Question

我正在尝试编写自定义丢失函数：我想将categorical_crossentropy应用于输入向量的各个部分，然后求和。

假设y_true，y_pred是1D向量。

代码：

def custom_loss(y_true, y_pred):

    loss_sum= 0.0
    for i in range(0,y_true.shape[0],dictionary_dims):
        loss_sum+= keras.backend.categorical_crossentropy(y_true[i*dictionary_dims:(i+1)*dictionary_dims], y_pred[i*dictionary_dims:(i+1)*dictionary_dims])

    return loss_sum

但是我收到了一个错误：

    for i in range(0,y_true.shape[0],dictionary_dims):
TypeError: __index__ returned non-int (type NoneType)

那么如何访问输入张量的形状以得到张量的子集？

更新还尝试直接通过tensorflow写丢失：

def custom_loss_tf(y_true, y_pred):

    print('tf.shape(y_true)',tf.shape(y_true)) #
    print('type(tf.shape(y_true))',type(tf.shape(y_true))) #

    sys.exit()

    loss_sum= 0.0
    for i in range(0,y_true.shape[0],dictionary_dims):
        loss_sum+= keras.backend.categorical_crossentropy(y_true[i*dictionary_dims:(i+1)*dictionary_dims], y_pred[i*dictionary_dims:(i+1)*dictionary_dims])

    return loss_sum

输出：

tf.shape(y_true) Tensor("Shape:0", shape=(2,), dtype=int32)
type(tf.shape(y_true)) <class 'tensorflow.python.framework.ops.Tensor'>

不确定shape=(2,)是什么意思，但这不是我所期望的，因为model.summary()显示最后一层是(None, 26)：

_________________________________________________________________
Layer (type)                 Output Shape              Param #
=================================================================
input_1 (InputLayer)         (None, 80, 120, 3)        0
_________________________________________________________________
conv2d_1 (Conv2D)            (None, 80, 120, 32)       896
_________________________________________________________________
max_pooling2d_1 (MaxPooling2 (None, 40, 60, 32)        0
_________________________________________________________________
activation_1 (Activation)    (None, 40, 60, 32)        0
_________________________________________________________________
conv2d_2 (Conv2D)            (None, 40, 60, 32)        9248
_________________________________________________________________
max_pooling2d_2 (MaxPooling2 (None, 20, 30, 32)        0
_________________________________________________________________
activation_2 (Activation)    (None, 20, 30, 32)        0
_________________________________________________________________
conv2d_3 (Conv2D)            (None, 20, 30, 64)        18496
_________________________________________________________________
max_pooling2d_3 (MaxPooling2 (None, 10, 15, 64)        0
_________________________________________________________________
activation_3 (Activation)    (None, 10, 15, 64)        0
_________________________________________________________________
conv2d_4 (Conv2D)            (None, 10, 15, 64)        36928
_________________________________________________________________
max_pooling2d_4 (MaxPooling2 (None, 5, 7, 64)          0
_________________________________________________________________
activation_4 (Activation)    (None, 5, 7, 64)          0
_________________________________________________________________
flatten_1 (Flatten)          (None, 2240)              0
_________________________________________________________________
head (Dense)                 (None, 26)                58266
=================================================================

Answer 1

这里有两件事：

如果您想获得张量形状，则应使用keras.backend中的int_shape函数。
第一个维度设置为批量维度，因此int_shape(y_true)[0]将返回批量大小。您应该使用int_shape(y_true)[1]。

Keras：如何在自定义损失中获得张量尺寸？

1 个答案: