Common pitfalls in interpretation of coefficients of linear models

In linear models, the target value is modeled as a linear combination of the features (see the Linear Models User Guide section for a description of a set of linear models available in scikit-learn). Coefficients in multiple linear models represent the relationship between the given feature, \(X_i\) and the target, \(y\), assuming that all the other features remain constant (conditional dependence). This is different from plotting \(X_i\) versus \(y\) and fitting a linear relationship: in that case all possible values of the other features are taken into account in the estimation (marginal dependence).

This example will provide some hints in interpreting coefficient in linear models, pointing at problems that arise when either the linear model is not appropriate to describe the dataset, or when features are correlated.

We will use data from the “Current Population Survey” from 1985 to predict wage as a function of various features such as experience, age, or education.

print(__doc__)

import numpy as np
import scipy as sp
import pandas as pd
import matplotlib.pyplot as plt
import seaborn as sns

The dataset: wages

We fetch the data from OpenML. Note that setting the parameter as_frame to True will retrieve the data as a pandas dataframe.

from sklearn.datasets import fetch_openml

survey = fetch_openml(data_id=534, as_frame=True)

Then, we identify features X and targets y: the column WAGE is our target variable (i.e., the variable which we want to predict).

X = survey.data[survey.feature_names]
X.describe(include="all")
EDUCATION SOUTH SEX EXPERIENCE UNION AGE RACE OCCUPATION SECTOR MARR
count 534.000000 534 534 534.000000 534 534.000000 534 534 534 534
unique NaN 2 2 NaN 2 NaN 3 6 3 2
top NaN no male NaN not_member NaN White Other Other Married
freq NaN 378 289 NaN 438 NaN 440 156 411 350
mean 13.018727 NaN NaN 17.822097 NaN 36.833333 NaN NaN NaN NaN
std 2.615373 NaN NaN 12.379710 NaN 11.726573 NaN NaN NaN NaN
min 2.000000 NaN NaN 0.000000 NaN 18.000000 NaN NaN NaN NaN
25% 12.000000 NaN NaN 8.000000 NaN 28.000000 NaN NaN NaN NaN
50% 12.000000 NaN NaN 15.000000 NaN 35.000000 NaN NaN NaN NaN
75% 15.000000 NaN NaN 26.000000 NaN 44.000000 NaN NaN NaN NaN
max 18.000000 NaN NaN 55.000000 NaN 64.000000 NaN NaN NaN NaN


Note that the dataset contains categorical and numerical variables. We will need to take this into account when preprocessing the dataset thereafter.

X.head()
EDUCATION SOUTH SEX EXPERIENCE UNION AGE RACE OCCUPATION SECTOR MARR
0 8.0 no female 21.0 not_member 35.0 Hispanic Other Manufacturing Married
1 9.0 no female 42.0 not_member 57.0 White Other Manufacturing Married
2 12.0 no male 1.0 not_member 19.0 White Other Manufacturing Unmarried
3 12.0 no male 4.0 not_member 22.0 White Other Other Unmarried
4 12.0 no male 17.0 not_member 35.0 White Other Other Married


Our target for prediction: the wage. Wages are described as floating-point number in dollars per hour.

y = survey.target.values.ravel()
survey.target.head()

Out:

0    5.10
1    4.95
2    6.67
3    4.00
4    7.50
Name: WAGE, dtype: float64

We split the sample into a train and a test dataset. Only the train dataset will be used in the following exploratory analysis. This is a way to emulate a real situation where predictions are performed on an unknown target, and we don’t want our analysis and decisions to be biased by our knowledge of the test data.

from sklearn.model_selection import train_test_split

X_train, X_test, y_train, y_test = train_test_split(
    X, y, random_state=42
)

First, let’s get some insights by looking at the variable distributions and at the pairwise relationships between them. Only numerical variables will be used. In the following plot, each dot represents a sample.

train_dataset = X_train.copy()
train_dataset.insert(0, "WAGE", y_train)
_ = sns.pairplot(train_dataset, kind='reg', diag_kind='kde')
plot linear model coefficient interpretation

Looking closely at the WAGE distribution reveals that it has a long tail. For this reason, we should take its logarithm to turn it approximately into a normal distribution (linear models such as ridge or lasso work best for a normal distribution of error).

