K折交叉验证

import pandas as pd
#读取数据集
datas = pd.read_csv('wdbc.csv',header=None)
#使用LabelEncoder类将类标转换为整数
from sklearn.preprocessing import LabelEncoder

X = datas.loc[:,2:].values
Y = datas.loc[:,1].values

le = LabelEncoder()
y = le.fit_transform(Y)

#划分训练数据集
from sklearn.model_selection import train_test_split

x_train,x_test,y_train,y_test = train_test_split(X,y,test_size=0.2)

from sklearn.preprocessing import StandardScaler
from sklearn.decomposition import PCA
from sklearn.linear_model import LogisticRegression
from sklearn.pipeline import  Pipeline
#n_components参数为所要保留的主成分个数，=2降到2维。
pipe = Pipeline([('standard',StandardScaler()),
                 ('PCA',PCA(n_components=2)),
                 ('clf',LogisticRegression(random_state=1,solver='liblinear'))])
# pipe.fit(x_train,y_train)
# print('Test:%.3f' %pipe.score(x_test,y_test))

#分层交叉验证
from sklearn.model_selection import StratifiedKFold
#n_folds交叉验证的份数，10交叉验证法

import numpy as np#n_splits分割数量
kfold = StratifiedKFold(n_splits=10)
kfold = kfold.split(x_train,y_train)
scores = []

for k , (train,test) in enumerate(kfold):
    pipe.fit(x_train[train],y_train[train])
    score = pipe.score(x_train[test],y_train[test])
    scores.append(score);
    #bincount每个索引出现的次数(0和1出现的个数，这里指类型出现个数)

    print('fold: %s,class num: %s,ACC: %.3f'%(k+1,np.bincount(y_train[train]),score))#计算平均值
print('mean acc:%.3f'%np.mean(scores))

#2

from sklearn.model_selection import cross_val_score

scoress = cross_val_score(estimator=pipe,
                          X=x_train,
                          y=y_train,
                          cv=10,
                          n_jobs=1)

print('ACC:%s'%scoress)

来源：CSDN

作者：hamimelon2020

链接：https://blog.csdn.net/weixin_40945354/article/details/104268892