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Match a string and replace remaning characters

I would like to extract the server type from the hostname and replace the remaining characters with underscores so I can then use it with the LIKE pattern match in SQLite
My initial approach was something like this (this is the excepted output):
>>> host = 'webus01'
>>> location = 'us'
>>> parts = list(host.partition(location))
>>> parts
['web', 'us', '01']
>>> parts[1] = "_" * len(parts[1])
>>> parts[2] = "_" * len(parts[2])
>>> "".join(parts) + ".%"
'web____.%'
But this won't work if the hostname starts with or contains the location name:
>>> host = 'utilityit01'
>>> pod = 'it'
>>> parts = list(host.partition(location))
>>> parts
['utilityit01', '', '']
>>> parts[1] = "_" * len(parts[1])
>>> parts[2] = "_" * len(parts[2])
>>> "".join(parts) + ".%"
'utilityit01.%'
Then I though it will be better to use RegEx for this to match only the location before digits.
The re.sub function seems to be a good candidate for this task but I'm not sure how to replace all characters instead of the match group as a whole:
>>> import re
>>> re.sub(r'it\d+.*', '_', 'utilityit01a')
'utility_'
The output in this case should be: utility_____.%.

Looks like you need.
import re
print(re.sub(r'(it\d+.*)', lambda x: '_'*len(x.group(1))+".%" , 'utilityit01a'))
print(re.sub(r'(us\d+.*)', lambda x: '_'*len(x.group(1))+".%" , 'webus01'))
Output:
utility_____.%
web____.%

You can find the substring that you want to replace first, and then replace it based on its length:
>>> import re
>>> host = 'utilityit01'
>>> substr = re.findall(r'it\d+.*', host)
>>> substr
['it01a']
>>> host.replace(substr[0], len(substr)*'_')
'utility_____'

Is there a Python equivalent to the Perl /e modifier for regular expressions? [duplicate]

Greetings all,
I'm not sure if this is possible but I'd like to use matched groups in a regex substitution to call variables.
a = 'foo'
b = 'bar'
text = 'find a replacement for me [[:a:]] and [[:b:]]'
desired_output = 'find a replacement for me foo and bar'
re.sub('\[\[:(.+):\]\]',group(1),text) #is not valid
re.sub('\[\[:(.+):\]\]','\1',text) #replaces the value with 'a' or 'b', not var value
thoughts?

You can specify a callback when using re.sub, which has access to the groups:
http://docs.python.org/library/re.html#text-munging
a = 'foo'
b = 'bar'
text = 'find a replacement for me [[:a:]] and [[:b:]]'
desired_output = 'find a replacement for me foo and bar'
def repl(m):
contents = m.group(1)
if contents == 'a':
return a
if contents == 'b':
return b
print re.sub('\[\[:(.+?):\]\]', repl, text)
Also notice the extra ? in the regular expression. You want non-greedy matching here.
I understand this is just sample code to illustrate a concept, but for the example you gave, simple string formatting is better.

Sounds like overkill. Why not just do something like
text = "find a replacement for me %(a)s and %(b)s"%dict(a='foo', b='bar')
?

>>> d={}
>>> d['a'] = 'foo'
>>> d['b'] = 'bar'
>>> text = 'find a replacement for me [[:a:]] and [[:b:]]'
>>> t=text.split(":]]")
>>> for n,item in enumerate(t):
... if "[[:" in item:
... t[n]=item[: item.rindex("[[:") +3 ] + d[ item.split("[[:")[-1]]
...
>>> print ':]]'.join( t )
'find a replacement for me [[:foo:]] and [[:bar:]]'

Find a regular extension of domains on a string python Regex

I would like to find extension "COM" from a sentence using regex in python.
>>> import re
>>> str = 'finding exstention from string on http://domain.coms/index/page/2'
>>> pattern = re.compile(r'([^\s.\s\:]+\.[^\.\s\:]*)')
>>>
Result:
domain : 'domain.com' ### notes: not domain.coms
url : 'http://domain.coms/index/page/2'

may be you are looking for this:
>>> import re
>>> str = 'finding exstention from string on http://domain.coms/index/page/2'
>>> pattern = re.compile(r'([^\/]*\.(?:com|en|org))')
>>> m = pattern.search(str)
>>> print m.group(1)
domain.com

((?:https?:\/\/)?(?:([^\s.\s\:]+\.[^\/]*)(?:\/|$)[^\.\s\:]*))
Try this.Group 1 will be the url.Group 2 will be the domain.
See demo.
http://regex101.com/r/sK8oK9/1

