I need to perform multiple polynomial regression and obtain statistics, p value, AIC etc.
As far as I understood I can do that with OLS, however I found only a way to produce a formula using one independent variable, like this:
model = 'act_hours ~ h_hours + I(h_hours**2)'
hours_model = smf.ols(formula = model, data = df)
I tried to define a formula using two independent variable, however I could not understand if that is the correct way and if the results are reasonable. The line that I doubt is model = 'Height ~ Diamet + I(Diamet**2) + area + I(area**2). The full code is this one:
import pandas as pd
import statsmodels.formula.api as smf
train = pd.read_csv(r'W:\...file.csv')
model = 'Height ~ Diamet + I(Diamet**2) + area + I(area**2)'
hours_model = smf.ols(formula = model, data = train).fit()
print(hours_model.summary())
The summary of the regression is here:
OLS Regression Results
==============================================================================
Dep. Variable: Height R-squared: 0.611
Model: OLS Adj. R-squared: 0.609
Method: Least Squares F-statistic: 376.0
Date: Fri, 04 Feb 2022 Prob (F-statistic): 1.33e-194
Time: 08:50:17 Log-Likelihood: -5114.6
No. Observations: 963 AIC: 1.024e+04
Df Residuals: 958 BIC: 1.026e+04
Df Model: 4
Covariance Type: nonrobust
==================================================================================
coef std err t P>|t| [0.025 0.975]
----------------------------------------------------------------------------------
Intercept 13.9287 60.951 0.229 0.819 -105.684 133.542
Diamet 0.6027 0.340 1.770 0.077 -0.066 1.271
I(Diamet ** 2) 0.0004 0.002 0.262 0.794 -0.003 0.004
area 3.3553 5.307 0.632 0.527 -7.060 13.771
I(area** 2) 0.2519 0.108 2.324 0.020 0.039 0.465
==============================================================================
Omnibus: 60.996 Durbin-Watson: 1.889
Prob(Omnibus): 0.000 Jarque-Bera (JB): 86.039
Skew: 0.528 Prob(JB): 2.07e-19
Kurtosis: 4.015 Cond. No. 4.45e+05
==============================================================================
Notes:
[1] Standard Errors assume that the covariance matrix of the errors is correctly specified.
[2] The condition number is large, 4.45e+05. This might indicate that there are
strong multicollinearity or other numerical problems.
I am running rolling regressions using the RollingOLS function on statsmodels.api, and wondering if its possible to get the summary statistics (betas, r^2, etc.) out for each regression done in the rolling regression.
Using a single OLS regression, you can get the summary information like such,
X_opt = X[:, [0,1,2,3]]
regressor_OLS = sm.OLS(endog= y, exog= X_opt).fit()
regressor_OLS.summary()
OLS Regression Results
==============================================================================
Dep. Variable: y R-squared: 0.951
Model: OLS Adj. R-squared: 0.948
Method: Least Squares F-statistic: 296.0
Date: Wed, 08 Aug 2018 Prob (F-statistic): 4.53e-30
Time: 00:46:48 Log-Likelihood: -525.39
No. Observations: 50 AIC: 1059.
Df Residuals: 46 BIC: 1066.
Df Model: 3
Covariance Type: nonrobust
==============================================================================
coef std err t P>|t| [0.025 0.975]
------------------------------------------------------------------------------
const 5.012e+04 6572.353 7.626 0.000 3.69e+04 6.34e+04
x1 0.8057 0.045 17.846 0.000 0.715 0.897
x2 -0.0268 0.051 -0.526 0.602 -0.130 0.076
x3 0.0272 0.016 1.655 0.105 -0.006 0.060
==============================================================================
Omnibus: 14.838 Durbin-Watson: 1.282
Prob(Omnibus): 0.001 Jarque-Bera (JB): 21.442
Skew: -0.949 Prob(JB): 2.21e-05
Kurtosis: 5.586 Cond. No. 1.40e+06
==============================================================================
is there a way to get this information for the regression run on each window for a rolling regression?
