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Dual-regularized one-class collaborative filtering with implicit feedback


Collaborative filtering plays a central role in many recommender systems. While most of the existing collaborative filtering methods are proposed for the explicit, multi-class settings (e.g., 1-5 stars in movie recommendation), many real-world applications actually belong to the one-class setting where user feedback is implicitly expressed (e.g., views in news recommendation and video recommendation). In this article, we propose dual-regularized one-class collaborative filtering models for implicit feedback. In particular, by dividing existing methods into point-wise class and pair-wise class, we first propose a point-wise model by integrating two existing methods and further exploiting the side information from both users and items. Next, we propose to add dual regularization into an existing pair-wise method with a different treatment of the side information. We also propose efficient algorithms to solve the proposed models. Extensive experimental evaluations on three real data sets demonstrate the effectiveness and efficiency of the proposed methods.

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This work is supported by the National Key Research and Development Program of China (No. 2017YFB1001801), the National Natural Science Foundation of China (No. 61690204, 61672274, 61702252), and the Collaborative Innovation Center of Novel Software Technology and Industrialization. Hanghang Tong is partially supported by NSF (IIS-1651203, IIS-1715385, CNS-1629888 and IIS-1743040), DTRA (HDTRA1-16-0017), ARO (W911NF-16-1-0168), and gifts from Huawei and Baidu.

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Correspondence to Yuan Yao.

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This article belongs to the Topical Collection: Special Issue on Geo-Social Computing

Guest Editors: Guandong Xu, Wen-Chih Peng, Hongzhi Yin, Zi (Helen) Huang



Proof Proof of Theorem 2

By ignoring constant terms, we can re-write (12) as

$$\begin{array}{@{}rcl@{}} J(\textbf{F}) &=& -2 \text{tr}[(\textbf{W}\odot\textbf{W}\odot(\textbf{R}+\textbf{P}))\textbf{G}\textbf{F}^{\prime}] + \text{tr}[(\textbf{W}\odot\textbf{W}\odot(\textbf{F}\textbf{G}^{\prime}))\textbf{G}\textbf{F}^{\prime}] \\ & & + \,\lambda_{r} \text{tr}(\textbf{F}\textbf{F}^{\prime}) + \lambda_{F} \text{tr}(\textbf{F}^{\prime}\textbf{D}_{M} \textbf{F}) - \lambda_{F} \text{tr}(\textbf{F}^{\prime}\textbf{M} \textbf{F}) \end{array} $$

Following the auxiliary function approach [25], an auxiliary function \(H(\textbf {F}, \tilde {\textbf {F}})\) of \(J(\textbf {F})\) must satisfy

$$\begin{array}{@{}rcl@{}} H(\textbf{F}, \textbf{F}) = J(\textbf{F}), ~~~~~~~~H(\textbf{F}, \tilde{\textbf{F}}) \geqslant J(\textbf{F}) \end{array} $$

We define

$$\begin{array}{@{}rcl@{}} \textbf{F}^{(t + 1)} = {\arg}\min\limits_{\textbf{F}} H(\textbf{F}, \textbf{F}^{(t)}) \end{array} $$

Then, by construction, we have

$$\begin{array}{@{}rcl@{}} J(\textbf{F}^{(t)}) = H(\textbf{F}^{(t)}, \textbf{F}^{(t)}) \geqslant H(\textbf{F}^{(t + 1)}, \textbf{F}^{(t)}) \geqslant J(\textbf{F}^{(t + 1)}) \end{array} $$

This would prove that \(J(\textbf {F}^{(t)})\) is monotonically decreasing.

In the remainder of proof, we need to find 1) an appropriate auxiliary function, and 2) the global minimum solution of the auxiliary function.

We start with the auxiliary function, and show that the following equation is one of the auxiliary functions for (24)

$$\begin{array}{@{}rcl@{}} H(\textbf{F}, \tilde{\textbf{F}}) &=& -2 \sum\limits_{u = 1}^{m} \sum\limits_{k = 1}^{r} [(\textbf{W}\odot\textbf{W}\odot(\textbf{R}+\textbf{P}))\textbf{G}](u,k) \tilde{\textbf{F}}(u,k) \\ & & (1+\log(\frac{\textbf{F}(u,k)}{\tilde{\textbf{F}}(u,k)})) \\ & & - \sum\limits_{u = 1}^{m} \sum\limits_{v = 1}^{m} \sum\limits_{k = 1}^{r} \lambda_{F} \textbf{M}(u,v) \tilde{\textbf{F}}(v,k) \tilde{\textbf{F}}(u,k) \\ & & (1+\log(\frac{\textbf{F}(v,k)\textbf{F}(u,k)}{\tilde{\textbf{F}}(v,k)\tilde{\textbf{F}}(u,k)})) \\ & & + \sum\limits_{u = 1}^{m} \sum\limits_{k = 1}^{r} \lambda_{r} \textbf{F}^{2}(u,k) \\ & & + \sum\limits_{u = 1}^{m} \sum\limits_{k = 1}^{r} \frac{[(\textbf{W}\odot\textbf{W}\odot(\tilde{\textbf{F}}\textbf{G}^{\prime}))\textbf{G}](u,k) \textbf{F}^{2}(u,k)}{\tilde{\textbf{F}}(u,k)} \\ & & + \sum\limits_{u = 1}^{m} \sum\limits_{k = 1}^{r} \frac{[\lambda_{F} \textbf{D}_{M} \tilde{\textbf{F}}](u,k) \textbf{F}^{2}(u,k)}{\tilde{\textbf{F}}(u,k)} \end{array} $$

For convenience, we name the five terms in (28) as \(E1\), \(E2\), \(E3\), \(E4\) and \(E5\), respectively. Then, for \(E3\) we have

$$\begin{array}{@{}rcl@{}} E3 &=& \lambda_{r} \text{tr}(\textbf{F}\textbf{F}^{\prime}) \end{array} $$

Using the inequality \(z \geqslant 1+\log z\), we have

$$\begin{array}{@{}rcl@{}} E1 &\geqslant& -2 \sum\limits_{u = 1}^{m} \sum\limits_{k = 1}^{r} [(\textbf{W}\odot\textbf{W}\odot(\textbf{R}+\textbf{P}))\textbf{G}](u,k) \textbf{F}(u,k) \\ &=& -2 \text{tr}[(\textbf{W}\odot\textbf{W}\odot(\textbf{R}+\textbf{P}))\textbf{G}\textbf{F}^{\prime}] \end{array} $$


$$\begin{array}{@{}rcl@{}} E2 \geqslant -\sum\limits_{u = 1}^{m} \sum\limits_{v = 1}^{m} \sum\limits_{k = 1}^{r} \lambda_{F} \textbf{M}(u,v) \textbf{F}(v,k) \textbf{F}(u,k) = - \lambda_{F} \text{tr}(\textbf{F}^{\prime}\textbf{M} \textbf{F}) \end{array} $$

For \(E5\), we use the following inequality [7]

$$\begin{array}{@{}rcl@{}} {\sum}_{i = 1}^{n} \sum\limits_{p = 1}^{k} \frac{(\textbf{A}\textbf{S}^{*}\textbf{B})\textbf{S}^{2}(i,p)}{\textbf{S}(i,p)} \geqslant \text{tr}(\textbf{S}^{*}\textbf{A}\textbf{S}\textbf{B}) \end{array} $$

where \(\textbf {A}_{n \times n}\), \(\textbf {B}_{k \times k}\), \(\textbf {S}_{n \times k}\), and \(\textbf {S}^{*}_{n \times k}\) are non-negative matrices, and \(\textbf {A}\) and \(\textbf {B}\) are symmetric. Therefore, we have

$$\begin{array}{@{}rcl@{}} E5 \geqslant \lambda_{F} \text{tr}(\textbf{F}^{\prime}\textbf{D}_{M} \textbf{F}) \end{array} $$

Finally, for \(E4\), let \(\textbf {F}(u,k) = \tilde {\textbf {F}}(u,k) \textbf {Q}(u,k)\) we have

$$\begin{array}{@{}rcl@{}} E4 &=& \sum\limits_{u = 1}^{m} \sum\limits_{i = 1}^{n} \sum\limits_{k = 1}^{r} \sum\limits_{l = 1}^{r} \frac{\tilde{\textbf{F}}(u,l) \textbf{G}^{\prime}(l,i) \textbf{W}^{2}(u,i) \textbf{G}(i,k)\textbf{F}^{2}(u,k)}{\tilde{\textbf{F}}(u,k)} \\ &=& \sum\limits_{u = 1}^{m} \sum\limits_{i = 1}^{n} \sum\limits_{k = 1}^{r} \sum\limits_{l = 1}^{r} \tilde{\textbf{F}}(u,l) \textbf{G}^{\prime}(l,i) \textbf{W}^{2}(u,i) \textbf{G}(i,k) \tilde{\textbf{F}}(u,k) \textbf{Q}^{2}(u,k) \\ &=& \sum\limits_{u = 1}^{m} \sum\limits_{i = 1}^{n} \sum\limits_{k = 1}^{r} \sum\limits_{l = 1}^{r} \tilde{\textbf{F}}(u,l) \textbf{G}^{\prime}(l,i) \textbf{W}^{2}(u,i) \textbf{G}(i,k) \tilde{\textbf{F}}(u,k) \\ & & (\frac{\textbf{Q}^{2}(u,k)+\textbf{Q}^{2}(u,l)}{2}) \\ &\geqslant& \sum\limits_{u = 1}^{m} \sum\limits_{i = 1}^{n} \sum\limits_{k = 1}^{r} \sum\limits_{l = 1}^{r} \tilde{\textbf{F}}(u,l) \textbf{G}^{\prime}(l,i) \textbf{W}^{2}(u,i) \textbf{G}(i,k) \tilde{\textbf{F}}(u,k) \\ & & (\textbf{Q}(u,k)\textbf{Q}(u,l)) \\ &=& \sum\limits_{u = 1}^{m} \sum\limits_{i = 1}^{n} \sum\limits_{k = 1}^{r} \sum\limits_{l = 1}^{r} \textbf{F}(u,l) \textbf{G}^{\prime}(l,i) \textbf{W}^{2}(u,i) \textbf{G}(i,k) \textbf{F}(u,k) \\ &=& \text{tr}[(\textbf{W}\odot\textbf{W}\odot(\textbf{F}\textbf{G}^{\prime}))\textbf{G}\textbf{F}^{\prime}] \end{array} $$

