Slope safety, stability evaluation, and protective measures based on machine learning
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摘要: 為了更加快捷、高效地判定邊坡穩定與否，基于機器學習，融合主成分分析法（PCA）、參數調整、影響因素權重分析等，建立了一種邊坡安全穩定性評價體系。研究發現，運用PCA可以在保留80%數據原信息的前提下將輸入變量維度從六維降至三維，但此時模型效果有所下降；隨機森林及梯度提升(XGBoost) 兩種學習算法均可搭建有效的邊坡安全穩定性評估模型，通過對其預測效果的對比分析，確定XGBoost為最佳評價模型。與此同時，采取卡方檢驗、F檢驗以及互信息法3種相關性檢驗手段，并通過計算評價因子的重要程度且加以可視化展示，明確了容重、坡高、內摩擦角以及內聚力4個內在因素的重要性，最終將評估結果與實際結合提出了邊坡安全防護措施。Abstract: In recent years, the slope instability has brought immeasurable costs to production and life of human. As a result, it is essential to correctly understand, analyze, and design the slope reasonably, and implement appropriate protective measures to minimize the loss and harm caused by its instability. By far, slope stability can be investigated using theoretical analysis, numerical modeling and machine learning prediction, among them machine learning prediction has been the most encouraging one. Many studies have been performed using machine learning algorithms to predict the slope stability. However, these methods suffers from poor accuracy and poor generalisation capbility, so its real-life application has been limited. In the current study, a machine learning-based slope safety and stability evaluation system is established by integrating principal component analysis, parameter adjustment, and influence factor weight analysis. It is shown that PCA can reduce the dimensions of the input variables from six to three while retaining 80% of the information; however, at the cost of the model’s effectiveness. The random forest and XGBoost (eXtreme Gradient Boosting) learning algorithms can both be employed to develop effective evaluation models for slope safety and stability. The comparative analysis of algorithms’ prediction effects established XGBoost as the best evaluation model, which can achieve the average accuracy of 92%, precision of 91%, recall of 96%, and the area under the receiver operating characteristic curve (AUC) of 0.95. In addition, this study employs three types of test methods: the chi-square test, F test correlation, and mutual information method, meanwhile by calculating and visualizing the importance of influencing factors, the influence of unit weight, slope height, internal friction angle and cohesion on slope stability is demonstrated. It has been shown that the unit weight is the most influencing factor for the slope stability. Finally, the slope safety protection measures are proposed by combining the evaluation results with the actual project.
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Key words:
- slope stability evaluation /
- machine learning /
- random forest /
- XGBoost /
- protective measures
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圖 1 影響因素的特征統計。（a）坡高分布；（b）坡角分布；（c）孔隙壓力比分布；（d）容重分布；（e）內聚力分布；（f）內摩擦角分布
Figure 1. Characteristic statistics of influencing factors: (a) distribution of slope height; (b) slope angle distribution; (c) pore pressure ratio distribution; (d) unit weight distribution; (e) distribution of cohesion; (f) internal friction angle distribution
圖 3 降維后的模型表現。（a）隨機森林模型；（b）XGBoost模型
Figure 3. Model performance after dimensionality reduction: (a) random forest model; (b) XGBoost model
圖 4 訓練集和驗證集的結果對比。（a）隨機森林模型；（b）XGBoost模型
Figure 4. Comparison of the results between the training set and the verification set: (a) random forest model; (b) XGBoost model
圖 5 測試集的10次預測準確率
Figure 5. Ten times prediction accuracy of the test set using random forest and XGBoost models
圖 6 調整后的ROC與AUC展示。（a）隨機森林模型；（b）XGBoost模型
Figure 6. ROC and AUC display after adjustment: (a) random forest model; (b) XGBoost model
圖 7 特征重要性排序。（a）隨機森林模型；（b）XGBoost模型
Figure 7. Rank of feature importance: (a) Random forest model; (b) XGBoost model
表 1 兩種算法評估結果差異表
Table 1. Different evaluation results obtained from the two algorithms
Indicators Random forest XGBoost Accuracy 0.86 0.92 Precision 0.90 0.91 Recall 0.85 0.96 AUC 0.92 0.95 
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表 2 三種特征選擇方法下的評價因子重要性
Table 2. Importance of evaluation factors under the three feature selection methods
Feature selection method Important evaluation factors and corresponding index scores Slope height Unit weight Internal friction angle Chi-square (${\chi ^2}$) test ${\chi ^2}$=6.9396 P=0.0084 ${\chi ^2}$=3.5136 P= 0.0609 ${\chi ^2}$=2.5307 P =0.1117 F test F =22.5560 P =0.00000441 F=49.7141 P=0 F =34.8498 P=0.00000002 Mutual information method 0.2354 0.2173 0.2791 (Cohesion)
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