Can you convert it into a equal join problem? That's the case mapreduce
can handle efficiently. Not sure if it address your problem but provide
a sample script.
a = load 'A' as (a0:chararray);
b = foreach a generate LOWER(a0) as b0;
c = load 'B' as (c0:chararray);
d = foreach c generate LOWER(c0) as d0;
e = foreach d generate d0, flatten(STRSPLIT(d0)) as e0;
f = join b by b0, e by e0;
g = foreach f generate d0, b0;
dump g;
Daniel
Arun A K wrote:
Hello
I have this problem to solve using Pig.
*Input*
1. Relation A which has only one field of type chararray. Sample of A
follows:
*abc*
*xyz gh*
*zzz yy*
*red*
Approximate numbers of rows in A = 10,000
2. Relation B which has only one field of type chararray. Sample of B
follows:
*red car*
*red ferrari*
*abc*
*abcd*
*xyz ghis*
Approximate numbers of rows in B = 1 billion
*Problem to be solved* I need to find all case-insensitive variants of each
term in relation A existing in relation B. For example: Term 'red' from A
would have variants 'red car' and 'red ferrari' in B.
I was able to get variants of one term in A from B using matches operator
i.e. matches '.*red.*' How to go about creating a complete solution for this
problem? Should I use a UDF or go for native Map Reduce? Am a bit confused
on how to proceed on this. I would really appreciate any help on this.
Thanks much.
Regards
Arun A K
