I'm a total amateur as far as data science goes, and I'm trying to figure out a way to do some string comparison on a large dataset.

I've a Google BigQuery table storing merchant transactions, but the store names are all over the board. For example, there can be 'Wal-Mart Super Center' and 'Wal-Mart SC #1234', or 'McDonalds F2222' and 'McDonalds #321'.

What I need to do is group ALL 'Wal-mart' and 'McDonalds' and whatever else. My first approach was doing a recursive reg-ex check, but that took forever and eventually timed-out.

What's the best approach for doing that with a table of 20 million+ rows? I'm open to trying out any technology that would fit this job.


This is an entity resolution aka record linkage aka data matching problem.

I would solve this by removing all of the non-alphabetical characters including numbers, casting into all uppercase and then employing a hierarchical match. First match up the exact cases and then move to a Levenshtein scoring between the fields. Make some sort of a decision about how large you will allow the Levenshtein or normalized Levenshtein score to get before you declare something a non-match.

Assign every row an id and when you have a match, reassign the lower of the IDs to both members of the match.

The Levenshtein distance algorithm is simple but brilliant (taken from here):

def levenshtein(a,b):
    "Calculates the Levenshtein distance between a and b."
    n, m = len(a), len(b)
    if n > m:
        # Make sure n <= m, to use O(min(n,m)) space
        a,b = b,a
        n,m = m,n

    current = range(n+1)
    for i in range(1,m+1):
        previous, current = current, [i]+[0]*n
        for j in range(1,n+1):
            add, delete = previous[j]+1, current[j-1]+1
            change = previous[j-1]
            if a[j-1] != b[i-1]:
                change = change + 1
            current[j] = min(add, delete, change)

    return current[n]

This Data Matching book is a good resource and is free for seven days on Amazon.

Nominally, this is an $n^2$ algorithm without exploiting some sorting efficiencies, so I would expect to have to use multiple cores on $2\times10^7$ rows. But this should run just fine on an 8 core AWS instance. It will eventually finish on a single core, but might take several hours.

Hope this helps!

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  • 2
    $\begingroup$ Thanks! Levenshtein is causing some issues because there are merchants like 'Google Marketplace' and 'Goodwill Marketplace' which have a lower score than even 'Wal-mart' and 'walgreens'. I think I can probably whitelist the outliers though. And double thanks for the book recommendation. $\endgroup$ – TerryMatula Aug 21 '15 at 15:57
  • 1
    $\begingroup$ That's a tough one as normalized Levenshtein is less than $.16$ which is a cutoff I often use. I guess you could also try extracting the words through space delineation and then comparing all words to all words in some smart way. This is a tough problem if you are going for high confidence! Another option is see where you get with exact match and then use a mechanical turk (or yourself) for the last couple million. $\endgroup$ – AN6U5 Aug 21 '15 at 16:07
  • $\begingroup$ Based on the examples it seems like it should be front-weighted in some way, as names that should be considered equivalent differ near the end but never im the beginning. $\endgroup$ – Adam Bittlingmayer Oct 18 '15 at 19:13
  • $\begingroup$ For example I would not remove a number from the beginning. Or at least look to see all the data with numbers in the beginning. $\endgroup$ – Adam Bittlingmayer Oct 18 '15 at 19:18
  • $\begingroup$ @Adam M. B., Levenshtein is a "base algorithm" and there are many derivations of it that have been tailored for specific advantages and cases. I haven't found prioritizing beginnings of words over their ends to be effective because my data often contains human error and typos can occur anywhere, but some datasets may benefit. There are a number of modifications to Levenshtein in the book that I referenced. I personally use an algo that only allows for an edit distance error of 1 or less. A key advantage being that it requires a high level of match and scales like $n$ rather than $n^2$. $\endgroup$ – AN6U5 Oct 19 '15 at 16:46

I'd be really tempted to be lazy and apply some old technology for a quick and dirty solution, with no programming, using the linux sort command. This will give you a lexicographically sorted list.

If the store names are not the first field, if just reorder them or tell sort to use a different field via the -k switch.

Save the data to a plain CSV text file and then sort them:

$sort myStores.csv > sortedByStore.csv

You can give sort a hand by allocating it plenty of memory, 16GB in this case:

$sort -S16G myStores.csv > sortedByStore.csv

You could go further and produce a list of unique store names and counts of instances for them to help you get a handle on what the data looks like:

$sort -S16G myStores.csv  | cut -f1 -d, | uniq -c > storeIdsAndCounts.csv

Or to avoid resorting and have only the unique IDs:

$cat sortedByStore.csv   | cut -f1 -d, | uniq  > storeIds.csv
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  • $\begingroup$ This might not be what the OP is looking for, but +1 for laziness! $\endgroup$ – eliasah Oct 18 '15 at 16:41

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