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 pd
from sklearn.model_selection import train_test_split
from sklearn.ensemble import RandomForestClassifier
from sklearn.metrics import accuracy_score, recall_score, f1_score, confusion_matrix
from sklearn.preprocessing import StandardScaler, OneHotEncoder
from sklearn.compose import ColumnTransformer
import matplotlib.pyplot as plt
data = 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}));
end
X = 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 plt
from sklearn.ensemble import RandomForestRegressor
from sklearn.cluster import KMeans
from sklearn.preprocessing import MinMaxScaler
data = 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));
figure
histogram(R_norm(prio==0), 'BinWidth',0.05,'FaceColor','g'); hold on
histogram(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');