Yup! The last two lines break the backlash-right clause.
Cool tool! Thanks.
On Wed, Aug 31, 2011 at 8:57 PM, Ted Dunning wrote:
> Hmm... I see this:
>
>
> http://latex.codecogs.com/gif.latex?F(x,y)=0%20~~\mbox{and}~~%20\left|%20\begin{array}{ccc}%20F''_{xx}%20&%20F''_{xy}%20&%20F'_x%20\\%20F''_
Hmm... I see this:
http://latex.codecogs.com/gif.latex?F(x,y)=0%20~~\mbox{and}~~%20\left|%20\begin{array}{ccc}%20F''_{xx}%20&%20F''_{xy}%20&%20F'_x%20\\%20F''_{yx}%20&%20F''_{yy}%20&%20F'_y%20\\%20F'_x%20&%20F'_y%20&%200%20\end{array}\right|%20=%200
Must be a cut and paste kind of thing.
On Wed,
On Aug 31, 2011, at 8:32pm, Ted Dunning wrote:
> The basics of latex notation are
>
> ^ for superscript
> _ for subscript
> {} for grouping
> \sum for summation
> \log for logs
> \Omega for upper case greek letter omega
> \alpha for lower case greek letter beta
> \int for integral.
>
> Se
The basics of latex notation are
^ for superscript
_ for subscript
{} for grouping
\sum for summation
\log for logs
\Omega for upper case greek letter omega
\alpha for lower case greek letter beta
\int for integral.
See http://www.codecogs.com/latex/eqneditor.php for a playground where you
This is a very good point that seems very likely to be the source of the
confusion.
On Wed, Aug 31, 2011 at 6:06 PM, Dmitriy Lyubimov wrote:
> Perhaps confusion may be stemming from the fact that inverse equal
> transpose if matrices are orthogonal (or even orthonormal), so
> sometimes Q^{-1}\eq
PS
> On Wed, Aug 31, 2011 at 3:57 PM, Lance Norskog wrote:
>>
>> I see a fair amount of stuff here in what I think is MathML, but is displays
>> raw in gmail.
>>
this is usually tex. I am not familiar with mathml that close but i
think it is fundamentally different
we usually denote inverse , A^{-1} or just A^-1
Apostrophe, superscript star or {top} always mean transpose. I never
saw apostrophe to be used for inverses.
Perhaps confusion may be stemming from the fact that inverse equal
transpose if matrices are orthogonal (or even orthonormal), so
sometimes
In this text-only notation, I though apostrophe meant inverse. What then is
matrix inversion?
I see a fair amount of stuff here in what I think is MathML, but is displays
raw in gmail.
On Wed, Aug 31, 2011 at 8:04 AM, Ted Dunning wrote:
> Uhh...
>
> A' is the transpose of A. Not the inverse.
>
Mathematically speaking, random sampling is just fine. Stratifying based on
various criteria can help avoid loss of accuracy so if you had several
clusters then down sampling heavily represented clusters might work, but the
accurate definition of clusters is harder than the cooccurrence analysis
t
Uhh...
A' is the transpose of A. Not the inverse.
A' A *is* the summation version.
On Wed, Aug 31, 2011 at 1:24 AM, Lance Norskog wrote:
> "Also, if your original matrix is A, then it is usually a shorter path to
> results to analyze the word (item) cooccurrence matrix A'A. The methods
> bel
"If you have a document (user) and a word (item), then you
have a joint probability that any given interaction will be between this
document and word. We pretend in this case that each interaction is
independent of every other which is patently not true, but very helpful."
So if you subsample ran
"Also, if your original matrix is A, then it is usually a shorter path to
results to analyze the word (item) cooccurrence matrix A'A. The methods
below work either way."
The cooccurrence definitions I'm finding only use the summation-based one in
wikipedia. Are there any involving inverting the m
Jeff,
I think that this is a much simpler exposition:
http://tdunning.blogspot.com/2008/03/surprise-and-coincidence.html
It makes the connection with entropy clear and allows a very simple
implementation for more than 2x2 situations.
More comments in-line:
On Mon, Aug 29, 2011 at 1:34 PM, Jeff
Friday I finally got around to reading Ted's paper "accurate methods for
statistics of surprise and coincidence" for a better understanding of how to
apply log likelihood.
Can somebody validate if I'm understanding/applying the idea correctly in
this case?
If we have a item/feature matrix (docume
http://www.nytimes.com/interactive/2010/01/10/nyregion/20100110-netflix-map.html
Do not fear demographics. Yes, some people rent movies with all-black casts,
and other people rent movies with all-white casts. And the Walmarts in the
SF East Bay have palettes full of Tyler Perry videos, while most
On Fri, Aug 26, 2011 at 8:29 AM, Jeff Hansen wrote:
> Thanks for the math Ted -- that was very helpful.
>
NP.
> ... I've been playing with smaller matrices
> mainly for my own learning purposes -- it's much easier to read through 200
> movies (most of which I've heard of) and get a gut feel,
Thanks for the math Ted -- that was very helpful.
