I have two scatterplots that I've placed on one plot. I want to find the linear regression line for the points of y1 and y2 combined (as in the regression between x and (y1 and y2) ), but I'm having difficulty since I usually only find the regression line for y1 or y2 separately. I also want to find the r^2 value (for the combined y1 and y2). I would appreciate any help I can get!
df1 = pd.DataFrame(np.random.randint(0,100,size=(15, 2)), columns=list('AB'))
y1 = df1['A']
y2 = df1['B']
plt.scatter(df1.index, y1)
plt.scatter(df1.index, y2)
plt.show()
Sounds like you want to 'stack' columns A and B together; many ways to do it, here is one using stack:
df2 = df1.stack().rename('A_and_B').reset_index(level = 1, drop = True).to_frame()
Then df.head() looks like this:
A_and_B
0 35
0 58
1 49
1 73
2 44
and the scatter plot:
plt.scatter(df2.index, df2['A_and_B'])
looks like
I don't know how you do regressions, you can apply your method to df2 now. For example:
import statsmodels.api as sm
res = sm.OLS(df2['A_and_B'], df2.index).fit()
res.summary()
output:
OLS Regression Results
=======================================================================================
Dep. Variable: A_and_B R-squared (uncentered): 0.517
Model: OLS Adj. R-squared (uncentered): 0.501
Method: Least Squares F-statistic: 31.10
Date: Mon, 14 Mar 2022 Prob (F-statistic): 5.11e-06
Time: 23:02:47 Log-Likelihood: -152.15
No. Observations: 30 AIC: 306.3
Df Residuals: 29 BIC: 307.7
Df Model: 1
Covariance Type: nonrobust
==============================================================================
coef std err t P>|t| [0.025 0.975]
------------------------------------------------------------------------------
x1 4.8576 0.871 5.577 0.000 3.076 6.639
==============================================================================
Omnibus: 3.466 Durbin-Watson: 1.244
Prob(Omnibus): 0.177 Jarque-Bera (JB): 1.962
Skew: -0.371 Prob(JB): 0.375
Kurtosis: 1.990 Cond. No. 1.00
==============================================================================
Notes:
[1] R² is computed without centering (uncentered) since the model does not contain a constant.
[2] Standard Errors assume that the covariance matrix of the errors is correctly specified.
Related
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've looked through the documentation and still can't figure this out. I want to run a WLS with multiple regressions.
statsmodels.api is imported as sm
Example of single variable.
X = Height
Y = Weight
res = sm.OLS(Y,X,).fit()
res.summary()
Say I also have:
X2 = Age
How do I add this into my regresssion?
You can put them into a data.frame and call out the columns (this way the output looks nicer too):
import statsmodels.api as sm
import pandas as pd
import numpy as np
Height = np.random.uniform(0,1,100)
Weight = np.random.uniform(0,1,100)
Age = np.random.uniform(0,30,100)
df = pd.DataFrame({'Height':Height,'Weight':Weight,'Age':Age})
res = sm.OLS(df['Height'],df[['Weight','Age']]).fit()
In [10]: res.summary()
Out[10]:
<class 'statsmodels.iolib.summary.Summary'>
"""
OLS Regression Results
=======================================================================================
Dep. Variable: Height R-squared (uncentered): 0.700
Model: OLS Adj. R-squared (uncentered): 0.694
Method: Least Squares F-statistic: 114.3
Date: Mon, 24 Aug 2020 Prob (F-statistic): 2.43e-26
Time: 15:54:30 Log-Likelihood: -28.374
No. Observations: 100 AIC: 60.75
Df Residuals: 98 BIC: 65.96
Df Model: 2
Covariance Type: nonrobust
==============================================================================
coef std err t P>|t| [0.025 0.975]
------------------------------------------------------------------------------
Weight 0.1787 0.090 1.988 0.050 0.000 0.357
Age 0.0229 0.003 8.235 0.000 0.017 0.028
==============================================================================
Omnibus: 2.938 Durbin-Watson: 1.813
Prob(Omnibus): 0.230 Jarque-Bera (JB): 2.223
Skew: -0.211 Prob(JB): 0.329
Kurtosis: 2.404 Cond. No. 49.7
==============================================================================
I use a 2nd order polynomial to predict how height and age affect weight for a soldier. You can pick up ansur_2_m.csv on my GitHub.
df=pd.read_csv('ANSUR_2_M.csv', encoding = "ISO-8859-1", usecols=['Weightlbs','Heightin','Age'], dtype={'Weightlbs':np.integer,'Heightin':np.integer,'Age':np.integer})
df=df.dropna()
df.reset_index()
df['Heightin2']=df['Heightin']**2
df['Age2']=df['Age']**2
formula="Weightlbs ~ Heightin+Heightin2+Age+Age2"
model_ols = smf.ols(formula,data=df).fit()
minHeight=df['Heightin'].min()
maxHeight=df['Heightin'].max()
avgAge = df['Age'].median()
print(minHeight,maxHeight,avgAge)
df2=pd.DataFrame()
df2['Heightin']=np.linspace(60,100,50)
df2['Heightin2']=df2['Heightin']**2
df2['Age']=28
df2['Age2']=df['Age']**2
df3=pd.DataFrame()
df3['Heightin']=np.linspace(60,100,50)
df3['Heightin2']=df2['Heightin']**2
df3['Age']=45
df3['Age2']=df['Age']**2
prediction28=model_ols.predict(df2)
prediction45=model_ols.predict(df3)
plt.clf()
plt.plot(df2['Heightin'],prediction28,label="Age 28")
plt.plot(df3['Heightin'],prediction45,label="Age 45")
plt.ylabel="Weight lbs"
plt.xlabel="Height in"
plt.legend()
plt.show()
print('A 45 year old soldier is more probable to weight more than an 28 year old soldier')
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!
