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核岭回归和支持向量回归的比较#

核岭回归（KRR）和支持向量回归（SVR）都通过使用核技巧来学习非线性函数，即它们在由相应核引导的空间中学习线性函数，这对应于原始空间中的非线性函数。它们在损失函数（岭损失与ε-不敏感损失）上有所不同。与SVR相比，KRR的拟合可以通过闭式解完成，并且对于中等规模的数据集通常更快。另一方面，所学习的模型是非稀疏的，因此在预测时比SVR更慢。

这个示例在一个人工数据集上展示了这两种方法，该数据集由一个正弦目标函数和每五个数据点添加的强噪声组成。

作者：scikit-learn 开发者 SPDX-License-Identifier: BSD-3-Clause

生成示例数据#

import numpy as np

rng = np.random.RandomState(42)

X = 5 * rng.rand(10000, 1)
y = np.sin(X).ravel()

# 添加噪声到目标
y[::5] += 3 * (0.5 - rng.rand(X.shape[0] // 5))

X_plot = np.linspace(0, 5, 100000)[:, None]

构建基于核的回归模型#

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)},
)

比较SVR和核岭回归的时间#

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))

Best SVR with params: {'C': 1.0, 'gamma': np.float64(0.1)} and R2 score: 0.737
SVR complexity and bandwidth selected and model fitted in 0.266 s
Best KRR with params: {'alpha': 0.1, 'gamma': np.float64(0.1)} and R2 score: 0.723
KRR complexity and bandwidth selected and model fitted in 0.076 s
Support vector ratio: 0.340
SVR prediction for 100000 inputs in 0.060 s
KRR prediction for 100000 inputs in 0.066 s

查看结果#

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()

前面的图比较了在使用网格搜索优化RBF核的复杂度/正则化和带宽时，KRR和SVR的学习模型。学习到的函数非常相似；然而，拟合KRR的速度大约是拟合SVR（均使用网格搜索）的3-4倍。

理论上，预测100000个目标值时，SVR可能会快大约三倍，因为它仅使用大约1/3的训练数据点作为支持向量学习了一个稀疏模型。然而，实际上，由于每个模型计算核函数的实现细节，KRR模型可能同样快甚至更快，尽管它需要进行更多的算术运算。

可视化训练和预测时间#

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")

此图比较了KRR和SVR在不同训练集规模下的拟合和预测时间。对于中等规模的训练集（少于几千个样本），KRR的拟合速度比SVR快；然而，对于较大的训练集，SVR的扩展性更好。关于预测时间，由于学习到的稀疏解，SVR在所有训练集规模下都应该比KRR快，但由于实现细节，这在实际中不一定成立。请注意，稀疏度以及预测时间取决于SVR的参数epsilon和C。

可视化学习曲线#

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()