I am running the following source code:
import statsmodels.formula.api as sm
# Add one column of ones for the intercept term
X = np.append(arr= np.ones((50, 1)).astype(int), values=X, axis=1)
regressor_OLS = sm.OLS(endog=y, exog=X).fit()
print(regressor_OLS.summary())
where
X is an 50x5 (before adding the intercept term) numpy array which looks like this:
[[0 1 165349.20 136897.80 471784.10]
[0 0 162597.70 151377.59 443898.53]...]
and y is a a 50x1 numpy array with float values for the dependent variable.
The first two columns are for a dummy variable with three different values. The rest of the columns are three different indepedent variables.
Although, it is said that the statsmodels.formula.api.OLS adds automatically an intercept term (see #stellacia's answer here: OLS using statsmodel.formula.api versus statsmodel.api) its summary does not show the statistical values of the intercept term as it evident below in my case:
OLS Regression Results
==============================================================================
Dep. Variable: Profit R-squared: 0.988
Model: OLS Adj. R-squared: 0.986
Method: Least Squares F-statistic: 727.1
Date: Sun, 01 Jul 2018 Prob (F-statistic): 7.87e-42
Time: 21:40:23 Log-Likelihood: -545.15
No. Observations: 50 AIC: 1100.
Df Residuals: 45 BIC: 1110.
Df Model: 5
Covariance Type: nonrobust
==============================================================================
coef std err t P>|t| [0.025 0.975]
------------------------------------------------------------------------------
x1 3464.4536 4905.406 0.706 0.484 -6415.541 1.33e+04
x2 5067.8937 4668.238 1.086 0.283 -4334.419 1.45e+04
x3 0.7182 0.066 10.916 0.000 0.586 0.851
x4 0.3113 0.035 8.885 0.000 0.241 0.382
x5 0.0786 0.023 3.429 0.001 0.032 0.125
==============================================================================
Omnibus: 1.355 Durbin-Watson: 1.288
Prob(Omnibus): 0.508 Jarque-Bera (JB): 1.241
Skew: -0.237 Prob(JB): 0.538
Kurtosis: 2.391 Cond. No. 8.28e+05
==============================================================================
For this reason, I added to my source code the line:
X = np.append(arr= np.ones((50, 1)).astype(int), values=X, axis=1)
as you can see at the beginning of my post and the statistical values of the intercept/constant are shown as below:
OLS Regression Results
==============================================================================
Dep. Variable: Profit R-squared: 0.951
Model: OLS Adj. R-squared: 0.945
Method: Least Squares F-statistic: 169.9
Date: Sun, 01 Jul 2018 Prob (F-statistic): 1.34e-27
Time: 20:25:21 Log-Likelihood: -525.38
No. Observations: 50 AIC: 1063.
Df Residuals: 44 BIC: 1074.
Df Model: 5
Covariance Type: nonrobust
==============================================================================
coef std err t P>|t| [0.025 0.975]
------------------------------------------------------------------------------
const 5.013e+04 6884.820 7.281 0.000 3.62e+04 6.4e+04
x1 198.7888 3371.007 0.059 0.953 -6595.030 6992.607
x2 -41.8870 3256.039 -0.013 0.990 -6604.003 6520.229
x3 0.8060 0.046 17.369 0.000 0.712 0.900
x4 -0.0270 0.052 -0.517 0.608 -0.132 0.078
x5 0.0270 0.017 1.574 0.123 -0.008 0.062
==============================================================================
Omnibus: 14.782 Durbin-Watson: 1.283
Prob(Omnibus): 0.001 Jarque-Bera (JB): 21.266
Skew: -0.948 Prob(JB): 2.41e-05
Kurtosis: 5.572 Cond. No. 1.45e+06
==============================================================================
Why the statistical values of the intercept are not showing when I do not add my myself an intercept term even though it is said that statsmodels.formula.api.OLS is adding this automatically?
"No constant is added by the model unless you are using formulas."
Therefore try something like below example. Variable names should be defined according to your data set.
