Delete all characters in a multline string up to a given pattern

Question

Using Python I need to delete all charaters in a multiline string up to the first occurrence of a given pattern. In Perl this can be done using regular expressions with something like:

#remove all chars up to first occurrence of cat or dog or rat
$pattern = 'cat|dog|rat' 
$pagetext =~ s/(.*?)($pattern)/$2/xms; 

What's the best way to do it in Python?

Answer 1

>>> import re
>>> s = 'hello cat!'
>>> m = re.search('cat|dog|rat', s)
>>> s[m.start():]
'cat!'

Of course you'll need to account for the case where there's no match in a real solution.

Or, more cleanly:

>>> import re
>>> s = 'hello cat!'
>>> p = 'cat|dog|rat'
>>> re.sub('.*?(?=%s)' % p, '', s, 1)
'cat!'

For multiline, use the re.DOTALL flag.

Answer 2

Something like this should do what you want:

import re
text = '   sdfda  faf foo zing baz bar'
match = re.search('foo|bar', text)
if match:
  print text[match.start():] # ==>  'foo zing baz bar'

Answer 3

non regex way

>>> s='hello cat!'
>>> pat=['cat','dog','rat']
>>> for n,i in enumerate(pat):
...     m=s.find(i)
...     if m != -1: print s[m:]
...
cat!

Answer 4

You want to delete all characters preceding the first occurrence of a pattern; as an example, you give "cat|dog|rat".

Code that achieves this using re:

re.sub("(?s).*?(cat|dog|rat)", "\\1", input_text, 1)

or, if you'll be using again this regular expression:

rex= re.compile("(?s).*?(cat|dog|rat)")
result= rex.sub("\\1", input_text, 1)

Note the non-greedy .*?. The initial (?s) allows to match newline characters too, before the word matching.

Examples:

>>> input_text= "I have a dog and a cat"
>>> re.sub(".*?(cat|dog|rat)", "\\1", input_text, 1)
'dog and a cat'

>>> re.sub("(?s).*?(cat|dog|rat)", "\\1", input_text, 1)
'I have no animals!'

>>> input_text= "This is irrational"
>>> re.sub("(?s).*?(cat|dog|rat)", "\\1", input_text, 1)
'rational'

In case you want to do the conversion only for the words cat, dog and rat, you'll have to change the regex into:

>>> re.sub(r"(?s).*?\b(cat|dog|rat)\b", "\\1", input_text, 1)
'This is irrational'

Answer 5

Another option is to use look ahead s/.*?(?=$pattern)//xs:

re.sub(r'(?s).*?(?=cat|dog|rat)', '', text, 1)

Non-regex way:

for option in 'cat dog rat'.split():
    index = text.find(option)
    if index != -1: # found
       text = text[index:]
       break

Non-regex way is almost 5 times faster (for some input):

$ python -mtimeit -s'from drop_until_word import drop_re, text, options;' \
> 'drop_re(text, options)'
1000 loops, best of 3: 1.06 msec per loop

$ python -mtimeit -s'from drop_until_word import drop_search, text, options;'\
> 'drop_search(text, options)'
10000 loops, best of 3: 184 usec per loop

$ python -mtimeit -s'from drop_until_word import drop_find, text, options;' \
> 'drop_find(text, options)'
1000 loops, best of 3: 207 usec per loop

Where drop_until_word.py is:

import re

def drop_re(text, options):
    return re.sub(r'(?s).*?(?='+'|'.join(map(re.escape, options))+')', '',
                  text, 1)

def drop_re2(text, options):
    return re.sub(r'(?s).*?('+'|'.join(map(re.escape, options))+')', '\\1',
                  text, 1)

def drop_search(text, options):
    m = re.search('|'.join(map(re.escape, options)), text)
    return text[m.start():] if m else text

def drop_find(text, options):
    indexes = [i for i in (text.find(option) for option in options) if i != -1]
    return text[min(indexes):] if indexes else text

text = open('/usr/share/dict/words').read()
options = 'cat dog rat'.split()

def test():
    assert drop_find(text, options) == drop_re(text, options) \
        == drop_re2(text, options) == drop_search(text, options)

    txt = 'dog before cat'
    r = txt
    for f in [drop_find, drop_re, drop_re2, drop_search]:
        assert r == f(txt, options), f.__name__


if __name__=="__main__":
    test()