We Keep Coding

match multiple substrings using findall from re library

I have a large array that contains strings with the following format in Python
some_array = ['MATH_SOME_TEXT_AND_NUMBER MORE_TEXT SOME_VALUE',
'SCIENCE_SOME_TEXT_AND_NUMBER MORE_TEXT SOME_VALUE',
'ART_SOME_TEXT_AND_NUMBER MORE_TEXT SOME_VALUE]
I just need to extract the substrings that start with MATH, SCIENCE and ART. So what I'm currently using
my_str = re.findall('MATH_.*? ', some_array )
if len(my_str) > 0:
print(my_str)
my_str = re.findall('SCIENCE_.*? ', some_array )
if len(my_str) !=0:
print(my_str)
my_str = re.findall('ART_.*? ', some_array )
if len(my_str) > 0:
print(my_str)
It seems to work, but I was wondering if the findall function can look for more than one substring in the same line or maybe there is a cleaner way of doing it with another function.

You can use | to match multiple different strings in a regular expression.
re.findall('(?:MATH|SCIENCE|ART)_.*? ', ...)
You could also use str.startswith along with a list comprehension.
res = [x for x in some_array if any(x.startswith(prefix)
for prefix in ('MATH', 'SCIENCE', 'ART'))]

You could also match optional non whitespace characters after one of the alternations, start with a word boundary to prevent a partial word match and match the trailing single space:
\b(?:MATH|SCIENCE|ART)_\S*
Regex demo
Or if only word characters \w:
\b(?:MATH|SCIENCE|ART)_\w*
Example
import re
some_array = ['MATH_SOME_TEXT_AND_NUMBER MORE_TEXT SOME_VALUE',
'SCIENCE_SOME_TEXT_AND_NUMBER MORE_TEXT SOME_VALUE',
'ART_SOME_TEXT_AND_NUMBER MORE_TEXT SOME_VALUE']
pattern = re.compile(r"\b(?:MATH|SCIENCE|ART)_\S* ")
for s in some_array:
print(pattern.findall(s))
Output
['MATH_SOME_TEXT_AND_NUMBER ']
['SCIENCE_SOME_TEXT_AND_NUMBER ']
['ART_SOME_TEXT_AND_NUMBER ']

split string on any special character using python

currently I can have many dynamic separators in string like
new_123_12313131
new$123$12313131
new#123#12313131
etc etc . I just want to check if there is a special character in string then just get value after last separator like in this example just want 12313131

This is a good use case for isdigit():
l = [
'new_123_12313131',
'new$123$12313131',
'new#123#12313131',
]
output = []
for s in l:
temp = ''
for char in s:
if char.isdigit():
temp += char
output.append(temp)
print(output)
Result: ['12312313131', '12312313131', '12312313131']

Assuming you define 'special character' as anything thats not alphanumeric, you can use the str.isalnum() function to determine the first special character and leverage it something like this:
def split_non_special(input) -> str:
"""
Find first special character starting from the end and get the last piece
"""
for i in reversed(input):
if not i.isalnum():
return input.split(i)[-1] # return as soon as a separator is found
return '' # no separator found
# inputs = ['new_123_12313131', 'new$123$12313131', 'new#123#12313131', 'eefwfwrfwfwf3243']
# outputs = [split_non_special(input) for input in inputs]
# ['12313131', '12313131', '12313131', ''] # outputs

just get value after last separator
the more obvious way is using re.findall:
from re import findall
findall(r'\d+$',text) # ['12313131']

Python supplies what seems to be what you consider "special" characters using the string library as string.punctuation. Which are these characters:
!"#$%&'()*+,-./:;<=>?#[\]^_`{|}~
Using that in conjunction with the re module you can do this:
from string import punctuation
import re
re.split(f"[{punctuation}]", my_string)
my_string being the string you want to split.
Results for your examples
['new', '123', '12313131']
To get just digits you can use:
re.split("\d", my_string)
Results:
['123', '12313131']

Splitting a string using re module of python

I have a string
s = 'count_EVENT_GENRE in [1,2,3,4,5]'
#I have to capture only the field 'count_EVENT_GENRE'
field = re.split(r'[(==)(>=)(<=)(in)(like)]', s)[0].strip()
#o/p is 'cou'
# for s = 'sum_EVENT_GENRE in [1,2,3,4,5]' o/p = 'sum_EVENT_GENRE'
which is fine
My doubt is for any character in (in)(like) it is splitting the string s at that character and giving me first slice.(as after "cou" it finds one matching char i:e n). It's happening for any string that contains any character from (in)(like).
Ex : 'percentage_AMOUNT' o/p = 'p'
as it finds a matching char as 'e' after p.
So i want some advice how to treat (in)(like) as words not as characters , when splitting occurs/matters.
please suggest a syntax.

