Python中的多元线性回归

我似乎找不到任何进行多元回归的python库。我发现的唯一的东西只是做简单的回归。我需要针对几个自变量（x1，x2，x3等）对我的因变量（y）进行回归。

例如，使用以下数据：

print 'y        x1      x2       x3       x4      x5     x6       x7'
for t in texts:
    print "{:>7.1f}{:>10.2f}{:>9.2f}{:>9.2f}{:>10.2f}{:>7.2f}{:>7.2f}{:>9.2f}" /
   .format(t.y,t.x1,t.x2,t.x3,t.x4,t.x5,t.x6,t.x7)

（以上输出：）

      y        x1       x2       x3        x4     x5     x6       x7
   -6.0     -4.95    -5.87    -0.76     14.73   4.02   0.20     0.45
   -5.0     -4.55    -4.52    -0.71     13.74   4.47   0.16     0.50
  -10.0    -10.96   -11.64    -0.98     15.49   4.18   0.19     0.53
   -5.0     -1.08    -3.36     0.75     24.72   4.96   0.16     0.60
   -8.0     -6.52    -7.45    -0.86     16.59   4.29   0.10     0.48
   -3.0     -0.81    -2.36    -0.50     22.44   4.81   0.15     0.53
   -6.0     -7.01    -7.33    -0.33     13.93   4.32   0.21     0.50
   -8.0     -4.46    -7.65    -0.94     11.40   4.43   0.16     0.49
   -8.0    -11.54   -10.03    -1.03     18.18   4.28   0.21     0.55

我将如何在python中进行回归，以获得线性回归公式：

Y = a1x1 + a2x2 + a3x3 + a4x4 + a5x5 + a6x6 + + a7x7 + c

sklearn.linear_model.LinearRegression 会做的：

from sklearn import linear_model
clf = linear_model.LinearRegression()
clf.fit([[getattr(t, 'x%d' % i) for i in range(1, 8)] for t in texts],
        [t.y for t in texts])

然后clf.coef_将具有回归系数。

sklearn.linear_model 也具有类似的接口，可以对回归进行各种正则化。

这是我创建的一些解决方法。我用R检查了它，它可以正常工作。

import numpy as np
import statsmodels.api as sm

y = [1,2,3,4,3,4,5,4,5,5,4,5,4,5,4,5,6,5,4,5,4,3,4]

x = [
     [4,2,3,4,5,4,5,6,7,4,8,9,8,8,6,6,5,5,5,5,5,5,5],
     [4,1,2,3,4,5,6,7,5,8,7,8,7,8,7,8,7,7,7,7,7,6,5],
     [4,1,2,5,6,7,8,9,7,8,7,8,7,7,7,7,7,7,6,6,4,4,4]
     ]

def reg_m(y, x):
    ones = np.ones(len(x[0]))
    X = sm.add_constant(np.column_stack((x[0], ones)))
    for ele in x[1:]:
        X = sm.add_constant(np.column_stack((ele, X)))
    results = sm.OLS(y, X).fit()
    return results

结果：

print reg_m(y, x).summary()

输出：

                            OLS Regression Results                            
==============================================================================
Dep. Variable:                      y   R-squared:                       0.535
Model:                            OLS   Adj. R-squared:                  0.461
Method:                 Least Squares   F-statistic:                     7.281
Date:                Tue, 19 Feb 2013   Prob (F-statistic):            0.00191
Time:                        21:51:28   Log-Likelihood:                -26.025
No. Observations:                  23   AIC:                             60.05
Df Residuals:                      19   BIC:                             64.59
Df Model:                           3                                         
==============================================================================
                 coef    std err          t      P>|t|      [95.0% Conf. Int.]
------------------------------------------------------------------------------
x1             0.2424      0.139      1.739      0.098        -0.049     0.534
x2             0.2360      0.149      1.587      0.129        -0.075     0.547
x3            -0.0618      0.145     -0.427      0.674        -0.365     0.241
const          1.5704      0.633      2.481      0.023         0.245     2.895

