1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
57
58
59
60
61
62
63
64
65
66
67
68
69
70
71
72
73
74
75
76
77
78
79
80
81
82
83
84
85
86
87
88
89
90
91
92
93
94
95
96
97
98
99
100
101
102
103
104
105
106
107
108
109
110
111
112
113
114
115
116
117
118
119
120
121
122
123
124
125
126
127
128
129
130
131
132
133
134
135
136
137
138
139
140
141
142
143
144
145
146
147
148
149
150
151
152
153
154
155
156
157
158
159
160
161
162
163
164
165
166
167
168
169
170
171
172
|
r"""
==============================================
Scaling the regularization parameter for SVCs
==============================================
The following example illustrates the effect of scaling the
regularization parameter when using :ref:`svm` for
:ref:`classification <svm_classification>`.
For SVC classification, we are interested in a risk minimization for the
equation:
.. math::
C \sum_{i=1, n} \mathcal{L} (f(x_i), y_i) + \Omega (w)
where
- :math:`C` is used to set the amount of regularization
- :math:`\mathcal{L}` is a `loss` function of our samples
and our model parameters.
- :math:`\Omega` is a `penalty` function of our model parameters
If we consider the loss function to be the individual error per
sample, then the data-fit term, or the sum of the error for each sample, will
increase as we add more samples. The penalization term, however, will not
increase.
When using, for example, :ref:`cross validation <cross_validation>`, to
set the amount of regularization with `C`, there will be a
different amount of samples between the main problem and the smaller problems
within the folds of the cross validation.
Since our loss function is dependent on the amount of samples, the latter
will influence the selected value of `C`.
The question that arises is "How do we optimally adjust C to
account for the different amount of training samples?"
In the remainder of this example, we will investigate the effect of scaling
the value of the regularization parameter `C` in regards to the number of
samples for both L1 and L2 penalty. We will generate some synthetic datasets
that are appropriate for each type of regularization.
"""
# Author: Andreas Mueller <amueller@ais.uni-bonn.de>
# Jaques Grobler <jaques.grobler@inria.fr>
# License: BSD 3 clause
# %%
# L1-penalty case
# ---------------
# In the L1 case, theory says that prediction consistency (i.e. that under
# given hypothesis, the estimator learned predicts as well as a model knowing
# the true distribution) is not possible because of the bias of the L1. It
# does say, however, that model consistency, in terms of finding the right set
# of non-zero parameters as well as their signs, can be achieved by scaling
# `C`.
#
# We will demonstrate this effect by using a synthetic dataset. This
# dataset will be sparse, meaning that only a few features will be informative
# and useful for the model.
from sklearn.datasets import make_classification
n_samples, n_features = 100, 300
X, y = make_classification(
n_samples=n_samples, n_features=n_features, n_informative=5, random_state=1
)
# %%
# Now, we can define a linear SVC with the `l1` penalty.
from sklearn.svm import LinearSVC
model_l1 = LinearSVC(penalty="l1", loss="squared_hinge", dual=False, tol=1e-3)
# %%
# We will compute the mean test score for different values of `C`.
import numpy as np
import pandas as pd
from sklearn.model_selection import validation_curve, ShuffleSplit
Cs = np.logspace(-2.3, -1.3, 10)
train_sizes = np.linspace(0.3, 0.7, 3)
labels = [f"fraction: {train_size}" for train_size in train_sizes]
results = {"C": Cs}
for label, train_size in zip(labels, train_sizes):
cv = ShuffleSplit(train_size=train_size, test_size=0.3, n_splits=50, random_state=1)
train_scores, test_scores = validation_curve(
model_l1, X, y, param_name="C", param_range=Cs, cv=cv
)
results[label] = test_scores.mean(axis=1)
results = pd.DataFrame(results)
# %%
import matplotlib.pyplot as plt
fig, axes = plt.subplots(nrows=1, ncols=2, sharey=True, figsize=(12, 6))
# plot results without scaling C
results.plot(x="C", ax=axes[0], logx=True)
axes[0].set_ylabel("CV score")
axes[0].set_title("No scaling")
# plot results by scaling C
for train_size_idx, label in enumerate(labels):
results_scaled = results[[label]].assign(
C_scaled=Cs * float(n_samples * train_sizes[train_size_idx])
)
results_scaled.plot(x="C_scaled", ax=axes[1], logx=True, label=label)
axes[1].set_title("Scaling C by 1 / n_samples")
_ = fig.suptitle("Effect of scaling C with L1 penalty")
# %%
# Here, we observe that the cross-validation-error correlates best with the
# test-error, when scaling our `C` with the number of samples, `n`.
#
# L2-penalty case
# ---------------
# We can repeat a similar experiment with the `l2` penalty. In this case, we
# don't need to use a sparse dataset.
#
# In this case, the theory says that in order to achieve prediction
# consistency, the penalty parameter should be kept constant as the number of
# samples grow.
#
# So we will repeat the same experiment by creating a linear SVC classifier
# with the `l2` penalty and check the test score via cross-validation and
# plot the results with and without scaling the parameter `C`.
rng = np.random.RandomState(1)
y = np.sign(0.5 - rng.rand(n_samples))
X = rng.randn(n_samples, n_features // 5) + y[:, np.newaxis]
X += 5 * rng.randn(n_samples, n_features // 5)
# %%
model_l2 = LinearSVC(penalty="l2", loss="squared_hinge", dual=True)
Cs = np.logspace(-4.5, -2, 10)
labels = [f"fraction: {train_size}" for train_size in train_sizes]
results = {"C": Cs}
for label, train_size in zip(labels, train_sizes):
cv = ShuffleSplit(train_size=train_size, test_size=0.3, n_splits=50, random_state=1)
train_scores, test_scores = validation_curve(
model_l2, X, y, param_name="C", param_range=Cs, cv=cv
)
results[label] = test_scores.mean(axis=1)
results = pd.DataFrame(results)
# %%
import matplotlib.pyplot as plt
fig, axes = plt.subplots(nrows=1, ncols=2, sharey=True, figsize=(12, 6))
# plot results without scaling C
results.plot(x="C", ax=axes[0], logx=True)
axes[0].set_ylabel("CV score")
axes[0].set_title("No scaling")
# plot results by scaling C
for train_size_idx, label in enumerate(labels):
results_scaled = results[[label]].assign(
C_scaled=Cs * float(n_samples * train_sizes[train_size_idx])
)
results_scaled.plot(x="C_scaled", ax=axes[1], logx=True, label=label)
axes[1].set_title("Scaling C by 1 / n_samples")
_ = fig.suptitle("Effect of scaling C with L2 penalty")
# %%
# So or the L2 penalty case, the best result comes from the case where `C` is
# not scaled.
plt.show()
|
