Hierarchical Compound Poisson Factorization

Authors: Mehmet Basbug, Barbara Engelhardt

ICML 2016 | Conference PDF | Archive PDF | Plain Text | LLM Run Details

Reproducibility Variable Result LLM Response
Research Type Experimental We compare HCPF with HPF on nine discrete and three continuous data sets and conclude that HCPF captures the relationship between sparsity and response better than HPF.
Researcher Affiliation Academia Mehmet E. Basbug MEHMETBASBUG@YAHOO.COM Princeton University, 35 Olden St., Princeton, NJ 08540 USA Barbara E. Engelhardt BEE@PRINCETON.EDU Princeton University, 35 Olden St., Princeton, NJ 08540 USA
Pseudocode Yes Algorithm 1 SVI for HCPF
Open Source Code No The paper does not provide a link to its source code or explicitly state that the code is publicly available.
Open Datasets Yes The rating data sets include amazon fine food ratings (Mc Auley & Leskovec, 2013), movielens (Harper & Konstan, 2015), netflix (Bell & Koren, 2007) and yelp... social media activity data sets (wordpress and tencent) (Niu et al., 2012)... biochemistry data set (merck) (Ma et al., 2015)... echonest (Bertin-Mahieux et al., 2011)... genomics data set (geuvadis) (Lappalainen et al., 2013).
Dataset Splits Yes We held out 20% and 1% of the non-missing entries for testing (Ytest NM) and validation, respectively.
Hardware Specification No The paper does not provide any specific details about the hardware used to run the experiments (e.g., CPU, GPU models, or memory).
Software Dependencies No The paper does not provide specific version numbers for any software dependencies or libraries used in the experiments.
Experiment Setup Yes In HCPF, we fix K = 160, ξ = 0.7 and τ = 10, 000 after an empirical study on smaller data sets. To set hyperparameters θ and κ, we use the maximum likelihood estimates of the element distribution parameters on the non-missing entries. ... We then used E[nui] to set the factorization hyperparameters η, ζ, ρ, ϱ, ω, ϖ. To create heavy tails and uninformative gamma priors, we set ϖ = ϱ = 0.1 and ω = ρ = 0.01.