==============================================================================
Omnibus:                        6.904   Durbin-Watson:                   1.905
Prob(Omnibus):                  0.032   Jarque-Bera (JB):                4.708
Skew:                          -0.849   Prob(JB):                       0.0950
Kurtosis:                       4.426   Cond. No.                         38.6

pandas 提供了运行此答案中给出的OLS的便捷方法：

使用Pandas Data Frame运行OLS回归

为了澄清起见，您给出的示例是多元线性回归，而不是多元线性回归。区别：

单个标量预测变量x和单个标量响应变量y的最简单情况就是简单线性回归。对多个和/或向量值的预测变量（用大写的X表示）的扩展被称为多元线性回归，也称为多元线性回归。几乎所有现实世界中的回归模型都涉及多个预测变量，而线性回归的基本描述通常用多元回归模型来表述。但是请注意，在这些情况下，响应变量y仍然是标量。另一个变量多元线性回归是指y是向量的情况，即与一般线性回归相同。

简而言之：

• 多元线性回归：响应y是一个标量。
• 多元线性回归：响应y是向量。

（另一个来源。）

您可以使用numpy.linalg.lstsq：

import numpy as np
y = np.array([-6,-5,-10,-5,-8,-3,-6,-8,-8])
X = np.array([[-4.95,-4.55,-10.96,-1.08,-6.52,-0.81,-7.01,-4.46,-11.54],[-5.87,-4.52,-11.64,-3.36,-7.45,-2.36,-7.33,-7.65,-10.03],[-0.76,-0.71,-0.98,0.75,-0.86,-0.50,-0.33,-0.94,-1.03],[14.73,13.74,15.49,24.72,16.59,22.44,13.93,11.40,18.18],[4.02,4.47,4.18,4.96,4.29,4.81,4.32,4.43,4.28],[0.20,0.16,0.19,0.16,0.10,0.15,0.21,0.16,0.21],[0.45,0.50,0.53,0.60,0.48,0.53,0.50,0.49,0.55]])
X = X.T # transpose so input vectors are along the rows
X = np.c_[X, np.ones(X.shape[0])] # add bias term
beta_hat = np.linalg.lstsq(X,y)[0]
print beta_hat

结果：

[ -0.49104607   0.83271938   0.0860167    0.1326091    6.85681762  22.98163883 -41.08437805 -19.08085066]

您可以通过以下方式查看估计的输出：

print np.dot(X,beta_hat)

结果：

[ -5.97751163,  -5.06465759, -10.16873217,  -4.96959788,  -7.96356915,  -3.06176313,  -6.01818435,  -7.90878145,  -7.86720264]

使用scipy.optimize.curve_fit。而且不仅适用于线性拟合。

from scipy.optimize import curve_fit
import scipy

def fn(x, a, b, c):
    return a + b*x[0] + c*x[1]

# y(x0,x1) data:
#    x0=0 1 2
# ___________
# x1=0 |0 1 2
# x1=1 |1 2 3
# x1=2 |2 3 4

x = scipy.array([[0,1,2,0,1,2,0,1,2,],[0,0,0,1,1,1,2,2,2]])
y = scipy.array([0,1,2,1,2,3,2,3,4])
popt, pcov = curve_fit(fn, x, y)
print popt

将数据转换为熊猫数据框（df）后，

import statsmodels.formula.api as smf
lm = smf.ols(formula='y ~ x1 + x2 + x3 + x4 + x5 + x6 + x7', data=df).fit()
print(lm.params)

默认情况下包括拦截项。

有关更多示例，请参见此笔记本。

我认为这可能是完成这项工作的最简单方法：

from random import random
from pandas import DataFrame
from statsmodels.api import OLS
lr = lambda : [random() for i in range(100)]
x = DataFrame({'x1': lr(), 'x2':lr(), 'x3':lr()})
x['b'] = 1
y = x.x1 + x.x2 * 2 + x.x3 * 3 + 4

print x.head()

         x1        x2        x3  b
0  0.433681  0.946723  0.103422  1
1  0.400423  0.527179  0.131674  1
2  0.992441  0.900678  0.360140  1
3  0.413757  0.099319  0.825181  1
4  0.796491  0.862593  0.193554  1

print y.head()

0    6.637392
1    5.849802
2    7.874218
3    7.087938
4    7.102337
dtype: float64

model = OLS(y, x)
result = model.fit()
print result.summary()