Use,
regressor_OLS = smf.ols(formula='Y_variable ~ X_variable', data=df).fit()
instead of,
regressor_OLS = sm.OLS(endog=y, exog=X).fit()
Can use this
X = sm.add_constant(X)
I'm doing a multiple linear regression, and trying to select the best subset of a number of independent variables. I would like to try to do all 1024 possible combinations in a "for" loop and save the best results based on condition number and r squared. I know it calculates both, giving results like:
model = sm.OLS(salarray, narraycareer)
results = model.fit()
print results.summary()
OLS Regression Results
==============================================================================
Dep. Variable: y R-squared: 0.425
Model: OLS Adj. R-squared: 0.409
Method: Least Squares F-statistic: 26.89
Date: Sat, 23 Sep 2017 Prob (F-statistic): 1.69e-27
Time: 00:58:14 Log-Likelihood: -1907.4
No. Observations: 263 AIC: 3831.
Df Residuals: 255 BIC: 3859.
Df Model: 7
Covariance Type: nonrobust
==============================================================================
coef std err t P>|t| [95.0% Conf. Int.]
------------------------------------------------------------------------------
const 535.9259 21.387 25.058 0.000 493.808 578.044
x1 -675.5296 302.245 -2.235 0.026 -1270.744 -80.315
x2 182.7168 436.493 0.419 0.676 -676.874 1042.307
x3 -48.2603 126.141 -0.383 0.702 -296.671 200.151
x4 445.0863 218.373 2.038 0.043 15.043 875.130
x5 344.0092 219.896 1.564 0.119 -89.035 777.053
x6 -41.5168 71.925 -0.577 0.564 -183.159 100.126
x7 96.5430 30.595 3.156 0.002 36.293 156.793
==============================================================================
Omnibus: 96.442 Durbin-Watson: 1.973
Prob(Omnibus): 0.000 Jarque-Bera (JB): 440.598
Skew: 1.438 Prob(JB): 2.11e-96
Kurtosis: 8.651 Cond. No. 61.7
==============================================================================
Warnings:
[1] Standard Errors assume that the covariance matrix of the errors is correctly specified.
>>>
But I can't find any documentation on how to get out condition number or r squared.
Thanks!
I found it, or the Spyder ide found it for me in the interpreter window.
>>> results.rsquared
0.42465891683421031
>>> results.condition_number
61.715714331759621
>>> >
When I typed "results." it gave a bunch of suggestions. Something vim doesn't do!
The following code is from
"Python for Data Analysis",chp 11,group transforms and analysis.
I show the version of each library as below.
# -*- coding: utf-8 -*-
""" Created on Sun Jun 4 13:33:47 2017
"Python for Data Analysis",chp 11,group transforms and analysis.
"""
import numpy as np # np.__version__'1.12.1'
import pandas as pd # pd.__version__ '0.20.2'
import random; random.seed(a=0,version=2)
import statsmodels.api as sm # statsmodels.__version__ '0.8.0'
import string
# generate tickers from random
N=1000
def rands(n):
choices=string.ascii_uppercase
return (''.join([random.choice(choices) for _ in range(n)]))
tickers=np.array([rands(5) for _ in range(N)])
# generate data for tickers
M=500
df=pd.DataFrame({'Momentum': np.random.randn(M)/200+0.03,
'Value':np.random.randn(M)/200+0.08,
'ShortInterest':np.random.randn(M)/200-0.02},
index=tickers[:M])
# create industry
ind_names=np.array(['Financial','Tech'])
sampler=np.random.randint(low=0,high=len(ind_names),size=N, dtype='l')
industries=pd.Series(ind_names[sampler],index=tickers,
name='industry')
#%% factor analysis
fac1,fac2,fac3=np.random.rand(3,1000)
ticker_subset=tickers.take(np.random.permutation(N)[:1000])
port=pd.Series(0.7*fac1-1.2*fac2+0.3*fac3+np.random.rand(1000),
index=ticker_subset)
factors=pd.DataFrame({'f1':fac1,'f2':fac2,'f3':fac3},
index=ticker_subset)
by_ind=port.groupby(industries)
This part is from the book, while pd.ols has been depreciated.
#%% use pd.ols, which is depreciated.
# AttributeError: module 'pandas' has no attribute 'ols'
def beta_exposure(chuck,factors=None):
return pd.ols(y=chuck, x=factors).beta
exposures_pd=by_ind.apply(beta_exposure,factors=factors)
print('\nexposures_pd\n',exposures_pd.unstack())
I would like to use sm.OLS, while I have trouble in selecting corresponding rows for x. How should I deal with it?
