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 
re.match(..) way of doing this. But yea. Seems like I'll just stick to using the parse module. +1Use 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] 
split() will continue to work just fine. (If the target portion itself contains dashes, you can tell split() to only split on the first dash.)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 += on a string is bad for performance, as is using re.match in a loop. +1 for the answer for the effort, though.
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.
hh-mm