The WAGE is increasing when EDUCATION is increasing. Note that the dependence between WAGE and EDUCATION represented here is a marginal dependence, i.e., it describes the behavior of a specific variable without keeping the others fixed.

Also, the EXPERIENCE and AGE are strongly linearly correlated.

The machine-learning pipeline

To design our machine-learning pipeline, we first manually check the type of data that we are dealing with:

survey.data.info()

Out:

<class 'pandas.core.frame.DataFrame'>
RangeIndex: 534 entries, 0 to 533
Data columns (total 10 columns):
 #   Column      Non-Null Count  Dtype
---  ------      --------------  -----
 0   EDUCATION   534 non-null    float64
 1   SOUTH       534 non-null    category
 2   SEX         534 non-null    category
 3   EXPERIENCE  534 non-null    float64
 4   UNION       534 non-null    category
 5   AGE         534 non-null    float64
 6   RACE        534 non-null    category
 7   OCCUPATION  534 non-null    category
 8   SECTOR      534 non-null    category
 9   MARR        534 non-null    category
dtypes: category(7), float64(3)
memory usage: 17.1 KB

As seen previously, the dataset contains columns with different data types and we need to apply a specific preprocessing for each data types. In particular categorical variables cannot be included in linear model if not coded as integers first. In addition, to avoid categorical features to be treated as ordered values, we need to one-hot-encode them. Our pre-processor will

  • one-hot encode (i.e., generate a column by category) the categorical columns;

  • as a first approach (we will see after how the normalisation of numerical values will affect our discussion), keep numerical values as they are.

from sklearn.compose import make_column_transformer
from sklearn.preprocessing import OneHotEncoder

categorical_columns = ['RACE', 'OCCUPATION', 'SECTOR',
                       'MARR', 'UNION', 'SEX', 'SOUTH']
numerical_columns = ['EDUCATION', 'EXPERIENCE', 'AGE']

preprocessor = make_column_transformer(
    (OneHotEncoder(drop='if_binary'), categorical_columns),
    remainder='passthrough'
)

To describe the dataset as a linear model we use a ridge regressor with a very small regularization and to model the logarithm of the WAGE.

from sklearn.pipeline import make_pipeline
from sklearn.linear_model import Ridge
from sklearn.compose import TransformedTargetRegressor

model = make_pipeline(
    preprocessor,
    TransformedTargetRegressor(
        regressor=Ridge(alpha=1e-10),
        func=np.log10,
        inverse_func=sp.special.exp10
    )
)

Processing the dataset

First, we fit the model.

_ = model.fit(X_train, y_train)

Then we check the performance of the computed model plotting its predictions on the test set and computing, for example, the median absolute error of the model.

from sklearn.metrics import median_absolute_error

y_pred = model.predict(X_train)

mae = median_absolute_error(y_train, y_pred)
string_score = f'MAE on training set: {mae:.2f} $/hour'
y_pred = model.predict(X_test)
mae = median_absolute_error(y_test, y_pred)
string_score += f'\nMAE on testing set: {mae:.2f} $/hour'
fig, ax = plt.subplots(figsize=(5, 5))
plt.scatter(y_test, y_pred)
ax.plot([0, 1], [0, 1], transform=ax.transAxes, ls="--", c="red")
plt.text(3, 20, string_score)
plt.title('Ridge model, small regularization')
plt.ylabel('Model predictions')
plt.xlabel('Truths')
plt.xlim([0, 27])
_ = plt.ylim([0, 27])
Ridge model, small regularization

The model learnt is far from being a good model making accurate predictions: this is obvious when looking at the plot above, where good predictions should lie on the red line.

In the following section, we will interpret the coefficients of the model. While we do so, we should keep in mind that any conclusion we draw is about the model that we build, rather than about the true (real-world) generative process of the data.

