Section 4.6 — Generalized linear models

Section 4.6 — Generalized linear models#

This notebook contains the code examples from Section 4.6 Generalized linear models from the No Bullshit Guide to Statistics.

Notebook setup#

# load Python modules
import os
import numpy as np
import pandas as pd
import seaborn as sns

# Figures setup
import matplotlib.pyplot as plt
plt.clf()  # needed otherwise `sns.set_theme` doesn't work
from plot_helpers import RCPARAMS
RCPARAMS.update({"figure.figsize": (5, 3)})   # good for screen
# RCPARAMS.update({'figure.figsize': (10, 3)})   # good for screen
# RCPARAMS.update({'figure.figsize': (4, 2)})  # good for print
sns.set_theme(
    context="paper",
    style="whitegrid",
    palette="colorblind",
    rc=RCPARAMS,
)

# High-resolution please
%config InlineBackend.figure_format = 'retina'

# Where to store figures
DESTDIR = "figures/lm/generalized"

<Figure size 640x480 with 0 Axes>

from ministats.utils import savefigure

#######################################################

Definitions#

Probability representations and link functions#

Odds#

0.5/(1-0.5), 0.9/(1-0.9), 0.2/(1-0.2)

(1.0, 9.000000000000002, 0.25)

Log-odds#

np.log(0.5/(1-0.5)), np.log(0.9/(1-0.9)), np.log(0.2/(1-0.2))

(0.0, 2.1972245773362196, -1.3862943611198906)

The logit function#

def logit(p):
    x = np.log(p / (1-p))
    return x

# ALT. import the function from `scipy.special`
from scipy.special import logit

logit(0.5), logit(0.9), logit(0.2)

(0.0, 2.1972245773362196, -1.3862943611198906)

The logistic function#

def expit(x):
    p = 1 / (1 + np.exp(-x))
    return p

# ALT. import the function from `scipy.special`
from scipy.special import expit

expit(0), expit(2), expit(-2)

(0.5, 0.8807970779778823, 0.11920292202211755)

# FIGURES ONLY
fig, (ax1, ax2) = plt.subplots(1, 2, figsize=(6, 2))

# plot the expit function, a.k.a. the logistic function
xs = np.linspace(-6, 6, 500)
sns.lineplot(x=xs, y=expit(xs), ax=ax1, label="$\\mathrm{expit}(x)$")
ax1.set_xlabel("$x$")
ax1.set_ylabel("$p$")

# plot the logit function
ps = np.linspace(0, 1, 600)
sns.lineplot(x=ps, y=logit(ps), ax=ax2, label="$\\mathrm{logit}(p)$", color="C1")
ax2.set_xlabel("$p$")
ax2.set_ylabel("$x$")


filename = os.path.join(DESTDIR, "logistic_and_logit_functions.pdf")
savefigure(fig, filename, tight_layout_kwargs=dict(w_pad=3))

Saved figure to figures/lm/generalized/logistic_and_logit_functions.pdf

Saved figure to figures/lm/generalized/logistic_and_logit_functions.png

../_images/1dfe18d065c7cd84fa9f8c77c33fcd0a44aa19182ee4620439b51efec08f3fd4.png

The logistic and logit functions are inverses#

expit(logit(0.2))

0.2

logit(expit(3))

3.000000000000003

Logistic regression#

TODO FORMULA

# FIGURES ONLY
from scipy.stats import bernoulli
from scipy.special import expit

# Define the logistic regression model function
def expit_model(x):
    p = expit(-10 + 2*x)
    return p

xlims = [0, 10]

stem_half_width = 0.03

with sns.axes_style("ticks"):
    fig, ax = plt.subplots(figsize=(5, 3))

    # Plot the logistic regression model
    xs = np.linspace(xlims[0], xlims[1], 200)
    ps = expit_model(xs)
    sns.lineplot(x=xs, y=ps, ax=ax, label=r"$p(x) = \mathrm{expit}(\beta_0 + \beta_1x)$", linewidth=2)

    # Plot Bernoulli distributions at specified x positions
    x_positions = [2,4,5,6,8,10]
    for x_pos in x_positions:
        p_pos = expit_model(x_pos)
        ys = [0,1]
        pmf = bernoulli(p=p_pos).pmf(ys)
        ys_plot = [p_pos-stem_half_width, p_pos+stem_half_width]
        ax.stem(ys_plot, x_pos - pmf, bottom=x_pos, orientation='horizontal')

    # Figure setup
    ax.set_xlabel("$x$")
    ax.set_ylabel("$p$")
    ax.legend(loc="upper left")

    filename = os.path.join(DESTDIR, "logistic_regression_xy_with_stemplots.pdf")
    savefigure(fig, filename)

Saved figure to figures/lm/generalized/logistic_regression_xy_with_stemplots.pdf
Saved figure to figures/lm/generalized/logistic_regression_xy_with_stemplots.png

../_images/7037635c81105b3b63caa1767e533482a4d9d79863285f5b1b3effbb51090082.png

expit(-6)

0.0024726231566347743

expit(10)

0.9999546021312976

Example 1: hiring student interns#

interns = pd.read_csv("../datasets/interns.csv")
print(interns.shape)
interns.head(3)

(100, 2)

	work	hired
0	42.5	1
1	39.3	0
2	43.2	1

import statsmodels.formula.api as smf

lr1 = smf.logit("hired ~ 1 + work", data=interns).fit()
print(lr1.params)

Optimization terminated successfully.
         Current function value: 0.138101
         Iterations 10
Intercept   -78.693205
work          1.981458
dtype: float64

lr1.summary()

Logit Regression Results
Dep. Variable:	hired	No. Observations:	100
Model:	Logit	Df Residuals:	98
Method:	MLE	Df Model:	1
Date:	Thu, 05 Sep 2024	Pseudo R-squ.:	0.8005
Time:	15:49:30	Log-Likelihood:	-13.810
converged:	True	LL-Null:	-69.235
Covariance Type:	nonrobust	LLR p-value:	6.385e-26

	coef	std err	z	P>\|z\|	[0.025	0.975]
Intercept	-78.6932	19.851	-3.964	0.000	-117.600	-39.787
work	1.9815	0.500	3.959	0.000	1.001	2.962

