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.. "auto_examples/miscellaneous/plot_kernel_ridge_regression.py"
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.. only:: html
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.. rst-class:: sphx-glr-example-title
.. _sphx_glr_auto_examples_miscellaneous_plot_kernel_ridge_regression.py:
=============================================
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.
.. GENERATED FROM PYTHON SOURCE LINES 21-23
Authors: Jan Hendrik Metzen <jhm@informatik.uni-bremen.de>
License: BSD 3 clause
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Generate sample data
--------------------
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.. code-block:: Python
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]
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Construct the kernel-based regression models
--------------------------------------------
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.. code-block:: Python
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)},
)
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Compare times of SVR and Kernel Ridge Regression
------------------------------------------------
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.. code-block:: Python
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))
.. rst-class:: sphx-glr-script-out
.. code-block:: none
Best SVR with params: {'C': 1.0, 'gamma': 0.09999999999999999} and R2 score: 0.737
SVR complexity and bandwidth selected and model fitted in 0.494 s
Best KRR with params: {'alpha': 0.1, 'gamma': 0.09999999999999999} and R2 score: 0.723
KRR complexity and bandwidth selected and model fitted in 0.192 s
Support vector ratio: 0.340
SVR prediction for 100000 inputs in 0.132 s
KRR prediction for 100000 inputs in 0.092 s
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Look at the results
-------------------
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.. code-block:: Python
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()
.. image-sg:: /auto_examples/miscellaneous/images/sphx_glr_plot_kernel_ridge_regression_001.png
:alt: SVR versus Kernel Ridge
:srcset: /auto_examples/miscellaneous/images/sphx_glr_plot_kernel_ridge_regression_001.png
:class: sphx-glr-single-img
.. GENERATED FROM PYTHON SOURCE LINES 122-134
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.
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Visualize training and prediction times
---------------------------------------
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.. code-block:: Python
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")
.. image-sg:: /auto_examples/miscellaneous/images/sphx_glr_plot_kernel_ridge_regression_002.png
:alt: Execution Time
:srcset: /auto_examples/miscellaneous/images/sphx_glr_plot_kernel_ridge_regression_002.png
:class: sphx-glr-single-img
.. GENERATED FROM PYTHON SOURCE LINES 180-188
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.
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Visualize the learning curves
-----------------------------
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.. code-block:: Python
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()
.. image-sg:: /auto_examples/miscellaneous/images/sphx_glr_plot_kernel_ridge_regression_003.png
:alt: Learning curves
:srcset: /auto_examples/miscellaneous/images/sphx_glr_plot_kernel_ridge_regression_003.png
:class: sphx-glr-single-img
.. rst-class:: sphx-glr-timing
**Total running time of the script:** (0 minutes 8.155 seconds)
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