ALS model - how to generate full_u * v^t * v?

Question

I'm trying to figure out how an ALS model can predict values for new users in between it being updated by a batch process. In my search, I came across this stackoverflow answer. I've copied the answer below for the reader's convenience:

You can get predictions for new users using the trained model (without updating it):

To get predictions for a user in the model, you use its latent representation (vector u of size f (number of factors)), which is multiplied by the product latent factor matrix (matrix made of the latent representations of all products, a bunch of vectors of size f) and gives you a score for each product. For new users, the problem is that you don't have access to their latent representation (you only have the full representation of size M (number of different products), but what you can do is use a similarity function to compute a similar latent representation for this new user by multiplying it by the transpose of the product matrix.

i.e. if you user latent matrix is u and your product latent matrix is v, for user i in the model, you get scores by doing: u_i * v for a new user, you don't have a latent representation, so take the full representation full_u and do: full_u * v^t * v This will approximate the latent factors for the new users and should give reasonable recommendations (if the model already gives reasonable recommendations for existing users)

To answer the question of training, this allows you to compute predictions for new users without having to do the heavy computation of the model which you can now do only once in a while. So you have you batch processing at night and can still make prediction for new user during the day.

Note: MLLIB gives you access to the matrix u and v

The quoted text above is an excellent answer, however, I'm struggling to understand how to programmatically implement this solution. For example, the matrix u and v can be obtained with:

# pyspark example

# ommitted for brevity ... loading movielens 1M ratings

model = ALS.train(ratings, rank, numIterations, lambdaParam)

matrix_u = model.userFeatures()

print(matrix_u.take(2)) # take a look at the dataset

This returns:

[
  (2, array('d', [0.26341307163238525, 0.1650490164756775, 0.118405282497406, -0.5976635217666626, -0.3913084864616394, -0.1379186064004898, -0.3866392970085144, -0.1768060326576233, -0.38342711329460144, 0.48550787568092346, -0.18867433071136475, -0.02757863700389862, 0.1410026103258133, 0.11498363316059113, 0.03958914801478386, 0.034536730498075485, 0.08427099883556366, 0.46969038248062134, -0.8230801224708557, -0.15124185383319855, 0.2566414773464203, 0.04326820373535156, 0.19077207148075104, 0.025207923725247383, -0.02030213735997677, 0.1696728765964508, 0.5714617967605591, -0.03885050490498543, -0.09797532111406326, 0.29186877608299255, -0.12768596410751343, -0.1582849770784378, 0.01933656632900238, -0.09131495654582977, 0.26577943563461304, -0.4543033838272095, -0.11789630353450775, 0.05775507912039757, 0.2891307771205902, -0.2147761881351471, -0.011787488125264645, 0.49508437514305115, 0.5610293745994568, 0.228189617395401, 0.624510645866394, -0.009683617390692234, -0.050237834453582764, -0.07940001785755157, 0.4686132073402405, -0.02288617007434368])), 
  (4, array('d', [-0.001666820957325399, -0.12487432360649109, 0.1252429485321045, -0.794727087020874, -0.3804478347301483, -0.04577340930700302, -0.42346617579460144, -0.27448347210884094, -0.25846347212791443, 0.5107921957969666, 0.04229479655623436, -0.10212298482656479, -0.13407345116138458, -0.2059325873851776, 0.12777331471443176, -0.318756639957428, 0.129398375749588, 0.4351944327354431, -0.9031049013137817, -0.29211774468421936, -0.02933369390666485, 0.023618215695023537, 0.10542935132980347, -0.22032295167446136, -0.1861676126718521, 0.13154461979866028, 0.6130356192588806, -0.10089754313230515, 0.13624103367328644, 0.22037173807621002, -0.2966669499874115, -0.34058427810668945, 0.37738317251205444, -0.3755438029766083, -0.2408779263496399, -0.35355791449546814, 0.05752146989107132, -0.15478627383708954, 0.3418906629085541, -0.6939512491226196, 0.4279302656650543, 0.4875738322734833, 0.5659542083740234, 0.1479463279247284, 0.5280753970146179, -0.24357643723487854, 0.14329688251018524, -0.2137598991394043, 0.011986476369202137, -0.015219110995531082]))
]

I can also do similar to get the v matrix:

matrix_v = model.productFeatures()

print(matrix_v.take(2)) # take a look at the dataset

This results in:

