2025年“创新杯”(原钉钉杯) 建模思路
B题 道路路面维护需求综合预测
2025钉钉杯 B题解题思路
任务A:道路维护需求预测(二分类)
1 问题分析
- 特征多样:数值型(PCI、AADT)+ 分类型(道路类型、沥青类型)。
- 样本不平衡:需维护路段占少数。
- 可解释性:需量化关键特征对维护需求的影响。
- 解决方案:随机森林——支持混合数据、鲁棒、自带特征重要性。
2 Python 代码
import pandas as pdfrom sklearn.model_selection import train_test_splitfrom sklearn.ensemble import RandomForestClassifierfrom sklearn.metrics import accuracy_score, recall_score, f1_score, confusion_matrixfrom sklearn.preprocessing import StandardScaler, OneHotEncoderfrom sklearn.compose import ColumnTransformerimport matplotlib.pyplot as pltdata = pd.read_csv(\'road_maintenance.csv\')X = data[[\'PCI\',\'Road_Type\',\'AADT\',\'Asphalt_Type\', \'Last_Maintenance\',\'Average_Rainfall\',\'Rutting\',\'IRI\']]y = data[\'Needs_Maintenance\']pre = ColumnTransformer([ (\'cat\', OneHotEncoder(), [\'Road_Type\',\'Asphalt_Type\']), (\'num\', StandardScaler(), [\'PCI\',\'AADT\',\'Last_Maintenance\', \'Average_Rainfall\',\'Rutting\',\'IRI\']) ])X_proc = pre.fit_transform(X)X_train, X_test, y_train, y_test = train_test_split( X_proc, y, test_size=0.2, random_state=42)clf = RandomForestClassifier(n_estimators=100, max_depth=10, min_samples_split=5, random_state=42)clf.fit(X_train, y_train)y_pred = clf.predict(X_test)print(\"Accuracy:\", accuracy_score(y_test, y_pred))print(\"Recall: \", recall_score(y_test, y_pred))print(\"F1: \", f1_score(y_test, y_pred))names = pre.get_feature_names_out()imp = clf.feature_importances_plt.barh(names, imp)plt.title(\'Feature Importance\')plt.show()
3 MATLAB 代码
data = readtable(\'road_maintenance.csv\');cat_vars = {\'Road_Type\',\'Asphalt_Type\'};for v = cat_vars data.(v{1}) = categorical(data.(v{1}));endX = data(:, {\'PCI\',\'Road_Type\',\'AADT\',\'Asphalt_Type\', ... \'Last_Maintenance\',\'Average_Rainfall\',\'Rutting\',\'IRI\'});y = data.Needs_Maintenance;Xenc = onehotencode(X, cat_vars);Xnorm = normalize(Xenc);rng(1)cv = cvpartition(size(Xnorm,1), \'HoldOut\', 0.2);Xtr = Xnorm(cv.training,:); ytr = y(cv.training);Xte = Xnorm(cv.test,:); yte = y(cv.test);model = TreeBagger(100, Xtr, ytr, ... \'Method\',\'classification\', ... \'MaxDepth\',10, \'MinParentSize\',5);y_hat = str2double(predict(model, Xte));cm = confusionmat(yte, y_hat);acc = sum(diag(cm))/sum(cm(:));rec = cm(2,2)/sum(cm(2,:));f1 = 2*rec*cm(2,2)/sum(cm(:,2))/(rec+cm(2,2)/sum(cm(:,2)));fprintf(\'Accuracy: %.4f\\nRecall: %.4f\\nF1: %.4f\\n\', acc, rec, f1);imp = model.OOBPermutedPredictorDeltaError;bar(imp)xticklabels(X.Properties.VariableNames)title(\'Feature Importance\')
任务B:维护紧急程度评分与优先级划分
1 思路
- 输出连续评分:将任务A的随机森林改为回归模型,输出
[0,1]
区间的紧急程度 R
。
- 无监督聚类:使用 K-means 把
R
划分为高、中、低三个优先级。
- 可解释验证:检查高优先级路段的 PCI、IRI 等核心指标,确保策略合理。
2 Python 代码
import numpy as np, pandas as pd, matplotlib.pyplot as pltfrom sklearn.ensemble import RandomForestRegressorfrom sklearn.cluster import KMeansfrom sklearn.preprocessing import MinMaxScalerdata = pd.read_csv(\'road_maintenance.csv\')X = data.drop(\'Needs_Maintenance\', axis=1)y = data[\'Needs_Maintenance\']rf = RandomForestRegressor(n_estimators=100, random_state=42)rf.fit(X, y)R = rf.predict(X)R_norm = MinMaxScaler().fit_transform(R.reshape(-1,1)).flatten()km = KMeans(n_clusters=3, random_state=42)clusters = km.fit_predict(R_norm.reshape(-1,1))centers = km.cluster_centers_.flatten()order = np.argsort(centers)prio_map = {order[0]:0, order[1]:1, order[2]:2}priorities = np.array([prio_map[c] for c in clusters])print(pd.Series(priorities).value_counts().sort_index())plt.hist(R_norm[priorities==0], bins=30, alpha=.7, color=\'green\', label=\'Low\')plt.hist(R_norm[priorities==1], bins=30, alpha=.7, color=\'blue\', label=\'Medium\')plt.hist(R_norm[priorities==2], bins=30, alpha=.7, color=\'red\', label=\'High\')plt.xlabel(\'Maintenance Urgency Score\')plt.ylabel(\'Number of Segments\')plt.title(\'Priority Distribution via K-means\')plt.legend(); plt.show()
3 MATLAB 代码
load(\'rf_regression_model.mat\'); data = readtable(\'road_maintenance.csv\');X = data{:, {\'PCI\',\'Road_Type\',\'AADT\',\'Asphalt_Type\', ... \'Last_Maintenance\',\'Average_Rainfall\',\'Rutting\',\'IRI\'}};R = predict(model, X);R_norm = (R - min(R)) / (max(R) - min(R));rng(1)[idx, centers] = kmeans(R_norm, 3);[~, order] = sort(centers);prio = zeros(size(idx));prio(idx==order(1)) = 0; prio(idx==order(2)) = 1; prio(idx==order(3)) = 2; fprintf(\'High: %d, Medium: %d, Low: %d\\n\', ... sum(prio==2), sum(prio==1), sum(prio==0));figurehistogram(R_norm(prio==0), \'BinWidth\',0.05,\'FaceColor\',\'g\'); hold onhistogram(R_norm(prio==1), \'BinWidth\',0.05,\'FaceColor\',\'b\');histogram(R_norm(prio==2), \'BinWidth\',0.05,\'FaceColor\',\'r\');xlabel(\'Maintenance Urgency Score\'); ylabel(\'Count\');title(\'Priority Distribution\'); legend(\'Low\',\'Medium\',\'High\');
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