我似乎无法找到任何进行多重回归的python库。我发现的唯一的东西只做简单的回归。我需要对几个自变量（x1，x2，x3等）回归我的因变量（y）。
例如，使用此数据：
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
如何在python中对这些进行回归，以获得线性回归公式：
Y = a1x1 + a2x2 + a3x3 + a4x4 + a5x5 + a6x6 + + a7x7 + c
答案 0 :(得分：90)
sklearn.linear_model.LinearRegression会这样做：
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也有类似的接口来对回归进行各种规范化。
答案 1 :(得分：57)
这是我创建的一个小工作。我用R检查了它，它的工作正常。
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提供了一种运行OLS的便捷方式，如下面的答案所示：
答案 2 :(得分：44)
为了澄清，您提供的示例是多个线性回归，而不是多变量线性回归参考。 Difference：
单个标量预测变量x和单个标量响应变量y的最简单情况称为简单线性回归。多个和/或向量值预测变量的扩展（用大写X表示）称为多元线性回归，也称为多变量线性回归。几乎所有现实世界的回归模型都涉及多个预测因子，线性回归的基本描述通常用多元回归模型来表达。但请注意，在这些情况下，响应变量y仍然是标量。另一个术语多元线性回归指的是y是矢量的情况，即与一般线性回归相同的情况。应强调多元线性回归与多变量线性回归之间的差异，因为它会在文献中引起很多混淆和误解。
简而言之：
（另一个source。）
答案 3 :(得分：26)
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算输出：
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]
答案 4 :(得分：11)
使用scipy.optimize.curve_fit。而且不仅仅是线性拟合。
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
答案 5 :(得分：8)
将数据转换为pandas数据帧（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)
默认情况下包含拦截术语。
有关更多示例，请参阅this notebook。
答案 6 :(得分：4)
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
==============================================================================
答案 7 :(得分：4)
可以使用上面引用的sklearn库处理多元线性回归。我正在使用Anaconda安装的Python 3.6。
按如下方式创建模型：
from sklearn.linear_model import LinearRegression
regressor = LinearRegression()
regressor.fit(X, y)
# display coefficients
print(regressor.coef_)
答案 8 :(得分：3)
您可以使用numpy.linalg.lstsq
答案 9 :(得分：1)
您可以使用以下功能并将其传递给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
答案 10 :(得分：1)
Scikit-learn是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查找用于多元线性回归的代码模板：
select act.ea_guid AS CLASSGUID, act.Object_ID as CLASSTYPE, act.Name, act.Stereotype,
o.Name as Lane, gen.Name as GenericElement
from (((t_object o
inner join t_objectproperties tv on (tv.Object_ID = o.Object_ID
and tv.Property = 'PartitionElementRef'))
left join t_object gen on gen.ea_guid = tv.Value)
left join t_object act on (act.ParentID = o.Object_ID
and act.Stereotype = 'Activity'))
where (o.Stereotype = 'Lane' and act.Stereotype = 'Activity')
就是这样。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：Linear Regression with an example
答案 11 :(得分：0)
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())
除了sm.OLS之外，您还可以使用sm.Logit或sm.Probit等。
答案 12 :(得分：0)
可以使用OpenTURNS处理找到这种线性模型。
在OpenTURNS中，这是通过LinearModelAlgorithm类完成的，该类根据数值样本创建线性模型。更具体地说，它构建以下线性模型：
Y = a0 + a1.X1 + ... + anXn + epsilon，
其中误差epsilon是高斯，均值和单位方差为零。假设您的数据在一个csv文件中，这是一个简单的脚本来获取回归系数ai：
from __future__ import print_function
import pandas as pd
import openturns as ot
# Assuming the data is a csv file with the given structure
# Y X1 X2 .. X7
df = pd.read_csv("./data.csv", sep="\s+")
# Build a sample from the pandas dataframe
sample = ot.Sample(df.values)
# The observation points are in the first column (dimension 1)
Y = sample[:, 0]
# The input vector (X1,..,X7) of dimension 7
X = sample[:, 1::]
# Build a Linear model approximation
result = ot.LinearModelAlgorithm(X, Y).getResult()
# Get the coefficients ai
print("coefficients of the linear regression model = ", result.getCoefficients())
您可以通过以下调用轻松获得置信区间：
# Get the confidence intervals at 90% of the ai coefficients
print(
"confidence intervals of the coefficients = ",
ot.LinearModelAnalysis(result).getCoefficientsConfidenceInterval(0.9),
)
您可以在OpenTURNS示例中找到更详细的示例。
答案 13 :(得分：0)
尝试使用高斯族的广义线性模型
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=zip(*reversed(X))
df=pd.DataFrame({'X':X,'y':y})
columns=7
for i in range(0,columns):
df['X'+str(i)]=df.apply(lambda row: row['X'][i],axis=1)
df=df.drop('X',axis=1)
print(df)
#model_formula='y ~ X0+X1+X2+X3+X4+X5+X6'
model_formula='y ~ X0'
model_family = sm.families.Gaussian()
model_fit = glm(formula = model_formula,
data = df,
family = model_family).fit()
print(model_fit.summary())
# Extract coefficients from the fitted model wells_fit
#print(model_fit.params)
intercept, slope = model_fit.params
# Print coefficients
print('Intercept =', intercept)
print('Slope =', slope)
# Extract and print confidence intervals
print(model_fit.conf_int())
df2=pd.DataFrame()
df2['X0']=np.linspace(0.50,0.70,50)
df3=pd.DataFrame()
df3['X1']=np.linspace(0.20,0.60,50)
prediction0=model_fit.predict(df2)
#prediction1=model_fit.predict(df3)
plt.plot(df2['X0'],prediction0,label='X0')
plt.ylabel("y")
plt.xlabel("X0")
plt.show()