I've been using sparseMatrix() from libray(Matrix) -- largely based on your
response to somebody elses email. I've been playing with smaller matrices
mainly for my own learning purposes -- it's much easier to read through 200
movies (most of which
I got this "axis of interest" concept from a presentation by one of the
Netflix team runner-ups, I don't know which one. He did not give a name for
it. Is there a standard term? I hate just making up new words.
Also, there are clusters of items at both ends, but there are also items
along the axis
That's correct. Well you just have to recompose the user row you are
interested in. It will no longer be sparse, at all. Those new values are
your estimated ratings.
On Fri, Aug 26, 2011 at 12:07 AM, Jeff Hansen wrote:
>
> I also think I may have missed a big step of the puzzle. For some reason
This is a little meditation on user v.s. item matrix density. The
heavy users and heavy items can be subsampled, once they are
identified. Hadoop's built-in sort does give a very simple
"map-increase" way to do this sort.
http://ultrawhizbang.blogspot.com/2011/08/sorted-recommender-data.html
On T
In matrix terms the binary user x item matrix maps a set of items to users
(A h = users who interacted with items in h). Similarly A' maps users to
items. Thus A' (A h) is the classic "users who x'ed this also x'ed that"
sort of operation. This can be rearranged to be (A'A) h.
This is where the
One thing I found interesting (but not particularly surprising) is that the
biggest singular value/vector was pretty much tied directly to volume. That
makes sense because the best predictor of whether a given fields value was 1
was whether it belonged to a row with lots of 1s or a column with lot
If you take the item vector an existing user, multiply that by the
left-hand SVD matrix, and multthe resulting vector[i] *
1/singularvalues[i], you should get the item's row in the left-hand
column. So, the left-hand column times 1/singular values gives you the
projection for a new user's item vect
The 200x10 matrix is indeed a matrix of 10 singular vectors, which are
eigenvectors of AA'. It's the columns, not rows, that are
eigenvectors.
The rows do mean something. I think it's fair to interpret the 10
singular values / vectors as corresponding to some underlying features
of tastes. The row
Well, I think my problem may have had more to do with what I was calling the
eigenvector... I was referring to the rows rather than the columns of U and
V. While the columns may be characteristic of the overall matrix, the rows
are characteristic of the user or item (in that they are a rank reduc
On Thu, Aug 25, 2011 at 1:53 PM, Jeff Hansen wrote:
> By the way, please ignore my use of the term eigenvector -- I have a
> feeling
> I completely misused it. I've never quite understood the concept, but to
> me
> that truncated 10 value long vector that corresponds to a movie seems to be
> "ch
By the way, please ignore my use of the term eigenvector -- I have a feeling
I completely misused it. I've never quite understood the concept, but to me
that truncated 10 value long vector that corresponds to a movie seems to be
"characteristic" of it (which is what the language eigen was always i
I've been playing around with this problem for the last week or so (or at
least this problem as I understood it based on your initial commentary
Lance) -- but purely in R using smaller data so I can 1. get my head wrapped
around the problem, and 2. get more familiar with R.
To make the problem a l
Sharpened:
http://ultrawhizbang.blogspot.com/2011/08/singular-vectors-for-recommendations.html
On Wed, Aug 10, 2011 at 11:53 PM, Sean Owen wrote:
> You may need to sharpen your terms / problem statement here :
>
> What is a geometric value -- just mean a continuous real value?
> So these are ite
You may need to sharpen your terms / problem statement here :
What is a geometric value -- just mean a continuous real value?
So these are item-feature vectors?
The middle bit of the output of an SVD is not a singular vector -- it's a
diagonal matrix containing singular values on the diagonal.
Th
A picture that might help explain the problem:
http://www.flickr.com/photos/54866255@N00/6031564308/in/photostream
On 8/10/11, Lance Norskog wrote:
> Zeroing in on the topic:
>
> I have:
> 1) a set of raw input vectors of a given length, one for each item.
> Each value in the vectors are geomet
Zeroing in on the topic:
I have:
1) a set of raw input vectors of a given length, one for each item.
Each value in the vectors are geometric, not bag-of-words or other.
The matrix is [# items , # dimensions].
2) An SVD of same:
left matrix of [ # items, #d features per item] * singular
vector[
SVDRecommender is intriguing, thanks for the pointer.
On Sun, Jul 10, 2011 at 12:15 PM, Ted Dunning wrote:
> Also, item-item similarity is often (nearly) the result of a matrix product.
> If yours is, then you can decompose the user x item matrix and the desired
> eigenvalues are the singular va
Also, item-item similarity is often (nearly) the result of a matrix product.
If yours is, then you can decompose the user x item matrix and the desired
eigenvalues are the singular values squared and the eigen vectors are the
right singular vectors for the decomposition.
On Sun, Jul 10, 2011 at 2
So it sounds like you want the SVD of the item-item similarity matrix? Sure,
you can use Mahout for that. If you are not in Hadoop land then look at
SVDRecomnender to crib some related code. It is decomposing the user item
matrix though.
But for this special case of a symmetric matrix your singula
I would like to find the singular vectors of an item-item data model.
That is, the largest singular vector has many items "close" to it, and
items at the ends are polar opposites in popularity. For example, the
Netflix dataset yielded chick flicks v.s. Star Trek movies and Harry
Potter v.s. (Stanle
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