[
  (2, array('d', [0.019985994324088097, 0.0673416256904602, -0.05697149783372879, -0.5434763431549072, -0.40705952048301697, -0.18632276356220245, -0.30776089429855347, -0.13178342580795288, -0.27466219663619995, 0.4183739423751831, -0.24422742426395416, -0.24130797386169434, 0.24116989970207214, 0.06833088397979736, -0.01750543899834156, 0.03404173627495766, 0.04333991929888725, 0.3577033281326294, -0.7044714689254761, 0.1438472419977188, 0.06652364134788513, -0.029888223856687546, -0.16717877984046936, 0.1027146726846695, -0.12836599349975586, 0.10197233408689499, 0.5053384900093079, 0.019304445013403893, -0.21254844963550568, 0.2705852687358856, -0.04169371724128723, -0.24098040163516998, -0.0683765709400177, -0.09532768279314041, 0.1006036177277565, -0.08682398498058319, -0.13584329187870026, -0.001340558985248208, 0.20587041974067688, -0.14007550477981567, -0.1831497997045517, 0.5021498203277588, 0.3049483597278595, 0.11236990243196487, 0.15783801674842834, -0.044139936566352844, -0.14372406899929047, 0.058535050600767136, 0.3777201473712921, -0.045475270599126816])), 
  (4, array('d', [0.10334215313196182, 0.1881643384695053, 0.09297363460063934, -0.457258403301239, -0.5272660255432129, -0.0989445373415947, -0.2053477019071579, -0.1644461452960968, -0.3771175146102905, 0.21405018866062164, -0.18553146719932556, 0.011830524541437626, 0.29562288522720337, 0.07959598302841187, -0.035378433763980865, -0.11786794662475586, -0.11603366583585739, 0.3776192367076874, -0.5124108791351318, 0.03971947357058525, -0.03365595266222954, 0.023278912529349327, 0.17436474561691284, -0.06317273527383804, 0.05118614062666893, 0.4375131130218506, 0.3281322419643402, 0.036590900272130966, -0.3759073317050934, 0.22429685294628143, -0.0728025734424591, -0.10945595055818558, 0.0728464275598526, 0.014129920862615108, -0.10701996833086014, -0.2496117204427719, -0.09409723430871964, -0.11898282915353775, 0.18940524756908417, -0.3211393356323242, -0.035668935626745224, 0.41765937209129333, 0.2636736035346985, -0.01290816068649292, 0.2824321389198303, 0.021533429622650146, -0.08053319901227951, 0.11117415875196457, 0.22975310683250427, 0.06993964314460754]))
]

However, I'm not sure how to progress from this to full_u * v^t * v

Coparcenary

If you already knew the full ratings, why do you need ALS model to predict them? In other words the post approach may not help. – Gramicidin

Answer 1

This new user is not the the matrix U, so you don't have its latent representation in 'k' factors, you only know its full representation, i.e., all its ratings. full_u here means all of the new user ratings in a dense format (not the sparse format ratings are) e.g.:

[0 2 0 0 0 1 0] if user u has rated item 2 with a 2 and item 6 with a 1.

then you can get v pretty much like you did and turning it to a matrix in numpy for instance:

pf = model.productFeatures()
Vt = np.matrix(np.asarray(pf.values().collect()))

then is is just a matter of multiplying: full_u*Vt*Vt.T

Vt and V are transposed compared to the other answer but that's just a matter of convention.

Note that the Vt*Vt.T product is fixed, so if you are going to use this for multiple new users it will be computationally more efficient to pre-compute it. Actually for more than one user it would be better to put all their ratings in bigU (in the same format as my one new user example) and just do the matrix product: bigU*Vt*Vt.T to get all the ratings for all the new users. Might still be worth checking that the product is done in the most efficient way in terms of number of operations.

Brocade

If you are interested, there is a follow-on question with a bounty: #41569474 – Coparcenary

Answer 2

Just a word of warning. People talk about the user and product matrices like they are left and right singular vectors. But as far as I understand, the method used to find U and V is an optimization of a straight squared error cost function, which makes none of the orthogonality guarantees of SVD.

In other words, think algebraically about what the above answer claims. We have a full ratings matrix R, an n by p matrix of ratings for n users over p products. We decompose it with k latent factors to approximate R = UV, where the rows of U, an n by k matrix, are the latent user representations, and the columns of V, a k by p matrix, are the latent product representations. In order to find latent user representations for a matrix R of entirely new users without refitting the model, we need:

       R = U V  
R V^{-1} = U V V^{-1}  
R V^{-1} = U I_{k}  
R V^{-1} = U  

where I_{k} is the k dimensional identity matrix and V^{-1} is the p by k right inverse of V. The tip above assumes that V^{T} = V^{-1}. This is not guaranteed. And in general there is no guarantee that assuming this is true will give you anything but nonsense answers.

Let me know if I'm missing something in the optimization method behind MLLib's CF implementation. Is there a trick in the ALS model that guarantees orthogonality that I'm missing?

Facility