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sklearn/examples/model_selection/plot_grid_search_stats.py

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"""
==================================================
Statistical comparison of models using grid search
==================================================
This example illustrates how to statistically compare the performance of models
trained and evaluated using :class:`~sklearn.model_selection.GridSearchCV`.
"""
# %%
# We will start by simulating moon shaped data (where the ideal separation
# between classes is non-linear), adding to it a moderate degree of noise.
# Datapoints will belong to one of two possible classes to be predicted by two
# features. We will simulate 50 samples for each class:
import matplotlib.pyplot as plt
import seaborn as sns
from sklearn.datasets import make_moons
X, y = make_moons(noise=0.352, random_state=1, n_samples=100)
sns.scatterplot(
x=X[:, 0], y=X[:, 1], hue=y, marker="o", s=25, edgecolor="k", legend=False
).set_title("Data")
plt.show()
# %%
# We will compare the performance of :class:`~sklearn.svm.SVC` estimators that
# vary on their `kernel` parameter, to decide which choice of this
# hyper-parameter predicts our simulated data best.
# We will evaluate the performance of the models using
# :class:`~sklearn.model_selection.RepeatedStratifiedKFold`, repeating 10 times
# a 10-fold stratified cross validation using a different randomization of the
# data in each repetition. The performance will be evaluated using
# :class:`~sklearn.metrics.roc_auc_score`.
from sklearn.model_selection import GridSearchCV, RepeatedStratifiedKFold
from sklearn.svm import SVC
param_grid = [
{"kernel": ["linear"]},
{"kernel": ["poly"], "degree": [2, 3]},
{"kernel": ["rbf"]},
]
svc = SVC(random_state=0)
cv = RepeatedStratifiedKFold(n_splits=10, n_repeats=10, random_state=0)
search = GridSearchCV(estimator=svc, param_grid=param_grid, scoring="roc_auc", cv=cv)
search.fit(X, y)
# %%
# We can now inspect the results of our search, sorted by their
# `mean_test_score`:
import pandas as pd
results_df = pd.DataFrame(search.cv_results_)
results_df = results_df.sort_values(by=["rank_test_score"])
results_df = results_df.set_index(
results_df["params"].apply(lambda x: "_".join(str(val) for val in x.values()))
).rename_axis("kernel")
results_df[["params", "rank_test_score", "mean_test_score", "std_test_score"]]
# %%
# We can see that the estimator using the `'rbf'` kernel performed best,
# closely followed by `'linear'`. Both estimators with a `'poly'` kernel
# performed worse, with the one using a two-degree polynomial achieving a much
# lower performance than all other models.
#
# Usually, the analysis just ends here, but half the story is missing. The
# output of :class:`~sklearn.model_selection.GridSearchCV` does not provide
# information on the certainty of the differences between the models.
# We don't know if these are **statistically** significant.
# To evaluate this, we need to conduct a statistical test.
# Specifically, to contrast the performance of two models we should
# statistically compare their AUC scores. There are 100 samples (AUC
# scores) for each model as we repreated 10 times a 10-fold cross-validation.
#
# However, the scores of the models are not independent: all models are
# evaluated on the **same** 100 partitions, increasing the correlation
# between the performance of the models.
# Since some partitions of the data can make the distinction of the classes
# particularly easy or hard to find for all models, the models scores will
# co-vary.
#
# Let's inspect this partition effect by plotting the performance of all models
# in each fold, and calculating the correlation between models across folds:
# create df of model scores ordered by performance
model_scores = results_df.filter(regex=r"split\d*_test_score")
# plot 30 examples of dependency between cv fold and AUC scores
fig, ax = plt.subplots()
sns.lineplot(
data=model_scores.transpose().iloc[:30],
dashes=False,
palette="Set1",
marker="o",
alpha=0.5,
ax=ax,
)
ax.set_xlabel("CV test fold", size=12, labelpad=10)
ax.set_ylabel("Model AUC", size=12)
ax.tick_params(bottom=True, labelbottom=False)
plt.show()
# print correlation of AUC scores across folds
print(f"Correlation of models:\n {model_scores.transpose().corr()}")
