Develop Reference

Regular expression to retrieve string parts within parentheses separated by commas

I have a String from which I want to take the values within the parenthesis. Then, get the values that are separated from a comma.
Example: x(142,1,23ERWA31)
I would like to get:
142
1
23ERWA31
Is it possible to get everything with one regex?
I have found a method to do so, but it is ugly.
This is how I did it in python:
import re
string = "x(142,1,23ERWA31)"
firstResult = re.search("\((.*?)\)", string)
secondResult = re.search("(?<=\()(.*?)(?=\))", firstResult.group(0))
finalResult = [x.strip() for x in secondResult.group(0).split(',')]
for i in finalResult:
print(i)
142
1
23ERWA31

This works for your example string:
import re
string = "x(142,1,23ERWA31)"
l = re.findall (r'([^(,)]+)(?!.*\()', string)
print (l)
Result: a plain list
['142', '1', '23ERWA31']
The expression matches a sequence of characters not in (,,,) and – to prevent the first x being picked up – may not be followed by a ( anywhere further in the string. This makes it also work if your preamble x consists of more than a single character.
findall rather than search makes sure all items are found, and as a bonus it returns a plain list of the results.

You can make this a lot simpler. You are running your first Regex but then not taking the result. You want .group(1) (inside the brackets), not .group(0) (the whole match). Once you have that you can just split it on ,:
import re
string = "x(142,1,23ERWA31)"
firstResult = re.search("\((.*?)\)", string)
for e in firstResult.group(1).split(','):
print(e)

A little wonky looking, and also assuming there's always going to be a grouping of 3 values in the parenthesis - but try this regex
\((.*?),(.*?),(.*?)\)
To extract all the group matches to a single object - your code would then look like
import re
string = "x(142,1,23ERWA31)"
firstResult = re.search("\((.*?),(.*?),(.*?)\)", string).groups()
You can then call the firstResult object like a list
>> print(firstResult[2])
23ERWA31

How do you find all instances of a substring, followed by a certain number of dynamic characters?

I'm trying to find all instances of a specific substring(a!b2 as an example) and return them with the 4 characters that follow after the substring match. These 4 following characters are always dynamic and can be any letter/digit/symbol.
I've tried searching, but it seems like the similar questions that are asked are requesting help with certain characters that can easily split a substring, but since the characters I'm looking for are dynamic, I'm not sure how to write the regex.

When using regex, you can use "." to dynamically match any character. Use {number} to specify how many characters to match, and use parentheses as in (.{number}) to specify that the match should be captured for later use.
>>> import re
>>> s = "a!b2foobar a!b2bazqux a!b2spam and eggs"
>>> print(re.findall("a!b2(.{4})", s))
['foob', 'bazq', 'spam']

import re
print (re.search(r'a!b2(.{4})')).group(1))
.{4} matches any 4 characters except special characters.
group(0) is the complete match of the searched string. You can read about group id here.

If you're only looking for how to grab the following 4 characters using Regex, what you are probably looking to use is the curly brace indicator for quantity to match: '{}'.
They go into more detail in the post here, but essentially you would do [a-Z][0-9]{X,Y} or (.{X,Y}), where X to Y is the number of characters you're looking for (in your case, you would only need {4}).
A more Pythonic way to solve this problem would be to make use of string slicing, and the index function however.
Eg. given an input_string, when you find the substring at index i using index, then you could use input_string[i+len(sub_str):i+len(sub_str)+4] to grab those special characters.
As an example,
input_string = 'abcdefg'
sub_str = 'abcd'
found_index = input_string.index(sub_str)
start_index = found_index + len(sub_str)
symbol = input_string[start_index: start_index + 4]
Outputs (to show it works with <4 as well): efg
Index also allows you to give start and end indexes for the search, so you could also use it in a loop if you wanted to find it for every sub string, with the start of the search index being the previous found index + 1.

How to parse values appear after the same string in python?

I have a input text like this (actual text file contains tons of garbage characters surrounding these 2 string too.)
(random_garbage_char_here)**value=xxx**;(random_garbage_char_here)**value=yyy**;(random_garbage_char_here)
I am trying to parse the text to store something like this:
value1="xxx" and value2="yyy".
I wrote python code as follows:
value1_start = content.find('value')
value1_end = content.find(';', value1_start)
value2_start = content.find('value')
value2_end = content.find(';', value2_start)
print "%s" %(content[value1_start:value1_end])
print "%s" %(content[value2_start:value2_end])
But it always returns:
value=xxx
value=xxx
Could anyone tell me how can I parse the text so that the output is:
value=xxx
value=yyy

Use a regex approach:
re.findall(r'\bvalue=[^;]*', s)
Or - if value can be any 1+ word (letter/digit/underscore) chars:
re.findall(r'\b\w+=[^;]*', s)
See the regex demo
Details:
\b - word boundary
value= - a literal char sequence value=
[^;]* - zero or more chars other than ;.
See the Python demo:
import re
rx = re.compile(r"\bvalue=[^;]*")
s = "$%$%&^(&value=xxx;$%^$%^$&^%^*value=yyy;%$#^%"
res = rx.findall(s)
print(res)

