GridSearchCV 在 scikit-learn 中的小数据集上速度极慢

2024-04-11

这很奇怪。我可以成功运行该示例grid_search_digits.py。但是，我无法对自己的数据进行网格搜索。

我有以下设置：

import sklearn
from sklearn.svm import SVC
from sklearn.grid_search import GridSearchCV
from sklearn.cross_validation import LeaveOneOut
from sklearn.metrics import auc_score

# ... Build X and y ....

tuned_parameters = [{'kernel': ['rbf'], 'gamma': [1e-3, 1e-4],
                     'C': [1, 10, 100, 1000]},
                    {'kernel': ['linear'], 'C': [1, 10, 100, 1000]}]

loo = LeaveOneOut(len(y))
clf = GridSearchCV(SVC(C=1), tuned_parameters, score_func=auc_score)
clf.fit(X, y, cv=loo)
....
print clf.best_estimator_
....

但我从来没有通过clf.fit（我让它运行约 1 小时）。

我也尝试过

clf.fit(X, y, cv=10)

and with

skf = StratifiedKFold(y,2)
clf.fit(X, y, cv=skf)

并遇到了同样的问题（它永远不会完成 clf.fit 语句）。我的数据很简单：

> X.shape
(27,26)

> y.shape
27

> numpy.sum(y)
5

> y.dtype
dtype('int64')


>?y
Type:       ndarray
String Form:[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1]
Length:     27
File:       /home/jacob04/opt/python/numpy/numpy-1.7.1/lib/python2.7/site-
packages/numpy/__init__.py                                                
Docstring:  <no docstring>
Class Docstring:
ndarray(shape, dtype=float, buffer=None, offset=0,
        strides=None, order=None)

> ?X
Type:       ndarray
String Form:
       [[ -3.61238468e+03  -3.61253920e+03  -3.61290196e+03  -3.61326679e+03
           7.84590361e+02   0.0000 <...> 0000e+00   2.22389150e+00   2.53252959e+00 
           2.11606216e+00  -1.99613432e+05  -1.99564828e+05]]
Length:     27
File:       /home/jacob04/opt/python/numpy/numpy-1.7.1/lib/python2.7/site-
packages/numpy/__init__.py                                                
Docstring:  <no docstring>
Class Docstring:
ndarray(shape, dtype=float, buffer=None, offset=0,
        strides=None, order=None)

这一切都与最新版本的 scikit-learn (0.13.1) 相关，并且：

$ pip freeze
Cython==0.19.1
PIL==1.1.7
PyXB==1.2.2
PyYAML==3.10
argparse==1.2.1
distribute==0.6.34
epc==0.0.5
ipython==0.13.2
jedi==0.6.0
matplotlib==1.3.x
nltk==2.0.4
nose==1.3.0
numexpr==2.1
numpy==1.7.1
pandas==0.11.0
pyparsing==1.5.7
python-dateutil==2.1
pytz==2013b
rpy2==2.3.1
scikit-learn==0.13.1
scipy==0.12.0
sexpdata==0.0.3
six==1.3.0
stemming==1.0.1
-e git+https://github.com/PyTables/PyTables.git@df7b20444b0737cf34686b5d88b4e674ec85575b#egg=tables-dev
tornado==3.0.1
wsgiref==0.1.2

奇怪的是，拟合单个 SVM 的速度非常快：

>  %timeit clf2 = svm.SVC(); clf2.fit(X,y)                                                                                                             
1000 loops, best of 3: 328 us per loop

Update

我注意到，如果我使用以下方法预先缩放数据：

from sklearn import preprocessing
X = preprocessing.scale(X)

网格搜索速度非常快。

为什么？为什么GridSearchCV对缩放非常敏感，而常规svm.SVC().fit is not?

正如已经指出的，为了SVM基于分类器（如y == np.int* ) 预处理是必须的，否则 ML 估计器的预测能力会因倾斜特征对决策函数的影响而丧失。

正如所反对的处理时间：

尝试更好地了解您的 AI/ML 模型过度拟合/泛化是什么[C,gamma]景观
尝试添加冗长进入初始 AI/ML 流程调整
尝试添加n_jobs进入数字运算
如果规模需要，尝试将网格计算添加到您的计算方法中

aGrid = aML_GS.GridSearchCV( aClassifierOBJECT,
                                    param_grid = aGrid_of_parameters,
                                    cv         = cv,
                                    n_jobs     = n_JobsOnMultiCpuCores,
                                    verbose    = 5 )