By substituting (29)-(33) into (28), we have \(H(\textbf {F}, \tilde {\textbf {F}}) \geqslant J(\textbf {F})\).

Next, we need to find the global minimum solution of \(H(\textbf {F}, \tilde {\textbf {F}})\). The gradient is

$$\begin{array}{@{}rcl@{}} \frac{1}{2} \frac{\partial{H(\textbf{F}, \tilde{\textbf{F}})}}{\partial{\textbf{F}(u,k)}} &=& - \frac{[(\textbf{W}\odot\textbf{W}\odot(\textbf{R}+ \textbf{P}))\textbf{G}](u,k) \tilde{\textbf{F}}(u,k)}{\textbf{F}(u,k)} \\ & & - \frac{[\lambda_{F} \textbf{M} \tilde{\textbf{F}}](u,k) \tilde{\textbf{F}}(u,k)}{\textbf{F}(u,k)} + \frac{[\lambda_{r} \tilde{\textbf{F}}](u,k) \textbf{F}(u,k)}{\tilde{\textbf{F}}(u,k)} \\ & & + \frac{[(\textbf{W}\odot\textbf{W}\odot(\tilde{\textbf{F}}\textbf{G}^{\prime}))\textbf{G}](u,k) \textbf{F}(u,k)}{\tilde{\textbf{F}}(u,k)} \\ & & + \frac{[\lambda_{F} \textbf{D}_{M} \tilde{\textbf{F}}](u,k) \textbf{F}(u,k)}{\tilde{\textbf{F}}(u,k)} \\ &=& - \frac{[(\textbf{W}\odot\textbf{W}\odot(\textbf{R}+ \textbf{P}))\textbf{G} + \lambda_{F} \textbf{M} \tilde{\textbf{F}}](u,k) \tilde{\textbf{F}}(u,k)}{\textbf{F}(u,k)} \\ & & + \frac{[(\textbf{W}\odot\textbf{W}\odot(\tilde{\textbf{F}}\textbf{G}^{\prime}))\textbf{G} + \lambda_{r} \tilde{\textbf{F}} + \lambda_{F} \textbf{D}_{M} \tilde{\textbf{F}}](u,k) \textbf{F}(u,k)}{\tilde{\textbf{F}}(u,k)} \\ \end{array} $$

We can further show that the Hessian matrix of \(H(\textbf {F}, \tilde {\textbf {F}})\) is a diagonal matrix with positive diagonal elements. Therefore, the global minimum can be obtained by setting (34) as zero, which results in

$$\begin{array}{@{}rcl@{}} \textbf{F}^{2}(u,k) = \tilde{\textbf{F}}^{2}(u,k) \frac{[(\textbf{W}\odot\textbf{W}\odot(\textbf{R}+ \textbf{P}))\textbf{G} + \lambda_{F} \textbf{M} \tilde{\textbf{F}}](u,k)}{[(\textbf{W}\odot\textbf{W}\odot(\tilde{\textbf{F}}\textbf{G}^{\prime}))\textbf{G} + \lambda_{r} \tilde{\textbf{F}} + \lambda_{F} \textbf{D}_{M} \tilde{\textbf{F}}](u,k)} \end{array} $$

Back to (26), \(\textbf {F}^{(t + 1)} = \textbf {F}\) and \(\textbf {F}^{(t)} = \tilde {\textbf {F}}\). Therefore, the updating rule in (14) decreases monotonically. Further, with equivalence between (14) and (16) as shown in the proof of Theorem 1, we have that (12) decreases monotonically under the updating rule of (16), which completes the proof. □

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Yao, Y., Tong, H., Yan, G. et al. Dual-regularized one-class collaborative filtering with implicit feedback. World Wide Web 22, 1099–1129 (2019).

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  • Recommender systems
  • One-class collaborative filtering
  • Implicit feedback
  • Dual regularization