《Python数据分析和应用》实验四-使用scikit-learn-构建模型【精选推荐】

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　实验四

　使用 scikit- -n learn 构建模型

　教材 P196

　实训 1- 实训 4 1 、实训 1 # 读数据 import pandas as pd wine=pd.read_csv("D:\\桌面\\实验四\\data\\wine.csv") winequality=pd.read_csv("D:\\桌面\\实验四\\data\\winequality.csv",sep=";") # 数据和标签拆分开 wine_data=wine.iloc[:,1:] wine_target=wine["Class"] print("wine 数据集的数据为：\n",wine_data) print("wine 数据集的标签为：\n",wine_target)

　winequality_data=winequality.iloc[:,:-1] winequality_target=winequality["quality"] print("winequality 数据集的数据为：\n",winequality_data) print("winequality 数据集的标签为：\n",winequality_target)

　 # 划分训练集和测试集 from sklearn.model_selection import train_test_split wine_data_train, wine_data_test, \ wine_target_train, wine_target_test = \ train_test_split(wine_data, wine_target, \ test_size=0.1, random_state=6) winequality_data_train, winequality_data_test, \ winequality_target_train, winequality_target_test = \ train_test_split(winequality_data, winequality_target, \ test_size=0.1, random_state=6) # 标准化数据集 from sklearn.preprocessing import StandardScaler stdScale = StandardScaler().fit(wine_data_train)

　wine_trainScaler = stdScale.transform(wine_data_train) wine_testScaler = stdScale.transform(wine_data_test)

　stdScale = StandardScaler().fit(winequality_data_train)

　winequality_trainScaler = stdScale.transform(winequality_data_train) winequality_testScaler = stdScale.transform(winequality_data_test) #PCA 降维 from sklearn.decomposition import PCA pca = PCA(n_components=5).fit(wine_trainScaler)

　wine_trainPca = pca.transform(wine_trainScaler)

　wine_testPca = pca.transform(wine_testScaler)

　pca = PCA(n_components=5).fit(winequality_trainScaler) winequality_trainPca = pca.transform(winequality_trainScaler)

　winequality_testPca = pca.transform(winequality_testScaler) 2 、实训 2 # 根据实训 1 的 的 wine 数据集处理的结果, 构建聚类数目为 3 的 K-Means 模型 from sklearn.cluster import Kmeans kmeans = KMeans(n_clusters = 3,random_state=1).fit(wine_trainScaler) print("构建的 KMeans 模型为：\n",kmeans)

　# 对比真实标签和聚类标签求取 FMI from sklearn.metrics import fowlkes_mallows_score score=fowlkes_mallows_score(wine_target_train,kmeans.labels_) print("wine 数据集的 FMI:%f"%(score))

　# 在聚类数目为 2~10 类时, 确定最优聚类数目 for i in range(2,11):

　 kmeans = KMeans(n_clusters = i,random_state=123).fit(wine_trainScaler)

　 score = fowlkes_mallows_score(wine_target_train,kmeans.labels_) print("iris 数据聚%d 类 FMI 评价分值为：%f" %(i,score))

　# 求取模型的轮廓系数, 绘制轮廓系数折线图, 确定最 优聚类数目 from sklearn.metrics import silhouette_score import matplotlib.pyplot as plt silhouettteScore = [] for i in range(2,11):

　 kmeans = KMeans(n_clusters = i,random_state=1).fit(wine)

　 score = silhouette_score(wine,kmeans.labels_)

　 silhouettteScore.append(score) plt.figure(figsize=(10,6)) plt.plot(range(2,11),silhouettteScore,linewidth=1.5, linestyle="-") plt.show()

　 # 求取 Calinski-Harabasz 指数, 确定最优聚类数 from sklearn.metrics import calinski_harabaz_score for i in range(2,11):

　 kmeans = KMeans(n_clusters = i,random_state=1).fit(wine)

　 score = calinski_harabaz_score(wine,kmeans.labels_) print("seeds 数据聚%d 类 calinski_harabaz 指数为：%f"%(i,score))

　出现错误，代码没有问题。换了个电脑出现结果;