# %%
# We can observe that the performance of the models highly depends on the fold.
#
# As a consequence, if we assume independence between samples we will be
# underestimating the variance computed in our statistical tests, increasing
# the number of false positive errors (i.e. detecting a significant difference
# between models when such does not exist) [1]_.
#
# Several variance-corrected statistical tests have been developed for these
# cases. In this example we will show how to implement one of them (the so
# called Nadeau and Bengio's corrected t-test) under two different statistical
# frameworks: frequentist and Bayesian.
# %%
# Comparing two models: frequentist approach
# ------------------------------------------
#
# We can start by asking: "Is the first model significantly better than the
# second model (when ranked by `mean_test_score`)?"
#
# To answer this question using a frequentist approach we could
# run a paired t-test and compute the p-value. This is also known as
# Diebold-Mariano test in the forecast literature [5]_.
# Many variants of such a t-test have been developed to account for the
# 'non-independence of samples problem'
# described in the previous section. We will use the one proven to obtain the
# highest replicability scores (which rate how similar the performance of a
# model is when evaluating it on different random partitions of the same
# dataset) while maintaining a low rate of false positives and false negatives:
# the Nadeau and Bengio's corrected t-test [2]_ that uses a 10 times repeated
# 10-fold cross validation [3]_.
#
# This corrected paired t-test is computed as:
#
# .. math::
# t=\frac{\frac{1}{k \cdot r}\sum_{i=1}^{k}\sum_{j=1}^{r}x_{ij}}
# {\sqrt{(\frac{1}{k \cdot r}+\frac{n_{test}}{n_{train}})\hat{\sigma}^2}}
#
# where :math:`k` is the number of folds,
# :math:`r` the number of repetitions in the cross-validation,
# :math:`x` is the difference in performance of the models,
# :math:`n_{test}` is the number of samples used for testing,
# :math:`n_{train}` is the number of samples used for training,
# and :math:`\hat{\sigma}^2` represents the variance of the observed
# differences.
#
# Let's implement a corrected right-tailed paired t-test to evaluate if the
# performance of the first model is significantly better than that of the
# second model. Our null hypothesis is that the second model performs at least
# as good as the first model.
import numpy as np
from scipy.stats import t
def corrected_std(differences, n_train, n_test):
"""Corrects standard deviation using Nadeau and Bengio's approach.
Parameters
----------
differences : ndarray of shape (n_samples,)
Vector containing the differences in the score metrics of two models.
n_train : int
Number of samples in the training set.
n_test : int
Number of samples in the testing set.
Returns
-------
corrected_std : float
Variance-corrected standard deviation of the set of differences.
"""
# kr = k times r, r times repeated k-fold crossvalidation,
# kr equals the number of times the model was evaluated
kr = len(differences)
corrected_var = np.var(differences, ddof=1) * (1 / kr + n_test / n_train)
corrected_std = np.sqrt(corrected_var)
return corrected_std
def compute_corrected_ttest(differences, df, n_train, n_test):
"""Computes right-tailed paired t-test with corrected variance.
Parameters
----------
differences : array-like of shape (n_samples,)
Vector containing the differences in the score metrics of two models.
df : int
Degrees of freedom.
n_train : int
Number of samples in the training set.
n_test : int
Number of samples in the testing set.
Returns
-------
t_stat : float
Variance-corrected t-statistic.
p_val : float
Variance-corrected p-value.
"""
mean = np.mean(differences)
std = corrected_std(differences, n_train, n_test)
t_stat = mean / std
p_val = t.sf(np.abs(t_stat), df) # right-tailed t-test
return t_stat, p_val
# %%
model_1_scores = model_scores.iloc[0].values # scores of the best model
model_2_scores = model_scores.iloc[1].values # scores of the second-best model
differences = model_1_scores - model_2_scores
n = differences.shape[0] # number of test sets
df = n - 1
n_train = len(list(cv.split(X, y))[0][0])
n_test = len(list(cv.split(X, y))[0][1])
t_stat, p_val = compute_corrected_ttest(differences, df, n_train, n_test)
print(f"Corrected t-value: {t_stat:.3f}\nCorrected p-value: {p_val:.3f}")
# %%
# We can compare the corrected t- and p-values with the uncorrected ones:
t_stat_uncorrected = np.mean(differences) / np.sqrt(np.var(differences, ddof=1) / n)
p_val_uncorrected = t.sf(np.abs(t_stat_uncorrected), df)
print(
f"Uncorrected t-value: {t_stat_uncorrected:.3f}\n"
f"Uncorrected p-value: {p_val_uncorrected:.3f}"
)
