"""
=====================================================
Prediction Intervals for Gradient Boosting Regression
=====================================================

This example shows how quantile regression can be used to create prediction
intervals. See :ref:`sphx_glr_auto_examples_ensemble_plot_hgbt_regression.py`
for an example showcasing some other features of
:class:`~ensemble.HistGradientBoostingRegressor`.

"""

# %%
# Generate some data for a synthetic regression problem by applying the
# function f to uniformly sampled random inputs.
import numpy as np

from sklearn.model_selection import train_test_split


def f(x):
    """The function to predict."""
    return x * np.sin(x)


rng = np.random.RandomState(42)
X = np.atleast_2d(rng.uniform(0, 10.0, size=1000)).T
expected_y = f(X).ravel()

# %%
# To make the problem interesting, we generate observations of the target y as
# the sum of a deterministic term computed by the function f and a random noise
# term that follows a centered `log-normal
# <https://en.wikipedia.org/wiki/Log-normal_distribution>`_. To make this even
# more interesting we consider the case where the amplitude of the noise
# depends on the input variable x (heteroscedastic noise).
#
# The lognormal distribution is non-symmetric and long tailed: observing large
# outliers is likely but it is impossible to observe small outliers.
sigma = 0.5 + X.ravel() / 10
noise = rng.lognormal(sigma=sigma) - np.exp(sigma**2 / 2)
y = expected_y + noise

# %%
# Split into train, test datasets:
X_train, X_test, y_train, y_test = train_test_split(X, y, random_state=0)

# %%
# Fitting non-linear quantile and least squares regressors
# --------------------------------------------------------
#
# Fit gradient boosting models trained with the quantile loss and
# alpha=0.05, 0.5, 0.95.
#
# The models obtained for alpha=0.05 and alpha=0.95 produce a 90% confidence
# interval (95% - 5% = 90%).
#
# The model trained with alpha=0.5 produces a regression of the median: on
# average, there should be the same number of target observations above and
# below the predicted values.
from sklearn.ensemble import GradientBoostingRegressor
from sklearn.metrics import mean_pinball_loss, mean_squared_error

all_models = {}
common_params = dict(
    learning_rate=0.05,
    n_estimators=200,
    max_depth=2,
    min_samples_leaf=9,
    min_samples_split=9,
)
for alpha in [0.05, 0.5, 0.95]:
    gbr = GradientBoostingRegressor(loss="quantile", alpha=alpha, **common_params)
    all_models["q %1.2f" % alpha] = gbr.fit(X_train, y_train)

# %%
# Notice that :class:`~sklearn.ensemble.HistGradientBoostingRegressor` is much
# faster than :class:`~sklearn.ensemble.GradientBoostingRegressor` starting with
# intermediate datasets (`n_samples >= 10_000`), which is not the case of the
# present example.
#
# For the sake of comparison, we also fit a baseline model trained with the
# usual (mean) squared error (MSE).
gbr_ls = GradientBoostingRegressor(loss="squared_error", **common_params)
all_models["mse"] = gbr_ls.fit(X_train, y_train)

# %%
# Create an evenly spaced evaluation set of input values spanning the [0, 10]
# range.
xx = np.atleast_2d(np.linspace(0, 10, 1000)).T

# %%
# Plot the true conditional mean function f, the predictions of the conditional
# mean (loss equals squared error), the conditional median and the conditional
# 90% interval (from 5th to 95th conditional percentiles).
import matplotlib.pyplot as plt

y_pred = all_models["mse"].predict(xx)
y_lower = all_models["q 0.05"].predict(xx)
y_upper = all_models["q 0.95"].predict(xx)
y_med = all_models["q 0.50"].predict(xx)

fig = plt.figure(figsize=(10, 10))
plt.plot(xx, f(xx), "g:", linewidth=3, label=r"$f(x) = x\,\sin(x)$")
plt.plot(X_test, y_test, "b.", markersize=10, label="Test observations")
plt.plot(xx, y_med, "r-", label="Predicted median")
plt.plot(xx, y_pred, "r-", label="Predicted mean")
plt.plot(xx, y_upper, "k-")
plt.plot(xx, y_lower, "k-")
plt.fill_between(
    xx.ravel(), y_lower, y_upper, alpha=0.4, label="Predicted 90% interval"
)
plt.xlabel("$x$")
plt.ylabel("$f(x)$")
plt.ylim(-10, 25)
plt.legend(loc="upper left")
plt.show()

# %%
# Comparing the predicted median with the predicted mean, we note that the
# median is on average below the mean as the noise is skewed towards high
# values (large outliers). The median estimate also seems to be smoother
# because of its natural robustness to outliers.
#
# Also observe that the inductive bias of gradient boosting trees is
# unfortunately preventing our 0.05 quantile to fully capture the sinoisoidal
# shape of the signal, in particular around x=8. Tuning hyper-parameters can
# reduce this effect as shown in the last part of this notebook.
#
# Analysis of the error metrics
# -----------------------------
#
# Measure the models with :func:`~sklearn.metrics.mean_squared_error` and
# :func:`~sklearn.metrics.mean_pinball_loss` metrics on the training dataset.
import pandas as pd


def highlight_min(x):
    x_min = x.min()
    return ["font-weight: bold" if v == x_min else "" for v in x]


results = []
for name, gbr in sorted(all_models.items()):
    metrics = {"model": name}
    y_pred = gbr.predict(X_train)
    for alpha in [0.05, 0.5, 0.95]:
        metrics["pbl=%1.2f" % alpha] = mean_pinball_loss(y_train, y_pred, alpha=alpha)
    metrics["MSE"] = mean_squared_error(y_train, y_pred)
    results.append(metrics)

pd.DataFrame(results).set_index("model").style.apply(highlight_min)

# %%
# One column shows all models evaluated by the same metric. The minimum number
# on a column should be obtained when the model is trained and measured with
# the same metric. This should be always the case on the training set if the
# training converged.
#
# Note that because the target distribution is asymmetric, the expected
# conditional mean and conditional median are significantly different and
# therefore one could not use the squared error model get a good estimation of
# the conditional median nor the converse.
#
# If the target distribution were symmetric and had no outliers (e.g. with a
# Gaussian noise), then median estimator and the least squares estimator would
# have yielded similar predictions.
#
# We then do the same on the test set.
results = []
for name, gbr in sorted(all_models.items()):
    metrics = {"model": name}
    y_pred = gbr.predict(X_test)
    for alpha in [0.05, 0.5, 0.95]:
        metrics["pbl=%1.2f" % alpha] = mean_pinball_loss(y_test, y_pred, alpha=alpha)
    metrics["MSE"] = mean_squared_error(y_test, y_pred)
    results.append(metrics)

pd.DataFrame(results).set_index("model").style.apply(highlight_min)


# %%
# Errors are higher meaning the models slightly overfitted the data. It still
# shows that the best test metric is obtained when the model is trained by
# minimizing this same metric.
#
# Note that the conditional median estimator is competitive with the squared
# error estimator in terms of MSE on the test set: this can be explained by
# the fact the squared error estimator is very sensitive to large outliers
# which can cause significant overfitting. This can be seen on the right hand
# side of the previous plot. The conditional median estimator is biased
# (underestimation for this asymmetric noise) but is also naturally robust to
# outliers and overfits less.
#
# .. _calibration-section:
#
# Calibration of the confidence interval
# --------------------------------------
#
# We can also evaluate the ability of the two extreme quantile estimators at
# producing a well-calibrated conditional 90%-confidence interval.
#
# To do this we can compute the fraction of observations that fall between the
# predictions:
def coverage_fraction(y, y_low, y_high):
    return np.mean(np.logical_and(y >= y_low, y <= y_high))


coverage_fraction(
    y_train,
    all_models["q 0.05"].predict(X_train),
    all_models["q 0.95"].predict(X_train),
)

# %%
# On the training set the calibration is very close to the expected coverage
# value for a 90% confidence interval.
coverage_fraction(
    y_test, all_models["q 0.05"].predict(X_test), all_models["q 0.95"].predict(X_test)
)