Interpreting coefficients: scale matters

First of all, we can take a look to the values of the coefficients of the regressor we have fitted.

feature_names = (model.named_steps['columntransformer']
                      .named_transformers_['onehotencoder']
                      .get_feature_names(input_features=categorical_columns))
feature_names = np.concatenate(
    [feature_names, numerical_columns])

coefs = pd.DataFrame(
    model.named_steps['transformedtargetregressor'].regressor_.coef_,
    columns=['Coefficients'], index=feature_names
)

coefs
Coefficients
RACE_Hispanic -0.013564
RACE_Other -0.009120
RACE_White 0.022549
OCCUPATION_Clerical 0.000048
OCCUPATION_Management 0.090530
OCCUPATION_Other -0.025099
OCCUPATION_Professional 0.071966
OCCUPATION_Sales -0.046634
OCCUPATION_Service -0.091051
SECTOR_Construction -0.000183
SECTOR_Manufacturing 0.031270
SECTOR_Other -0.031011
MARR_Unmarried -0.032405
UNION_not_member -0.117154
SEX_male 0.090808
SOUTH_yes -0.033823
EDUCATION 0.054699
EXPERIENCE 0.035005
AGE -0.030867


The AGE coefficient is expressed in “dollars/hour per living years” while the EDUCATION one is expressed in “dollars/hour per years of education”. This representation of the coefficients has the benefit of making clear the practical predictions of the model: an increase of \(1\) year in AGE means a decrease of \(0.030867\) dollars/hour, while an increase of \(1\) year in EDUCATION means an increase of \(0.054699\) dollars/hour. On the other hand, categorical variables (as UNION or SEX) are adimensional numbers taking either the value 0 or 1. Their coefficients are expressed in dollars/hour. Then, we cannot compare the magnitude of different coefficients since the features have different natural scales, and hence value ranges, because of their different unit of measure. This is more visible if we plot the coefficients.

coefs.plot(kind='barh', figsize=(9, 7))
plt.title('Ridge model, small regularization')
plt.axvline(x=0, color='.5')
plt.subplots_adjust(left=.3)
Ridge model, small regularization

Indeed, from the plot above the most important factor in determining WAGE appears to be the variable UNION, even if our intuition might tell us that variables like EXPERIENCE should have more impact.

Looking at the coefficient plot to gauge feature importance can be misleading as some of them vary on a small scale, while others, like AGE, varies a lot more, several decades.

This is visible if we compare the standard deviations of different features.

X_train_preprocessed = pd.DataFrame(
    model.named_steps['columntransformer'].transform(X_train),
    columns=feature_names
)

X_train_preprocessed.std(axis=0).plot(kind='barh', figsize=(9, 7))
plt.title('Features std. dev.')
plt.subplots_adjust(left=.3)
Features std. dev.

Multiplying the coefficients by the standard deviation of the related feature would reduce all the coefficients to the same unit of measure. As we will see after this is equivalent to normalize numerical variables to their standard deviation, as \(y = \sum{coef_i \times X_i} = \sum{(coef_i \times std_i) \times (X_i / std_i)}\).

In that way, we emphasize that the greater the variance of a feature, the larger the weight of the corresponding coefficient on the output, all else being equal.

coefs = pd.DataFrame(
    model.named_steps['transformedtargetregressor'].regressor_.coef_ *
    X_train_preprocessed.std(axis=0),
    columns=['Coefficient importance'], index=feature_names
)
coefs.plot(kind='barh', figsize=(9, 7))
plt.title('Ridge model, small regularization')
plt.axvline(x=0, color='.5')
plt.subplots_adjust(left=.3)
Ridge model, small regularization

Now that the coefficients have been scaled, we can safely compare them.

Warning

Why does the plot above suggest that an increase in age leads to a decrease in wage? Why the initial pairplot is telling the opposite?

The plot above tells us about dependencies between a specific feature and the target when all other features remain constant, i.e., conditional dependencies. An increase of the AGE will induce a decrease of the WAGE when all other features remain constant. On the contrary, an increase of the EXPERIENCE will induce an increase of the WAGE when all other features remain constant. Also, AGE, EXPERIENCE and EDUCATION are the three variables that most influence the model.