# %%
# Using the conventional significance alpha level at `p=0.05`, we observe that
# the uncorrected t-test concludes that the first model is significantly better
# than the second.
#
# With the corrected approach, in contrast, we fail to detect this difference.
#
# In the latter case, however, the frequentist approach does not let us
# conclude that the first and second model have an equivalent performance. If
# we wanted to make this assertion we need to use a Bayesian approach.
# %%
# Comparing two models: Bayesian approach
# ---------------------------------------
# We can use Bayesian estimation to calculate the probability that the first
# model is better than the second. Bayesian estimation will output a
# distribution followed by the mean :math:`\mu` of the differences in the
# performance of two models.
#
# To obtain the posterior distribution we need to define a prior that models
# our beliefs of how the mean is distributed before looking at the data,
# and multiply it by a likelihood function that computes how likely our
# observed differences are, given the values that the mean of differences
# could take.
#
# Bayesian estimation can be carried out in many forms to answer our question,
# but in this example we will implement the approach suggested by Benavoli and
# colleagues [4]_.
#
# One way of defining our posterior using a closed-form expression is to select
# a prior conjugate to the likelihood function. Benavoli and colleagues [4]_
# show that when comparing the performance of two classifiers we can model the
# prior as a Normal-Gamma distribution (with both mean and variance unknown)
# conjugate to a normal likelihood, to thus express the posterior as a normal
# distribution.
# Marginalizing out the variance from this normal posterior, we can define the
# posterior of the mean parameter as a Student's t-distribution. Specifically:
#
# .. math::
# St(\mu;n-1,\overline{x},(\frac{1}{n}+\frac{n_{test}}{n_{train}})
# \hat{\sigma}^2)
#
# where :math:`n` is the total number of samples,
# :math:`\overline{x}` represents the mean difference in the scores,
# :math:`n_{test}` is the number of samples used for testing,
# :math:`n_{train}` is the number of samples used for training,
# and :math:`\hat{\sigma}^2` represents the variance of the observed
# differences.
#
# Notice that we are using Nadeau and Bengio's corrected variance in our
# Bayesian approach as well.
#
# Let's compute and plot the posterior:
# initialize random variable
t_post = t(
df, loc=np.mean(differences), scale=corrected_std(differences, n_train, n_test)
)
# %%
# Let's plot the posterior distribution:
x = np.linspace(t_post.ppf(0.001), t_post.ppf(0.999), 100)
plt.plot(x, t_post.pdf(x))
plt.xticks(np.arange(-0.04, 0.06, 0.01))
plt.fill_between(x, t_post.pdf(x), 0, facecolor="blue", alpha=0.2)
plt.ylabel("Probability density")
plt.xlabel(r"Mean difference ($\mu$)")
plt.title("Posterior distribution")
plt.show()
# %%
# We can calculate the probability that the first model is better than the
# second by computing the area under the curve of the posterior distribution
# from zero to infinity. And also the reverse: we can calculate the probability
# that the second model is better than the first by computing the area under
# the curve from minus infinity to zero.
better_prob = 1 - t_post.cdf(0)
print(
f"Probability of {model_scores.index[0]} being more accurate than "
f"{model_scores.index[1]}: {better_prob:.3f}"
)
print(
f"Probability of {model_scores.index[1]} being more accurate than "
f"{model_scores.index[0]}: {1 - better_prob:.3f}"
)
