回归评价指标MSE、RMSE、MAE、MAPE及python实现

回归评价指标公式

假设：
预测值：$\hat{y}=\{\widehat{y_1},\widehat{y_2},\dots ,\widehat{y_n}\}$
真实值：$y=\{y_1,y_2,\dots ,y_n\}$

MSE

均方误差 – Mean Square Error
$$MSE=\frac{1}{n}\sum _{i=1}^n(\widehat{y_i}-y_i{)}^2$$

RMSE

均方根误差 – Root Mean Square Error
$$RMSE=\sqrt{\frac{1}{n}\sum _{i=1}^n(\widehat{y_i}-y_i{)}^2}$$

MAE

平均绝对误差 – Mean Absolute Error
$$MAE=\frac{1}{n}\sum _{i=1}^n|\widehat{y_i}-y_i|$$

MAPE

平均绝对百分比误差 – Mean Absolute Percentage Error
$$MAPE=\frac{100\%}{n}\sum _{i=1}^n|\frac{\widehat{y_i}-y_i}{y_i}|$$

python实现

numpy

numpy的范数可以实现 MSE、RMSE、MAE，也可以都自己写。

numpy范数说明：
x_norm = np.linalg.norm(x, ord=None, axis=None, keepdims=False)

import numpy as np
y = np.array([1,1])
y_hat = np.array([2,3])
## 利用范数实现
## MSE--相当于y-y_hat的二阶范数的平方/n
MSE = np.linalg.norm(y-y_hat, ord=2)**2/len(y)
## RMSE--相当于y-y_hat的二阶范数/根号n
RMSE = np.linalg.norm(y-y_hat, ord=2)/len(y)**0.5
## MAE--相当于y-y_hat的一阶范数/n
MAE = np.linalg.norm(y-y_hat, ord=1)/len(y)

## 自己写
MSE = np.mean(np.square(y - y_hat))
RMSE = np.sqrt(np.mean(np.square(y - y_hat)))
MAE = np.mean(np.abs(y-y_hat))
MAPE = np.mean(np.abs((y - y_hat) / y)) * 100

sklearn

sklearn.metrics 可以实现 MSE、RMSE、MAE

from sklearn import metrics
y = np.array([1,1])
y_hat = np.array([2,3])
MSE = metrics.mean_squared_error(y, y_hat)
RMSE = metrics.mean_squared_error(y, y_hat)**0.5
MAE = metrics.mean_absolute_error(y, y_hat)
MAPE = metrics.mean_absolute_percentage_error(y, y_hat)