我的r平方得分为负,但我使用k折交叉验证的准确度评分达到92%左右

Question

对于下面的代码,我的r平方得分是负面的,但我使用k折交叉验证的准确度得分为92%.这怎么可能？我使用随机森林回归算法来预测一些数据.数据集的链接在以下链接中给出:https: //www.kaggle.com/ludobenistant/hr-analytics

import numpy as np
import pandas as pd
from sklearn.preprocessing import LabelEncoder,OneHotEncoder

dataset = pd.read_csv("HR_comma_sep.csv")
x = dataset.iloc[:,:-1].values   ##Independent variable
y = dataset.iloc[:,9].values     ##Dependent variable

##Encoding the categorical variables

le_x1 = LabelEncoder()
x[:,7] = le_x1.fit_transform(x[:,7])
le_x2 = LabelEncoder()
x[:,8] = le_x1.fit_transform(x[:,8])
ohe = OneHotEncoder(categorical_features = [7,8])
x = ohe.fit_transform(x).toarray()


##splitting the dataset in training and testing data

from sklearn.cross_validation import train_test_split
y = pd.factorize(dataset['left'].values)[0].reshape(-1, 1)
x_train, x_test, y_train, y_test = train_test_split(x, y, test_size = 0.2, random_state = 0)

from sklearn.preprocessing import StandardScaler
sc_x = StandardScaler()
x_train = sc_x.fit_transform(x_train)
x_test = sc_x.transform(x_test)
sc_y = StandardScaler()
y_train = sc_y.fit_transform(y_train)

from sklearn.ensemble import RandomForestRegressor
regressor = RandomForestRegressor(n_estimators = 10, random_state = 0)
regressor.fit(x_train, y_train)

y_pred = regressor.predict(x_test)
print(y_pred)
from sklearn.metrics import r2_score
r2_score(y_test , y_pred)

from sklearn.model_selection import cross_val_score
accuracies = cross_val_score(estimator = regressor, X = x_train, y = y_train, cv = 10)
accuracies.mean()
accuracies.std()

Answer 1

你的问题有几个问题......

对于初学者来说,你正在做一个很基本的错误:你认为你正在使用精度指标,而你在回归设置,下面实际使用的度量是均方误差(MSE).

准确性是分类中使用的度量,它与正确分类的示例的百分比有关 - 请查看Wikipedia条目以获取更多详细信息.

您选择的回归程序(随机森林)内部使用的度量标准包含在regressor.fit(x_train, y_train)命令的详细输出中- 请注意criterion='mse'参数:

RandomForestRegressor(bootstrap=True, criterion='mse', max_depth=None,
           max_features='auto', max_leaf_nodes=None,
           min_impurity_split=1e-07, min_samples_leaf=1,
           min_samples_split=2, min_weight_fraction_leaf=0.0,
           n_estimators=10, n_jobs=1, oob_score=False, random_state=0,
           verbose=0, warm_start=False)

MSE是一个正的连续数量,它不是上限为1,即如果你得到0.92的值,这意味着......好吧,0.92,而不是 92%.

知道这一点,最好明确包含MSE作为交叉验证的评分函数:

cv_mse = cross_val_score(estimator = regressor, X = x_train, y = y_train, cv = 10, scoring='neg_mean_squared_error')
cv_mse.mean()
# -2.433430574463703e-28

出于所有实际目的,这是零 - 你几乎完美地适应训练集; 为了确认,这里是你训练集上的(完美再次)R平方分数:

train_pred = regressor.predict(x_train)
r2_score(y_train , train_pred)
# 1.0

但是,与往常一样,当您在测试集上应用模型时,真实的时刻就来了; 你的第二个错误是,既然你用缩放训练你的回归量y_train,你也应该y_test在评估之前进行缩放:

y_test = sc_y.fit_transform(y_test)
r2_score(y_test , y_pred)
# 0.9998476914664215

你在测试集中得到一个非常好的R平方(接近1).

那么MSE呢？

from sklearn.metrics import mean_squared_error
mse_test = mean_squared_error(y_test, y_pred)
mse_test
# 0.00015230853357849051