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I'm looking for a partner. I have class at the following times:
M 10:30-11:30
T, Th 10:30-1:30
W, F 2:30-5:30

-James

I get out of class at 1:30, so I have lots of free time.

e-mail: gdemart@cs

ok, the assignment discription asks us to:
"implement tf-idf as discribed in class slides"
therefore i've been wondering:
1. are we really supposed to return the vector space similarity of hte query term doc
and the current doc as the "relevence score"?

2. since the normalizer is defined to be (sqr (tf-idf)squred) shouldn't each individual
term have its own normalizer? in stead of a whole document sharing 1 normalizer?

3. if the query term were to be treated as a doc, then the Tf would be "1" for each
term, and since idf/normalizer is a constant then the whole "vector space similarity"
deal would just be changing the fromular of weight to weight=tf*(idf/normalizer)^2?

sorry to have ranted on like this. alot of my questions sound silly, but i am just really confused …
any explaination is appreciated.
cheers, jiatao

Hi everyone,

Someone asked a couple of questions that I'll cite here and
answer:

Question1: Treating Query as a document

According to slide 44 IR, we did in class yesterday , you suppose to treat your
query as a document and calcualte its W (for each term). Do the same for the
current document you are trying to rank(for each term). Then do that cross
product thing with the array of W from both current doc and query to get how
close together are they are

It is unclear to me that how do you calculate W (normalize) for the query?
Formular for W(not normalize) is tf * log(N/nk) where nk is number of times
term appear in all documents. Is nk including the one in query as well, since
you are counting query as a document?

One generally does not include the query document itself in this
count. The reason is that people like to precompute the idf term,
and obviously we can't do that if the query contents can change it.
In practice, the number of documents is so large that this isn't a
real concern regarding the idf value; the difference between including
the query or not will usually be vanishingly small.

What about when we are calculating W for the document we are trying to rank?,
do you need to count in words inside the query to a document space of
log(N/nk)? (thus N would increase by 1, and nk would increase by the number of
times the term appear in the query?).

One does not consider internal-doc frequency to compute this. It's
just the number of docs in which the count is >= 1.


Question2: getNormalize()?

Upon talking to professor, he also want this thing normalize~. I try writting
it out, and it is very long (2 loop for calculating dominator for each the
current document and the query. Another 2 loops to calcualte W (not normalize)
for document, and the query. you need to go though query k^2 times to find out
how many times each term occur in the query as well since you can not convery
query array to Document. Its long such that I think I am doing the wrong thing.
So I was wondering what is "getNormalizer()" function in Document class do?

I mentioned in a previous email what the getNormalizer() result
actually contains. While you don't have to incorporate the query,
you're right that it takes a loop through the whole corpus to compute.
That's a little burdensome for >1M docs, so we've precomputed it
for you.

Thanks,
—Mike

[CSE454] some more questions by gdemartgdemart, 21 Oct 2006 21:10

Hi everyone,

A few people have asked for a more precise description of the
result from the getNormalizer() call in Document.

It's the per-document length normalizer. It's not computed
per-term-weight. There's just one unique value per document.

—Mike

[CSE454] getNormalizer() by gdemartgdemart, 21 Oct 2006 21:09

Hi everyone,

There's an inconsistency in the assignment about whether your
DocumentRanker code should inherit from "IRanker" or "IExtendedRanker".

Sorry about that; I made a versioning error in the assignment text.
You should inherit from "IExtendedRanker". Its methods are exactly the
same as the "IRanker" interface.

Thanks,
—Mike

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