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PORegular expressions in Python. Can anybody help a newbie?
 

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  1. PORegular expressions in Python. Can anybody help a newbie?
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    <p>This simple problem is kiling me. I posted something earlier about trying to clean up a database of addresses and somebody suggested GeoPy to check the validity of the addresses. Great tool which I did not know, but before doing that, I need to clean up the database a little bit, since geopy will not deal with messy formatting. The solution is to use regular expressions, which I think I have sort of fixed for most of the types of addresses I seen in the database. Nevertheless, I am having problems with the last RegExp I defined (called r4 in the code), because it is retuning part of the first parenthesis which I don't need, and I don't know why I have a extra white spaces when it returns the last group (City: London, Country: England). Can anybody help?</p> <pre><code>import re r1 = '\s*ForeignZip.*--\s*([\d\.]+)' r2 = '(\w+)\W*,\W*(\w*)' r3 = '(?&lt;=\().*?(?=\))' r4 = '(\w+\W\()' Location = [' ForeignZip (xxx) -- 734.450','Washington, DC.','London (England)'] for item in Location: print item match1 = re.search(r1,item) match2 = re.search(r2,item) match3 = re.search(r3,item) match4 = re.search(r4,item) if match1: print 'pattern 1 found:', match1.group(1) elif match2: print 'pattern 2 found: City :' + match2.group(1) + ", State :" + match2.group(2) elif match3: print 'pattern 3 found: City: ', match4.group() + ", Country :" + match3.group(0) else: print 'no match' </code></pre> <p>This returns</p> <pre><code> ForeignZip (xxx) -- 734.450 pattern 1 found: 734.50 Washington, DC. pattern 2 found: City :Washington, State :DC London (England) pattern 3 found: City: London (, Country :England </code></pre>
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    1. USLuis Miguel
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    1. COYour main problem is that `\(` is *inside* your capture group. If you don't want to include it in the capture group, but still want to use it to match, place it outside of the parentheses in your regex. Also, It's inefficient to evaluate all 4 patterns. Why don't you check to see if a match is found after each regex is run, that way, if you find a match with the first pattern, you can avoid evaluating all the rest of the patterns.
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      1. PORegular expressions in Python. Can anybody help a newbie?
      1. USJoel Cornett
 

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