Checking the variability of the coefficients

We can check the coefficient variability through cross-validation: it is a form of data perturbation (related to resampling).

If coefficients vary significantly when changing the input dataset their robustness is not guaranteed, and they should probably be interpreted with caution.

from sklearn.model_selection import cross_validate
from sklearn.model_selection import RepeatedKFold

cv_model = cross_validate(
    model, X, y, cv=RepeatedKFold(n_splits=5, n_repeats=5),
    return_estimator=True, n_jobs=-1
)
coefs = pd.DataFrame(
    [est.named_steps['transformedtargetregressor'].regressor_.coef_ *
     X_train_preprocessed.std(axis=0)
     for est in cv_model['estimator']],
    columns=feature_names
)
plt.figure(figsize=(9, 7))
sns.swarmplot(data=coefs, orient='h', color='k', alpha=0.5)
sns.boxplot(data=coefs, orient='h', color='cyan', saturation=0.5)
plt.axvline(x=0, color='.5')
plt.xlabel('Coefficient importance')
plt.title('Coefficient importance and its variability')
plt.subplots_adjust(left=.3)
Coefficient importance and its variability

The problem of correlated variables

The AGE and EXPERIENCE coefficients are affected by strong variability which might be due to the collinearity between the 2 features: as AGE and EXPERIENCE vary together in the data, their effect is difficult to tease apart.

To verify this interpretation we plot the variability of the AGE and EXPERIENCE coefficient.

plt.ylabel('Age coefficient')
plt.xlabel('Experience coefficient')
plt.grid(True)
plt.xlim(-0.4, 0.5)
plt.ylim(-0.4, 0.5)
plt.scatter(coefs["AGE"], coefs["EXPERIENCE"])
_ = plt.title('Co-variations of coefficients for AGE and EXPERIENCE '
              'across folds')
Co-variations of coefficients for AGE and EXPERIENCE across folds

Two regions are populated: when the EXPERIENCE coefficient is positive the AGE one is negative and viceversa.

To go further we remove one of the 2 features and check what is the impact on the model stability.

column_to_drop = ['AGE']

cv_model = cross_validate(
    model, X.drop(columns=column_to_drop), y,
    cv=RepeatedKFold(n_splits=5, n_repeats=5),
    return_estimator=True, n_jobs=-1
)
coefs = pd.DataFrame(
    [est.named_steps['transformedtargetregressor'].regressor_.coef_ *
     X_train_preprocessed.drop(columns=column_to_drop).std(axis=0)
     for est in cv_model['estimator']],
    columns=feature_names[:-1]
)
plt.figure(figsize=(9, 7))
sns.swarmplot(data=coefs, orient='h', color='k', alpha=0.5)
sns.boxplot(data=coefs, orient='h', color='cyan', saturation=0.5)
plt.axvline(x=0, color='.5')
plt.title('Coefficient importance and its variability')
plt.xlabel('Coefficient importance')
plt.subplots_adjust(left=.3)
Coefficient importance and its variability

The estimation of the EXPERIENCE coefficient is now less variable and remain important for all models trained during cross-validation.

Preprocessing numerical variables

As said above (see “The machine-learning pipeline”), we could also choose to scale numerical values before training the model. This can be useful to apply a similar amount regularization to all of them in the Ridge. The preprocessor is redefined in order to subtract the mean and scale variables to unit variance.

from sklearn.preprocessing import StandardScaler

preprocessor = make_column_transformer(
    (OneHotEncoder(drop='if_binary'), categorical_columns),
    (StandardScaler(), numerical_columns),
    remainder='passthrough'
)

The model will stay unchanged.

model = make_pipeline(
    preprocessor,
    TransformedTargetRegressor(
        regressor=Ridge(alpha=1e-10),
        func=np.log10,
        inverse_func=sp.special.exp10
    )
)

_ = model.fit(X_train, y_train)

Again, we check the performance of the computed model using, for example, the median absolute error of the model and the R squared coefficient.

y_pred = model.predict(X_train)
mae = median_absolute_error(y_train, y_pred)
string_score = f'MAE on training set: {mae:.2f} $/hour'
y_pred = model.predict(X_test)
mae = median_absolute_error(y_test, y_pred)
string_score += f'\nMAE on testing set: {mae:.2f} $/hour'
fig, ax = plt.subplots(figsize=(6, 6))
plt.scatter(y_test, y_pred)
ax.plot([0, 1], [0, 1], transform=ax.transAxes, ls="--", c="red")

plt.text(3, 20, string_score)

plt.title('Ridge model, small regularization, normalized variables')
plt.ylabel('Model predictions')
plt.xlabel('Truths')
plt.xlim([0, 27])
_ = plt.ylim([0, 27])
Ridge model, small regularization, normalized variables

For the coefficient analysis, scaling is not needed this time.

coefs = pd.DataFrame(
    model.named_steps['transformedtargetregressor'].regressor_.coef_,
    columns=['Coefficients'], index=feature_names
)
coefs.plot(kind='barh', figsize=(9, 7))
plt.title('Ridge model, small regularization, normalized variables')
plt.axvline(x=0, color='.5')
plt.subplots_adjust(left=.3)
Ridge model, small regularization, normalized variables

We now inspect the coefficients across several cross-validation folds.

cv_model = cross_validate(
    model, X, y, cv=RepeatedKFold(n_splits=5, n_repeats=5),
    return_estimator=True, n_jobs=-1
)
coefs = pd.DataFrame(
    [est.named_steps['transformedtargetregressor'].regressor_.coef_
     for est in cv_model['estimator']],
    columns=feature_names
)
plt.figure(figsize=(9, 7))
sns.swarmplot(data=coefs, orient='h', color='k', alpha=0.5)
sns.boxplot(data=coefs, orient='h', color='cyan', saturation=0.5)
plt.axvline(x=0, color='.5')
plt.title('Coefficient variability')
plt.subplots_adjust(left=.3)
Coefficient variability

The result is quite similar to the non-normalized case.

Linear models with regularization

In machine-learning practice, Ridge Regression is more often used with non-negligible regularization.

Above, we limited this regularization to a very little amount. Regularization improves the conditioning of the problem and reduces the variance of the estimates. RidgeCV applies cross validation in order to determine which value of the regularization parameter (alpha) is best suited for prediction.

from sklearn.linear_model import RidgeCV

model = make_pipeline(
    preprocessor,
    TransformedTargetRegressor(
        regressor=RidgeCV(alphas=np.logspace(-10, 10, 21)),
        func=np.log10,
        inverse_func=sp.special.exp10
    )
)

_ = model.fit(X_train, y_train)

First we check which value of \(\alpha\) has been selected.

model[-1].regressor_.alpha_

Out:

10.0

Then we check the quality of the predictions.

y_pred = model.predict(X_train)
mae = median_absolute_error(y_train, y_pred)
string_score = f'MAE on training set: {mae:.2f} $/hour'
y_pred = model.predict(X_test)
mae = median_absolute_error(y_test, y_pred)
string_score += f'\nMAE on testing set: {mae:.2f} $/hour'

fig, ax = plt.subplots(figsize=(6, 6))
plt.scatter(y_test, y_pred)
ax.plot([0, 1], [0, 1], transform=ax.transAxes, ls="--", c="red")

plt.text(3, 20, string_score)

plt.title('Ridge model, regularization, normalized variables')
plt.ylabel('Model predictions')
plt.xlabel('Truths')
plt.xlim([0, 27])
_ = plt.ylim([0, 27])
Ridge model, regularization, normalized variables

The ability to reproduce the data of the regularized model is similar to the one of the non-regularized model.

coefs = pd.DataFrame(
    model.named_steps['transformedtargetregressor'].regressor_.coef_,
    columns=['Coefficients'], index=feature_names
)
coefs.plot(kind='barh', figsize=(9, 7))
plt.title('Ridge model, regularization, normalized variables')
plt.axvline(x=0, color='.5')
plt.subplots_adjust(left=.3)
Ridge model, regularization, normalized variables

The coefficients are significantly different. AGE and EXPERIENCE coefficients are both positive but they now have less influence on the prediction.