　3 、实训 3 # 读取 wine 数据集, 区分标签和数据 import pandas as pd wine = pd.read_csv ("D:\\桌面\\实验四\\data\\wine.csv") wine_data=wine.iloc[:,1:] wine_target=wine["Class"] #将 将 wine 数据集划分为训练集和测试集 from sklearn.model_selection import train_test_split wine_data_train, wine_data_test, \ wine_target_train, wine_target_test = \ train_test_split(wine_data, wine_target, \

　test_size=0.1, random_state=6) # 使用离差标准化方法标准化 wine 数据集 from sklearn.preprocessing import MinMaxScaler stdScale = MinMaxScaler().fit(wine_data_train)

　wine_trainScaler = stdScale.transform(wine_data_train) wine_testScaler = stdScale.transform(wine_data_test) # 构建 SVM 模型, 并预测测试集结果。

　from sklearn.svm import SVC

　svm = SVC().fit(wine_trainScaler,wine_target_train) print("建立的 SVM 模型为：\n",svm) wine_target_pred = svm.predict(wine_testScaler) print("预测前 10 个结果为：\n",wine_target_pred[:10])

　#打印出分类报告,评价分类模型性能 from sklearn.metrics import classification_report print("使用 SVM 预测 iris 数据的分类报告为：","\n",

　 classification_report(wine_target_test,

　 wine_target_pred))

　4 、实训 4 # 根据 wine_quality 数据集处理的结果, 构建线性回归模型。

　from sklearn.linear_model import LinearRegression clf = LinearRegression().fit(winequality_trainPca,winequality_target_train) y_pred = clf.predict(winequality_testPca) print("线性回归模型预测前 10 个结果为：","\n",y_pred[:10])

　# 根据 wine_quality 数据集处理的结果, 构建梯度提升回归模型 from sklearn.ensemble import GradientBoostingRegressor GBR_wine = GradientBoostingRegressor().\

　fit(winequality_trainPca,winequality_target_train) wine_target_pred = GBR_wine.predict(winequality_testPca) print("梯度提升回归模型预测前 10 个结果为：","\n",wine_target_pred[:10]) print("真实标签前十个预测结果为：","\n",list(winequality_target_test[:10]))

　# 结合真实评分和预测评分, 计算均方误差、中值绝对误差、可解释方差值 # 根据得分, 判定模型的性能优劣 from sklearn.metrics import mean_absolute_error from sklearn.metrics import mean_squared_error from sklearn.metrics import median_absolute_error from sklearn.metrics import explained_variance_score from sklearn.metrics import

　 r2_score print("线性回归模型评价结果：") print("winequality 数据线性回归模型的平均绝对误差为：",

　mean_absolute_error(winequality_target_test,y_pred)) print("winequality 数据线性回归模型的均方误差为：",

　mean_squared_error(winequality_target_test,y_pred)) print("winequality 数据线性回归模型的中值绝对误差为：",

　median_absolute_error(winequality_target_test,y_pred)) print("winequality 数据线性回归模型的可解释方差值为：",

　explained_variance_score(winequality_target_test,y_pred)) print("winequality 数据线性回归模型的 R 方值为：",

　r2_score(winequality_target_test,y_pred)) print("梯度提升回归模型评价结果：") from sklearn.metrics import explained_variance_score,\ mean_absolute_error,mean_squared_error,median_absolute_error,r2_score print("winequality 数据梯度提升回归树模型的平均绝对误差为：",

　mean_absolute_error(winequality_target_test,wine_target_pred)) print("winequality 数据梯度提升回归树模型的均方误差为：",

　mean_squared_error(winequality_target_test,wine_target_pred)) print("winequality 数据梯度提升回归树模型的中值绝对误差为：",

　median_absolute_error(winequality_target_test,wine_target_pred)) print("winequality 数据梯度提升回归树模型的可解释方差值为：",

　explained_variance_score(winequality_target_test,wine_target_pred)) print("winequality 数据梯度提升回归树模型的 R 方值为：",

　 r2_score(winequality_target_test,wine_target_pred))

　通过对比两者的回归评价指标发现，线性回归模型的平均绝对误差、均方误差、中值绝对误差都要大于梯度回归模型，而可解释方差和 R2 值都要小于梯度回归模型。对于一个回归模型来说，平均绝对误差、均方误差、中值绝对误差越接近 0 越好，可解释方差和 R2 值越接近 1 越好，因此可知，梯度回归模型在五个指标上都要优于线性回归模型，故在本题中，梯度回归模型性能更优。

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