Question

我有一个问题，我正在尝试实施KFold和cross_val_score。我的目标是计算mean_squared_error，为此我使用了以下代码：

from sklearn import linear_model
import numpy as np
from sklearn.metrics import mean_squared_error
from sklearn.model_selection import KFold, cross_val_score

x = np.random.random((10000,20))
y = np.random.random((10000,1))

x_train = x[7000:]
y_train = y[7000:]

x_test = x[:7000]
y_test = y[:7000]

Model = linear_model.LinearRegression()
Model.fit(x_train,y_train)

y_predicted  = Model.predict(x_test)

MSE = mean_squared_error(y_test,y_predicted)
print(MSE)

kfold = KFold(n_splits = 100, random_state = None, shuffle = False)

results = cross_val_score(Model,x,y,cv=kfold, scoring='neg_mean_squared_error')
print(results.mean())

我认为没关系，我得到了以下结果：

结果：0.0828856459279和 - 0.083069435946

但是当我尝试在其他一些例子（来自Kaggle House价格的数据）上这样做时，它不能正常工作，至少我是这么认为的。

train = pd.read_csv('train.csv')

Insert missing values...
...

train = pd.get_dummies(train)
y = train['SalePrice']
train = train.drop(['SalePrice'], axis = 1)

x_train = train[:1000].values.reshape(-1,339)
y_train = y[:1000].values.reshape(-1,1)
y_train_normal = np.log(y_train)

x_test = train[1000:].values.reshape(-1,339)
y_test = y[1000:].values.reshape(-1,1)

Model = linear_model.LinearRegression()
Model.fit(x_train,y_train_normal)

y_predicted = Model.predict(x_test)
y_predicted_transform = np.exp(y_predicted)

MSE = mean_squared_error(y_test, y_predicted_transform)
print(MSE)

kfold = KFold(n_splits = 10, random_state = None, shuffle = False)

results = cross_val_score(Model,train,y, cv = kfold, scoring = "neg_mean_squared_error")
print(results.mean())

我在这里得到以下结果：0.912874946869和-6.16986926564e+16

显然，mean_squared_error计算的'手动'与mean_squared_error在KFold的帮助下计算的不同。

我对我犯错误的地方感兴趣？

Answer 1

差异是因为，与您的第一种方法（训练/测试集）相比，在您的简历方法中，您使用非标准化 y数据来拟合回归，因此您的MSE。要获得可比较的结果，您应该执行以下操作：

y_normal = np.log(y)
y_test_normal = np.log(y_test)

MSE = mean_squared_error(y_test_normal, y_predicted) # NOT y_predicted_transform
results = cross_val_score(Model, train, y_normal, cv = kfold, scoring = "neg_mean_squared_error")

Python手动和cross_val_score预测的结果不同

1 个答案: