Plot Kernel Ridge Regression¶
============================================= Comparison of kernel ridge regression and SVR¶
Both kernel ridge regression (KRR) and SVR learn a non-linear function by employing the kernel trick, i.e., they learn a linear function in the space induced by the respective kernel which corresponds to a non-linear function in the original space. They differ in the loss functions (ridge versus epsilon-insensitive loss). In contrast to SVR, fitting a KRR can be done in closed-form and is typically faster for medium-sized datasets. On the other hand, the learned model is non-sparse and thus slower than SVR at prediction-time.
This example illustrates both methods on an artificial dataset, which consists of a sinusoidal target function and strong noise added to every fifth datapoint.
Imports for Comparing Kernel Ridge Regression and Support Vector Regression¶
KernelRidge and SVR both learn non-linear functions via the kernel trick but differ fundamentally in their loss functions and solution sparsity: Kernel Ridge Regression (KRR) minimizes squared error with L2 regularization in the kernel-induced feature space, yielding a closed-form solution that requires inverting an n-by-n kernel matrix – O(n^3) training but using all training points for prediction. SVR minimizes epsilon-insensitive loss (ignoring errors smaller than epsilon), producing a sparse solution where only the support vectors (typically a fraction of training points) contribute to predictions. This sparsity makes SVR prediction O(n_sv * d) rather than O(n * d), which becomes significant when the support vector ratio is low.
GridSearchCV jointly tunes the regularization strength (C for SVR, alpha for KRR) and the RBF kernel bandwidth (gamma), revealing the performance-complexity tradeoff between the two methods: The sinusoidal target with heavy-tailed noise (every 5th point corrupted) tests robustness: SVR’s epsilon-insensitive loss naturally ignores small residuals, while KRR’s squared loss is sensitive to outliers. The timing comparison shows that KRR trains faster for small-to-medium datasets (closed-form solution) while SVR scales better for large datasets (the quadratic programming problem benefits from the sparse solution). The LearningCurveDisplay further reveals how test error decreases as training set size grows, with both methods converging to similar performance but at different computational costs.
# Authors: The scikit-learn developers
# SPDX-License-Identifier: BSD-3-Clause
# %%
# Generate sample data
# --------------------
import numpy as np
rng = np.random.RandomState(42)
X = 5 * rng.rand(10000, 1)
y = np.sin(X).ravel()
# Add noise to targets
y[::5] += 3 * (0.5 - rng.rand(X.shape[0] // 5))
X_plot = np.linspace(0, 5, 100000)[:, None]
# %%
# Construct the kernel-based regression models
# --------------------------------------------
from sklearn.kernel_ridge import KernelRidge
from sklearn.model_selection import GridSearchCV
from sklearn.svm import SVR
train_size = 100
svr = GridSearchCV(
SVR(kernel="rbf", gamma=0.1),
param_grid={"C": [1e0, 1e1, 1e2, 1e3], "gamma": np.logspace(-2, 2, 5)},
)
kr = GridSearchCV(
KernelRidge(kernel="rbf", gamma=0.1),
param_grid={"alpha": [1e0, 0.1, 1e-2, 1e-3], "gamma": np.logspace(-2, 2, 5)},
)
# %%
# Compare times of SVR and Kernel Ridge Regression
# ------------------------------------------------
import time
t0 = time.time()
svr.fit(X[:train_size], y[:train_size])
svr_fit = time.time() - t0
print(f"Best SVR with params: {svr.best_params_} and R2 score: {svr.best_score_:.3f}")
print("SVR complexity and bandwidth selected and model fitted in %.3f s" % svr_fit)
t0 = time.time()
kr.fit(X[:train_size], y[:train_size])
kr_fit = time.time() - t0
print(f"Best KRR with params: {kr.best_params_} and R2 score: {kr.best_score_:.3f}")
print("KRR complexity and bandwidth selected and model fitted in %.3f s" % kr_fit)
sv_ratio = svr.best_estimator_.support_.shape[0] / train_size