The regularization reduces the influence of correlated variables on the model because the weight is shared between the two predictive variables, so neither alone would have strong weights.

On the other hand, the weights obtained with regularization are more stable (see the Ridge regression and classification User Guide section). This increased stability is visible from the plot, obtained from data perturbations, in a cross validation. This plot can be compared with the previous one.

cv_model = cross_validate(
    model, X, y, cv=RepeatedKFold(n_splits=5, n_repeats=5),
    return_estimator=True, n_jobs=-1
)
coefs = pd.DataFrame(
    [est.named_steps['transformedtargetregressor'].regressor_.coef_ *
     X_train_preprocessed.std(axis=0)
     for est in cv_model['estimator']],
    columns=feature_names
)

plt.ylabel('Age coefficient')
plt.xlabel('Experience coefficient')
plt.grid(True)
plt.xlim(-0.4, 0.5)
plt.ylim(-0.4, 0.5)
plt.scatter(coefs["AGE"], coefs["EXPERIENCE"])
_ = plt.title('Co-variations of coefficients for AGE and EXPERIENCE '
              'across folds')
Co-variations of coefficients for AGE and EXPERIENCE across folds

Linear models with sparse coefficients

Another possibility to take into account correlated variables in the dataset, is to estimate sparse coefficients. In some way we already did it manually when we dropped the AGE column in a previous Ridge estimation.

Lasso models (see the Lasso User Guide section) estimates sparse coefficients. LassoCV applies cross validation in order to determine which value of the regularization parameter (alpha) is best suited for the model estimation.

from sklearn.linear_model import LassoCV

model = make_pipeline(
    preprocessor,
    TransformedTargetRegressor(
        regressor=LassoCV(alphas=np.logspace(-10, 10, 21), max_iter=100000),
        func=np.log10,
        inverse_func=sp.special.exp10
    )
)

_ = model.fit(X_train, y_train)

First we verify which value of \(\alpha\) has been selected.

model[-1].regressor_.alpha_

Out:

0.001

Then we check the quality of the predictions.

y_pred = model.predict(X_train)
mae = median_absolute_error(y_train, y_pred)
string_score = f'MAE on training set: {mae:.2f} $/hour'
y_pred = model.predict(X_test)
mae = median_absolute_error(y_test, y_pred)
string_score += f'\nMAE on testing set: {mae:.2f} $/hour'

fig, ax = plt.subplots(figsize=(6, 6))
plt.scatter(y_test, y_pred)
ax.plot([0, 1], [0, 1], transform=ax.transAxes, ls="--", c="red")

plt.text(3, 20, string_score)

plt.title('Lasso model, regularization, normalized variables')
plt.ylabel('Model predictions')
plt.xlabel('Truths')
plt.xlim([0, 27])
_ = plt.ylim([0, 27])
Lasso model, regularization, normalized variables

For our dataset, again the model is not very predictive.

coefs = pd.DataFrame(
    model.named_steps['transformedtargetregressor'].regressor_.coef_,
    columns=['Coefficients'], index=feature_names
)
coefs.plot(kind='barh', figsize=(9, 7))
plt.title('Lasso model, regularization, normalized variables')
plt.axvline(x=0, color='.5')
plt.subplots_adjust(left=.3)
Lasso model, regularization, normalized variables

A Lasso model identifies the correlation between AGE and EXPERIENCE and suppresses one of them for the sake of the prediction.

It is important to keep in mind that the coefficients that have been dropped may still be related to the outcome by themselves: the model chose to suppress them because they bring little or no additional information on top of the other features. Additionnaly, this selection is unstable for correlated features, and should be interpreted with caution.

Lessons learned

  • Coefficients must be scaled to the same unit of measure to retrieve feature importance. Scaling them with the standard-deviation of the feature is a useful proxy.

  • Coefficients in multivariate linear models represent the dependency between a given feature and the target, conditional on the other features.

  • Correlated features induce instabilities in the coefficients of linear models and their effects cannot be well teased apart.

  • Different linear models respond differently to feature correlation and coefficients could significantly vary from one another.

  • Inspecting coefficients across the folds of a cross-validation loop gives an idea of their stability.

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