Keras自定义损失

Question

我对keras有点陌生。我设法建立了一个具有两个输出的网络：

q_dot_P : <tf.Tensor 'concatenate_1/concat:0' shape=(?, 7) dtype=float32>
q_dot_N : <tf.Tensor 'concatenate_2/concat:0' shape=(?, 10) dtype=float32>

我希望计算上述表达式，q_dot_P是\ delta ^ {q} _P，而q_dot_N是\ delta ^ {q} _P。

这是我的尝试：

nN = 10 
nP = 7    
__a = keras.layers.RepeatVector(nN)( q_dot_P ) #OK, same as 1 . q_dot_P
__b = keras.layers.RepeatVector(nP)( q_dot_N ) #OK, same as 1 . q_dot_N
minu = keras.layers.Subtract()( [keras.layers.Permute( (2,1) )( __b ), __a ] )   
minu = keras.layers.Lambda( lambda x: x + 0.1)( minu )      
minu = keras.layers.Maximum()( [ minu, K.zeros(nN, nP) ] ) #this fails

keras.layers.Maximum()失败。

Traceback (most recent call last):
  File "noveou_train_netvlad.py", line 226, in <module>
    minu = keras.layers.Maximum()( [ minu, K.zeros(nN, nP) ] )
  File "/usr/local/lib/python2.7/dist-packages/keras/engine/base_layer.py", line 457, in __call__
    output = self.call(inputs, **kwargs)
  File "/usr/local/lib/python2.7/dist-packages/keras/layers/merge.py", line 115, in call
    return self._merge_function(reshaped_inputs)
  File "/usr/local/lib/python2.7/dist-packages/keras/layers/merge.py", line 301, in _merge_function
    output = K.maximum(output, inputs[i])
  File "/usr/local/lib/python2.7/dist-packages/keras/backend/tensorflow_backend.py", line 1672, in maximum
    return tf.maximum(x, y)
  File "/usr/local/lib/python2.7/dist-packages/tensorflow/python/ops/gen_math_ops.py", line 4707, in maximum
    "Maximum", x=x, y=y, name=name)
  File "/usr/local/lib/python2.7/dist-packages/tensorflow/python/framework/op_def_library.py", line 546, in _apply_op_helper
    inferred_from[input_arg.type_attr]))
TypeError: Input 'y' of 'Maximum' Op has type string that does not match type float32 of argument 'x'.

实现此目标的最简单方法是什么？

---

遵循@rvinas的建议

我在喀拉拉邦有一个时间分布模型。参见Keras TimeDistributed layer without LSTM

def custom_loss(y_true, y_pred):
    nP = 2
    nN = 2

    # y_pred.shape = shape=(?, 5, 512)
    q = y_pred[:,0:1,:]    # shape=(?, 1, 512)
    P = y_pred[:,1:1+nP,:] # shape=(?, 2, 512)
    N = y_pred[:,1+nP:,:]  # shape=(?, 2, 512)
    q_dot_P = keras.layers.dot( [q,P], axes=-1 )  # shape=(?, 1, 2)
    q_dot_N = keras.layers.dot( [q,N], axes=-1 )  # shape=(?, 1, 2)

    epsilon = 0.1  # Your epsilon here

    zeros = K.zeros((nP, nN), dtype='float32')
    ones_m = K.ones(nP, dtype='float32')
    ones_n = K.ones(nN, dtype='float32')
    code.interact( local=locals() , banner='custom_loss')
    aux = ones_m[None, :, None] * q_dot_N[:, None, :] \
          - q_dot_P[:, :, None] * ones_n[None, None, :] \
          + epsilon * ones_m[:, None] * ones_n[None, :]

    return K.maximum(zeros, aux)

主要内容：

# In __main__
#---------------------------------------------------------------------------
# Setting Up core computation
#---------------------------------------------------------------------------
input_img = Input( shape=(image_nrows, image_ncols, image_nchnl ) )
cnn = make_vgg( input_img )    
out = NetVLADLayer(num_clusters = 16)( cnn )
model = Model( inputs=input_img, outputs=out )

#--------------------------------------------------------------------------
# TimeDistributed
#--------------------------------------------------------------------------
t_input = Input( shape=(1+nP+nN, image_nrows, image_ncols, image_nchnl ) )
t_out = TimeDistributed( model )( t_input )
t_model = Model( inputs=t_input, outputs=t_out )

t_model.compile( loss=custom_loss, optimizer='sgd' )

Answer 1

您可以如下定义损失函数：

import keras.backend as K

nN = 10
nP = 7


def custom_loss(y_true, y_pred):
    q_dot_P = ...  # Extract q_dot_P from y_pred
    q_dot_N = ...  # Extract q_dot_N from y_pred
    epsilon = ...  # Your epsilon here

    zeros = K.zeros((nP, nN), dtype='float32')
    ones_m = K.ones(nP, dtype='float32')
    ones_n = K.ones(nN, dtype='float32')
    aux = ones_m[None, :, None] * q_dot_N[:, None, :] \
          - q_dot_P[:, :, None] * ones_n[None, None, :] \
          + epsilon * ones_m[:, None] * ones_n[None, :]

    return K.maximum(zeros, aux)

并将此函数传递给model.compile()。

注意：未经测试。