交叉验证决策树

编程问答 2022-06-01

问题描述

在创建一个决策树函数后，我决定检查树的准确度，并确认如果我用相同的数据制作另一棵树，至少第一次分割是相同的

from sklearn.model_selection import train_test_split
import pandas as pd
import numpy as np
import os
from sklearn import tree
from sklearn import preprocessing
import sys
from sklearn.tree import DecisionTreeClassifier
from sklearn.model_selection import cross_val_score
from sklearn.model_selection import KFold

.....

def desicion_tree(data_set:pd.DataFrame,val_1 : str,val_2 : str):
    #Encoder  -- > fit doesn't accept strings
    feature_cols = data_set.columns[0:-1]
    X = data_set[feature_cols] # Independent variables
    y = data_set.Mut #class
    y = y.to_list()
    le = preprocessing.LabelBinarizer()
    y = le.fit_transform(y)
    # Split data set into training set and test set
    X_train,X_test,y_train,y_test = train_test_split(X,y,test_size=0.25,random_state=1) # 75% 
    # Create Decision Tree classifer object
    clf = DecisionTreeClassifier(max_depth= 4,criterion= 'entropy')
    # Train Decision Tree Classifer
    clf.fit(X_train,y_train)
    # Predict the response for test dataset
    y_pred = clf.predict(X_test)
    #Perform cross validation
    for i in range(2,8):
        plt.figure(figsize=(14,7))
        # Perform Kfold cross validation
        #cv = ShuffleSplit(test_size=0.25,random_state=0)
        kf = KFold(n_splits=5,shuffle= True)
        scores = cross_val_score(estimator=clf,X=X,y=y,n_jobs=4,cv=kf)
        print("%0.2f accuracy with a standard deviation of %0.2f" % (scores.mean(),scores.std()))
        tree.plot_tree(clf,filled = True,feature_names=feature_cols,class_names=[val_1,val_2])
        plt.show()
desicion_tree(car_rep_sep_20,'Categorial','Non categorial')

Down ，我编写了一个循环，以便使用 Kfold 重新创建具有拆分值的树。准确率在变化（大约 90%）但树是一样的，我哪里弄错了？

解决方法

cross_val_score 克隆估计器，以便在各种折叠上进行拟合和评分，因此 clf 对象保持与循环前将其拟合到整个数据集时相同，因此绘制的树是那个树，而不是任何交叉验证的树。

为了得到您想要的结果，我认为您可以将 cross_validate 与选项 return_estimator=True 一起使用。如果您的 cv 对象具有所需的拆分数量，您也不应该需要循环：

kf = KFold(n_splits=5,shuffle=True)
cv_results = cross_validate(
    estimator=clf,X=X,y=y,n_jobs=4,cv=kf,return_estimator=True,)
print("%0.2f accuracy with a standard deviation of %0.2f" % (
    cv_results['test_score'].mean(),cv_results['test_score'].std(),))
for est in cv_results['estimator']:
    tree.plot_tree(est,filled=True,feature_names=feature_cols,class_names=[val_1,val_2])
    plt.show();

或者，在折叠（或其他 cv 迭代）上手动循环，拟合模型并在循环中绘制其树。

for-loopmachine-learningpythontraining-data