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
         


        

Answer 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')

Answer 2

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 df: the pandas dataframe containing only the predictor features, not the response variable
    :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 3

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 4

Although it is already late, I am adding some modifications from the given answer. To get the best set after removing multicollinearity if we use @Chef1075 solution then we will lose the variables which are correlated. We have to remove only one of them. To do this I came with the following solution using @steve answer:

import pandas as pd
from sklearn.linear_model import LinearRegression

def sklearn_vif(exogs, data):
    '''
    This function calculates variance inflation function in sklearn way. 
     It is a comparatively faster process.

    '''
    # 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
df = pd.DataFrame(
{'a': [1, 1, 2, 3, 4,1],
 'b': [2, 2, 3, 2, 1,3],
 'c': [4, 6, 7, 8, 9,5],
 'd': [4, 3, 4, 5, 4,6],
 'e': [8,8,14,15,17,20]}
  )

df_vif= sklearn_vif(exogs=df.columns, data=df).sort_values(by='VIF',ascending=False)
while (df_vif.VIF>5).any() ==True:
    red_df_vif= df_vif.drop(df_vif.index[0])
    df= df[red_df_vif.index]
    df_vif=sklearn_vif(exogs=df.columns,data=df).sort_values(by='VIF',ascending=False)




print(df)

   d  c  b
0  4  4  2
1  3  6  2
2  4  7  3
3  5  8  2
4  4  9  1
5  6  5  3

Answer 5

here code using dataframe python:

To create data

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]

To create dataframe

import pandas as pd
data = pd.DataFrame()
data["a"] = a
data["b"] = b
data["c"] = c
data["d"] = d

Calculate VIF

cc = np.corrcoef(data, rowvar=False)
VIF = np.linalg.inv(cc)
VIF.diagonal()

Result

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

Answer 6

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.