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Home/ Questions/Q 9120879
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Editorial Team
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Editorial Team
Asked: June 17, 20262026-06-17T05:43:39+00:00 2026-06-17T05:43:39+00:00

I have the following two lists: input = [‘MAPLEWOOD AVE’, ‘LYNNDALE ‘, ‘SUGAR DR’]

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I have the following two lists:

input = ['MAPLEWOOD AVE', 'LYNNDALE ', 'SUGAR DR']

ref = ['LYNNDALE (?:RD)?', 'HOMAN (?:AVE)?', 'MAPLEWOOD (?:AVE)?', 'LYNNDALE (?:LN)?']

I would like to look for all matches for each element within input with ref. The output would be a dictionary with each key being an input element, and each value being a ref element matched to the corresponding input element, like the following:

{'MAPLEWOOD AVE' : 'MAPLEWOOD AVE', 'LYNNDALE ' : 'LYNNDALE RD', 'LYNNDALE LN', 'SUGAR DR':}

The following allows me to iterate over input in search of a findall match within ref (which contains embedded regex groupings). However, I cannot retrieve the corresponding match element(s) from ref as values alongside each input element:

combined = "(" + ")|(".join(ref) + ")"

l = []

for i in input:
    if re.findall(combined,i):
         l.append(i)
...
MAPLEWOOD AVE
LYNNDALE
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  1. Editorial Team
    Editorial Team
    2026-06-17T05:43:40+00:00Added an answer on June 17, 2026 at 5:43 am

    Try:

    import re
    
    input = ['MAPLEWOOD AVE', 'LYNNDALE ', 'SUGAR DR']
    ref = ['LYNNDALE (?:RD)?', 'HOMAN (?:AVE)?', 'MAPLEWOOD (?:AVE)?', 'LYNNDALE (?:LN)?']
    output = dict([ (i, [ r for r in ref if re.match(r, i) ]) for i in input ])
    

    Or if you’re using Python 3:

    output = { i : [ r for r in ref if re.match(r, i) ] for i in input }
    

    Also you could compile your regexs to speed them up a little:

    ref_re = [ re.compile(r) for r in ref ]
    output = { i : [ r.pattern for r in ref_re if r.match(i) ] for i in input }
    

    UPD:
    Maybe you want to use matched part as values, not patterns:

    output = { i : [ r.match(i).group(0) for r in ref_re if r.match(i) ] for i in input }
    
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