za-z0-9]+))')
>>> x = RE_SID.search('sid:I118uailfriedx151201005423521">>')
>>> x.group('sid')
'I118uailfriedx151201005423521'
and another like so:
>>> RE_SID = re.compile(r'(?P<sid>(?<=sid:<<")([A-Za-z0-9]+))')
>>> x = RE_SID.search('sid:<<"I118uailfriedx151201005423521')
>>> x.group('sid')
'I118uailfriedx151201005423521'
How can I combine these two patterns in a way that, after parsing these two different lines,:
sid:A111uancalual2626x151130185758596
sid:<<"I118uailfriedx151201005423521">>
returns only the corresponding id to me.

RE_SID = re.compile(r'sid:(<<")?(?P<sid>([A-Za-z0-9]+))')
Use this, I've just tested and it is working for me. I've moved some part out.

Instead of tweaking your regex, you can make your strings easier to parse by just removing any characters except alphanumeric and a colon. Then, just split by colon and get the last item:
>>> import re
>>>
>>> test_strings = ['sid:I118uailfriedx151201005423521">>', 'sid:<<"I118uailfriedx151201005423521']
>>> pattern = re.compile(r"[^A-Za-z0-9:]")
>>> for test_string in test_strings:
... print(pattern.sub("", test_string).split(":")[-1])
...
I118uailfriedx151201005423521
I118uailfriedx151201005423521

You can achieve what you want with a single regex:
\bsid:\W*(?P<sid>\w+)
See the regex demo
The regex breakdown:
\bsid - whole word sid
: - a literal colon
\W* - zero or more non-word characters
(?P<sid>\w+) - one or more word characters captured into a group named "sid"
Python demo:
import re
p = re.compile(r'\bsid:\W*(?P<sid>\w+)')
#test_str = "sid:I118uailfriedx151201005423521\">>" # => I118uailfriedx151201005423521
test_str = "sid:<<\"I118uailfriedx151201005423521" # => I118uailfriedx151201005423521
m = p.search(test_str)
if m:
print(m.group("sid"))

python 3 regular expression match string meta-character

I want to write a line of regular expression that can match strings like "(2000)" with years in parentheses. then I can check if any string contains the substring "2000".
for example, I want the regex to match (2000) not 2000, or (20000),or (200).
That is to say: they have to have exactly four digits, the first digit between 1 and 2; they have to include the parentheses.
also 2000 is just an example I use but really I want to the regex to include all the possible years.

You have to escape the open and close paranthesis,
>>> import re
>>> str = """foo(2000)bar(1000)foobar2000"""
>>> regex = r'\(2000\)'
>>> result = re.findall(regex, str)
>>> print(result)
['(2000)']
OR
>>> import re
>>> str = """foo(2000)bar(1000)foobar(2014)barfoo(2020)"""
>>> regex = r'\([0-9]{4}\)'
>>> result = re.findall(regex, str)
>>> print(result)
['(2000)', '(1000)', '(2014)', '(2020)']
It matches all the four digit numbers(year's) present within the paranthesis.

Special characters need to be escaped with a backslash. A parenthesis ( becomes \(. Therefore (2000) becomes \(2000\).
Then you can do something like:
if re.search(r"\(2000\)", subject):
# Successful match
else:
# Match attempt failed

>>> import re
>>> x = re.match(r'\((\d*?)\)', "(2000)")
>>> x.group(1)
'2000'

How do I extract certain parts of strings in Python?

Say I have three strings:
abc534loif
tvd645kgjf
tv96fjbd_gfgf
and three lists:
beginning captures just the first part of the string "the name"
middle captures just the number
end contains only the rest of the characters that are after the number portion
How do I accomplish this in the most efficent way?

Use regular expressions?
>>> import re
>>> strings = 'abc534loif tvd645kgjf tv96fjbd_gfgf'.split()
>>> for s in strings:
... for match in re.finditer(r'\b([a-z]+)(\d+)(.+?)\b', s):
... print match.groups()
...
('abc', '534', 'loif')
('tvd', '645', 'kgjf')
('tv', '96', 'fjbd_gfgf')

This is language agnostic approach that aims at higher efficiency:
find first digit in the string and save its position p0
find last digit in the string and save its position p1
extract substring from 0 to p0-1 into beginning
extract substring from p0 to p1 into middle
extract substring from p1+1 to length-1 into end

I guess you're looking for re.findall:
strs = """
abc534loif
tvd645kgjf
tv96fjbd_gfgf
"""
import re
print re.findall(r'\b(\w+?)(\d+)(\w+)', strs)
>> [('abc', '534', 'loif'), ('tvd', '645', 'kgjf'), ('tv', '96', 'fjbd_gfgf')]

>>> import itertools as it
>>> s="abc534loif"
>>> [''.join(j) for i,j in it.groupby(s, key=str.isdigit)]
['abc', '534', 'loif']

I'd something like this:
>>> import re
>>> l = ['abc534loif', 'tvd645kgjf', 'tv96fjbd_gfgf']
>>> regex = re.compile('([a-z_]+)(\d+)([a-z_]+)')
>>> beginning, middle, end = zip(*[regex.match(s).groups() for s in l])
>>> beginning
('abc', 'tvd', 'tv')
>>> middle
('534', '645', '96')
>>> end
('loif', 'kgjf', 'fjbd_gfgf')

I wouls use regualar expressions like:
(?P<beginning>[^0-9]*)(?P<middle>[^0-9]*)(?P<end>[^0-9]*)
and pull out the three matching sections.
import re
m = re.match(r"(?P<beginning>[^0-9]*)(?P<middle>[^0-9]*)(?P<end>[^0-9]*)", "abc534loif")
m.group('beginning')
m.group('middle')
m.group('end')

import re #You want to match a string against a pattern so you import the regular expressions module 're'
mystring = "abc1234def" #Just a string to test with
match = re.match(r"^(\D+)([0)9]+](\D+)$") #Our regular expression. Everything between brackets is 'captured', meaning that it is accessible as one of the 'groups' in the returned match object. The ^ sign matches at the beginning of a string, while the $ matches the end. the characters in between the square brackets [0-9] are character ranges, so [0-9] matches any digit character, \D is any non-digit character.
if match: # match will be None if the string didn't match the pattern, so we need to check for that, as None.group doesn't exist.
beginning = match.group(1)
middle = match.group(2)
end = match.group(3)