使用 python 进行线性回归的简单预测答案

【问题标题】：Simple prediction using linear regression with python使用 python 进行线性回归的简单预测
【发布时间】：2015-06-19 19:36:20
【问题描述】：

data2 = pd.DataFrame(data1['kwh'])
data2
                          kwh
date    
2012-04-12 14:56:50     1.256400
2012-04-12 15:11:55     1.430750
2012-04-12 15:27:01     1.369910
2012-04-12 15:42:06     1.359350
2012-04-12 15:57:10     1.305680
2012-04-12 16:12:10     1.287750
2012-04-12 16:27:14     1.245970
2012-04-12 16:42:19     1.282280
2012-04-12 16:57:24     1.365710
2012-04-12 17:12:28     1.320130
2012-04-12 17:27:33     1.354890
2012-04-12 17:42:37     1.343680
2012-04-12 17:57:41     1.314220
2012-04-12 18:12:44     1.311970
2012-04-12 18:27:46     1.338980
2012-04-12 18:42:51     1.357370
2012-04-12 18:57:54     1.328700
2012-04-12 19:12:58     1.308200
2012-04-12 19:28:01     1.341770
2012-04-12 19:43:04     1.278350
2012-04-12 19:58:07     1.253170
2012-04-12 20:13:10     1.420670
2012-04-12 20:28:15     1.292740
2012-04-12 20:43:15     1.322840
2012-04-12 20:58:18     1.247410
2012-04-12 21:13:20     0.568352
2012-04-12 21:28:22     0.317865
2012-04-12 21:43:24     0.233603
2012-04-12 21:58:27     0.229524
2012-04-12 22:13:29     0.236929
2012-04-12 22:28:34     0.233806
2012-04-12 22:43:38     0.235618
2012-04-12 22:58:43     0.229858
2012-04-12 23:13:43     0.235132
2012-04-12 23:28:46     0.231863
2012-04-12 23:43:55     0.237794
2012-04-12 23:59:00     0.229634
2012-04-13 00:14:02     0.234484
2012-04-13 00:29:05     0.234189
2012-04-13 00:44:09     0.237213
2012-04-13 00:59:09     0.230483
2012-04-13 01:14:10     0.234982
2012-04-13 01:29:11     0.237121
2012-04-13 01:44:16     0.230910
2012-04-13 01:59:22     0.238406
2012-04-13 02:14:21     0.250530
2012-04-13 02:29:24     0.283575
2012-04-13 02:44:24     0.302299
2012-04-13 02:59:25     0.322093
2012-04-13 03:14:30     0.327600
2012-04-13 03:29:31     0.324368
2012-04-13 03:44:31     0.301869
2012-04-13 03:59:42     0.322019
2012-04-13 04:14:43     0.325328
2012-04-13 04:29:43     0.306727
2012-04-13 04:44:46     0.299012
2012-04-13 04:59:47     0.303288
2012-04-13 05:14:48     0.326205
2012-04-13 05:29:49     0.344230
2012-04-13 05:44:50     0.353484
...

65701 rows × 1 columns

我有这个索引和 1 列的数据框。我想使用线性回归和 sklearn 进行简单的预测。我很困惑，我不知道如何设置 X 和 y（我希望 x 值是时间和 y 值 kwh...)。我是 Python 新手，所以每一个帮助都很有价值。谢谢。

【问题讨论】：

标签： python scikit-learn linear-regression

【解决方案1】：

您要做的第一件事是将数据拆分为两个数组，X 和 y。 X 的每个元素都是一个日期，y 的对应元素是相关的 kwh。

一旦你有了它，你会想要使用 sklearn.linear_model.LinearRegression 来做回归。文档是here。

对于每个 sklearn 模型，有两个步骤。首先，您必须适合您的数据。然后，将要预测kwh的日期放入另一个数组X_predict中，并使用predict方法预测kwh。

from sklearn.linear_model import LinearRegression

X = []  # put your dates in here
y = []  # put your kwh in here

model = LinearRegression()
model.fit(X, y)

X_predict = []  # put the dates of which you want to predict kwh here
y_predict = model.predict(X_predict)

【讨论】：

预测给出了什么？结果数组中的数字是多少？

【解决方案2】：

Predict() 函数将二维数组作为参数。所以，如果你想预测简单线性回归的值，那么你必须在二维数组内发出预测值，例如，

model.predict([[2012-04-13 05:55:30]]);

如果是多元线性回归，那么，

model.predict([[2012-04-13 05:44:50,0.327433]])

【讨论】：

【解决方案3】：

线性回归：

import pandas as pd  
import numpy as np  
import matplotlib.pyplot as plt  
data=pd.read_csv('Salary_Data.csv')  
X=data.iloc[:,:-1].values  
y=data.iloc[:,1].values  

