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

Regex in python: combining 2 regex expressions into one

Suppose I have the following list:
a = ['35','years','opened','7,000','churches','rev.','mr.','brandt','said','adding','denomination','national','goal','one','church','every','10,000','persons']
I want to remove all elements, that contain numbers and elements, that end with dots.
So I want to delete '35','7,000','10,000','mr.','rev.'
I can do it separately using the following regex:
regex = re.compile('[a-zA-Z\.]')
regex2 = re.compile('[0-9]')
But when I try to combine them I delete either all elements or nothing.
How can I combine two regex correctly?

This should work:
reg = re.compile('[a-zA-Z]+\.|[0-9,]+')
Note that your first regex is wrong because it deletes any string within a dot inside it.
To avoid this, I included [a-zA-Z]+\. in the combined regex.
Your second regex is also wrong as it misses a "+" and a ",", which I included in the above solution.
Here a demo.
Also, if you assume that elements which end with a dot might contain some numbers the complete solution should be:
reg = re.compile('[a-zA-Z0-9]+\.|[0-9,]+')

If you don't need to capture the result, this matches any string with a dot at the end, or any with a number in it.
\.$|\d

You could use:
(?:[^\d\n]*\d)|.*\.$
See a demo on regex101.com.

Here is a way to do the job:
import re
a = ['35','years','opened','7,000','churches','rev.','mr.','brandt','said','adding','denomination','national','goal','one','church','every','10,000','per.sons']
b = []
for s in a:
if not re.search(r'^(?:[\d,]+|.*\.)$', s):
b.append(s)
print b
Output:
['years', 'opened', 'churches', 'brandt', 'said', 'adding', 'denomination', 'national', 'goal', 'one', 'church', 'every', 'per.sons']
Demo & explanation

Python - How to use regex to find multiple words and extract them at the same time

Using Regular Expression, I want to find all the match words in a sentence and extract the wanted part in the matches words at the same time.
I use the API "findall" from "re" module to find the match words and plus the brackets to extract the parts I want.
For example I have a string "0xQQ1A, 0xWW2B, 0xEE3C, 0xQQ4C".
I only want the remaining two words after "0xQQ" or "0xWW", which will result in a list ["1A", "2B, "4C"].
Here is my code:
import re
MyString = "0xQQ1A, 0xWW2B, 0xEE3C, 0xQQ4C"
MySearch = re.compile("0xQQ(\w{2})|0xWW(\w{2})")
MyList = MySearch.findall(MyString)
print MyList
So my expected result is ["1A", "2B, "4C"].
But the actual result is [('1A', ''), ('', '2B'), ('4C', '')]
I think I might have used the combination of "()" and "|" in the wrong way.
Thx for the help!

Two different capturing groups will result in two items in the output (whatever matched each).
Instead, use a single capturing group and put your | (OR) earlier:
re.compile("0x(?:QQ|WW)(\w{2})")
((?:...) is a non-capturing group that matches ... - used to limit the effects of the | to only the QQ/WW split, without adding another capture to the output.)

You can try this:
import re
string = "0xQQ1A, 0xWW2B, 0xEE3C, 0xQQ4C"
pattern = re.compile(r"(0xQQ|0xWW)(\w{2})")
result = [match[2] for match in pattern.finditer(string)]
result will be:
['1A', '2B', '4C']

Cutting string in python

I did some search but couldn't find any useful information.
s = ['33PM']
My aim is to cut 'PM' from s[0] and append it as s[1].

You can use re.findall to extract continuous range of numbers and characters. \d+ would extract all numbers and \w+ would extract all character ranges
>>> import re
>>> s = re.findall(r'\d+|\w+', s[0])
>>> s
['33', 'PM']

Here is a method that uses simple Python code, avoiding the complications of regular expressions. This is designed for when you know that 'PM' is in the string, and if there is any text in the string after that it will be moved to the second list item together with the 'PM. This code also assumes that you care only about the first item in the list--any later items will be dropped.
s = ['33PM']
string0 = s[0]
loc = string0.find('PM')
s = [string0[:loc], string0[loc:]]
If you now print s the result is
['33', 'PM']

Why is the split() returning list objects that are empty? [duplicate]

I have the following file names that exhibit this pattern:
000014_L_20111007T084734-20111008T023142.txt
000014_U_20111007T084734-20111008T023142.txt
...
I want to extract the middle two time stamp parts after the second underscore '_' and before '.txt'. So I used the following Python regex string split:
time_info = re.split('^[0-9]+_[LU]_|-|\.txt$', f)
But this gives me two extra empty strings in the returned list:
time_info=['', '20111007T084734', '20111008T023142', '']
How do I get only the two time stamp information? i.e. I want:
time_info=['20111007T084734', '20111008T023142']

I'm no Python expert but maybe you could just remove the empty strings from your list?
str_list = re.split('^[0-9]+_[LU]_|-|\.txt$', f)
time_info = filter(None, str_list)

Don't use re.split(), use the groups() method of regex Match/SRE_Match objects.
>>> f = '000014_L_20111007T084734-20111008T023142.txt'
>>> time_info = re.search(r'[LU]_(\w+)-(\w+)\.', f).groups()
>>> time_info
('20111007T084734', '20111008T023142')
You can even name the capturing groups and retrieve them in a dict, though you use groupdict() rather than groups() for that. (The regex pattern for such a case would be something like r'[LU]_(?P<groupA>\w+)-(?P<groupB>\w+)\.')

If the timestamps are always after the second _ then you can use str.split and str.strip:
>>> strs = "000014_L_20111007T084734-20111008T023142.txt"
>>> strs.strip(".txt").split("_",2)[-1].split("-")
['20111007T084734', '20111008T023142']

Since this came up on google and for completeness, try using re.findall as an alternative!
This does require a little re-thinking, but it still returns a list of matches like split does. This makes it a nice drop-in replacement for some existing code and gets rid of the unwanted text. Pair it with lookaheads and/or lookbehinds and you get very similar behavior.
Yes, this is a bit of a "you're asking the wrong question" answer and doesn't use re.split(). It does solve the underlying issue- your list of matches suddenly have zero-length strings in it and you don't want that.

>>> f='000014_L_20111007T084734-20111008T023142.txt'
>>> f[10:-4].split('-')
['0111007T084734', '20111008T023142']
or, somewhat more general:
>>> f[f.rfind('_')+1:-4].split('-')
['20111007T084734', '20111008T023142']