Variance Inflation Factor in Python
I'm trying to calculate the variance inflation factor (VIF) for each column in a simple dataset in python:
a b c d
1 2 4 4
1 2 6 3
2 3 7 4
3 2 8 5
4 1 9 4
I have already done this in R using the vif function from the usdm library which gives the following results:
a <- c(1, 1, 2, 3, 4)
b <- c(2, 2, 3, 2, 1)
c <- c(4, 6, 7, 8, 9)
d <- c(4, 3, 4, 5, 4)
df <- data.frame(a, b, c, d)
vif_df <- vif(df)
print(vif_df)
Variables VIF
a 22.95
b 3.00
c 12.95
d 3.00
However, when I do the same in python using the statsmodel vif function, my results are:
a = [1, 1, 2, 3, 4]
b = [2, 2, 3, 2, 1]
c = [4, 6, 7, 8, 9]
d = [4, 3, 4, 5, 4]
ck = np.column_stack([a, b, c, d])
vif = [variance_inflation_factor(ck, i) for i in range(ck.shape[1])]
print(vif)
Variables VIF
a 47.136986301369774
b 28.931506849315081
c 80.31506849315096
d 40.438356164383549
The results are vastly different, even though the inputs are the same. In general, results from the statsmodel VIF function seem to be wrong, but I'm not sure if this is because of the way I am calling it or if it is an issue with the function itself.
I was hoping someone could help me figure out whether I was incorrectly calling the statsmodel function or explain the discrepancies in the results. If it's an issue with the function then are there any VIF alternatives in python?
Solution 1:
As mentioned by others and in this post by Josef Perktold, the function's author, variance_inflation_factor
expects the presence of a constant in the matrix of explanatory variables. One can use add_constant
from statsmodels to add the required constant to the dataframe before passing its values to the function.
from statsmodels.stats.outliers_influence import variance_inflation_factor
from statsmodels.tools.tools import add_constant
df = pd.DataFrame(
{'a': [1, 1, 2, 3, 4],
'b': [2, 2, 3, 2, 1],
'c': [4, 6, 7, 8, 9],
'd': [4, 3, 4, 5, 4]}
)
X = add_constant(df)
>>> pd.Series([variance_inflation_factor(X.values, i)
for i in range(X.shape[1])],
index=X.columns)
const 136.875
a 22.950
b 3.000
c 12.950
d 3.000
dtype: float64
I believe you could also add the constant to the right most column of the dataframe using assign
:
X = df.assign(const=1)
>>> pd.Series([variance_inflation_factor(X.values, i)
for i in range(X.shape[1])],
index=X.columns)
a 22.950
b 3.000
c 12.950
d 3.000
const 136.875
dtype: float64
The source code itself is rather concise:
def variance_inflation_factor(exog, exog_idx):
"""
exog : ndarray, (nobs, k_vars)
design matrix with all explanatory variables, as for example used in
regression
exog_idx : int
index of the exogenous variable in the columns of exog
"""
k_vars = exog.shape[1]
x_i = exog[:, exog_idx]
mask = np.arange(k_vars) != exog_idx
x_noti = exog[:, mask]
r_squared_i = OLS(x_i, x_noti).fit().rsquared
vif = 1. / (1. - r_squared_i)
return vif
It is also rather simple to modify the code to return all of the VIFs as a series:
from statsmodels.regression.linear_model import OLS
from statsmodels.tools.tools import add_constant
def variance_inflation_factors(exog_df):
'''
Parameters
----------
exog_df : dataframe, (nobs, k_vars)
design matrix with all explanatory variables, as for example used in
regression.
Returns
-------
vif : Series
variance inflation factors
'''
exog_df = add_constant(exog_df)
vifs = pd.Series(
[1 / (1. - OLS(exog_df[col].values,
exog_df.loc[:, exog_df.columns != col].values).fit().rsquared)
for col in exog_df],
index=exog_df.columns,
name='VIF'
)
return vifs
>>> variance_inflation_factors(df)
const 136.875
a 22.950
b 3.000
c 12.950
Name: VIF, dtype: float64
Per the solution of @T_T, one can also simply do the following:
vifs = pd.Series(np.linalg.inv(df.corr().to_numpy()).diagonal(),
index=df.columns,
name='VIF')
Solution 2:
I believe the reason for this is due to a difference in Python's OLS. OLS, which is used in the python variance inflation factor calculation, does not add an intercept by default. You definitely want an intercept in there however.
What you'd want to do is add one more column to your matrix, ck, filled with ones to represent a constant. This will be the intercept term of the equation. Once this is done, your values should match out properly.
Edited: replaced zeroes with ones