Possibly complete quasi-separation: A fraction 0.32 of observations can be
perfectly predicted. This might indicate that there is complete
quasi-separation. In this case some parameters will not be identified.

ax = sns.scatterplot(data=interns, x="work", y="hired")
wgrid = np.linspace(27, 50, 100)
hired_preds = lr1.predict({"work": wgrid})
sns.lineplot(x=wgrid, y=hired_preds, ax=ax);

../_images/177d412a04ceb10fa9c484d9973a2e95022cecb8b2182df55c77a05d85cdff8b.png

from ministats import plot_reg
plot_reg(lr1)

# FIGURES ONLY
filename = os.path.join(DESTDIR, "logistic_regression_interns_hired_vs_work.pdf")
savefigure(plt.gcf(), filename)

Saved figure to figures/lm/generalized/logistic_regression_interns_hired_vs_work.pdf
Saved figure to figures/lm/generalized/logistic_regression_interns_hired_vs_work.png

../_images/77b8dddda53b94cfbb8c1ab358072b2fbff2eb97114c00aa7b281f5ed3b0ebc3.png

lr1.summary()

Logit Regression Results
Dep. Variable:	hired	No. Observations:	100
Model:	Logit	Df Residuals:	98
Method:	MLE	Df Model:	1
Date:	Thu, 05 Sep 2024	Pseudo R-squ.:	0.8005
Time:	15:49:31	Log-Likelihood:	-13.810
converged:	True	LL-Null:	-69.235
Covariance Type:	nonrobust	LLR p-value:	6.385e-26

	coef	std err	z	P>\|z\|	[0.025	0.975]
Intercept	-78.6932	19.851	-3.964	0.000	-117.600	-39.787
work	1.9815	0.500	3.959	0.000	1.001	2.962

Possibly complete quasi-separation: A fraction 0.32 of observations can be
perfectly predicted. This might indicate that there is complete
quasi-separation. In this case some parameters will not be identified.

Interpreting the model parameters#

Parameters as changes in the log-odds#

lr1.params["work"]

1.9814577697476699

Parameters as ratios of odds#

np.exp(lr1.params["work"])

7.25330893626573

Differences in probabilities#

What is the marginal effect of the predictor work for an intern who invests 40 hours of effort?

# using `statsmodels`
lr1.get_margeff(atexog={1:40}).summary_frame()

	dy/dx	Std. Err.	z	Pr(>\|z\|)	Conf. Int. Low	Cont. Int. Hi.
work	0.45783	0.112623	4.065157	0.000048	0.237093	0.678567

lr1.get_margeff(atexog={1:42}).summary_frame()

	dy/dx	Std. Err.	z	Pr(>\|z\|)	Conf. Int. Low	Cont. Int. Hi.
work	0.020949	0.021358	0.98084	0.326672	-0.020912	0.06281

# # ALT. manual calculation plugging into derivative of `expit`
# p40 = lr1.predict({"work":40}).item()
# marg_effect_at_40 = p40 * (1 - p40) * lr1.params['work']
# marg_effect_at_40

Prediction#

p42 = lr1.predict({"work":42})[0]
p42

0.9893134055105761

Poisson regression#

# FIGURES ONLY
from scipy.stats import poisson

# Define the linear model function
def exp_model(x):
    lam = np.exp(1 + 0.2*x)
    return lam

onepixel = 0.07

xlims = [0, 20]
ylims = [0, 100]

with sns.axes_style("ticks"):
    fig, ax = plt.subplots(figsize=(5, 3))

    # Plot the linear model
    xs = np.linspace(xlims[0], xlims[1], 200)
    lams = exp_model(xs)
    sns.lineplot(x=xs, y=lams, ax=ax, label=r"$\mu_Y(x) = \exp(\beta_0 + \beta_1x)$", linewidth=2)
    
    # Plot Gaussian distributions at specified x positions and add sigma lines
    x_positions = range(2, xlims[1]-1, 3)
    for x_pos in x_positions:
        lam_pos = exp_model(x_pos)
        sigma = np.sqrt(lam_pos)
        ys_lower = int(lam_pos-2.5*sigma)
        ys_upper = int(lam_pos+3.4*sigma)
        ys = np.arange(ys_lower, ys_upper, 3)
        pmf = poisson(mu=lam_pos).pmf(ys)
        # ax.fill_betweenx(ys, x_pos - 2 * pmf * sigma, x_pos, color="grey", alpha=0.5)
        ax.stem(ys, x_pos- 2 * pmf * sigma, bottom=x_pos, orientation='horizontal')
        # Draw vertical sigma line and label it on the opposite side of the Gaussian shape
        # ax.plot([x_pos+onepixel, x_pos+onepixel], [lam_pos, lam_pos - sigma], "k", lw=1)
        # ax.text(x_pos + 0.1, lam_pos - sigma / 2 - 3*onepixel, r"$\sigma$", fontsize=12, va="center")

    # y-intercept
    ax.text(0 - 0.6, np.exp(1), r"$\exp(\beta_0)$", fontsize=10, va="center", ha="right")

    # Set up x-axis
    ax.set_xlim(xlims)
    ax.set_xlabel("$x$")
    ax.set_xticks(range(2, xlims[1], 3))
    ax.set_xticklabels([])
    
    # Set up y-axis
    ax.set_ylim([ylims[0]-4,ylims[1]])
    ax.set_ylabel("$y$")
    ax.set_yticks(list(range(ylims[0],ylims[1],20)) + [np.exp(1)] )
    ax.set_yticklabels([])
    
    ax.legend(loc="upper left")

    filename = os.path.join(DESTDIR, "poisson_regression_xy_with_stemplots.pdf")
    savefigure(fig, filename)