# %%
# In contrast with the frequentist approach, we can compute the probability
# that one model is better than the other.
#
# Note that we obtained similar results as those in the frequentist approach.
# Given our choice of priors, we are essentially performing the same
# computations, but we are allowed to make different assertions.
# %%
# Region of Practical Equivalence
# ^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^^
# Sometimes we are interested in determining the probabilities that our models
# have an equivalent performance, where "equivalent" is defined in a practical
# way. A naive approach [4]_ would be to define estimators as practically
# equivalent when they differ by less than 1% in their accuracy. But we could
# also define this practical equivalence taking into account the problem we are
# trying to solve. For example, a difference of 5% in accuracy would mean an
# increase of $1000 in sales, and we consider any quantity above that as
# relevant for our business.
#
# In this example we are going to define the
# Region of Practical Equivalence (ROPE) to be :math:`[-0.01, 0.01]`. That is,
# we will consider two models as practically equivalent if they differ by less
# than 1% in their performance.
#
# To compute the probabilities of the classifiers being practically equivalent,
# we calculate the area under the curve of the posterior over the ROPE
# interval:
rope_interval = [-0.01, 0.01]
rope_prob = t_post.cdf(rope_interval[1]) - t_post.cdf(rope_interval[0])
print(
f"Probability of {model_scores.index[0]} and {model_scores.index[1]} "
f"being practically equivalent: {rope_prob:.3f}"
)
# %%
# We can plot how the posterior is distributed over the ROPE interval:
x_rope = np.linspace(rope_interval[0], rope_interval[1], 100)
plt.plot(x, t_post.pdf(x))
plt.xticks(np.arange(-0.04, 0.06, 0.01))
plt.vlines([-0.01, 0.01], ymin=0, ymax=(np.max(t_post.pdf(x)) + 1))
plt.fill_between(x_rope, t_post.pdf(x_rope), 0, facecolor="blue", alpha=0.2)
plt.ylabel("Probability density")
plt.xlabel(r"Mean difference ($\mu$)")
plt.title("Posterior distribution under the ROPE")
plt.show()
# %%
# As suggested in [4]_, we can further interpret these probabilities using the
# same criteria as the frequentist approach: is the probability of falling
# inside the ROPE bigger than 95% (alpha value of 5%)? In that case we can
# conclude that both models are practically equivalent.
# %%
# The Bayesian estimation approach also allows us to compute how uncertain we
# are about our estimation of the difference. This can be calculated using
# credible intervals. For a given probability, they show the range of values
# that the estimated quantity, in our case the mean difference in
# performance, can take.
# For example, a 50% credible interval [x, y] tells us that there is a 50%
# probability that the true (mean) difference of performance between models is
# between x and y.
#
# Let's determine the credible intervals of our data using 50%, 75% and 95%:
cred_intervals = []
intervals = [0.5, 0.75, 0.95]
for interval in intervals:
cred_interval = list(t_post.interval(interval))
cred_intervals.append([interval, cred_interval[0], cred_interval[1]])
cred_int_df = pd.DataFrame(
cred_intervals, columns=["interval", "lower value", "upper value"]
).set_index("interval")
cred_int_df
# %%
# As shown in the table, there is a 50% probability that the true mean
# difference between models will be between 0.000977 and 0.019023, 70%
# probability that it will be between -0.005422 and 0.025422, and 95%
# probability that it will be between -0.016445 and 0.036445.
# %%
# Pairwise comparison of all models: frequentist approach
# -------------------------------------------------------
#
# We could also be interested in comparing the performance of all our models
# evaluated with :class:`~sklearn.model_selection.GridSearchCV`. In this case
# we would be running our statistical test multiple times, which leads us to
# the `multiple comparisons problem
# <https://en.wikipedia.org/wiki/Multiple_comparisons_problem>`_.
#
# There are many possible ways to tackle this problem, but a standard approach
# is to apply a `Bonferroni correction
# <https://en.wikipedia.org/wiki/Bonferroni_correction>`_. Bonferroni can be
# computed by multiplying the p-value by the number of comparisons we are
# testing.
#
# Let's compare the performance of the models using the corrected t-test:
from itertools import combinations
from math import factorial
n_comparisons = factorial(len(model_scores)) / (
factorial(2) * factorial(len(model_scores) - 2)
)
pairwise_t_test = []
for model_i, model_k in combinations(range(len(model_scores)), 2):
model_i_scores = model_scores.iloc[model_i].values
model_k_scores = model_scores.iloc[model_k].values
differences = model_i_scores - model_k_scores
t_stat, p_val = compute_corrected_ttest(differences, df, n_train, n_test)
p_val *= n_comparisons # implement Bonferroni correction
# Bonferroni can output p-values higher than 1
p_val = 1 if p_val > 1 else p_val
pairwise_t_test.append(
[model_scores.index[model_i], model_scores.index[model_k], t_stat, p_val]
)
pairwise_comp_df = pd.DataFrame(
pairwise_t_test, columns=["model_1", "model_2", "t_stat", "p_val"]
).round(3)
pairwise_comp_df
# %%
# We observe that after correcting for multiple comparisons, the only model
# that significantly differs from the others is `'2_poly'`.
# `'rbf'`, the model ranked first by
# :class:`~sklearn.model_selection.GridSearchCV`, does not significantly
# differ from `'linear'` or `'3_poly'`.
# %%
# Pairwise comparison of all models: Bayesian approach
# ----------------------------------------------------
#
# When using Bayesian estimation to compare multiple models, we don't need to
# correct for multiple comparisons (for reasons why see [4]_).
#
# We can carry out our pairwise comparisons the same way as in the first
# section:
pairwise_bayesian = []
for model_i, model_k in combinations(range(len(model_scores)), 2):
model_i_scores = model_scores.iloc[model_i].values
model_k_scores = model_scores.iloc[model_k].values
differences = model_i_scores - model_k_scores
t_post = t(
df, loc=np.mean(differences), scale=corrected_std(differences, n_train, n_test)
)
worse_prob = t_post.cdf(rope_interval[0])
better_prob = 1 - t_post.cdf(rope_interval[1])
rope_prob = t_post.cdf(rope_interval[1]) - t_post.cdf(rope_interval[0])
pairwise_bayesian.append([worse_prob, better_prob, rope_prob])
pairwise_bayesian_df = pd.DataFrame(
pairwise_bayesian, columns=["worse_prob", "better_prob", "rope_prob"]
).round(3)
pairwise_comp_df = pairwise_comp_df.join(pairwise_bayesian_df)
pairwise_comp_df
# %%
# Using the Bayesian approach we can compute the probability that a model
# performs better, worse or practically equivalent to another.
#
# Results show that the model ranked first by
# :class:`~sklearn.model_selection.GridSearchCV` `'rbf'`, has approximately a
# 6.8% chance of being worse than `'linear'`, and a 1.8% chance of being worse
# than `'3_poly'`.
# `'rbf'` and `'linear'` have a 43% probability of being practically
# equivalent, while `'rbf'` and `'3_poly'` have a 10% chance of being so.
#
# Similarly to the conclusions obtained using the frequentist approach, all
# models have a 100% probability of being better than `'2_poly'`, and none have
# a practically equivalent performance with the latter.
# %%
# Take-home messages
# ------------------
# - Small differences in performance measures might easily turn out to be
# merely by chance, but not because one model predicts systematically better
# than the other. As shown in this example, statistics can tell you how
# likely that is.
# - When statistically comparing the performance of two models evaluated in
# GridSearchCV, it is necessary to correct the calculated variance which
# could be underestimated since the scores of the models are not independent
# from each other.
# - A frequentist approach that uses a (variance-corrected) paired t-test can
# tell us if the performance of one model is better than another with a
# degree of certainty above chance.
# - A Bayesian approach can provide the probabilities of one model being
# better, worse or practically equivalent than another. It can also tell us
# how confident we are of knowing that the true differences of our models
# fall under a certain range of values.
# - If multiple models are statistically compared, a multiple comparisons
# correction is needed when using the frequentist approach.
# %%
# .. rubric:: References
#
# .. [1] Dietterich, T. G. (1998). `Approximate statistical tests for
# comparing supervised classification learning algorithms
# <http://web.cs.iastate.edu/~jtian/cs573/Papers/Dietterich-98.pdf>`_.
# Neural computation, 10(7).
# .. [2] Nadeau, C., & Bengio, Y. (2000). `Inference for the generalization
# error
# <https://papers.nips.cc/paper/1661-inference-for-the-generalization-error.pdf>`_.
# In Advances in neural information processing systems.
# .. [3] Bouckaert, R. R., & Frank, E. (2004). `Evaluating the replicability
# of significance tests for comparing learning algorithms
# <https://www.cms.waikato.ac.nz/~ml/publications/2004/bouckaert-frank.pdf>`_.
# In Pacific-Asia Conference on Knowledge Discovery and Data Mining.
# .. [4] Benavoli, A., Corani, G., Demšar, J., & Zaffalon, M. (2017). `Time
# for a change: a tutorial for comparing multiple classifiers through
# Bayesian analysis
# <http://www.jmlr.org/papers/volume18/16-305/16-305.pdf>`_.
# The Journal of Machine Learning Research, 18(1). See the Python
# library that accompanies this paper `here
# <https://github.com/janezd/baycomp>`_.
# .. [5] Diebold, F.X. & Mariano R.S. (1995). `Comparing predictive accuracy
# <http://www.est.uc3m.es/esp/nueva_docencia/comp_col_get/lade/tecnicas_prediccion/Practicas0708/Comparing%20Predictive%20Accuracy%20(Dielbold).pdf>`_
# Journal of Business & economic statistics, 20(1), 134-144.
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