You could try the below.
>>> s = "finding exstention from string on http://domain.coms/index/page/2"
>>> m = re.search(r'(\S+?([^/.]+\.[^/]+)\S+)', s).group(1)
>>> m
'http://domain.coms/index/page/2'
>>> m = re.search(r'(\S+?([^/.]+\.[^/]+)\S+)', s).group(2)
>>> m
'domain.coms'

python find digits with leading '_v'

Is there a better way of finding digits in a string which starts with '_v' which stands for version number? What I want is just '001'
filename = 'greatv02_v001_jam.mb'
parts = re.split('_v|\_',filename)
>>['greatv02', '001', 'jam.mb']
b = re.findall(r'\d+', filename)
>>['02', '001']
Is there a way to split a string with something along these lines?
parts = re.split('_v###_',filename)
or
parts = re.split('_v*_',filename)

You could use lookarounds:
>>> filename = 'greatv02_v001_jam.mb'
>>> import re
>>> re.findall(r'(?<=_v)\d+', filename)
['001']
>>>
>>> filename = 'greatv02_v001_av456jam.mb'
>>> re.findall(r'(?<=_v)\d+', filename)
['001']
>>> filename = 'greatv02_v001_v456jam.mb'
>>> re.findall(r'(?<=_v)\d+', filename)
['001', '456']
>>>

Ugly, but you could partition the file name twice
>>> filename.partition('_v')[2].partition('_')[0]
'001'

Use regex's grouping like this:
.*_v(\d+).*
Demo:
>>> filename = 'greatv02_v001_jam.mb'
>>> pattern = re.compile(r'.*_v(\d+).*')
>>> re.search(pattern, filename).group(1)
'001'

How about the regex _v(?P<version>\d+).*:
>>> regex = re.compile("_v(?P<version>\d+).*")
>>> r = regex.search(string)
# List the groups found
>>> r.groups()
(u'001',)
# List the named dictionary objects found
>>> r.groupdict()
{u'version': u'001'}
# Run findall
>>> regex.findall(string)
[u'001']
# Run timeit test
>>> setup = ur"import re; regex =re.compile("_v(?P<version>\d+).*");string="""greatv02_v00 ...
>>> t = timeit.Timer('regex.search(string)',setup)
>>> t.timeit(10000)
0.005126953125

How to search and export value with python and re?

I'm trying to export some value from the text to a txt file.
my text has this form:
"a='one' b='2' c='3' a='two' b='8' c='3'"
I want to export all the value of the key "a"
The result must be like
one
two

The other answers are correct for your particular case, but I think a regex with lookbehind/lookahead is a more general solution, i.e.:
import re
text = "a='one' b='2' c='3' a='two' b='8' c='3'"
expr = r"(?<=a=')[^']*(?=')"
matches = re.findall(expr,text)
for m in matches:
print m ##or whatever
This will match for any expression between single quotes preceded by a=, i.e. a='xyz', a='my#1.abcd' and a='a=5%' will all match

This regex is very easy to understand:
pattern = r"a='(.*?)'"
It doesn't use lookarounds (like (?<=a=')[^']*(?=') ) - so it's very simple ..
Whole program:
#!/usr/bin/python
import re
text = "a='one' b='2' c='3' a='two' b='8' c='3'"
pattern = r"a='(.*?)'"
for m in re.findall( pattern, text ):
print m

you can use something like this:
import re
r = re.compile(r"'([a-z]+)'")
f = open('input')
text = f.read()
m = r.finditer(text)
for mm in m:
print mm.group(1)

thought i would give a solution without re:
>>> text = "a='one' b='2' c='3' a='two' b='8' c='3'"
>>> step1 = text.split(" ")
>>> step1
["a='one'", "b='2'", "c='3'", "a='two'", "b='8'", "c='3'"]
>>> step2 = []
>>> for pair in step1:
split_pair = pair.split("=")
step2.append([split_pair[0],split_pair[1]])
>>> print step2
[['a', "'one'"], ['b', "'2'"], ['c', "'3'"], ['a', "'two'"], ['b', "'8'"], ['c', "'3'"]]
>>> results = []
>>> for split_pair in step2:
if split_pair[0] == "a":
results.append(split_pair[1])
>>> results
["'one'", "'two'"]
not the most elegant method, but it works.

Another non-regex solution: you could use the shlex module and the .partition method (or .split() with maxsplit=1):
>>> import shlex
>>> s = "a='one' b='2' c='3' a='two' b='8' c='3'"
>>> shlex.split(s)
['a=one', 'b=2', 'c=3', 'a=two', 'b=8', 'c=3']
>>> shlex.split(s)[0].partition("=")
('a', '=', 'one')
and so it's simply
>>> for group in shlex.split(s):
... key, eq, val = group.partition("=")
... if key == 'a':
... print val
...
one
two
with lots of variations of the same.