您用于预测的 gram 矩阵似乎是错误的。

一旦你拟合了 SVM，它的预测$x$是：$y = \text{sign} \langle w,\phi(x)\rangle$

肯定有可能无法计算$\phi(x)$但需要使用预先计算的 gram 矩阵。代替$w = \sum_i^m\alpha_i\phi(x_i)$ 进入上述预测表达式：

$$y = \text{sign } \langle \sum_i^m\alpha\phi(x_i),\phi(x) \rangle = \text{sign } \sum_i^m\alpha_i\langle\phi(x_i),\phi(x) \rangle = \text{sign } \sum_i^m\alpha_i K(x_i,x) \rangle = \text{sign } (Xx)^T\alpha = \text{sign } {G}\alpha$$

在哪里$X$是训练数据的设计矩阵（行$i$的$X$是$\phi(x_i)$) 和 gram 矩阵$G=(Xx)^T$（注意 G 不是对称的）

您可以将此作为参考：

from sklearn.datasets import load_digits
from sklearn.svm import SVC
from sklearn.utils import shuffle
from sklearn.metrics import accuracy_score
import numpy as np


digits = load_digits()
X, y = shuffle(digits.data, digits.target)
X_train, X_test = X[:1000, :], X[100:, :]
y_train, y_test = y[:1000], y[100:]

svc = SVC(kernel='precomputed')

kernel_train = np.dot(X_train, X_train.T)  # linear kernel

svc.fit(kernel_train, y_train)

#kernel_test = np.dot(X_test, X_train[svc.support_, :].T)
kernel_test = np.dot(X_test, X_train.T)
y_pred = svc.predict(kernel_test)
print 'accuracy score: %0.3f' % accuracy_score(y_test, y_pred)

我发现我通常会从 kfold 循环之外的训练/矢量化中得到该错误。

vec.fit_transform()、vec.fit_transform()、clf.fit()、clf.pred() - 所有这四个函数调用都必须在循环内部，以确保所有矩阵都获得正确的形状。

您的测试算法设置是这样的吗？：

k_folds = sklearn.cross_validation.StratifiedKFold(list_of_annotations, k=5)

for train_index, test_index in k_folds:
    X_train, X_test = X[train_index], X[test_index]
    y_train, y_test = y[train_index], y[test_index]

    vec = TfidfVectorizer()
--> X_train_matrix = vec.fit_transform(X_train)
--> X_test_matrix = vec.transform(X_test)

    clf = svm.SVC()
--> clf.fit(X_train_matrix, y_train)

--> y_pred = clf.predict(X_test_matrix)
    precision, recall, fscore, support = precision_recall_fscore_support(y_test, y_pred)
    # print/save accuracy scores