                            OLS Regression Results                            
==============================================================================
Dep. Variable:                      y   R-squared:                       1.000
Model:                            OLS   Adj. R-squared:                  1.000
Method:                 Least Squares   F-statistic:                 5.859e+30
Date:                Wed, 09 Dec 2015   Prob (F-statistic):               0.00
Time:                        15:17:32   Log-Likelihood:                 3224.9
No. Observations:                 100   AIC:                            -6442.
Df Residuals:                      96   BIC:                            -6431.
Df Model:                           3                                         
Covariance Type:            nonrobust                                         
==============================================================================
                 coef    std err          t      P>|t|      [95.0% Conf. Int.]
------------------------------------------------------------------------------
x1             1.0000   8.98e-16   1.11e+15      0.000         1.000     1.000
x2             2.0000   8.28e-16   2.41e+15      0.000         2.000     2.000
x3             3.0000   8.34e-16    3.6e+15      0.000         3.000     3.000
b              4.0000   8.51e-16    4.7e+15      0.000         4.000     4.000
==============================================================================
Omnibus:                        7.675   Durbin-Watson:                   1.614
Prob(Omnibus):                  0.022   Jarque-Bera (JB):                3.118
Skew:                           0.045   Prob(JB):                        0.210
Kurtosis:                       2.140   Cond. No.                         6.89
==============================================================================

可以使用上面提到的sklearn库处理多个线性回归。我正在使用Python 3.6的Anaconda安装。

如下创建模型：

from sklearn.linear_model import LinearRegression
regressor = LinearRegression()
regressor.fit(X, y)

# display coefficients
print(regressor.coef_)

您可以使用numpy.linalg.lstsq

您可以使用下面的函数并将其传递给DataFrame：

def linear(x, y=None, show=True):
    """
    @param x: pd.DataFrame
    @param y: pd.DataFrame or pd.Series or None
              if None, then use last column of x as y
    @param show: if show regression summary
    """
    import statsmodels.api as sm

    xy = sm.add_constant(x if y is None else pd.concat([x, y], axis=1))
    res = sm.OLS(xy.ix[:, -1], xy.ix[:, :-1], missing='drop').fit()

    if show: print res.summary()
    return res

Scikit-learn是一个适用于Python的机器学习库，可以为您完成这项工作。只需将sklearn.linear_model模块导入脚本即可。

在python中使用sklearn查找多重线性回归的代码模板：

import numpy as np
import matplotlib.pyplot as plt #to plot visualizations
import pandas as pd

# Importing the dataset
df = pd.read_csv(<Your-dataset-path>)
# Assigning feature and target variables
X = df.iloc[:,:-1]
y = df.iloc[:,-1]

# Use label encoders, if you have any categorical variable
from sklearn.preprocessing import LabelEncoder
labelencoder = LabelEncoder()
X['<column-name>'] = labelencoder.fit_transform(X['<column-name>'])

from sklearn.preprocessing import OneHotEncoder
onehotencoder = OneHotEncoder(categorical_features = ['<index-value>'])
X = onehotencoder.fit_transform(X).toarray()

# Avoiding the dummy variable trap
X = X[:,1:] # Usually done by the algorithm itself

#Spliting the data into test and train set
from sklearn.model_selection import train_test_split
X_train, X_test, y_train, y_test = train_test_split(X,y, random_state = 0, test_size = 0.2)

# Fitting the model
from sklearn.linear_model import LinearRegression
regressor = LinearRegression()
regressor.fit(X_train, y_train)

# Predicting the test set results
y_pred = regressor.predict(X_test)

而已。您可以将此代码用作在任何数据集中实现多元线性回归的模板。为了更好地理解示例，请访问：带有示例的线性回归

这是另一种基本方法：

from patsy import dmatrices
import statsmodels.api as sm

y,x = dmatrices("y_data ~ x_1 + x_2 ", data = my_data)
### y_data is the name of the dependent variable in your data ### 
model_fit = sm.OLS(y,x)
results = model_fit.fit()
print(results.summary())

代替sm.OLS您也可以使用sm.Logitor sm.Probit和等。