问题
I can't seem to find any python libraries that do multiple regression. The only things I find only do simple regression. I need to regress my dependent variable (y) against several independent variables (x1, x2, x3, etc.).
For example, with this data:
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)
(output for above:)
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
How would I regress these in python, to get the linear regression formula:
Y = a1x1 + a2x2 + a3x3 + a4x4 + a5x5 + a6x6 + +a7x7 + c
回答1:
sklearn.linear_model.LinearRegression will do it:
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])
Then clf.coef_ will have the regression coefficients.
sklearn.linear_model also has similar interfaces to do various kinds of regularizations on the regression.
回答2:
Here is a little work around that I created. I checked it with R and it works correct.
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
Result:
print reg_m(y, x).summary()
Output:
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 provides a convenient way to run OLS as given in this answer:
Run an OLS regression with Pandas Data Frame
回答3:
Just to clarify, the example you gave is multiple linear regression, not multivariate linear regression refer. Difference:
The very simplest case of a single scalar predictor variable x and a single scalar response variable y is known as simple linear regression. The extension to multiple and/or vector-valued predictor variables (denoted with a capital X) is known as multiple linear regression, also known as multivariable linear regression. Nearly all real-world regression models involve multiple predictors, and basic descriptions of linear regression are often phrased in terms of the multiple regression model. Note, however, that in these cases the response variable y is still a scalar. Another term multivariate linear regression refers to cases where y is a vector, i.e., the same as general linear regression. The difference between multivariate linear regression and multivariable linear regression should be emphasized as it causes much confusion and misunderstanding in the literature.
In short:
- multiple linear regression: the response y is a scalar.
- multivariate linear regression: the response y is a vector.
(Another source.)
回答4:
You can use 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
Result:
[ -0.49104607 0.83271938 0.0860167 0.1326091 6.85681762 22.98163883 -41.08437805 -19.08085066]
You can see the estimated output with:
print np.dot(X,beta_hat)
Result:
[ -5.97751163, -5.06465759, -10.16873217, -4.96959788, -7.96356915, -3.06176313, -6.01818435, -7.90878145, -7.86720264]
回答5:
Use scipy.optimize.curve_fit. And not only for linear fit.
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
回答6:
Once you convert your data to a pandas dataframe (df),
import statsmodels.formula.api as smf
lm = smf.ols(formula='y ~ x1 + x2 + x3 + x4 + x5 + x6 + x7', data=df).fit()
print(lm.params)
The intercept term is included by default.
See this notebook for more examples.
回答7:
I think this may the most easy way to finish this work:
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
==============================================================================
回答8:
Multiple Linear Regression can be handled using the sklearn library as referenced above. I'm using the Anaconda install of Python 3.6.
Create your model as follows:
from sklearn.linear_model import LinearRegression
regressor = LinearRegression()
regressor.fit(X, y)
# display coefficients
print(regressor.coef_)
回答9:
You can use numpy.linalg.lstsq
回答10:
You can use the function below and pass it a 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
回答11:
Here is an alternative and basic method:
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())
Instead of sm.OLS you can also use sm.Logit or sm.Probit and etc.
回答12:
Scikit-learn is a machine learning library for Python which can do this job for you. Just import sklearn.linear_model module into your script.
Find the code template for Multiple Linear Regression using sklearn in Python:
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)
That's it. You can use this code as a template for implementing Multiple Linear Regression in any dataset. For a better understanding with an example, Visit: Linear Regression with an example
来源：https://stackoverflow.com/questions/11479064/multiple-linear-regression-in-python