numpy:如何在np数组中选择特定索引进行k-fold交叉验证？

Question

我有一个尺寸为5000 x 3027(CIFAR-10数据集)的矩阵形式的训练数据集.在numpy中使用array_split,我将它分成5个不同的部分,我想只选择其中一个部分作为交叉验证折叠.然而,当我使用像XTrain [[Indexes]]之类的东西时,我的问题出现了,其中索引是像[0,1,2,3]这样的数组,因为这样做会给我一个尺寸为4 x 1000 x 3027的3D张量,而不是矩阵.如何将"4 x 1000"折叠成4000行,以获得4000 x 3027的矩阵？

for fold in range(len(X_train_folds)):
    indexes = np.delete(np.arange(len(X_train_folds)), fold) 
    XTrain = X_train_folds[indexes]
    X_cv = X_train_folds[fold]
    yTrain = y_train_folds[indexes]
    y_cv = y_train_folds[fold]

    classifier.train(XTrain, yTrain)
    dists = classifier.compute_distances_no_loops(X_cv)
    y_test_pred = classifier.predict_labels(dists, k)

    num_correct = np.sum(y_test_pred == y_test)
    accuracy = float(num_correct/num_test)
    k_to_accuracy[k] = accuracy

Answer 1

也许您可以试试看（numpy的新手，所以如果我做的事情效率低下/错误，将很乐意予以纠正）

X_train_folds = np.array_split(X_train, num_folds)
y_train_folds = np.array_split(y_train, num_folds)
k_to_accuracies = {}

for k in k_choices:
    k_to_accuracies[k] = []
    for i in range(num_folds):
        training_data, test_data = np.concatenate(X_train_folds[:i] + X_train_folds[i+1:]), X_train_folds[i]
        training_labels, test_labels = np.concatenate(y_train_folds[:i] + y_train_folds[i+1:]), y_train_folds[i]
        classifier.train(training_data, training_labels)
        predicted_labels = classifier.predict(test_data, k)
        k_to_accuracies[k].append(np.sum(predicted_labels == test_labels)/len(test_labels))

Answer 2

我建议使用scikit-learn包。它已经配备了大量常见的机器学习工具，例如K 折交叉验证生成器：

>>> from sklearn.cross_validation import KFold
>>> X = # your data [samples x features]
>>> y = # gt labels
>>> kf = KFold(X.shape[0], n_folds=5)

然后，迭代kf：

>>> for train_index, test_index in kf:
        X_train, X_test = X[train_index], X[test_index]
        y_train, y_test = y[train_index], y[test_index]
        # do something

上述循环将被执行n_folds多次，每次都有不同的训练和测试指标。