Use regex to filter the data you want from the "junk characters":
>>> import re
>>> _input = '#4#5%value=xxx38u952035983049;3^&^*(^%$3value=yyy#%$#^&*^%;$#%$#^'
>>> matches = re.findall(r'[a-zA-Z0-9]+=[a-zA-Z0-9]+', _input)
>>> matches
['value=xxx', 'value=yyy']
>>> for match in matches:
print(match)
value=xxx
value=yyy
>>>
Summary or the regular expression:
[a-zA-Z0-9]+: One or more alphanumeric characters
=: literal equal sign
[a-zA-Z0-9]+: One or more alphanumeric characters

For this input:
content = '(random_garbage_char_here)**value=xxx**;(random_garbage_char_here)**value=yyy**;(random_garbage_char_here)'
use a simple regex and manually strip off the first and last two characters:
import re
values = [x[2:-2] for x in re.findall(r'\*\*value=.*?\*\*', content)]
for value in values:
print(value)
Output:
value=xxx
value=yyy
Here the assumption is that there are always two leading and two trailing * as in **value=xxx**.

You already have good answers based on the re module. That would certainly be the simplest way.
If for any reason (perfs?) you prefere to use str methods, it is indeed possible. But you must search the second string past the end of the first one :
value2_start = content.find('value', value1_end)
value2_end = content.find(';', value2_start)

Slice substrings from long string to a list in python

In python I have long string like (of which I removed all breaks)
stringA = 'abcdefkey:12/eas9ghijklkey:43/e3mnop'
What I want to do is to search this string for all occurrences of "key:", then extract the "values" following "key:".
One further complication for me is that I don't know how long these values belonging to key are (e.g. key:12/eas9 and key:43/e3). All I do know is that they do have to end with a digit whereas the rest of the string does not contain any digits.
This is why my idea was to slice from the indices of key plus the next say 10 characters (e.g. key:12/eas9g) and then work backward until isdigit() is false.
I tried to split my initial string (that did contain breaks):
stringA_split = re.split("\n", stringA)
for linex in stringA_split:
index_start = linex.rfind("key:")
index_end = index_start + 8
print(linex[index_start:index_end]
#then work backward
However, inserting line breaks does not help in any way as they are meaningless from a pdf-to-txt conversion.
How would I then solve this (e.g. as a start with getting all indices of '"key:"' and slice this to a list)?

import re
>>> re.findall('key:(\d+[^\d]+[\d])', stringA)
['12/eas9', '43/e3']
\d+ # One or more digits.
[^\d]+ # Everything except a digit (equivalent to [\D]).
[\d] # The final digit
(\d+[^\d]+[\d]) # The group of the expression above
'key:(\d+[^\d]+[\d])' # 'key:' followed by the group expression
If you want key: in your result:
>>> re.findall('(key:\d+[^\d]+[\d])', stringA)
['key:12/eas9', 'key:43/e3']

I'm not 100% sure I understand your definition of what defines a value, but I think this will get you what you described
import re
stringA = 'abcdefkey:12/eas9ghijklkey:43/e3mnop'
for v in stringA.split('key:'):
ma = re.match(r'(\d+\/.*\d+)', v)
if ma:
print ma.group(1)
This returns:
12/eas9
43/e3

You can apply just one RE that gets all the keys into an array of tuples:
import re
p=re.compile('key\:(\d+)\/([^\d]+\d)')
ret=p.findall(stringA)
After the execution, you have:
ret
[('12', 'eas9'), ('43', 'e3')]

edit: a better answer was posted above. I misread the original question when proposing to reverse here, which really wasn't necessary. Good luck!
If you know that the format is always key:, what if you reversed the string and rex for :yek? You'd isolate all keys and then can reverse them back
import re
# \w is alphanumeric, you may want to add some symbols
rex = re.compile("\w*:yek")
word = 'abcdefkey:12/eas9ghijklkey:43/e3mnop'
matches = re.findall(rex, word[::-1])
matches = [match[::-1] for match in matches]

Remove -#### in zipcodes

How do I remove the +4 from zipcodes, in python?
I've got data like
85001
52804-3233
Winston-Salem
And I want that to become
85001
52804
Winston-Salem

>>> zip = '52804-3233'
>>> zip[:5]
'52804'
...and of course when you parse your lines from the original data you should insert some kind of rule to distinguish between zipcode to fix and other strings, but I don't know how your data looks like, so I can't help much (you could check if they are only digits and the '-' symbol, maybe?).

>>> import re
>>> s = "52804-3233"
>>> # regex to remove a dash and 4 digits after the dash after 5 digits:
>>> re.sub('(\d{5})-\d{4}', '\\1', s)
'52804'
The \\1 is a so called back reference and gets replaced by the first group, which would be the 5 digit zipcode in this case.

You could try something like this:
for input in inputs:
if input[:5].isnumeric():
input = input[:5]
# Takes the first 5 characters from the string
Just take away the first 5 characters of anything that is numbers in the first 5 positions.

re.sub('-\d{4}$', '', zipcode)

This grabs all items of the format 00000-0000 with a space or other word boundary before and after the number and replaces it with the first five digits. The other regex's posted will match some other number formats that you might not want.
re.sub('\b(\d{5})-\d{4}\b', '\\1', zipcode)

Or without regex:
output = [line[:5] if line[:5].isnumeric() and line[6:].isnumeric() else line for line in text if line]