Reconcile tickerplant feeds using longest common subsequence matching

Sanket Agrawal just posted some very cool code to the k4 mailing list for finding the longest common¬†sub-sequences. This actually solves a problem that some colleagues had where they wanted to combine two tickerplant feeds that never seemed to match up. Here’s an example:


q)/ assuming t is our perfect knowledge table

q)t
time sym size price
------------------------
09:00 A 8 7.343062
09:01 A 0 5.482385
09:02 B 5 8.847715
09:03 A 1 6.78881
09:04 B 5 3.432312
09:05 A 0 0.2801381
09:06 A 2 3.775222
09:07 B 3 1.676582
09:08 B 7 7.163578
09:09 B 4 3.300548

Let us now create two tables, u and v, neither of which contain all the data.

q)u:t except t 7 8
q)v:t except t 1 2 3 4
q)u
time sym size price
------------------------
09:00 A 8 7.343062
09:01 A 0 5.482385
09:02 B 5 8.847715
09:03 A 1 6.78881
09:04 B 5 3.432312
09:05 A 0 0.2801381
09:06 A 2 3.775222
09:09 B 4 3.300548
q)v
time sym size price
------------------------
09:00 A 8 7.343062
09:05 A 0 0.2801381
09:06 A 2 3.775222
09:07 B 3 1.676582
09:08 B 7 7.163578
09:09 B 4 3.300548

We can find the indices that differ using Sankets difftables function

q)p:diffTables[u;v;t `sym;`price`size]
q)show p 0; / rows in u that are not in v
1 2 3 4
q)show p 1; / rows in v that are not in u
3 4

/ combine together again

q)`time xasc (update src:`u from u),update src:`v from v p 1
time sym size price src
----------------------------
09:00 A 8 7.343062 u
09:01 A 0 5.482385 u
09:02 B 5 8.847715 u
09:03 A 1 6.78881 u
09:04 B 5 3.432312 u
09:05 A 0 0.2801381 u
09:06 A 2 3.775222 u
09:07 B 3 1.676582 v
09:08 B 7 7.163578 v
09:09 B 4 3.300548 u
q)t~`time xasc u,v p 1
1b

The code can be downloaded at:
http://code.kx.com/wsvn/code/contrib/sagrawal/lcs/miller.q

http://code.kx.com/wsvn/code/contrib/sagrawal/lcs/myers.q

Thanks Sanket.

Algorithm details:

Myers O(ND): http://www.xmailserver.org/diff2.pdf

Miller O(NP):  http://www.bookoff.co.jp/files/ir_pr/6c/np_diff.pdf

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