Answering your question, the [(==)(>=)(<=)(in)(like)] is a character class matching single characters you defined inside the class. To match sequences of characters, you need to remove [ and ] and use alternation:
r'==?|>=?|<=?|\b(?:in|like)\b'
or better:
r'[=><]=?|\b(?:in|like)\b'
You code would look like:
import re
ss = ['count_EVENT_GENRE in [1,2,3,4,5]','coint_EVENT_GENRE = "ROMANCE"']
for s in ss:
field = re.split(r'[=><]=?|\b(?:in|like)\b', s)[0].strip()
print(field)
However, there might be other (easier, or safer - depending on the actual specifications) ways to get what you want (splitting with space and getting the first item, use re.match with r'\w+' or r'[a-z]+(?:_[A-Z]+)+', etc.)
If your value is at the start of the string and starts with lowercase ASCII letters, and then can have any amount of sequences of _ followed with uppercase ASCII letters, use:
re.match(r'[a-z]+(?:_[A-Z]+)*', s)
Full demo code:
import re
ss = ['count_EVENT_GENRE in [1,2,3,4,5]','coint_EVENT_GENRE = "ROMANCE"']
for s in ss:
fieldObj = re.match(r'[a-z]+(?:_[A-Z]+)*', s)
if fieldObj:
print(fieldObj.group())

If you want only the first word of your string, then this should do the job:
import re
s = 'count_EVENT_GENRE in [1,2,3,4,5]'
field = re.split(r'\W', s)[0]
# count_EVENT_GENRE

Is there anything wrong with using split?
>>> s = 'count_EVENT_GENRE in [1,2,3,4,5]'
>>> s.split(' ')[0]
'count_EVENT_GENRE'
>>> s = 'coint_EVENT_GENRE = "ROMANCE"'
>>> s.split(' ')[0]
'coint_EVENT_GENRE'
>>>

How to extract just the characters "abc-3456" from the given text in python

i have this code
import re
text = "this is my desc abc-3456"
m = re.findall("\w+\\-\d+", text)
print m
This prints ['abc-3456'] but i want to get only abc-3456 (without the square brackets and the quotes].
How to do this?

import re
text = "this is my desc abc-3456"
m = re.findall("\w+\\-\d+", text)
print m[0]

re.findall(pattern, string, flags=0)
Return all non-overlapping matches of pattern in string, as a list of strings.
findall returns list of strings. If you want the first one then use m[0].
print m[0] will give string without [] and ''.

If you only want the first (or only) result, do this:
import re
text = "this is my desc abc-3456"
m = re.search("\w+\\-\d+", text)
print m.group()

re.findall retuns a list of matches. In that list the result is a string. You can use re.finditer if you want.
In python, a list's representation is in brackets: [member1, member2, ...].
A string ("somestring") representation is in quotes: 'somestring'.
This means the representation of a list of strings is:
['somestring1', 'somestring2', ...]
So you have a string in a list, the characters you want to remove are a part of python's representation and not a part of the data you have.
To get the string simply take the first element from the list:
mystring = m[0]

Search for quotes with regular expression

I'm looking for a way to search a text file for quotes made by author and then print them out. My script so far:
import re
#searches end of string
print re.search('"$', 'i am searching for quotes"')
#searches start of string
print re.search('^"' , '"i am searching for quotes"')
What I would like to do
import re
## load text file
quotelist = open('A.txt','r').read()
## search for strings contained with quotation marks
re.search ("-", quotelist)
## Store in list or Dict
Dict = quotelist
## Print quotes
print Dict
I also tried
import re
buffer = open('bbc.txt','r').read()
quotes = re.findall(r'.*"[^"].*".*', buffer)
for quote in quotes:
print quote
# Add quotes to list
l = []
for quote in quotes:
print quote
l.append(quote)

Develop a regular expression that matches all the expected characters you would expect to see inside of a quoted string. Then use the python method findall in re to find all occurrences of the match.
import re
buffer = open('file.txt','r').read()
quotes = re.findall(r'"[^"]*"',buffer)
for quote in quotes:
print quote
Searching between " and ” requires a unicode-regex search such as:
quotes = re.findall(ur'"[^\u201d]*\u201d',buffer)
And for a document that uses " and ” interchangeably for quotation termination
quotes = re.findall(ur'"[^"^\u201d]*["\u201d]', buffer)

You don't need regular expressions to find static strings. You should use this Python idiom for finding strings:
>>> haystack = 'this is the string to search!'
>>> needle = '!'
>>> if needle in haystack:
print 'Found', needle
Creating a list is easy enough -
>>> matches = []
Storing matches is easy too...
>>> matches.append('add this string to matches')
This should be enough to get you started. Good luck!
An addendum to address the comment below...
l = []
for quote in matches:
print quote
l.append(quote)