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Extract date string from (more) complex string (possibly a regex match)

I have a string template that looks like 'my_index-{year}'.
I do something like string_template.format(year=year) where year is some string. Result of this is some string that looks like my_index-2011.
Now. to my question. I have a string like my_index-2011 and my template 'my_index-{year}' What might be a slick way to extract the {year} portion?
[Note: I know of the existence of parse library]

There is this module called parse which provides an opposite to format() functionality:
Parse strings using a specification based on the Python format() syntax.
>>> from parse import parse
>>> s = "my_index-2011"
>>> f = "my_index-{year}"
>>> parse(f, s)['year']
'2011'
And, an alternative option and, since you are extracting a year, would be to use the dateutil parser in a fuzzy mode:
>>> from dateutil.parser import parse
>>> parse("my_index-2011", fuzzy=True).year
2011

Use the split() string function to split the string into two parts around the dash, then grab just the second part.
mystring = "my_index-2011"
year = mystring.split("-")[1]

I assume "year" is 4 digits and you have multiple indexes
import re
res = ''
patterns = [ '%s-[0-9]{4}'%index for index in idx ]
for index,pattern in zip(idx,patterns):
res +=' '.join( re.findall(pattern ,data) ).replace(index+'-','') + ' '
---update---
dummyString = 'adsf-1234 fsfdr lkjdfaif ln ewr-1234 adsferggs sfdgrsfgadsf-3456'
dummyIdx = ['ewr','adsf']
output
1234 1234 3456

Yes, a regex would be helpful here.
In [1]: import re
In [2]: s = 'my_string-2014'
In [3]: print( re.search('\d{4}', s).group(0) )
2014
Edit: I should have mentioned your regex can be more sophisticated. You can haul out a subcomponent of a more specific string, for example:
In [4]: print( re.search('my_string-(\d{4})$', s).group(1) )
2014
Given the problem you presented, I think any "find the year" formula should be expressible in terms of a regular expression.

You are going to want to use the string method split to split on "-", and then catch the last element as your year:
year = "any_index-2016".split("-")[-1]
Because you caught the last element (using -1 as the index), your index can have hyphens in them, and you will still extract the year appropriately.

Python - Most elegant way to extract a substring, being given left and right borders [duplicate]

This question already has answers here:
How to extract the substring between two markers?
(22 answers)
Closed 4 years ago.
I have a string - Python :
string = "/foo13546897/bar/Atlantis-GPS-coordinates/bar457822368/foo/"
Expected output is :
"Atlantis-GPS-coordinates"
I know that the expected output is ALWAYS surrounded by "/bar/" on the left and "/" on the right :
"/bar/Atlantis-GPS-coordinates/"
Proposed solution would look like :
a = string.find("/bar/")
b = string.find("/",a+5)
output=string[a+5,b]
This works, but I don't like it.
Does someone know a beautiful function or tip ?

You can use split:
>>> string.split("/bar/")[1].split("/")[0]
'Atlantis-GPS-coordinates'
Some efficiency from adding a max split of 1 I suppose:
>>> string.split("/bar/", 1)[1].split("/", 1)[0]
'Atlantis-GPS-coordinates'
Or use partition:
>>> string.partition("/bar/")[2].partition("/")[0]
'Atlantis-GPS-coordinates'
Or a regex:
>>> re.search(r'/bar/([^/]+)', string).group(1)
'Atlantis-GPS-coordinates'
Depends on what speaks to you and your data.

What you haven't isn't all that bad. I'd write it as:
start = string.find('/bar/') + 5
end = string.find('/', start)
output = string[start:end]
as long as you know that /bar/WHAT-YOU-WANT/ is always going to be present. Otherwise, I would reach for the regular expression knife:
>>> import re
>>> PATTERN = re.compile('^.*/bar/([^/]*)/.*$')
>>> s = '/foo13546897/bar/Atlantis-GPS-coordinates/bar457822368/foo/'
>>> match = PATTERN.match(s)
>>> match.group(1)
'Atlantis-GPS-coordinates'

import re
pattern = '(?<=/bar/).+?/'
string = "/foo13546897/bar/Atlantis-GPS-coordinates/bar457822368/foo/"
result = re.search(pattern, string)
print string[result.start():result.end() - 1]
# "Atlantis-GPS-coordinates"
That is a Python 2.x example. What it does first is:
1. (?<=/bar/) means only process the following regex if this precedes it (so that /bar/ must be before it)
2. '.+?/' means any amount of characters up until the next '/' char
Hope that helps some.
If you need to do this kind of search a bunch it is better to 'compile' this search for performance, but if you only need to do it once don't bother.