Saved figure to figures/lm/generalized/poisson_regression_xy_with_stemplots.pdf
Saved figure to figures/lm/generalized/poisson_regression_xy_with_stemplots.png

../_images/d7a4ae8a998f4cb89c56d709ecf787c03178bb6afea6843720bb0a8c4435568b.png

Example 2: hard disk failures over time#

hdisks = pd.read_csv("../datasets/hdisks.csv")
hdisks.head(3)

	age	failures
0	1.7	3
1	14.6	46
2	10.9	23

import statsmodels.formula.api as smf

pr2 = smf.poisson("failures ~ 1 + age", data=hdisks).fit()
pr2.params

Optimization terminated successfully.
         Current function value: 2.693129
         Iterations 6

Intercept    1.075999
age          0.193828
dtype: float64

pr2.summary()

Poisson Regression Results
Dep. Variable:	failures	No. Observations:	100
Model:	Poisson	Df Residuals:	98
Method:	MLE	Df Model:	1
Date:	Thu, 05 Sep 2024	Pseudo R-squ.:	0.6412
Time:	15:49:32	Log-Likelihood:	-269.31
converged:	True	LL-Null:	-750.68
Covariance Type:	nonrobust	LLR p-value:	2.271e-211

	coef	std err	z	P>\|z\|	[0.025	0.975]
Intercept	1.0760	0.076	14.114	0.000	0.927	1.225
age	0.1938	0.007	28.603	0.000	0.181	0.207

from ministats import plot_reg
plot_reg(pr2)

# FIGURES ONLY
filename = os.path.join(DESTDIR, "poisson_regression_hdisks_failures_vs_age.pdf")
savefigure(plt.gcf(), filename)

Saved figure to figures/lm/generalized/poisson_regression_hdisks_failures_vs_age.pdf
Saved figure to figures/lm/generalized/poisson_regression_hdisks_failures_vs_age.png

../_images/984dd7f8d91a8ed7563622bc958a47fb16fcb896b19c2135f4653157d253f488.png

pr2.summary()

Poisson Regression Results
Dep. Variable:	failures	No. Observations:	100
Model:	Poisson	Df Residuals:	98
Method:	MLE	Df Model:	1
Date:	Thu, 05 Sep 2024	Pseudo R-squ.:	0.6412
Time:	15:49:33	Log-Likelihood:	-269.31
converged:	True	LL-Null:	-750.68
Covariance Type:	nonrobust	LLR p-value:	2.271e-211

	coef	std err	z	P>\|z\|	[0.025	0.975]
Intercept	1.0760	0.076	14.114	0.000	0.927	1.225
age	0.1938	0.007	28.603	0.000	0.181	0.207

Interpreting the model parameters#

Log-counts#

pr2.params["age"]

0.19382784821454072

Incidence rate ratio (IRR)#

np.exp(pr2.params["age"])

1.213887292102993

Marginal effect#

What is the marginal effect of the predictor age for a 10 year old hard disk installation?

# using `statsmodels` .get_margeff() method
pr2.get_margeff(atexog={1:10}).summary_frame()

	dy/dx	Std. Err.	z	Pr(>\|z\|)	Conf. Int. Low	Cont. Int. Hi.
age	3.94912	0.151882	26.001165	4.804151e-149	3.651435	4.246804

# # ALT. manual calculation of the slope by evaluating the derivative
# b_0 = pr2.params['Intercept']
# b_age = pr2.params['age']
# np.exp(b_0 + b_age*10)*b_age

Predictions#

lam10 = pr2.predict({"age":10})[0]
lam10

20.374365915173986

from scipy.stats import poisson
Hhat = poisson(mu=lam10)
Hhat.ppf(0.05), Hhat.ppf(0.95)

(13.0, 28.0)

Explanations#

The exponential family of distributions#

exponential
Gaussian (normal)
Poisson
Binomial

The generalized linear model template#

choose

Generalized linear models using statsmodels#

import statsmodels.api as sm

Norm = sm.families.Gaussian()
Bin = sm.families.Binomial()
Pois = sm.families.Poisson()

Linear model#

students = pd.read_csv('../datasets/students.csv')
formula0 = "score ~ 1 + effort"
glm0 = smf.glm(formula0, data=students, family=Norm).fit()
glm0.params

Intercept    32.465809
effort        4.504850
dtype: float64

Logistic regression#

formula1 = "hired ~ 1 + work"
glm1 = smf.glm(formula1, data=interns, family=Bin).fit()
glm1.params
# glm1.summary()

Intercept   -78.693205
work          1.981458
dtype: float64

Poisson regression#

formula2 = "failures ~ 1 + age"
glm2 = smf.glm(formula2, data=hdisks, family=Pois).fit()
glm2.params
# glm2.summary()

Intercept    1.075999
age          0.193828
dtype: float64

Fitting generalized linear models#

Standardization of predictors#

from scipy.stats import zscore

efforts = students["effort"]
zefforts = zscore(efforts)
zefforts.head(3)

  1.092044
 -0.114057
 -0.161876
Name: effort, dtype: float64

# # ALT. define custom function equivalent to zscore(data)
# def standardize(data):
#     datamean = data.mean()
#     datastd = data.std(ddof=1)
#     zdata = (data - datamean) / datastd
#     return zdata

students["zeffort"] = zefforts
lm1s = smf.ols("score~1+zeffort", data=students).fit()
lm1s.params

Intercept    72.580000
zeffort       8.478568
dtype: float64

lm1s.params["zeffort"] / students["effort"].std(ddof=0)

4.504850344209072

glm0.params["effort"]

4.504850344209074

Marginal effects#

It can be difficult to interpret GLM parameters directly, but we can always ask the question about “slopes” the rate of change of interesting parameters.

import marginaleffects as me

---------------------------------------------------------------------------
ModuleNotFoundError                       Traceback (most recent call last)
Cell In[50], line 1
----> 1 import marginaleffects as me

ModuleNotFoundError: No module named 'marginaleffects'

Raw parameters#

lr1.params["work"], np.exp(lr1.params["work"])

(1.9814577697476699, 7.25330893626573)

The parameter tells us the log-odds ratio changes by 1.981, or equivalently, that the odd-ratio changes by a factor of 7.253 for each additional hour of work.

Marginal effect at a user-specified value#

Example To calculate the marginal effect of the predictor work for an intern who invests 40 hours of effort.

# using `statsmodels`
lr1.get_margeff(atexog={1:40}).summary_frame()

	dy/dx	Std. Err.	z	Pr(>\|z\|)	Conf. Int. Low	Cont. Int. Hi.
work	0.45783	0.112623	4.065157	0.000048	0.237093	0.678567

# ALT. using `marginaleffects`
dg40 = me.datagrid(lr1, work=[40])
me.slopes(lr1, newdata=dg40).to_pandas()

	term	contrast	estimate	std_error	statistic	p_value	s_value	conf_low	conf_high
0	work	dY/dX	0.45783	0.112263	4.078207	0.000045	14.427447	0.237799	0.677861

# ALT2. manual calculation plugging into derivative of `expit`
p40 = lr1.predict({"work":40}).item()
marg_effect_at_40 = p40 * (1 - p40) * lr1.params['work']
marg_effect_at_40

0.45782997989918905

Average marginal effect (AME)#

For each observation \((w_i,h_i)\), compute the marginal effect at \(w=w_i\), then average them together.

lr1.get_margeff().summary_frame()

	dy/dx	Std. Err.	z	Pr(>\|z\|)	Conf. Int. Low	Cont. Int. Hi.
work	0.08077	0.00422	19.139418	1.185874e-81	0.072499	0.089041

# ALT. using the `marginaleffects` package
me.avg_slopes(lr1).to_pandas()

	term	contrast	estimate	std_error	statistic	p_value	s_value	conf_low	conf_high
0	work	mean(dY/dX)	0.08077	0.00422	19.138224	0.0	inf	0.072498	0.089042

# ALT2. manual computation using a for-loop
meffects = []
for i, row in interns.iterrows():
    p = lr1.predict({"work":row["work"]}).item()
    meffect = p * (1-p) * lr1.params['work']
    meffects.append(meffect)
AME = np.mean(meffects)
AME

0.08076985840552156

Marginal effect at the mean (MEM)#

lr1.get_margeff(at="mean").summary_frame()

	dy/dx	Std. Err.	z	Pr(>\|z\|)	Conf. Int. Low	Cont. Int. Hi.
work	0.47034	0.121697	3.864838	0.000111	0.231818	0.708862

# ALT. using the `marginaleffects` package
me.slopes(lr1, newdata="mean").to_pandas()

	term	contrast	estimate	std_error	statistic	p_value	s_value	conf_low	conf_high
0	work	dY/dX	0.47034	0.121818	3.861021	0.000113	13.112488	0.231582	0.709098

# ALT2. manual computation
meanwork = interns["work"].mean()
p_at_mean = lr1.predict({"work":meanwork}).item()
MEM = p_at_mean * (1-p_at_mean) * lr1.params['work']
MEM

0.47034017780957577

The marginaleffects package provides some useful plots to visualize the predictions and slopes.

# me.plot_predictions(lr1, condition="work")
# me.plot_slopes(lr1, condition="work")

Links to learn more about marginal effects

Discussion#

Model diagnostics and validation#

# Dispersion from GLM attributes
# glm2.pearson_chi2 / glm2.df_resid

# Calculate Pearson chi-squared statistic
observed = hdisks['failures']
predicted = pr2.predict()
pearson_residuals = (observed - predicted) / np.sqrt(predicted)
pearson_chi2 = np.sum(pearson_residuals**2)
df_resid = pr2.df_resid
dispersion = pearson_chi2 / df_resid
print(f'Dispersion: {dispersion}')
# If dispersion > 1, consider Negative Binomial regression

Dispersion: 0.9869289289681199

ScikitLearn models#

# Cross check with sklearn
from sklearn.linear_model import LogisticRegression
X1_skl = interns[["work"]]
y1_skl = interns["hired"]
lr1_skl = LogisticRegression(penalty=None).fit(X1_skl, y1_skl)
lr1_skl.intercept_, lr1_skl.coef_

(array([-78.69320824]), array([[1.98145785]]))

lr1.params

Intercept   -78.693205
work          1.981458
dtype: float64

hdisks.dtypes

age         float64
failures      int64
dtype: object

from sklearn.linear_model import PoissonRegressor
X2_skl = hdisks[["age"]]
y2_skl = hdisks["failures"]
pr2_skl = PoissonRegressor(alpha=0).fit(X2_skl, y2_skl)
pr2_skl.intercept_, pr2_skl.coef_

(1.0759942750890905, array([0.19382823]))

pr2.params

Intercept    1.075999
age          0.193828
dtype: float64

Logistic regression as a building blocks for neural networks#

The operation of the perceptron, which is the basic building block of neural networks, is essentially the same as linear regression model:

constant intercept (bias term)
linear combination of inputs
nonlinear function used to force the output to be between 0 and 1

Limitations of GLMs#

GLMs assume observations are independent
Assumes distribution \(\mathcal{M}\) is one of the exponential family
Outliers can be problematic
Interpretability

Exercises#

Exercise 1: probabilities to odds and log-odds#

0.3/(1-0.3), 0.99/(1-0.99), 0.7/(1-0.7)

(0.4285714285714286, 98.99999999999991, 2.333333333333333)

logit(0.3), logit(0.99), logit(0.7)

(-0.8472978603872036, 4.595119850134589, 0.8472978603872034)

Exercise 2: log-odds to probabilities#

expit(-1), expit(1), expit(2)

(0.2689414213699951, 0.7310585786300049, 0.8807970779778823)

Exercise 3: students pass or fail#

a) Load the dataset students.csv and add a column passing that contains 1 or 0, based on the above threshold score of 70.

students = pd.read_csv('../datasets/students.csv')
students["passing"] = (students["score"] > 70).astype(int)
# students.head()

b) Fit a logistic regression model for passing based on effort variable.

lmpass = smf.logit("passing ~ 1 + effort", data=students).fit()
print(lmpass.params)

efforts = np.linspace(0, 13, 100)
passing_preds = lmpass.predict({"effort": efforts})
ax = sns.scatterplot(data=students, x="effort", y="passing", alpha=0.3)
sns.lineplot(x=efforts, y=passing_preds, ax=ax);

filename = os.path.join(DESTDIR, "logistic_regression_students_passing_vs_effort.pdf")
savefigure(plt.gcf(), filename)

Optimization terminated successfully.
         Current function value: 0.276583
         Iterations 8
Intercept   -16.257302
effort        2.047882
dtype: float64
Saved figure to figures/lm/generalized/logistic_regression_students_passing_vs_effort.pdf
Saved figure to figures/lm/generalized/logistic_regression_students_passing_vs_effort.png

../_images/e70db10c70033c34fc1facb75f3d6602d8207518bcee0cc9feea7b876fa2204c.png

c) Use your model to predict the probability of a new student with effort 10 will pass.

lmpass.predict({"effort":10})

0    0.985536
dtype: float64

# ALT. compute prediction manually
intercept, b_effort = lmpass.params
expit(intercept + b_effort*10)

0.9855358765361845

Exercise 4: titanic survival data#

Fit a logistic regression model that calculates the probability of survival for people who were on the Titanic, based on the data in datasets/exercises/titanic.csv. Use the variables age, sex, and pclass as predictors.

cf. Titanic_Logistic_Regression.ipynb

titanic = pd.read_csv('../datasets/exercises/titanic.csv')
formula = "survived ~ age + C(sex) + C(pclass)"
lrtitanic = smf.logit(formula, data=titanic).fit()
lrtitanic.params

Optimization terminated successfully.
         Current function value: 0.453279
         Iterations 6

Intercept         3.777013
C(sex)[T.M]      -2.522781
C(pclass)[T.2]   -1.309799
C(pclass)[T.3]   -2.580625
age              -0.036985
dtype: float64

Use your logistic regression model to estimate the probability of survival for a 30 year old female traveling in second class.

pass30Fpclass2 = {"age":30, "sex":"F", "pclass":2}
lrtitanic.predict(pass30Fpclass2)

0    0.795378
dtype: float64

# # Cross check with sklearn
# from sklearn.linear_model import LogisticRegression
# df = pd.get_dummies(titanic, columns=['sex', 'pclass'], drop_first=True)
# X, y = df.drop('survived', axis=1), df['survived']
# sktitanic = LogisticRegression(penalty=None)
# sktitanic.fit(X, y)
# sktitanic.intercept_, sktitanic.coef_

Exercise 5: asthma attacks#

Fit a Poisson regression model to the ../datasets/exercises/asthma.csv dataset.

data source drkamarul/multivar_data_analysis

asthma = pd.read_csv("../datasets/exercises/asthma.csv")
asthma

	gender	res_inf	ghq12	attack
0	female	yes	21	6
1	male	no	17	4
2	male	yes	30	8
3	female	yes	22	5
4	male	yes	27	2
...	...	...	...	...
115	male	yes	0	2
116	female	yes	31	2
117	female	yes	18	2
118	female	yes	21	3
119	female	yes	11	2

120 rows × 4 columns

formula_asthma = "attack ~ 1 + C(gender) + C(res_inf) + ghq12"
prasthma = smf.poisson(formula_asthma, data=asthma).fit()
prasthma.params

Optimization terminated successfully.
         Current function value: 1.707281
         Iterations 6

Intercept           -0.315387
C(gender)[T.male]   -0.041905
C(res_inf)[T.yes]    0.426431
ghq12                0.049508
dtype: float64

cf. https://bookdown.org/drki_musa/dataanalysis/poisson-regression.html#multivariable-analysis-1

Exercise 6: student admissions dataset#

The dataset datasets/exercises/binary.csv contains information about the acceptance decision for 400 students to a prestigious school. Try to fit a logistic regression model for the variable admit using the variables gre, gpa, and rank as predictors.

# ORIGINAL https://stats.idre.ucla.edu/stat/data/binary.csv
binary = pd.read_csv('../datasets/exercises/binary.csv')
binary.head(3)

	admit	gre	gpa	rank
0	0	380	3.61	3
1	1	660	3.67	3
2	1	800	4.00	1

lrbinary = smf.logit('admit ~ gre + gpa + C(rank)', data=binary).fit()
lrbinary.params

Optimization terminated successfully.
         Current function value: 0.573147
         Iterations 6

Intercept      -3.989979
C(rank)[T.2]   -0.675443
C(rank)[T.3]   -1.340204
C(rank)[T.4]   -1.551464
gre             0.002264
gpa             0.804038
dtype: float64

The above model uses the rank=1 as the reference category an the log odds reported are with respect to this catrgory

\[ \log p(accept|rank=1) / \log p(accept|rank=2) = \texttt{C(rank)rank[T.2]} = -0.675443 \]

etc. for others rank[T.3] -1.340204 rank[T.4] -1.551464

See LogisticRegressionChangeOfReferenceCategoricalValue.ipynb for exercise recodign relative to different refrence level.

