Variance Inflation Factor in Python

Question

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?

Answer 1

Example for Boston Data:

VIF is calculated by auxiliary regression, so not dependent on the actual fit.

See below:

from patsy import dmatrices
from statsmodels.stats.outliers_influence import variance_inflation_factor
import statsmodels.api as sm

# Break into left and right hand side; y and X
y, X = dmatrices(formula="medv ~ crim + zn + nox + ptratio + black + rm ", data=boston, return_type="dataframe")

# For each Xi, calculate VIF
vif = [variance_inflation_factor(X.values, i) for i in range(X.shape[1])]

# Fit X to y
result = sm.OLS(y, X).fit()

Answer 2

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

Answer 3

For future comers to this thread (like me):

import numpy as np
import scipy as sp

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])
cc = sp.corrcoef(ck, rowvar=False)
VIF = np.linalg.inv(cc)
VIF.diagonal()

This code gives

array([22.95,  3.  , 12.95,  3.  ])

[EDIT]

In response to a comment, I tried to use DataFrame as much as possible (numpy is required to invert a matrix).

import pandas as pd
import numpy as np

a = [1, 1, 2, 3, 4]
b = [2, 2, 3, 2, 1]
c = [4, 6, 7, 8, 9]
d = [4, 3, 4, 5, 4]

df = pd.DataFrame({'a':a,'b':b,'c':c,'d':d})
df_cor = df.corr()
pd.DataFrame(np.linalg.inv(df.corr().values), index = df_cor.index, columns=df_cor.columns)

The code gives

       a            b           c           d
a   22.950000   6.453681    -16.301917  -6.453681
b   6.453681    3.000000    -4.080441   -2.000000
c   -16.301917  -4.080441   12.950000   4.080441
d   -6.453681   -2.000000   4.080441    3.000000

The diagonal elements give VIF.

Answer 4

I wrote this function based on some other posts I saw on Stack and CrossValidated. It shows the features which are over the threshold and returns a new dataframe with the features removed.

from statsmodels.stats.outliers_influence import variance_inflation_factor 
from statsmodels.tools.tools import add_constant

def calculate_vif_(df, thresh=5):
    '''
    Calculates VIF each feature in a pandas dataframe
    A constant must be added to variance_inflation_factor or the results will be incorrect

    :param X: the pandas dataframe
    :param thresh: the max VIF value before the feature is removed from the dataframe
    :return: dataframe with features removed
    '''
    const = add_constant(df)
    cols = const.columns
    variables = np.arange(const.shape[1])
    vif_df = pd.Series([variance_inflation_factor(const.values, i) 
               for i in range(const.shape[1])], 
              index=const.columns).to_frame()

    vif_df = vif_df.sort_values(by=0, ascending=False).rename(columns={0: 'VIF'})
    vif_df = vif_df.drop('const')
    vif_df = vif_df[vif_df['VIF'] > thresh]

    print 'Features above VIF threshold:\n'
    print vif_df[vif_df['VIF'] > thresh]

    col_to_drop = list(vif_df.index)

    for i in col_to_drop:
        print 'Dropping: {}'.format(i)
        df = df.drop(columns=i)

    return df

Answer 5

In case you don't wanna deal with variance_inflation_factor and add_constant. Please consider the following two functions.

1. Use formula in statasmodels:

import pandas as pd
import statsmodels.formula.api as smf

def get_vif(exogs, data):
    '''Return VIF (variance inflation factor) DataFrame

    Args:
    exogs (list): list of exogenous/independent variables
    data (DataFrame): the df storing all variables

    Returns:
    VIF and Tolerance DataFrame for each exogenous variable

    Notes:
    Assume we have a list of exogenous variable [X1, X2, X3, X4].
    To calculate the VIF and Tolerance for each variable, we regress
    each of them against other exogenous variables. For instance, the
    regression model for X3 is defined as:
                        X3 ~ X1 + X2 + X4
    And then we extract the R-squared from the model to calculate:
                    VIF = 1 / (1 - R-squared)
                    Tolerance = 1 - R-squared
    The cutoff to detect multicollinearity:
                    VIF > 10 or Tolerance < 0.1
    '''

    # initialize dictionaries
    vif_dict, tolerance_dict = {}, {}

    # create formula for each exogenous variable
    for exog in exogs:
        not_exog = [i for i in exogs if i != exog]
        formula = f"{exog} ~ {' + '.join(not_exog)}"

        # extract r-squared from the fit
        r_squared = smf.ols(formula, data=data).fit().rsquared

        # calculate VIF
        vif = 1/(1 - r_squared)
        vif_dict[exog] = vif

        # calculate tolerance
        tolerance = 1 - r_squared
        tolerance_dict[exog] = tolerance

    # return VIF DataFrame
    df_vif = pd.DataFrame({'VIF': vif_dict, 'Tolerance': tolerance_dict})

    return df_vif

---

2. Use LinearRegression in sklearn:

# import warnings
# warnings.simplefilter(action='ignore', category=FutureWarning)
import pandas as pd
from sklearn.linear_model import LinearRegression

def sklearn_vif(exogs, data):

    # initialize dictionaries
    vif_dict, tolerance_dict = {}, {}

    # form input data for each exogenous variable
    for exog in exogs:
        not_exog = [i for i in exogs if i != exog]
        X, y = data[not_exog], data[exog]

        # extract r-squared from the fit
        r_squared = LinearRegression().fit(X, y).score(X, y)

        # calculate VIF
        vif = 1/(1 - r_squared)
        vif_dict[exog] = vif

        # calculate tolerance
        tolerance = 1 - r_squared
        tolerance_dict[exog] = tolerance

    # return VIF DataFrame
    df_vif = pd.DataFrame({'VIF': vif_dict, 'Tolerance': tolerance_dict})

    return df_vif

---

Example:

import seaborn as sns

df = sns.load_dataset('car_crashes')
exogs = ['alcohol', 'speeding', 'no_previous', 'not_distracted']

[In] %%timeit -n 100
get_vif(exogs=exogs, data=df)

[Out]
                      VIF   Tolerance
alcohol          3.436072   0.291030
no_previous      3.113984   0.321132
not_distracted   2.668456   0.374749
speeding         1.884340   0.530690

69.6 ms ± 8.96 ms per loop (mean ± std. dev. of 7 runs, 100 loops each)

[In] %%timeit -n 100
sklearn_vif(exogs=exogs, data=df)

[Out]
                      VIF   Tolerance
alcohol          3.436072   0.291030
no_previous      3.113984   0.321132
not_distracted   2.668456   0.374749
speeding         1.884340   0.530690

15.7 ms ± 1.4 ms per loop (mean ± std. dev. of 7 runs, 100 loops each)

Answer 6

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 the intercept term of the equation. Once this is done, your values should match out properly.

Edited: replaced zeroes with ones