Adversarial Surrogate Losses for Ordinal Regression
Authors: Rizal Fathony, Mohammad Ali Bashiri, Brian Ziebart
NeurIPS 2017 | Conference PDF | Archive PDF | Plain Text | LLM Run Details
| Reproducibility Variable | Result | LLM Response |
|---|---|---|
| Research Type | Experimental | We conduct our experiments on a benchmark dataset for ordinal regression [14], evaluate the performance using mean absolute error (MAE), and perform statistical tests on the results of different hinge loss surrogate methods. |
| Researcher Affiliation | Academia | Rizal Fathony Mohammad Bashiri Brian D. Ziebart Department of Computer Science University of Illinois at Chicago Chicago, IL 60607 {rfatho2, mbashi4, bziebart}@uic.edu |
| Pseudocode | No | The paper describes optimization approaches (e.g., 'stochastic optimization', 'quadratic program') but does not provide any structured pseudocode or algorithm blocks. |
| Open Source Code | No | The paper does not provide any explicit statement or link regarding the availability of its source code. |
| Open Datasets | Yes | We conduct our experiments on a benchmark dataset for ordinal regression [14], ... The benchmark contains datasets taken from the UCI Machine Learning repository [39]... |
| Dataset Splits | Yes | In the experiment, we first make 20 random splits of each dataset into training and testing sets. We performed two stages of five-fold cross validation on the first split training set for tuning each model s regularization constant λ. |
| Hardware Specification | No | The paper does not provide any specific details about the hardware (e.g., GPU, CPU models, memory, cloud instances) used for running the experiments. |
| Software Dependencies | No | The paper does not list specific software dependencies with version numbers (e.g., Python, library versions, or solver versions) required for replication. |
| Experiment Setup | Yes | In the first stage, the possible values for λ are 2 i, i = {1, 3, 5, 7, 9, 11, 13}. Using the best λ in the first stage, we set the possible values for λ in the second stage as 2 i 2 λ0, i = { 3, 2, 1, 0, 1, 2, 3}, where λ0 is the best parameter obtained in the first stage. ... we set C equals to 2i C0, i = { 2, 1, 0, 1, 2} and γ equals to 2iγ0, i = { 2, 1, 0, 1, 2}, where C0 and γ0 are the best parameters obtained in the first stage. |