Develop Reference

Python re to replace the subtitles with new formats

I have a file with the following format:
===Subtitle 1===
text ....
===Subtitle 2===
text ....
How could I replace ===Subtitle 1=== with Section Subtitle 1 using python?
I tried this:
import re
s = '===Subtitle 1==='
lst = re.findall('===[\S+a-zA-Z0-9]===', s)
print lst
But I cannot print out anything.

You don't need use regex here, just use str.replace() like this:
>>> a = '===Subtitle 1==='
>>> a.replace('=', '')
'Subtitle 1'
>>>
But if you'd like use regex...
>>> import re
>>> re.findall('=+(.+?)=+', a)
['Subtitle 1']
>>> re.findall('=+(.+?)=+', a)[0]
'Subtitle 1'
>>>

Extracting a substring of a string in Python based on presence of another string

common is always present regardless of string. Using that information, I'd like to grab the substring that comes just before it, in this case, "banana":
string = "apple_orange_banana_common_fruit"
In this case, "fruit":
string = "fruit_common_apple_banana_orange"
How would I go about doing this in Python?

You can use re.search() to extract the substring:
>>> import re
>>> s = 'apple_orange_banana_common_fruit'
>>> re.search(r'([a-zA-Z]+)_common', s).group(1)
'banana'

This will return a list of matches:
import re
string = "apple_orange_banana_common_fruit"
preceding_word = re.findall("[A-Za-z]+(?=_common)", string)
If common only occurs once per string, you might be better off using hwnd's solution.

import re
string = "apple_orange_bananna_common_fruit"
preceding_word = re.search('([a-zAZ]+)(?=_common)', string)
print (preceding_word.group(1))

>>> string = "fruit_common_apple_banana_orange"
>>> parts = string.split('_')
>>> print parts[parts.index('common') - 1]
fruit
>>> string = "apple_orange_banana_common_fruit"
>>> parts = string.split('_')
>>> print parts[parts.index('common') - 1]
banana

String to float and regex

I have a problem with regex and my string...
I need a solution in my regex code for take a float number of a string. I don't know why this code doesn't work.
from bs4 import BeautifulSoup
import urllib2
from re import sub
url = 'http://www.ebay.es/itm/PET-SHOP-BOYS-OFFICIAL-PROMO-BARCELONA-ELECTRIC-TOUR-BEER-CERVEZA-20cl-BOTTLE-/111116266655' #raw_input('Dime la url que deseas: ')
code = urllib2.urlopen(url).read();
soup = BeautifulSoup(code)
info = soup.find('span', id='v4-27').contents[0]
print info
info = sub("[\D]+,+[\D]", "", info)
i = float(info)
print i

\D means Non-digits. You need to use \d instead. Look here for details: http://en.wikipedia.org/wiki/Regular_expression#Character_classes
Updated
I see, that your approach is to replace all non-digits chars. To my mind, match needed information is more clear:
>>> import re
>>> s = "15,00 EUR"
>>> price_string = re.search('(\d+,\d+)', s).group(1)
>>> price_string
'15,00'
>>> float(price_string.replace(',', '.'))
15.0

regex to find postition between two markers in string

i need to find anything between
show_detail&
and
;session_id=1445045
in
https://www.site.gov.uk//search/cgi-bin/contract_search/contract_search.cgi?rm=show_detail&id=4035219;num=1;session_id=1445045;start=0;recs=20;subscription=1;value=0
using regex in python.
i know i need to use lookbehind/ahead but i can't seem to make it work!
please help!
thanks :)

Why use a regex?
>>>> url = 'https://ww.site.gov.....'
>>> start = url.index('show_detail&') + len('show_detail&')
>>> end = url.index(';session_id=')
>>> url[start:end]
'id=4035219;num=1'

>>> s= "https://www.site.gov.uk//search/cgi-bin/contract_search/contract_search.cgi?rm=show_detail&id=4035219;num=1;session_id=1445045;start=0;recs=20;subscription=1;value=0"
>>> s.split(";session_id=1445045")[0].split("show_detail&")[-1]
'id=4035219;num=1'
>>>

You can use a non greedy match (.*?) in between your markers.
>>> import re
>>> url = "https://www.site.gov.uk//search/cgi-bin/contract_search/contract_search.cgi?rm=show_detail&id=4035219;num=1;session_id=1445045;start=0;recs=20;subscription=1;value=0"
>>> m = re.search("show_detail&(.*?);session_id=1445045", url)
>>> m.group(1)
'id=4035219;num=1'

regex = re.compile(r"(?<=show_detail&).*?(?=;session_id=1445045)"
should work. See here for more info on lookaround assertions.

import re
url = "https://www.site.gov.uk//search/cgi-bin/contract_search/contract_search.cgi?rm=show_detail&id=4035219;num=1;session_id=1445045;start=0;recs=20;subscription=1;value=0"
pattern = "([^>].+)(show_detail&)([^>].+)(session_id=1445045)([^>].+)"
reg = re.compile(r''''''+pattern+'''''',flags = re.S)
match =reg.search(url)
print match.group(3)
this would work i think

python regex to get all text until a (, and get text inside brackets

I need help with two regex operations.
Get all text until an open bracket.
e.g. 'this is so cool (234)' => 'this is so cool'
Get the text inside the brackets, so the number '234'

Up until the paren: regex = re.compile("(.*?)\s*\(")
Inside the first set of parens: regex = re.compile(".*?\((.*?)\)")
Edit: Single regex version: regex = re.compile("(.*?)\s*\((.*?)\)")
Example output:
>>> import re
>>> r1 = re.compile("(.*?)\s*\(")
>>> r2 = re.compile(".*?\((.*?)\)")
>>> text = "this is so cool (234)"
>>> m1 = r1.match(text)
>>> m1.group(1)
'this is so cool'
>>> m2 = r2.match(text)
>>> m2.group(1)
'234'
>>> r3 = re.compile("(.*?)\s*\((.*?)\)")
>>> m3 = r3.match(text)
>>> m3.group(1)
'this is so cool'
>>> m3.group(2)
'234'
>>>
Note of course that this won't work right with multiple sets of parens, as it's only expecting one parenthesized block of text (as per your example). The language of matching opening/closing parens of arbitrary recurrence is not regular.

Sounds to me like you could just do this:
re.findall('[^()]+', mystring)
Splitting would work, too:
re.split('[()]', mystring)
Either way, the text before the first parenthesis will be the first item in the resulting array, and the text inside the first set of parens will be the second item.

No need for regular expression.
>>> s="this is so cool (234)"
>>> s.split("(")[0]
'this is so cool '
>>> s="this is so cool (234) test (123)"
>>> for i in s.split(")"):
... if "(" in i:
... print i.split("(")[-1]
...
234
123

Here is my own library function version without regex.
def between(left,right,s):
before,_,a = s.partition(left)
a,_,after = a.partition(right)
return before,a,after
s="this is so cool (234)"
print('\n'.join(between('(',')',s)))