使用 python 进行线性回归的简单预测
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【中文标题】使用 python 进行线性回归的简单预测【英文标题】：Simple prediction using linear regression with python 【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】：
【参考方案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？【参考方案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】：
【参考方案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？【参考方案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()
欲了解更多信息，请访问
https://github.com/wins999/Cricket_Chirps_Vs_Temprature--Simple-Linear-Regression-in-Python-
【讨论】：
【参考方案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】：
【参考方案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】：
【参考方案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】：
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