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操纵神经网络的输出

操纵神经网络的输出

holdtom 2022-05-11 16:31:04
我有一个神经网络,输入 (m, 2, 3, 96, 96) 和输出 (m, 2, 128)。我试图通过减去 output[m][0][0] - output[m][0][1] 将输出转换为 (m, 1, 128),然后通过输入1x128 输出到密集层我在网络和包装器中尝试了 Lambda 和 keras.backend.Subtract 层def faceRecoModel(input_shape):    """    Implementation of the Inception model used for FaceNet    Arguments:    input_shape -- shape of the images of the dataset    Returns:    model -- a Model() instance in Keras    """    # Define the input as a tensor with shape input_shape    X_input = Input(input_shape)    # Zero-Padding    X = ZeroPadding2D((3, 3))(X_input)    # First Block    X = Conv2D(64, (7, 7), strides=(2, 2), name='conv1')(X)    X = BatchNormalization(axis=1, name='bn1')(X)    X = Activation('relu')(X)    # Zero-Padding + MAXPOOL    X = ZeroPadding2D((1, 1))(X)    X = MaxPooling2D((3, 3), strides=2)(X)    # Second Block    X = Conv2D(64, (1, 1), strides=(1, 1), name='conv2')(X)    X = BatchNormalization(axis=1, epsilon=0.00001, name='bn2')(X)    X = Activation('relu')(X)    # Zero-Padding + MAXPOOL    X = ZeroPadding2D((1, 1))(X)    # Second Block    X = Conv2D(192, (3, 3), strides=(1, 1), name='conv3')(X)    X = BatchNormalization(axis=1, epsilon=0.00001, name='bn3')(X)    X = Activation('relu')(X)    # Zero-Padding + MAXPOOL    X = ZeroPadding2D((1, 1))(X)    X = MaxPooling2D(pool_size=3, strides=2)(X)    # Inception 1: a/b/c    X = inception_block_1a(X)    X = inception_block_1b(X)    X = inception_block_1c(X)    # Inception 2: a/b    X = inception_block_2a(X)    X = inception_block_2b(X)    # Inception 3: a/b    X = inception_block_3a(X)    X = inception_block_3b(X)    # Top layer    X = AveragePooling2D(pool_size=(3, 3), strides=(1, 1), data_format='channels_first')(X)    X = Flatten()(X)    X = Dense(128, name='dense_layer')(X)    # L2 normalization    X = Lambda(lambda x: K.l2_normalize(x, axis=1))(X)
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Qyouu

TA贡献1786条经验 获得超11个赞

X = Lambda(lambda x: return x[:,0] - x[:,1])(X)
X = Dense(...)(X)


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反对 回复 2022-05-11
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