# # Cross check with sklearn
# from sklearn.linear_model import LogisticRegression
# df = pd.get_dummies(binary, columns=['rank'], drop_first=True)
# X, y = df.drop("admit", axis=1), df["admit"]
# lr = LogisticRegression(solver="lbfgs", penalty=None, max_iter=1000)
# lr.fit(X, y)
# lr.intercept_,  lr.coef_

Exercise 7: ship accidents#

https://rdrr.io/cran/AER/man/ShipAccidents.html

https://pages.stern.nyu.edu/~wgreene/Text/tables/tablelist5.htm

https://pages.stern.nyu.edu/~wgreene/Text/tables/TableF21-3.txt

# TODO

Bonus Exercises#

Bonus exercise A: honors class#

https://stats.oarc.ucla.edu/other/mult-pkg/faq/general/faq-how-do-i-interpret-odds-ratios-in-logistic-regression/

honors = pd.read_csv("../datasets/exercises/honors.csv")
honors.sample(4)

	female	read	write	math	hon	femalexmath
128	1	50	49	56	0	56
41	0	55	59	62	0	0
182	1	52	67	57	1	57
103	1	63	52	54	0	54

Constant model#

lrhon1 = smf.logit("hon ~ 1", data=honors).fit()
lrhon1.params

Optimization terminated successfully.
         Current function value: 0.556775
         Iterations 5

Intercept   -1.12546
dtype: float64

expit(lrhon1.params["Intercept"])

0.24500000000000005

honors["hon"].value_counts(normalize=True)

hon
0    0.755
1    0.245
Name: proportion, dtype: float64

Using only a categorical variable#

lrhon2 = smf.logit("hon ~ 1 + female", data=honors).fit()
lrhon2.params

Optimization terminated successfully.
         Current function value: 0.549016
         Iterations 5

Intercept   -1.470852
female       0.592782
dtype: float64

pd.crosstab(honors["hon"], honors["female"], margins=True)

female	0	1	All
hon
0	74	77	151
1	17	32	49
All	91	109	200

b0 = lrhon2.params["Intercept"]
b_female = lrhon2.params["female"]

# male prob
expit(b0), 17/91

(0.18681318681318684, 0.18681318681318682)

# male odds
np.exp(b0), 17/74  # = (17/91) / (74/91)

(0.2297297297297298, 0.22972972972972974)

# male log-odds
b0, np.log(17/74)

(-1.4708517491479534, -1.4708517491479536)

# female prob
expit(b0 + b_female), 32/109

(0.2935779816513761, 0.29357798165137616)

# female odds
np.exp(b0 + b_female), 32/77 # = (32/109) / (77/109)

(0.4155844155844155, 0.4155844155844156)

b0 + b_female, np.log(32/77)

(-0.8780695190539575, -0.8780695190539572)

# odds female relative to male
np.exp(b_female)

1.809014514896867

Logistic regression with a single continuous predictor variable#

lrhon3 = smf.logit("hon ~ 1 + math", data=honors).fit()
lrhon3.params

Optimization terminated successfully.
         Current function value: 0.417683
         Iterations 7

Intercept   -9.793942
math         0.156340
dtype: float64

So the model equation is

\[ \log(p/(1-p)) = \texttt{logit}(p) = -9.793942 + .1563404 \cdot \texttt{math} \]

# Increase in log-odds between math=54 and math=55
p54 = lrhon3.predict({"math":[54]}).item()
p55 = lrhon3.predict({"math":[55]}).item()
logit(p55) - logit(p54), lrhon3.params["math"]

(0.1563403555859233, 0.15634035558592282)

We can say now that the coefficient for math is the difference in the log odds. In other words, for a one-unit increase in the math score, the expected change in log odds is .1563404.

# Increase (multiplicative) in odds for unit increase in math
np.exp(lrhon3.params["math"]),  (p55/(1-p55)) / (p54/(1-p54))

(1.1692240873242836, 1.1692240873242843)

So we can say for a one-unit increase in math score, we expect to see about 17% increase in the odds of being in an honors class. This 17% of increase does not depend on the value that math is held at.

Logistic regression with multiple predictor variables and no interaction terms#

lrhon4 = smf.logit("hon ~ 1 + math + female + read", data=honors).fit()
lrhon4.params

Optimization terminated successfully.
         Current function value: 0.390424
         Iterations 7

Intercept   -11.770246
math          0.122959
female        0.979948
read          0.059063
dtype: float64

Logistic regression with an interaction term of two predictor variables#

lrhon5 = smf.logit("hon ~ 1 + math + female + femalexmath", data=honors).fit()
lrhon5.params

Optimization terminated successfully.
         Current function value: 0.399417
         Iterations 7

Intercept     -8.745841
math           0.129378
female        -2.899863
femalexmath    0.066995
dtype: float64

# ALT. without using `femalexmath` column
# lrhon5 = smf.logit("hon ~ 1 + math + female + female*math", data=honors).fit()
# lrhon5.params

Bonus exercise: LA high schools (NOT A VERY GOOD FIT FOR POISSON MODEL)#

Dataset info: http://www.philender.com/courses/intro/assign/data.html

This dataset consists of data from computer exercises collected from two high school in the Los Angeles area.

http://www.philender.com/courses/intro/code.html

lahigh_raw = pd.read_stata("https://stats.idre.ucla.edu/stat/stata/notes/lahigh.dta")
lahigh = lahigh_raw.convert_dtypes()

lahigh["gender"] = lahigh["gender"].astype(object).replace({1:"F", 2:"M"})
lahigh["ethnic"] = lahigh["ethnic"].astype(object).replace({
    1:"Native American",
    2:"Asian",
    3:"African-American",
    4:"Hispanic",
    5:"White",
    6:"Filipino",
    7:"Pacific Islander"})
lahigh["school"] = lahigh["school"].astype(object).replace({1:"Alpha", 2:"Beta"})
lahigh.head()

---------------------------------------------------------------------------
gaierror                                  Traceback (most recent call last)
File ~/.pyenv/versions/3.9.4/lib/python3.9/urllib/request.py:1346, in AbstractHTTPHandler.do_open(self, http_class, req, **http_conn_args)
   1345 try:
-> 1346     h.request(req.get_method(), req.selector, req.data, headers,
   1347               encode_chunked=req.has_header('Transfer-encoding'))
   1348 except OSError as err: # timeout error

File ~/.pyenv/versions/3.9.4/lib/python3.9/http/client.py:1253, in HTTPConnection.request(self, method, url, body, headers, encode_chunked)
   1252 """Send a complete request to the server."""
-> 1253 self._send_request(method, url, body, headers, encode_chunked)