print("Support vector ratio: %.3f" % sv_ratio)
t0 = time.time()
y_svr = svr.predict(X_plot)
svr_predict = time.time() - t0
print("SVR prediction for %d inputs in %.3f s" % (X_plot.shape[0], svr_predict))
t0 = time.time()
y_kr = kr.predict(X_plot)
kr_predict = time.time() - t0
print("KRR prediction for %d inputs in %.3f s" % (X_plot.shape[0], kr_predict))
# %%
# Look at the results
# -------------------
import matplotlib.pyplot as plt
sv_ind = svr.best_estimator_.support_
plt.scatter(
X[sv_ind],
y[sv_ind],
c="r",
s=50,
label="SVR support vectors",
zorder=2,
edgecolors=(0, 0, 0),
)
plt.scatter(X[:100], y[:100], c="k", label="data", zorder=1, edgecolors=(0, 0, 0))
plt.plot(
X_plot,
y_svr,
c="r",
label="SVR (fit: %.3fs, predict: %.3fs)" % (svr_fit, svr_predict),
)
plt.plot(
X_plot, y_kr, c="g", label="KRR (fit: %.3fs, predict: %.3fs)" % (kr_fit, kr_predict)
)
plt.xlabel("data")
plt.ylabel("target")
plt.title("SVR versus Kernel Ridge")
_ = plt.legend()
# %%
# The previous figure compares the learned model of KRR and SVR when both
# complexity/regularization and bandwidth of the RBF kernel are optimized using
# grid-search. The learned functions are very similar; however, fitting KRR is
# approximately 3-4 times faster than fitting SVR (both with grid-search).
#
# Prediction of 100000 target values could be in theory approximately three
# times faster with SVR since it has learned a sparse model using only
# approximately 1/3 of the training datapoints as support vectors. However, in
# practice, this is not necessarily the case because of implementation details
# in the way the kernel function is computed for each model that can make the
# KRR model as fast or even faster despite computing more arithmetic
# operations.
# %%
# Visualize training and prediction times
# ---------------------------------------
plt.figure()
sizes = np.logspace(1, 3.8, 7).astype(int)
for name, estimator in {
"KRR": KernelRidge(kernel="rbf", alpha=0.01, gamma=10),
"SVR": SVR(kernel="rbf", C=1e2, gamma=10),
}.items():
train_time = []
test_time = []
for train_test_size in sizes:
t0 = time.time()
estimator.fit(X[:train_test_size], y[:train_test_size])
train_time.append(time.time() - t0)
t0 = time.time()
estimator.predict(X_plot[:1000])
test_time.append(time.time() - t0)
plt.plot(
sizes,
train_time,
"o-",
color="r" if name == "SVR" else "g",
label="%s (train)" % name,
)
plt.plot(
sizes,
test_time,
"o--",
color="r" if name == "SVR" else "g",
label="%s (test)" % name,
)
plt.xscale("log")
plt.yscale("log")
plt.xlabel("Train size")
plt.ylabel("Time (seconds)")
plt.title("Execution Time")
_ = plt.legend(loc="best")
# %%
# This figure compares the time for fitting and prediction of KRR and SVR for
# different sizes of the training set. Fitting KRR is faster than SVR for
# medium-sized training sets (less than a few thousand samples); however, for
# larger training sets SVR scales better. With regard to prediction time, SVR
# should be faster than KRR for all sizes of the training set because of the
# learned sparse solution, however this is not necessarily the case in practice
# because of implementation details. Note that the degree of sparsity and thus
# the prediction time depends on the parameters epsilon and C of the SVR.
# %%
# Visualize the learning curves
# -----------------------------
from sklearn.model_selection import LearningCurveDisplay
_, ax = plt.subplots()
svr = SVR(kernel="rbf", C=1e1, gamma=0.1)
kr = KernelRidge(kernel="rbf", alpha=0.1, gamma=0.1)
common_params = {
"X": X[:100],
"y": y[:100],
"train_sizes": np.linspace(0.1, 1, 10),
"scoring": "neg_mean_squared_error",
"negate_score": True,
"score_name": "Mean Squared Error",
"score_type": "test",
"std_display_style": None,
"ax": ax,
}
LearningCurveDisplay.from_estimator(svr, **common_params)
LearningCurveDisplay.from_estimator(kr, **common_params)
ax.set_title("Learning curves")
ax.legend(handles=ax.get_legend_handles_labels()[0], labels=["SVR", "KRR"])
plt.show()