回归评估指标-Python实现

常用回归模型评估指标的python计算逻辑。python和相关package版本：

import sys
import sklearn
import pandas as pd
import numpy as np
import math 
print("Python版本：",sys.version)
print("sklearn版本：",sklearn.__version__)
print("pandas：",pd.__version__)
print("numpy：",np.__version__)

Python版本： 3.7.0 (default, Jun 28 2018, 08:04:48) [MSC v.1912 64 bit (AMD64)]
sklearn版本： 0.22
pandas： 0.23.4
numpy： 1.17.4

MAE

from sklearn.metrics import mean_absolute_error
y_true = [3, -0.5, 2, 7]
y_pred = [2.5, 0.0, 2, 8]
mean_absolute_error(y_true, y_pred)

0.5

sum([abs(y_true[i]-y_pred[i]) for i in range(len(y_true))])/len(y_true)

0.5

MSE

from sklearn.metrics import mean_squared_error
y_true = [3, -0.5, 2, 7]
y_pred = [2.5, 0.0, 2, 8]
mean_squared_error(y_true, y_pred)

0.375

sum([abs(y_true[i]-y_pred[i])**2 for i in range(len(y_true))])/len(y_true)

0.375

RMSE

这个没啥说的，mse开方就行

math.sqrt(mean_squared_error(y_true, y_pred))

0.6123724356957945

R方

from sklearn.metrics import r2_score
r2_score(y_true, y_pred)

0.9486081370449679

1 - sum([(y_true[i] - y_pred[i])**2 for i in range(len(y_true))]) / sum(
    [(y_true[i] - sum(y_true) / len(y_true))**2 for i in range(len(y_true))])

0.9486081370449679

整合

  mape由于可能存在分母为0的情况，因此暂时没有统计。把其他指标整合一下，格式化输出：

from sklearn.metrics import mean_absolute_error
from sklearn.metrics import mean_squared_error
from sklearn.metrics import r2_score
def reg_metric(y_true, y_predict):
    mae = mean_absolute_error(y_true, y_pred)
    mse = mean_squared_error(y_true, y_pred)
    rmse = math.sqrt(mean_squared_error(y_true, y_pred))
    r2 = r2_score(y_true, y_pred)
    print("MAE:",mae)
    print("MSE:",mse)
    print("RMSE:",rmse)
    print("R Square:",r2)

reg_metric(y_true=y_true, y_predict=y_pred)

MAE: 0.5
MSE: 0.375
RMSE: 0.6123724356957945
R Square: 0.9486081370449679

                                2020-06-16、连绵梅雨季 于南京市江宁区九龙湖