File ~/.pyenv/versions/3.9.4/lib/python3.9/http/client.py:1299, in HTTPConnection._send_request(self, method, url, body, headers, encode_chunked)
   1298     body = _encode(body, 'body')
-> 1299 self.endheaders(body, encode_chunked=encode_chunked)

File ~/.pyenv/versions/3.9.4/lib/python3.9/http/client.py:1248, in HTTPConnection.endheaders(self, message_body, encode_chunked)
   1247     raise CannotSendHeader()
-> 1248 self._send_output(message_body, encode_chunked=encode_chunked)

File ~/.pyenv/versions/3.9.4/lib/python3.9/http/client.py:1008, in HTTPConnection._send_output(self, message_body, encode_chunked)
   1007 del self._buffer[:]
-> 1008 self.send(msg)
   1010 if message_body is not None:
   1011 
   1012     # create a consistent interface to message_body

File ~/.pyenv/versions/3.9.4/lib/python3.9/http/client.py:948, in HTTPConnection.send(self, data)
    947 if self.auto_open:
--> 948     self.connect()
    949 else:

File ~/.pyenv/versions/3.9.4/lib/python3.9/http/client.py:1415, in HTTPSConnection.connect(self)
   1413 "Connect to a host on a given (SSL) port."
-> 1415 super().connect()
   1417 if self._tunnel_host:

File ~/.pyenv/versions/3.9.4/lib/python3.9/http/client.py:919, in HTTPConnection.connect(self)
    918 """Connect to the host and port specified in __init__."""
--> 919 self.sock = self._create_connection(
    920     (self.host,self.port), self.timeout, self.source_address)
    921 self.sock.setsockopt(socket.IPPROTO_TCP, socket.TCP_NODELAY, 1)

File ~/.pyenv/versions/3.9.4/lib/python3.9/socket.py:822, in create_connection(address, timeout, source_address)
    821 err = None
--> 822 for res in getaddrinfo(host, port, 0, SOCK_STREAM):
    823     af, socktype, proto, canonname, sa = res

File ~/.pyenv/versions/3.9.4/lib/python3.9/socket.py:953, in getaddrinfo(host, port, family, type, proto, flags)
    952 addrlist = []
--> 953 for res in _socket.getaddrinfo(host, port, family, type, proto, flags):
    954     af, socktype, proto, canonname, sa = res

gaierror: [Errno 8] nodename nor servname provided, or not known

During handling of the above exception, another exception occurred:

URLError                                  Traceback (most recent call last)
Cell In[102], line 1
----> 1 lahigh_raw = pd.read_stata("https://stats.idre.ucla.edu/stat/stata/notes/lahigh.dta")
      2 lahigh = lahigh_raw.convert_dtypes()
      4 lahigh["gender"] = lahigh["gender"].astype(object).replace({1:"F", 2:"M"})

File ~/Projects/Minireference/STATSbook/noBSstatsnotebooks/venv/lib/python3.9/site-packages/pandas/io/stata.py:2109, in read_stata(filepath_or_buffer, convert_dates, convert_categoricals, index_col, convert_missing, preserve_dtypes, columns, order_categoricals, chunksize, iterator, compression, storage_options)
   2106     return reader
   2108 with reader:
-> 2109     return reader.read()

File ~/Projects/Minireference/STATSbook/noBSstatsnotebooks/venv/lib/python3.9/site-packages/pandas/io/stata.py:1683, in StataReader.read(self, nrows, convert_dates, convert_categoricals, index_col, convert_missing, preserve_dtypes, columns, order_categoricals)
   1671 @Appender(_read_method_doc)
   1672 def read(
   1673     self,
   (...)
   1681     order_categoricals: bool | None = None,
   1682 ) -> DataFrame:
-> 1683     self._ensure_open()
   1685     # Handle options
   1686     if convert_dates is None:

File ~/Projects/Minireference/STATSbook/noBSstatsnotebooks/venv/lib/python3.9/site-packages/pandas/io/stata.py:1175, in StataReader._ensure_open(self)
   1171 """
   1172 Ensure the file has been opened and its header data read.
   1173 """
   1174 if not hasattr(self, "_path_or_buf"):
-> 1175     self._open_file()

File ~/Projects/Minireference/STATSbook/noBSstatsnotebooks/venv/lib/python3.9/site-packages/pandas/io/stata.py:1188, in StataReader._open_file(self)
   1181 if not self._entered:
   1182     warnings.warn(
   1183         "StataReader is being used without using a context manager. "
   1184         "Using StataReader as a context manager is the only supported method.",
   1185         ResourceWarning,
   1186         stacklevel=find_stack_level(),
   1187     )
-> 1188 handles = get_handle(
   1189     self._original_path_or_buf,
   1190     "rb",
   1191     storage_options=self._storage_options,
   1192     is_text=False,
   1193     compression=self._compression,
   1194 )
   1195 if hasattr(handles.handle, "seekable") and handles.handle.seekable():
   1196     # If the handle is directly seekable, use it without an extra copy.
   1197     self._path_or_buf = handles.handle

File ~/Projects/Minireference/STATSbook/noBSstatsnotebooks/venv/lib/python3.9/site-packages/pandas/io/common.py:728, in get_handle(path_or_buf, mode, encoding, compression, memory_map, is_text, errors, storage_options)
    725     codecs.lookup_error(errors)
    727 # open URLs
--> 728 ioargs = _get_filepath_or_buffer(
    729     path_or_buf,
    730     encoding=encoding,
    731     compression=compression,
    732     mode=mode,
    733     storage_options=storage_options,
    734 )
    736 handle = ioargs.filepath_or_buffer
    737 handles: list[BaseBuffer]