I want to have a coefficient and Newey-West standard error associated with it.
I am looking for Python library (ideally, but any working solutions is fine) that can do what the following R code is doing:
library(sandwich)
library(lmtest)
a <- matrix(c(1,3,5,7,4,5,6,4,7,8,9))
b <- matrix(c(3,5,6,2,4,6,7,8,7,8,9))
temp.lm = lm(a ~ b)
temp.summ <- summary(temp.lm)
temp.summ$coefficients <- unclass(coeftest(temp.lm, vcov. = NeweyWest))
print (temp.summ$coefficients)
Result:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 2.0576208 2.5230532 0.8155281 0.4358205
b 0.5594796 0.4071834 1.3740235 0.2026817
I get the coefficients and associated with them standard errors.
I see statsmodels.stats.sandwich_covariance.cov_hac module, but I don't see how to make it work with OLS.
Edited (10/31/2015) to reflect preferred coding style for statsmodels as fall 2015.
In statsmodels version 0.6.1 you can do the following:
import pandas as pd
import numpy as np
import statsmodels.formula.api as smf
df = pd.DataFrame({'a':[1,3,5,7,4,5,6,4,7,8,9],
'b':[3,5,6,2,4,6,7,8,7,8,9]})
reg = smf.ols('a ~ 1 + b',data=df).fit(cov_type='HAC',cov_kwds={'maxlags':1})
print reg.summary()
OLS Regression Results
==============================================================================
Dep. Variable: a R-squared: 0.281
Model: OLS Adj. R-squared: 0.201
Method: Least Squares F-statistic: 1.949
Date: Sat, 31 Oct 2015 Prob (F-statistic): 0.196
Time: 03:15:46 Log-Likelihood: -22.603
No. Observations: 11 AIC: 49.21
Df Residuals: 9 BIC: 50.00
Df Model: 1
Covariance Type: HAC
==============================================================================
coef std err z P>|z| [95.0% Conf. Int.]
------------------------------------------------------------------------------
Intercept 2.0576 2.661 0.773 0.439 -3.157 7.272
b 0.5595 0.401 1.396 0.163 -0.226 1.345
==============================================================================
Omnibus: 0.361 Durbin-Watson: 1.468
Prob(Omnibus): 0.835 Jarque-Bera (JB): 0.331
Skew: 0.321 Prob(JB): 0.847
Kurtosis: 2.442 Cond. No. 19.1
==============================================================================
Warnings:
[1] Standard Errors are heteroscedasticity and autocorrelation robust (HAC) using 1 lags and without small sample correction
Or you can use the get_robustcov_results method after fitting the model:
reg = smf.ols('a ~ 1 + b',data=df).fit()
new = reg.get_robustcov_results(cov_type='HAC',maxlags=1)
print new.summary()
OLS Regression Results
==============================================================================
Dep. Variable: a R-squared: 0.281
Model: OLS Adj. R-squared: 0.201
Method: Least Squares F-statistic: 1.949
Date: Sat, 31 Oct 2015 Prob (F-statistic): 0.196
Time: 03:15:46 Log-Likelihood: -22.603
No. Observations: 11 AIC: 49.21
Df Residuals: 9 BIC: 50.00
Df Model: 1
Covariance Type: HAC
==============================================================================
coef std err z P>|z| [95.0% Conf. Int.]
------------------------------------------------------------------------------
Intercept 2.0576 2.661 0.773 0.439 -3.157 7.272
b 0.5595 0.401 1.396 0.163 -0.226 1.345
==============================================================================
Omnibus: 0.361 Durbin-Watson: 1.468
Prob(Omnibus): 0.835 Jarque-Bera (JB): 0.331
Skew: 0.321 Prob(JB): 0.847
Kurtosis: 2.442 Cond. No. 19.1
==============================================================================
Warnings:
[1] Standard Errors are heteroscedasticity and autocorrelation robust (HAC) using 1 lags and without small sample correction
The defaults for statsmodels are slightly different than the defaults for the equivalent method in R. The R method can be made equivalent to the statsmodels default (what I did above) by changing the vcov, call to the following:
temp.summ$coefficients <- unclass(coeftest(temp.lm,
vcov. = NeweyWest(temp.lm,lag=1,prewhite=FALSE)))
print (temp.summ$coefficients)
Estimate Std. Error t value Pr(>|t|)
(Intercept) 2.0576208 2.6605060 0.7733945 0.4591196
b 0.5594796 0.4007965 1.3959193 0.1962142
You can also still do Newey-West in pandas (0.17), although I believe the plan is to deprecate OLS in pandas:
print pd.stats.ols.OLS(df.a,df.b,nw_lags=1)
-------------------------Summary of Regression Analysis-------------------------
Formula: Y ~ <x> + <intercept>
Number of Observations: 11
Number of Degrees of Freedom: 2
R-squared: 0.2807
Adj R-squared: 0.2007
Rmse: 2.0880
F-stat (1, 9): 1.5943, p-value: 0.2384
Degrees of Freedom: model 1, resid 9
-----------------------Summary of Estimated Coefficients------------------------
Variable Coef Std Err t-stat p-value CI 2.5% CI 97.5%
--------------------------------------------------------------------------------
x 0.5595 0.4431 1.26 0.2384 -0.3090 1.4280
intercept 2.0576 2.9413 0.70 0.5019 -3.7073 7.8226
*** The calculations are Newey-West adjusted with lags 1
---------------------------------End of Summary---------------------------------