#split dataset in train and testing set   
from sklearn.cross_validation import train_test_split  
X_train,X_test,Y_train,Y_test=train_test_split(X,y,test_size=10,random_state=0)  

from sklearn.linear_model import LinearRegression  
regressor=LinearRegression()  
regressor.fit(X_train,Y_train)  
y_pre=regressor.predict(X_test)

【讨论】：

能否进一步解释如何选择数据，因为这也是问题的一部分？

【解决方案4】：

您可以查看我在 Github 上的代码，其中我使用带有简单线性回归模型的昆虫蟋蟀的啁啾声来预测温度。我已经用 cmets 解释了代码

#Import the libraries required
import numpy as np
import matplotlib.pyplot as plt
import pandas as pd

#Importing the excel data 
dataset = pd.read_excel('D:\MachineLearing\Machine Learning A-Z Template Folder\Part 2 - Regression\Section 4 - Simple Linear Regression\CricketChirpsVs.Temperature.xls')

x = dataset.iloc[:, :-1].values
y = dataset.iloc[:, 1].values

#Split the data into train and test dataset
from sklearn.cross_validation import train_test_split
x_train,x_test,y_train,y_test=train_test_split(x,y,test_size=1/3,random_state=42)

#Fitting Simple Linear regression data model to train data set
from sklearn.linear_model import LinearRegression
regressorObject=LinearRegression()
regressorObject.fit(x_train,y_train)

#predict the test set
y_pred_test_data=regressorObject.predict(x_test)


# Visualising the Training set results in a scatter plot
plt.scatter(x_train, y_train, color = 'red')
plt.plot(x_train, regressorObject.predict(x_train), color = 'blue')
plt.title('Cricket Chirps vs Temperature (Training set)')
plt.xlabel('Cricket Chirps (chirps/sec for the striped ground cricket) ')
plt.ylabel('Temperature (in degrees Fahrenheit)')
plt.show()

# Visualising the test set results in a scatter plot
plt.scatter(x_test, y_test, color = 'red')
plt.plot(x_train, regressorObject.predict(x_train), color = 'blue')
plt.title('Cricket Chirps vs Temperature (Test set)')
plt.xlabel('Cricket Chirps (chirps/sec for the striped ground cricket) ')
plt.ylabel('Temperature (in degrees Fahrenheit)')
plt.show()

【讨论】：

【解决方案5】：

将数据集拆分为训练集和测试集后

from sklearn.model_selection import train_test_split
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size = 0.2, random_state =0)

在训练集上训练您的简单线性回归模型

from sklearn.linear_model import LinearRegression
regressor = LinearRegression()
regressor.fit(X_train, y_train)

预测测试集结果

y_predict = regressor.predict(X_test)

【讨论】：

【解决方案6】：

您应该实现以下代码。

import pandas as pd
from sklearn.linear_model import LinearRegression # to build linear regression model
from sklearn.cross_validation import train_test_split # to split dataset

data2 = pd.DataFrame(data1['kwh'])
data2 = data2.reset_index() # will create new index (0 to 65700) so date column wont be an index now.
X = data2.iloc[:,0]   # date column
y = data2.iloc[:,-1]  # kwh column

Xtrain, Xtest, ytrain, ytest = train_test_split(X, y, train_size=0.80, random_state=20)  

linearModel = LinearRegression()
linearModel.fit(Xtrain, ytrain)
ypred = model.predict(Xtest)

这里 ypred 会给你概率。

【讨论】：

【解决方案7】：

以防万一有人正在寻找没有 sklearn 的解决方案

import numpy as np
import pandas as pd

def variance(values, mean):
    return sum([(val-mean)**2 for val in values])

def covariance(x, mean_x, y , mean_y):
    covariance = 0.0
    for r in range(len(x)):
        covariance = covariance + (x[r] - mean_x) * (y[r] - mean_y)
    return covariance

def get_coef(df):
    mean_x = sum(df['x']) / float(len(df['x']))
    mean_y = sum(df['y']) / float(len(df['y']))
    variance_x = variance(df['x'], mean_x)
    #variance_y = variance(df['y'], mean_y)
    covariance_x_y = covariance(df['x'],mean_x,df['y'],mean_y)
    m = covariance_x_y / variance_x
    c = mean_y - m * mean_x
    return m,c

def get_y(x,m,c):
    return m*x+c

灵感来自https://github.com/dhirajk100/Linear-Regression-from-Scratch-in-Python/blob/master/Linear%20Regression%20%20from%20Scratch%20Without%20Sklearn.ipynb

【讨论】：