我试图比较我拥有的数据集上的多个分类器 . 为了获得分类器的准确准确度分数，我现在对每个分类器执行10倍交叉验证 . 除了SVM（线性和rbf内核）之外，这对所有这些都很顺利 . 数据加载如下：
dataset = pd.read_csv("data/distance_annotated_indels.txt", delimiter="\t", header=None)
X = dataset.iloc[:, [5,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26]].values
y = dataset.iloc[:, 4].values
from sklearn.model_selection import train_test_split
X_train, X_test, y_train, y_test = train_test_split(X, y, test_size = 0.2)
from sklearn.preprocessing import StandardScaler
sc = StandardScaler()
X_train = sc.fit_transform(X_train)
X_test = sc.transform(X_test)
例如随机森林的交叉验证工作正常：
start = time.time()
classifier = RandomForestClassifier(n_estimators = 100, criterion = 'entropy')
classifier.fit(X_train, y_train)
y_pred = classifier.predict(X_test)
cv = ShuffleSplit(n_splits=10, test_size=0.2)
scores = cross_val_score(classifier, X, y, cv=10)
print(classification_report(y_test, y_pred))
print("Random Forest accuracy after 10 fold CV: %0.2f (+/- %0.2f)" % (scores.mean(), scores.std() * 2) + ", " + str(round(time.time() - start, 3)) + "s")
输出：
precision recall f1-score support
0 0.97 0.95 0.96 3427
1 0.95 0.97 0.96 3417
avg / total 0.96 0.96 0.96 6844
Random Forest accuracy after 10 fold CV: 0.92 (+/- 0.06), 90.842s
然而对于SVM来说，这个过程需要很长时间（等待2个小时，但仍然没有） . sklearn网站并没有让我更聪明 . 对于SVM分类器，我应该做些什么吗？ SVM代码如下：
start = time.time()
classifier = SVC(kernel = 'linear')
classifier.fit(X_train, y_train)
y_pred = classifier.predict(X_test)
scores = cross_val_score(classifier, X, y, cv=10)
print(classification_report(y_test, y_pred))
print("Linear SVM accuracy after 10 fold CV: %0.2f (+/- %0.2f)" % (scores.mean(), scores.std() * 2) + ", " + str(round(time.time() - start, 3)) + "s")
1 回答
如果您有大量样本，则问题的计算复杂性会受到影响，请参阅Training complexity of Linear SVM .
考虑使用
cross_val_score的verbose标志来查看有关进度的更多日志 . 此外，将n_jobs设置为值> 1（或者甚至使用n_jobs设置为-1的所有CPU，如果内存允许），您可以通过并行化加速计算 . http://scikit-learn.org/stable/modules/generated/sklearn.model_selection.cross_val_score.html可用于评估这些选项 .如果性能很差，我会考虑减少
cv的值（有关此问题的讨论，请参阅https://stats.stackexchange.com/questions/27730/choice-of-k-in-k-fold-cross-validation）