File ~/Projects/Minireference/STATSbook/noBSstatsnotebooks/venv/lib/python3.9/site-packages/pandas/io/common.py:384, in _get_filepath_or_buffer(filepath_or_buffer, encoding, compression, mode, storage_options)
    382 # assuming storage_options is to be interpreted as headers
    383 req_info = urllib.request.Request(filepath_or_buffer, headers=storage_options)
--> 384 with urlopen(req_info) as req:
    385     content_encoding = req.headers.get("Content-Encoding", None)
    386     if content_encoding == "gzip":
    387         # Override compression based on Content-Encoding header

File ~/Projects/Minireference/STATSbook/noBSstatsnotebooks/venv/lib/python3.9/site-packages/pandas/io/common.py:289, in urlopen(*args, **kwargs)
    283 """
    284 Lazy-import wrapper for stdlib urlopen, as that imports a big chunk of
    285 the stdlib.
    286 """
    287 import urllib.request
--> 289 return urllib.request.urlopen(*args, **kwargs)

File ~/.pyenv/versions/3.9.4/lib/python3.9/urllib/request.py:214, in urlopen(url, data, timeout, cafile, capath, cadefault, context)
    212 else:
    213     opener = _opener
--> 214 return opener.open(url, data, timeout)

File ~/.pyenv/versions/3.9.4/lib/python3.9/urllib/request.py:517, in OpenerDirector.open(self, fullurl, data, timeout)
    514     req = meth(req)
    516 sys.audit('urllib.Request', req.full_url, req.data, req.headers, req.get_method())
--> 517 response = self._open(req, data)
    519 # post-process response
    520 meth_name = protocol+"_response"

File ~/.pyenv/versions/3.9.4/lib/python3.9/urllib/request.py:534, in OpenerDirector._open(self, req, data)
    531     return result
    533 protocol = req.type
--> 534 result = self._call_chain(self.handle_open, protocol, protocol +
    535                           '_open', req)
    536 if result:
    537     return result

File ~/.pyenv/versions/3.9.4/lib/python3.9/urllib/request.py:494, in OpenerDirector._call_chain(self, chain, kind, meth_name, *args)
    492 for handler in handlers:
    493     func = getattr(handler, meth_name)
--> 494     result = func(*args)
    495     if result is not None:
    496         return result

File ~/.pyenv/versions/3.9.4/lib/python3.9/urllib/request.py:1389, in HTTPSHandler.https_open(self, req)
   1388 def https_open(self, req):
-> 1389     return self.do_open(http.client.HTTPSConnection, req,
   1390         context=self._context, check_hostname=self._check_hostname)

File ~/.pyenv/versions/3.9.4/lib/python3.9/urllib/request.py:1349, in AbstractHTTPHandler.do_open(self, http_class, req, **http_conn_args)
   1346         h.request(req.get_method(), req.selector, req.data, headers,
   1347                   encode_chunked=req.has_header('Transfer-encoding'))
   1348     except OSError as err: # timeout error
-> 1349         raise URLError(err)
   1350     r = h.getresponse()
   1351 except:

URLError: <urlopen error [Errno 8] nodename nor servname provided, or not known>

formula = "daysabs ~ 1 + mathnce + langnce + C(gender)"
prlahigh = smf.poisson(formula, data=lahigh).fit()
prlahigh.params

# IRR
np.exp(prlahigh.params[1:])

# CI for IRR F 
np.exp(prlahigh.conf_int().loc["C(gender)[T.M]"])

# prlahigh.summary()
# prlahigh.aic, prlahigh.bic

Diagnostics#

via https://www.statsmodels.org/dev/examples/notebooks/generated/postestimation_poisson.html

prdiag = prlahigh.get_diagnostic()
# Plot observed versus predicted frequencies for entire sample
# prdiag.plot_probs();

# Other:
# ['plot_probs',
#  'probs_predicted',
#  'results',
#  'test_chisquare_prob',
#  'test_dispersion',
#  'test_poisson_zeroinflation',
#  'y_max']

# Code to get exactly the same numbers as in
# https://stats.oarc.ucla.edu/stata/output/poisson-regression/  
formula2 = "daysabs ~ 1 + mathnce + langnce + C(gender, Treatment(1))"
prlahigh2 = smf.poisson(formula2, data=lahigh).fit()
prlahigh2.params

Links#

TODO

Section 4.6 — Generalized linear models

Contents

Section 4.6 — Generalized linear models#

Notebook setup#

Definitions#

Probability representations and link functions#

Odds#

Log-odds#

The logit function#

The logistic function#

The logistic and logit functions are inverses#

Logistic regression#

Example 1: hiring student interns#

Interpreting the model parameters#

Parameters as changes in the log-odds#

Parameters as ratios of odds#

Differences in probabilities#

Prediction#

Poisson regression#

Example 2: hard disk failures over time#

Interpreting the model parameters#

Log-counts#

Incidence rate ratio (IRR)#

Marginal effect#

Predictions#

Explanations#

The exponential family of distributions#

The generalized linear model template#

Generalized linear models using statsmodels#

Linear model#

Logistic regression#

Poisson regression#

Fitting generalized linear models#

Standardization of predictors#

Marginal effects#

Raw parameters#

Marginal effect at a user-specified value#

Average marginal effect (AME)#

Marginal effect at the mean (MEM)#

Discussion#

Model diagnostics and validation#

ScikitLearn models#

Logistic regression as a building blocks for neural networks#

Limitations of GLMs#

Exercises#

Exercise 1: probabilities to odds and log-odds#

Exercise 2: log-odds to probabilities#

Exercise 3: students pass or fail#

Exercise 4: titanic survival data#

Exercise 5: asthma attacks#

Exercise 6: student admissions dataset#

Exercise 7: ship accidents#

Bonus Exercises#

Bonus exercise A: honors class#

Constant model#

Using only a categorical variable#

Logistic regression with a single continuous predictor variable#

Logistic regression with multiple predictor variables and no interaction terms#

Logistic regression with an interaction term of two predictor variables#

Bonus exercise: LA high schools (NOT A VERY GOOD FIT FOR POISSON MODEL)#

Diagnostics#

Links#