Question

我正在尝试使用Keras构建一个多分类器。我不确定我是否正确实施了它。数据就是这样

标签时间序列变量[0：25728}

index 0  1   2   3   4            25728
  0   1  2.5 3.2 1.6 1.05 ........ 2.54
  1   5  3.2 1.6 1.5 1.49 ........ 1.41
  2   1  2.3 3.2 1.5 1.52 ........ 2.11
  3   3  0.2 3.1 1.5 1.89 ........ 0.81
  4   8  1.2 1.1 0.2 1.19 ........ 3.71
  .   5  .    .   .   .   ........   .
  .   7  .    .   .   .   ........   .
1323  5  .    .   .   .   ........   .

这是代码。我将数据分割了68％，然后将1D数组重塑为2D数组。如384 * 67 = 25728 因此，对于一个标签，通过67形成矢量384的图像

x_train =x_train.reshape(x_train.shape[0],384,67)
x_test =x_test.reshape(x_test.shape[0],384,67)

train_label_y = df_train_y[2].values
test_label_y = df_test_y[2].values

train_Y = np.array(train_label_y[:], dtype=int)
test_Y  = np.array(test_label_y[:], dtype=int)

Y_train = convert_to_one_hot(train_Y, C = 8)
Y_test  = convert_to_one_hot(test_Y,  C = 8)

x_train = x_train.reshape(x_train.shape + (1,))
x_test = x_test.reshape(x_test.shape + (1,))

input_shape = x_train.shape[1:]
model = Sequential()
model.add(Conv2D(2, kernel_size=(3, 3), padding='same', input_shape=input_shape))
model.add(Activation('relu'))
model.add(MaxPooling2D(pool_size=(2, 2)))

model.add(Conv2D(64, kernel_size=(3, 3), padding='same'))
model.add(Activation('relu'))
model.add(MaxPooling2D(pool_size=(2, 2)))

model.add(Conv2D(128, kernel_size=(3, 3), padding='same'))
model.add(Activation('relu'))
model.add(MaxPooling2D(pool_size=(2, 2)))

model.add(Conv2D(8, kernel_size=(3, 3), padding='same'))
model.add(Activation('relu'))
model.add(MaxPooling2D(pool_size=(2, 2)))

model.add(GlobalAveragePooling2D())
model.add(Dense(8, activation='sigmoid'))

model.compile(loss='categorical_crossentropy', optimizer='rmsprop', metrics=['accuracy'])
hist = model.fit(x_train, Y_train, batch_size=batch_size, epochs=nb_epochs, verbose=1)
score = model.evaluate(x_test, Y_test)
print("Accuracy: %.2f%%" % (score[1] * 100))

它提供96.16％的准确性，但我不相信这是真的。我想预测标签。

如何预测标签？
我做错了什么？

请帮助！谢谢。

Answer 1

获得怪异结果的原因是，您应该在最后一层使用“ softmax”激活。

将时间序列作为输入图像向量的Keras多标签时间序列分类

1 个答案: