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Python 如何从这个线性回归中得到r_值?_Python_Machine Learning_Scikit Learn_Linear Regression - Fatal编程技术网

Python 如何从这个线性回归中得到r_值?

Python 如何从这个线性回归中得到r_值?,python,machine-learning,scikit-learn,linear-regression,Python,Machine Learning,Scikit Learn,Linear Regression,我使用下面的python脚本使用线性回归预测一个数字 from selenium import webdriver from webdriver_manager.chrome import ChromeDriverManager from selenium.webdriver.common.by import By from selenium.webdriver.support.ui import WebDriverWait from selenium.webdriver.support imp

我使用下面的python脚本使用线性回归预测一个数字

from selenium import webdriver
from webdriver_manager.chrome import ChromeDriverManager
from selenium.webdriver.common.by import By
from selenium.webdriver.support.ui import WebDriverWait
from selenium.webdriver.support import expected_conditions as EC
from sklearn.linear_model import LinearRegression
import time
import numpy as np
from sklearn.svm import SVR
import pytz
from datetime import datetime
from sys import argv, exit
import os, psutil

################################################
if len(argv) != 5:
  print (argv[0] + '<train count> <timeout(s)> <predict date(Y/M/D)> <predict clock(H:M:S)>')
  exit(2)
X_predict = [(int(datetime.strptime(argv[3] + " " + argv[4], '%Y/%m/%d %H:%M:%S').timestamp()*(10000000)))]
################################################

X=[]
y=[]
chromeOptions = webdriver.ChromeOptions()
chromeOptions.add_argument("--headless")
chromeOptions.add_argument("--remote-debugging-port=2212")
chromeOptions.add_argument('--no-sandbox')
chromeOptions.add_argument('--disable-dev-shm-usage')
driver = webdriver.Chrome('/usr/bin/chromedriver',chrome_options=chromeOptions)
driver.get('https://sample.com/')

elem_xpath = '//div[contains(text(), "number")]/following-sibling::div'

for i in range(1, int(argv[1])):
    try:
        elem = WebDriverWait(driver, 10).until(EC.visibility_of_element_located((By.XPATH, elem_xpath)))
        print ("train => ", i)
        X.append(int(time.time()*(10000000)))
        y.append(int(elem.text.replace(',', '')))
        time.sleep(int(argv[2]))
    finally:
        driver.quit
        
##############################################
X = np.array(X).reshape(-1, 1)
y = np.array(y).reshape(-1, 1)
X_predict = np.array(X_predict).reshape(-1, 1)
##############################################
    
svr_rbf = LinearRegression()
y_rbf = svr_rbf.fit(X,y).predict(X_predict)


print ('y_rbf: {}'.format(int(y_rbf)))
print('memory usage: {} MB'.format(
int(psutil.Process(os.getpid()).memory_info().rss/1024/1024)
)) 

从selenium导入webdriver
从webdriver_manager.chrome导入ChromeDriverManager
从selenium.webdriver.common.by导入
从selenium.webdriver.support.ui导入WebDriverWait
从selenium.webdriver.support将预期的_条件导入为EC
从sklearn.linear\u模型导入线性回归
导入时间
将numpy作为np导入
从sklearn.svm导入SVR
进口皮茨
从日期时间导入日期时间
从系统导入argv中退出
导入操作系统,psutil
################################################
如果len(argv)!=5:
打印(argv[0]+'')
出口(2)
X_predict=[(int(datetime.strtime(argv[3]+“”+argv[4],“%Y/%m/%d%H:%m:%S”)。timestamp()*(10000000))]
################################################
X=[]
y=[]
chromeOptions=webdriver.chromeOptions()
chromeOptions.add_参数(“--headless”)
添加参数(“--远程调试端口=2212”)
chromeOptions.add_参数('--no sandbox')
chromeOptions.add_参数('--disable dev shm usage')
driver=webdriver.Chrome('/usr/bin/chromedriver',Chrome\u options=chromeOptions)
司机,上车https://sample.com/')
elem_xpath='//div[包含(text(),“number”)]/以下同级::div'
对于范围(1,int(argv[1])中的i:
尝试:
elem=WebDriverWait(driver,10).until(位于((By.XPATH,elem_XPATH))的元素的EC.visibility_)
打印(“列车=>”,i)
X.append(int(time.time()*(10000000)))
y、 追加(int(elem.text.replace(',',''))
睡眠时间(int(argv[2]))
最后:
司机,退出
##############################################
X=np.数组(X).重塑(-1,1)
y=np.数组(y).重塑(-1,1)
X_predict=np.数组(X_predict).重塑(-1,1)
##############################################
svr_rbf=线性回归()
y_rbf=svr_rbf.拟合(X,y).预测(X_预测)
打印('y_rbf:{}'。格式(int(y_rbf)))
打印('内存使用:{}MB'。格式(
int(psutil.Process(os.getpid()).memory_info().rss/1024/1024)
)) 
代码运行良好。 据我所知,我们可以从线性回归中得到一些输出,如斜率、截距、r_值、p_值和标准误差。r_值将告诉您线性回归的准确性。
在上面的脚本中,如何获得r_值?

这有点难说,但我认为您正在寻找类似的值

import numpy as np
import pandas as pd
import matplotlib.pyplot as plt

import statsmodels.api as sm
from statsmodels.sandbox.regression.predstd import wls_prediction_std

np.random.seed(9876789)

# OLS estimation¶
# Artificial data:

nsample = 100
x = np.linspace(0, 10, 100)
X = np.column_stack((x, x**2))
beta = np.array([1, 0.1, 10])
e = np.random.normal(size=nsample)

# Our model needs an intercept so we add a column of 1s:
X = sm.add_constant(X)
y = np.dot(X, beta) + e

# Fit and summary:
model = sm.OLS(y, X)
results = model.fit()
print(results.summary())
结果:


欢迎来到SO。强烈建议,在匆忙提出问题之前,您执行一些操作,始终从相关的开始。非常感谢,我知道如何正常获取这些信息,但我不知道如何为我上面提供的代码执行这些操作。请参阅下面的链接。