python多元线性回归_numpy-Python中的多元线性回归numpy - Python中的多元线性回归
我似乎⽆法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
我如何在python中回归这些，以获得线性回归公式：
Y = a1x1 + a2x2 + a3x3 + a4x4 + a5x5 + a6x6 + + a7x7 + c
11个解决⽅案
90 votes
sklearn.linear_model将做到这⼀点：
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])
那么sklearn.linear_model将具有回归系数。
sklearn.linear_model也有类似的接⼝，可以对回归进⾏各种规范化。
Dougal answered 2019-07-04T01:27:06Z
57 votes
这是我创建的⼀个⼩⼯作。 我⽤R检查它，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_lumn_stack((x[0], ones)))
for ele in x[1:]:
X = sm.add_lumn_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：
使⽤Pandas Dataframe运⾏OLS回归
Akavall answered 2019-07-04T01:27:49Z
44 votes
只是为了澄清，你给出的例⼦是多元线性回归，⽽不是多元线性回归参考。 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：
多元线性回归：响应y是标量。
多元线性回归：响应y是向量。
（另⼀个来源。）
Franck Dernoncourt answered 2019-07-04T01:28:46Z
26 votes
你可以使⽤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]
Imran answered 2019-07-04T01:29:21Z
11 votes
使⽤scipy.optimize.curve_fit.⽽不仅仅是线性拟合。
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
Volodimir Kopey answered 2019-07-04T01:29:45Z
8 votes
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)
截距项默认包含在内。
有关更多⽰例，请参阅此笔记本
canary_in_the_data_mine answered 2019-07-04T01:30:23Z
4 votes
我认为这可能是完成这项⼯作最简单的⽅法：
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: nonrobustvariable used in lambda
============================================================================== coef std err t P>|t| [95.0% Conf. Int.]

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