Why do I get only one parameter from a statsmodels OLS fit

Question

Here is what I am doing:

$ python
Python 2.7.6 (v2.7.6:3a1db0d2747e, Nov 10 2013, 00:42:54) 
[GCC 4.2.1 (Apple Inc. build 5666) (dot 3)] on darwin
>>> import statsmodels.api as sm
>>> statsmodels.__version__
'0.5.0'
>>> import numpy 
>>> y = numpy.array([1,2,3,4,5,6,7,8,9])
>>> X = numpy.array([1,1,2,2,3,3,4,4,5])
>>> res_ols = sm.OLS(y, X).fit()
>>> res_ols.params
array([ 1.82352941])

I had expected an array with two elements?!? The intercept and the slope coefficient?

[Docs](http://statsmodels.sourceforge.net/devel/generated/statsmodels.regression.linear_model.OLS.html): An interecept is not included by default and should be added by the user. See statsmodels.tools.add_constant. — alko, Dec 20 '13 at 10:33
What is the significance of add_constant() here. When I generate a model in linear reg., I would expect to have an intercept, y = mX + C. What's the intention to have someone do additional operation of adding constant on top of input vector. — Abhi, Sep 18 '16 at 22:57
Interestingly, if you use the R-like formula api in statsmodels that gives you the intercept by default. — MJMacarty, Jan 06 '19 at 00:43

score 85 · Accepted Answer · edited Aug 18 '18 at 00:53

85

Try this:

X = sm.add_constant(X)
sm.OLS(y,X)

as in the documentation:

An intercept is not included by default and should be added by the user

statsmodels.tools.tools.add_constant

edited Aug 18 '18 at 00:53

Max Ghenis

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answered Dec 20 '13 at 10:30

behzad.nouri

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I am quite puzzled by this. Why isn't an intercept added by default? Why do you want to run the linear regression *without* the bloody constant? It makes no sense to me. – FaCoffee Oct 16 '17 at 18:24
what does adding a column of ones to an array do to X? – Golden Lion Jan 31 '22 at 21:20

score 10 · Answer 2 · edited Sep 19 '16 at 12:05

10

Just to be complete, this works:

>>> import numpy 
>>> import statsmodels.api as sm
>>> y = numpy.array([1,2,3,4,5,6,7,8,9])
>>> X = numpy.array([1,1,2,2,3,3,4,4,5])
>>> X = sm.add_constant(X)
>>> res_ols = sm.OLS(y, X).fit()
>>> res_ols.params
array([-0.35714286,  1.92857143])

It does give me a different slope coefficient, but I guess that figures as we now do have an intercept.

edited Sep 19 '16 at 12:05

Community

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answered Dec 20 '13 at 10:44

Tom

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score 4 · Answer 3 · edited Oct 05 '18 at 19:22

4

Try this, it worked for me:

import statsmodels.formula.api as sm

from statsmodels.api import add_constant

X_train = add_constant(X_train)

X_test = add_constant(X_test)


model = sm.OLS(y_train,X_train)

results = model.fit()

y_pred=results.predict(X_test)

results.params

edited Oct 05 '18 at 19:22

Vishesh Shrivastav

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answered Oct 05 '18 at 17:22

sup

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use `import statsmodels.api as sm` instead. `formula.api` will not have `OLS` (capital case) in the next release, only `ols` (lower case for formula interface) – Josef Oct 05 '18 at 19:14

score 2 · Answer 4 · answered Sep 25 '15 at 05:28

2

I'm running 0.6.1 and it looks like the "add_constant" function has been moved into the statsmodels.tools module. Here's what I ran that worked:

res_ols = sm.OLS(y, statsmodels.tools.add_constant(X)).fit()

answered Sep 25 '15 at 05:28

lukearmistead

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score 1 · Answer 5 · answered Nov 04 '18 at 11:35

i did add the code X = sm.add_constant(X) but python did not return the intercept value so using a little algebra i decided to do it myself in code:

this code computes regression over 35 samples, 7 features plus one intercept value that i added as feature to the equation:

import statsmodels.api as sm
from sklearn import datasets ## imports datasets from scikit-learn
import numpy as np
import pandas as pd

x=np.empty((35,8)) # (numSamples, oneIntercept + numFeatures))
feature_names = np.empty((8,))
y = np.empty((35,))

dbfv = open("dataset.csv").readlines()


interceptConstant = 1;
i = 0
# reading data and writing in numpy arrays
while i<len(dbfv):
    cells = dbfv[i].split(",")
    j = 0
    x[i][j] = interceptConstant
    feature_names[j] = str(j)
    while j<len(cells)-1:
        x[i][j+1] = cells[j]
        feature_names[j+1] = str(j+1)
        j += 1
    y[i] = cells[len(cells)-1]
    i += 1
# creating dataframes
df = pd.DataFrame(x, columns=feature_names)

target = pd.DataFrame(y, columns=["TARGET"])

X = df
y = target["TARGET"]

model = sm.OLS(y, X).fit()

print(model.params)

# predictions = model.predict(X) # make the predictions by the model


# Print out the statistics
print(model.summary())

score 0 · Answer 6 · edited May 08 '22 at 00:55

0

Try this

X = sm.add_constant(X)
ols= sm.OLS(y,X)
res_ols= ols.fit()
res_ols.params
res_ols.params[0]
res_ols.params[1]
print(res_ols.summary())

edited May 08 '22 at 00:55

Marco Bonelli

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answered May 08 '22 at 00:35

kyramichel

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Why do I get only one parameter from a statsmodels OLS fit

6 Answers6

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