如何在Python的Randomforest模型中获得准确性

Question

我得到了这个脚本，该脚本用Randomforest和LinearRegression预测秒数据集的值。正常工作，线性回归的准确性为18％，太糟糕了。

。

所以我要尝试随机前景，但是我不知道如何计算该模型的准确性。

import pandas as pd
from sklearn.ensemble import RandomForestRegressor
from sklearn.datasets import make_regression
import numpy as np
import pandas as pd
import scipy
import matplotlib.pyplot as plt
from pylab import rcParams
import urllib
import sklearn
from sklearn.linear_model import RidgeCV, LinearRegression, Lasso
from sklearn.preprocessing import StandardScaler, PolynomialFeatures
from sklearn.model_selection import GridSearchCV
data = pd.read_csv('EncuestaVieja.csv')
X = data[['Edad','Sexo','v1','v2','v3']]
y = data['Alumna']
dataP = pd.read_csv('EncuestaVieja_test.csv')
X_p = dataP[['Edad','Sexo','v1','v2','v3']]
y_p = dataP['Alumna']
dataT = pd.read_csv('EncuestaVieja_test_2.csv')
X_t = dataT[['Edad','Sexo','v1','v2','v3']]
y_t = dataT['Alumna']
regr = linear_model.LinearRegression()
regr.fit(X, y)
lr = RandomForestRegressor(n_estimators=50)
lr.fit(X, y)
X_test = pd.read_csv('EncuestaNueva.csv')[['Edad','Sexo','v1','v2','v3']]
predictions = regr.predict(X_test)

predictions2 = lr.predict(X_test)
print( 'RandomForest Accuracy: ')
print(((predictions2)))
print( '')
from sklearn.linear_model import LinearRegression
regressor = LinearRegression()
regressor.fit(X_p,y_p)
accuracy = regressor.score(X_t,y_t)
print( 'Linear Regression Accuracy: ', accuracy*100,'%')
print(((predictions)))

输出：

RandomForest Accuracy: 
[ 1.64  2.54  2.6   2.38  1.64  1.32  1.68  2.56  3.    2.28  2.38  2.68
  2.9   2.5   2.78  1.96  1.56  2.6   2.12  2.76  2.74  1.66  1.68  2.12
  2.3   2.36  2.28  2.28  2.82  1.7   1.86  2.36  1.24]
Linear Regression Accuracy:  18.1336149086 %
[ 1.2681851   1.02802219  3.13377072  2.96885127  2.30808853  1.98814349
  2.39233726  2.8638321   1.86640316  2.63073399  2.21166731  2.25201016
  1.95065189  2.65360517  3.08855254  1.01229733  2.18225606  2.41802534
  2.43539473  2.50227407  1.71105799  1.88238089  2.12152321  3.33525397
  2.72820453  2.43241713  2.88757874  2.6242382   2.63087916  1.98379487
  2.25430306  1.96810279  0.8554685 ]

Answer 1

我认为这是用score((方法来处理的

lr.score(x_test, y_test)

这将返回您的模型的R^2值。好像在您的情况下，您只有x_test。请注意，这不是准确性。回归模型不使用诸如分类模型之类的精度。取而代之的是计算出不同的指标，例如均值误差或确定系数。这些指标可以表明准确预测值匹配已知值或回归模型如何拟合回归线。