Using re (slower than other solutions):
>>> import re
>>> string = "/foo13546897/bar/Atlantis-GPS-coordinates/bar457822368/foo/"
>>> re.search(r'(?<=/bar/)[^/]+(?=/)', string).group()
'Atlantis-GPS-coordinates'

python regex find characters from and end of the string

svn-backup-test,2014/09/24/18/Rev1223/FullSvnCheckout.tgz
from the following string I need to fetch Rev1233. So i was wondering if we can have any regexpression to do that. I like to do following string.search ("Rev" uptill next /)
so far I split this using split array
s1,s2,s3,s4,s5 = string ("/",4)

You don't need a regex to do this. It is as simple as:
str = 'svn-backup-test,2014/09/24/18/Rev1223/FullSvnCheckout.tgz'
str.split('/')[-2]

Here is a quick python example
>>> impot re
>>> s = 'svn-backup-test,2014/09/24/18/Rev1223/FullSvnCheckout.tgz'
>>> p = re.compile('.*/(Rev\d+)/.*')
>>> p.match(s).groups()[0]
'Rev1223'

Find second part from the end using regex, if preferred:
/(Rev\d+)/[^/]+$
http://regex101.com/r/cC6fO3/1
>>> import re
>>> m = re.search('/(Rev\d+)/[^/]+$', 'svn-backup-test,2014/09/24/18/Rev1223/FullSvnCheckout.tgz')
>>> m.groups()[0]
'Rev1223'

Regex to retrieve the last few characters of a string

Regex to retrieve the last portion of a string:
https://play.google.com/store/apps/details?id=com.lima.doodlejump
I'm looking to retrieve the string followed by id=
The following regex didn't seem to work in python
sampleURL = "https://play.google.com/store/apps/details?id=com.lima.doodlejump"
re.search("id=(.*?)", sampleURL).group(1)
The above should give me an output:
com.lima.doodlejump
Is my search group right?

Your regular expression
(.*?)
will not work because, it will match between zero and unlimited times, as few times as possible (becasue of the ?). So, you have the following choices of RegEx
(.*) # Matches the rest of the string
(.*?)$ # Matches till the end of the string
But, you don't need RegEx at all here, simply split the string like this
data = "https://play.google.com/store/apps/details?id=com.lima.doodlejump"
print data.split("id=", 1)[-1]
Output
com.lima.doodlejump
If you really have to use RegEx, you can do like this
data = "https://play.google.com/store/apps/details?id=com.lima.doodlejump"
import re
print re.search("id=(.*)", data).group(1)
Output
com.lima.doodlejump

I'm surprised that nobody has mentioned urlparse yet...
>>> s = "https://play.google.com/store/apps/details?id=com.lima.doodlejump"
>>> urlparse.urlparse(s)
ParseResult(scheme='https', netloc='play.google.com', path='/store/apps/details', params='', query='id=com.lima.doodlejump', fragment='')
>>> urlparse.parse_qs(urlparse.urlparse(s).query)
{'id': ['com.lima.doodlejump']}
>>> urlparse.parse_qs(urlparse.urlparse(s).query)['id']
['com.lima.doodlejump']
>>> urlparse.parse_qs(urlparse.urlparse(s).query)['id'][0]
'com.lima.doodlejump'
The HUGE advantage here is that if the url query string gets more components then it could easily break the other solutions which rely on a simple str.split. It won't confuse urlparse however :).

Just split it in the place you want:
id = url.split('id=')[1]
If you print id, you'll get:
com.lima.doodlejump
Regex isn't needed here :)
However, in case there are multiple id=s in your string, and you only wanted the last one:
id = url.split('id=')[-1]
Hope this helps!

This works:
>>> import re
>>> sampleURL = "https://play.google.com/store/apps/details?id=com.lima.doodlejump"
>>> re.search("id=(.+)", sampleURL).group(1)
'com.lima.doodlejump'
>>>
Instead of capturing non-greedily for zero or more characters, this code captures greedily for one or more.

Regular Expression to extract parts of Twitter query

I have the following string from which I want to extract the q and geocode values.
?since_id=261042755432763393&q=salvia&geocode=39.862712%2C-75.33958%2C10mi
I've tried the following regular expression.
expr = re.compile('\[\=\](.*?)\[\&\]')
vals = expr.match(str)
However, vals is None. I'm also not sure how to find something before, say, q= versus =.

No need for a regex (using Python 3):
>>> from urllib.parse import parse_qs
>>> query = parse_qs(str[1:])
>>> query
{'q': ['salvia'], 'geocode': ['39.862712,-75.33958,10mi'], 'since_id': ['261042755432763393']}
>>> query['q']
['salvia']
>>> query['geocode']
['39.862712,-75.33958,10mi']
Obviously, str contains your input.
Since (according to your tag) you are using Python 2.7, I think you need to change the import statement to this, though:
from urlparse import parse_qs
and if you were using Python before version 2.6, the import statement is
from cgi import parse_qs

I think this can be easily done without regex:
string = '?since_id=261042755432763393&q=salvia&geocode=39.862712%2C-75.33958%2C10mi'
parts = string[1:].split('&') # the [1:] is to leave out the '?'
pairs = {}
for part in parts:
try:
key, value = part.split('=')
pairs[key] = value
except:
pass
And pairs should contain all the key-value pairs of the string.