#%% use sm.OLS, which is not show in the book.
def exposure(chuck,factors):
y=np.array(chuck).reshape(len(chuck),1)
# The following code is wrong, as the rows number is not the corresponding rows as y
# I use [:len(chuck)] just to keep x have same rows number as y.
x=factors[['f1','f2','f3']][:len(chuck)]
print(x[:5])
print(x.shape)
sx=sm.OLS(y,x).fit()
print(sx.summary())
return sm.OLS(y,x).fit()
exposures_sm=exposure(port, factors)
after several try, I think maybe I can do it by combine Series and DataFrame.
factors_data['port']=port
def group_ols(fts):
results=[]
for ind, ft in fts:
y=ft.loc[:,'port']
x=ft.loc[:,['f1','f2','f3']]
result=sm.OLS(y,x).fit()
results.append((ind,result.summary()))
return results
exposures_sm=group_ols(factors_data.groupby(industries))
exposures_sm
the result is like this.
[('Financial', <class 'statsmodels.iolib.summary.Summary'>
"""
OLS Regression Results
==============================================================================
Dep. Variable: port R-squared: 0.746
Model: OLS Adj. R-squared: 0.744
Method: Least Squares F-statistic: 482.4
Date: Thu, 29 Jun 2017 Prob (F-statistic): 2.37e-146
Time: 17:13:34 Log-Likelihood: -134.55
No. Observations: 497 AIC: 275.1
Df Residuals: 494 BIC: 287.7
Df Model: 3
Covariance Type: nonrobust
==============================================================================
coef std err t P>|t| [0.025 0.975]
------------------------------------------------------------------------------
f1 1.0231 0.043 23.894 0.000 0.939 1.107
f2 -0.9639 0.042 -23.146 0.000 -1.046 -0.882
f3 0.6397 0.042 15.391 0.000 0.558 0.721
==============================================================================
Omnibus: 34.466 Durbin-Watson: 1.916
Prob(Omnibus): 0.000 Jarque-Bera (JB): 12.724
Skew: -0.063 Prob(JB): 0.00173
Kurtosis: 2.226 Cond. No. 3.24
==============================================================================
Warnings:
[1] Standard Errors assume that the covariance matrix of the errors is correctly specified.
"""), ('Tech', <class 'statsmodels.iolib.summary.Summary'>
"""
OLS Regression Results
==============================================================================
Dep. Variable: port R-squared: 0.738
Model: OLS Adj. R-squared: 0.736
Method: Least Squares F-statistic: 468.9
Date: Thu, 29 Jun 2017 Prob (F-statistic): 7.30e-145
Time: 17:13:34 Log-Likelihood: -172.76
No. Observations: 503 AIC: 351.5
Df Residuals: 500 BIC: 364.2
Df Model: 3
Covariance Type: nonrobust
==============================================================================
coef std err t P>|t| [0.025 0.975]
------------------------------------------------------------------------------
f1 1.0530 0.045 23.525 0.000 0.965 1.141
f2 -0.8811 0.045 -19.742 0.000 -0.969 -0.793
f3 0.5762 0.046 12.538 0.000 0.486 0.667
==============================================================================
Omnibus: 45.191 Durbin-Watson: 2.013
Prob(Omnibus): 0.000 Jarque-Bera (JB): 15.547
Skew: -0.123 Prob(JB): 0.000421
Kurtosis: 2.175 Cond. No. 3.29
==============================================================================
Warnings:
[1] Standard Errors assume that the covariance matrix of the errors is correctly specified.
""")]
I would like to suggest leverage apply function as it is more scalable if you happen deal with large data
first build a function
from sklearn.linear_model import LinearRegression
def reg_function(x, y):
regr = LinearRegression(normalize=True, fit_intercept = True)
regr.fit(y, x)
return(regr.coef_)
here you can use statmodel library if you prefer
df_ur.groupby([key_col]).apply(lambda x: reg_function(x[col_y].values.reshape(-1, 1), x[col_x].values.reshape(-1, 1)))
key_col is the column you would like to groupby - col_x and col_y are your dependent and independent variables
here because I am doing univarate regression, I have to use reshape otherwise, it won't be treated line nx1 matrix