有时，GridSearchCV()确实会占用大量的 CPU 时间/CPU 资源池，即使使用了上述所有技巧之后。

因此，如果您确定特征工程、数据完整性和特征域预处理正确完成，请保持冷静，不要惊慌。

[GridSearchCV] ................ C=16777216.0, gamma=0.5, score=0.761619 -62.7min
[GridSearchCV] C=16777216.0, gamma=0.5 .........................................
[GridSearchCV] ................ C=16777216.0, gamma=0.5, score=0.792793 -64.4min
[GridSearchCV] C=16777216.0, gamma=1.0 .........................................
[GridSearchCV] ............... C=16777216.0, gamma=1.0, score=0.793103 -116.4min
[GridSearchCV] C=16777216.0, gamma=1.0 .........................................
[GridSearchCV] ............... C=16777216.0, gamma=1.0, score=0.794603 -205.4min
[GridSearchCV] C=16777216.0, gamma=1.0 .........................................
[GridSearchCV] ............... C=16777216.0, gamma=1.0, score=0.771772 -200.9min
[GridSearchCV] C=16777216.0, gamma=2.0 .........................................
[GridSearchCV] ............... C=16777216.0, gamma=2.0, score=0.713643 -446.0min
[GridSearchCV] C=16777216.0, gamma=2.0 .........................................
[GridSearchCV] ............... C=16777216.0, gamma=2.0, score=0.743628 -184.6min
[GridSearchCV] C=16777216.0, gamma=2.0 .........................................
[GridSearchCV] ............... C=16777216.0, gamma=2.0, score=0.761261 -281.2min
[GridSearchCV] C=16777216.0, gamma=4.0 .........................................
[GridSearchCV] ............... C=16777216.0, gamma=4.0, score=0.670165 -138.7min
[GridSearchCV] C=16777216.0, gamma=4.0 .........................................
[GridSearchCV] ................ C=16777216.0, gamma=4.0, score=0.760120 -97.3min
[GridSearchCV] C=16777216.0, gamma=4.0 .........................................
[GridSearchCV] ................ C=16777216.0, gamma=4.0, score=0.732733 -66.3min
[GridSearchCV] C=16777216.0, gamma=8.0 .........................................
[GridSearchCV] ................ C=16777216.0, gamma=8.0, score=0.755622 -13.6min
[GridSearchCV] C=16777216.0, gamma=8.0 .........................................
[GridSearchCV] ................ C=16777216.0, gamma=8.0, score=0.772114 - 4.6min
[GridSearchCV] C=16777216.0, gamma=8.0 .........................................
[GridSearchCV] ................ C=16777216.0, gamma=8.0, score=0.717718 -14.7min
[GridSearchCV] C=16777216.0, gamma=16.0 ........................................
[GridSearchCV] ............... C=16777216.0, gamma=16.0, score=0.763118 - 1.3min
[GridSearchCV] C=16777216.0, gamma=16.0 ........................................
[GridSearchCV] ............... C=16777216.0, gamma=16.0, score=0.746627 -  25.4s
[GridSearchCV] C=16777216.0, gamma=16.0 ........................................
[GridSearchCV] ............... C=16777216.0, gamma=16.0, score=0.738739 -  44.9s
[Parallel(n_jobs=1)]: Done 2700 out of 2700 | elapsed: 5670.8min finished

正如上面所问的“......常规svm.SVC().fit” 请注意，它使用默认值[C,gamma]值，因此与您的模型/问题域的行为无关。

回复：更新

哦，是的，确实，SVM 输入的正则化/缩放是这个 AI/ML 工具的强制性任务。 scikit-learn 有很好的工具来生成和重用aScalerOBJECT对于先验缩放（之前aDataSET进入.fit()）和事后临时缩放，一旦您需要重新缩放新的example并将其发送给预测器来回答它的魔力通过请求
anSvmCLASSIFIER.predict( aScalerOBJECT.transform( aNewExampleX ) )

( Yes, aNewExampleX可能是一个矩阵，因此要求对多个答案进行“矢量化”处理）

Performance relief of O( M² . N¹ ) computational complexity

In contrast to the below posted guess, that the Problem-"width", measured as N == a number of SVM-Features in matrix X is to be blamed for an overall computing time, the SVM classifier with rbf-kernel is by-design an O( M² . N¹ ) problem.

因此，对观察总数（示例）存在二次依赖性，移入训练（.fit()）或交叉验证阶段，并且很难说，如果“减少”特征的（仅线性）“宽度”，监督学习分类器将获得更好的预测能力，这本身bear输入到 SVM 分类器构建的预测能力中，不是吗？

本文内容由网友自发贡献，版权归原作者所有，本站不承担相应法律责任。如您发现有涉嫌抄袭侵权的内容，请联系:hwhale#tublm.com(使用前将#替换为@)