# %%
# On the test set, the estimated confidence interval is slightly too narrow.
# Note, however, that we would need to wrap those metrics in a cross-validation
# loop to assess their variability under data resampling.
#
# Tuning the hyper-parameters of the quantile regressors
# ------------------------------------------------------
#
# In the plot above, we observed that the 5th percentile regressor seems to
# underfit and could not adapt to sinusoidal shape of the signal.
#
# The hyper-parameters of the model were approximately hand-tuned for the
# median regressor and there is no reason that the same hyper-parameters are
# suitable for the 5th percentile regressor.
#
# To confirm this hypothesis, we tune the hyper-parameters of a new regressor
# of the 5th percentile by selecting the best model parameters by
# cross-validation on the pinball loss with alpha=0.05:

# %%
from sklearn.experimental import enable_halving_search_cv  # noqa
from sklearn.model_selection import HalvingRandomSearchCV
from sklearn.metrics import make_scorer
from pprint import pprint

param_grid = dict(
    learning_rate=[0.05, 0.1, 0.2],
    max_depth=[2, 5, 10],
    min_samples_leaf=[1, 5, 10, 20],
    min_samples_split=[5, 10, 20, 30, 50],
)
alpha = 0.05
neg_mean_pinball_loss_05p_scorer = make_scorer(
    mean_pinball_loss,
    alpha=alpha,
    greater_is_better=False,  # maximize the negative loss
)
gbr = GradientBoostingRegressor(loss="quantile", alpha=alpha, random_state=0)
search_05p = HalvingRandomSearchCV(
    gbr,
    param_grid,
    resource="n_estimators",
    max_resources=250,
    min_resources=50,
    scoring=neg_mean_pinball_loss_05p_scorer,
    n_jobs=2,
    random_state=0,
).fit(X_train, y_train)
pprint(search_05p.best_params_)

# %%
# We observe that the hyper-parameters that were hand-tuned for the median
# regressor are in the same range as the hyper-parameters suitable for the 5th
# percentile regressor.
#
# Let's now tune the hyper-parameters for the 95th percentile regressor. We
# need to redefine the `scoring` metric used to select the best model, along
# with adjusting the alpha parameter of the inner gradient boosting estimator
# itself:
from sklearn.base import clone

alpha = 0.95
neg_mean_pinball_loss_95p_scorer = make_scorer(
    mean_pinball_loss,
    alpha=alpha,
    greater_is_better=False,  # maximize the negative loss
)
search_95p = clone(search_05p).set_params(
    estimator__alpha=alpha,
    scoring=neg_mean_pinball_loss_95p_scorer,
)
search_95p.fit(X_train, y_train)
pprint(search_95p.best_params_)

# %%
# The result shows that the hyper-parameters for the 95th percentile regressor
# identified by the search procedure are roughly in the same range as the hand-
# tuned hyper-parameters for the median regressor and the hyper-parameters
# identified by the search procedure for the 5th percentile regressor. However,
# the hyper-parameter searches did lead to an improved 90% confidence interval
# that is comprised by the predictions of those two tuned quantile regressors.
# Note that the prediction of the upper 95th percentile has a much coarser shape
# than the prediction of the lower 5th percentile because of the outliers:
y_lower = search_05p.predict(xx)
y_upper = search_95p.predict(xx)

fig = plt.figure(figsize=(10, 10))
plt.plot(xx, f(xx), "g:", linewidth=3, label=r"$f(x) = x\,\sin(x)$")
plt.plot(X_test, y_test, "b.", markersize=10, label="Test observations")
plt.plot(xx, y_upper, "k-")
plt.plot(xx, y_lower, "k-")
plt.fill_between(
    xx.ravel(), y_lower, y_upper, alpha=0.4, label="Predicted 90% interval"
)
plt.xlabel("$x$")
plt.ylabel("$f(x)$")
plt.ylim(-10, 25)
plt.legend(loc="upper left")
plt.title("Prediction with tuned hyper-parameters")
plt.show()

# %%
# The plot looks qualitatively better than for the untuned models, especially
# for the shape of the of lower quantile.
#
# We now quantitatively evaluate the joint-calibration of the pair of
# estimators:
coverage_fraction(y_train, search_05p.predict(X_train), search_95p.predict(X_train))
# %%
coverage_fraction(y_test, search_05p.predict(X_test), search_95p.predict(X_test))
# %%
# The calibration of the tuned pair is sadly not better on the test set: the
# width of the estimated confidence interval is still too narrow.
#
# Again, we would need to wrap this study in a cross-validation loop to
# better assess the variability of those estimates.