python实现K折交叉验证

 import numpy as np from sklearn import datasets from sklearn.neighbors import KNeighborsClassifier from sklearn.model_selection import KFold # 主要用于K折交叉验证 # 导入iris数据集 iris = datasets.load_iris() X = iris.data y = iris.target print(X.shape,y.shape) # 定义想要搜索的K值，这里定义8个不同的值 ks = [1,3,5,7,9,11,13,15] # 进行5折交叉验证，KFold返回的是每一折中训练数据和验证数据的index # 假设数据样本为：[1,3,5,6,11,12,43,12,44,2],总共10个样本 # 则返回的kf的格式为（前面的是训练数据，后面的验证集）： # [0,1,3,5,6,7,8,9],[2,4] # [0,1,2,4,6,7,8,9],[3,5] # [1,2,3,4,5,6,7,8],[0,9] # [0,1,2,3,4,5,7,9],[6,8] # [0,2,3,4,5,6,8,9],[1,7] kf = KFold(n_splits = 5, random_state=2001, shuffle=True) # 保存当前最好的k值和对应的准确率 best_k = ks[0] best_score = 0 # 循环每一个k值 for k in ks: curr_score = 0 for train_index,valid_index in kf.split(X): # 每一折的训练以及计算准确率 clf = KNeighborsClassifier(n_neighbors=k) cl<span style="color:transparent">来源gaodai#ma#com搞*!代#%^码$网</span>f.fit(X[train_index],y[train_index]) curr_score = curr_score + clf.score(X[valid_index],y[valid_index]) # 求一下5折的平均准确率 avg_score = curr_score/5 if avg_score > best_score: best_k = k best_score = avg_score print("current best score is :%.2f" % best_score,"best k:%d" %best_k) print("after cross validation, the final best k is :%d" %best_k)