Python 如何修改填充向量的seq2seq成本函数?
Tensorflow通过在构建RNN层时使用参数“sequence_length”来支持动态长度序列,其中模型在序列大小=“sequence_length”即返回零向量后不学习序列 然而,如何修改at的代价函数以遇到屏蔽序列,从而只对实际序列而不是整个填充序列计算代价和复杂度Python 如何修改填充向量的seq2seq成本函数?,python,dynamic,tensorflow,deep-learning,lstm,Python,Dynamic,Tensorflow,Deep Learning,Lstm,Tensorflow通过在构建RNN层时使用参数“sequence_length”来支持动态长度序列,其中模型在序列大小=“sequence_length”即返回零向量后不学习序列 然而,如何修改at的代价函数以遇到屏蔽序列,从而只对实际序列而不是整个填充序列计算代价和复杂度 def sequence_loss_by_example(logits, targets, weights, average_across_timesteps=True, softmax_loss_function=No
def sequence_loss_by_example(logits, targets, weights, average_across_timesteps=True, softmax_loss_function=None, name=None):
if len(targets) != len(logits) or len(weights) != len(logits):
raise ValueError("Lengths of logits, weights, and targets must be the same "
"%d, %d, %d." % (len(logits), len(weights), len(targets)))
with ops.op_scope(logits + targets + weights, name,
"sequence_loss_by_example"):
log_perp_list = []
for logit, target, weight in zip(logits, targets, weights):
if softmax_loss_function is None:
# TODO(irving,ebrevdo): This reshape is needed because
# sequence_loss_by_example is called with scalars sometimes, which
# violates our general scalar strictness policy.
target = array_ops.reshape(target, [-1])
crossent = nn_ops.sparse_softmax_cross_entropy_with_logits(
logit, target)
else:
crossent = softmax_loss_function(logit, target)
log_perp_list.append(crossent * weight)
log_perps = math_ops.add_n(log_perp_list)
if average_across_timesteps:
total_size = math_ops.add_n(weights)
total_size += 1e-12 # Just to avoid division by 0 for all-0 weights.
log_perps /= total_size
return log_perps
此函数已经支持通过使用权重计算动态序列长度的成本。只要确保“填充目标”的权重为0,这些步骤的交叉熵将被推到0:
log_perp_list.append(crossent * weight)
总大小也将仅反映非填充步骤:
total_size = math_ops.add_n(weights)
如果使用零填充,则导出权重的一种方法如下所示:
weights = tf.sign(tf.abs(model.targets))
(请注意,您可能需要将其转换为与目标相同的类型)我只是这样做了。然而,当我在句子层面使用ptb模型()进行序列学习时,我得到了不同的代价和困惑。这就是为什么,我在想我们是否需要定制任何东西。到目前为止,模型学习所有训练数据的连续序列。我只是把它改成在句子层次上学习序列,因此所有句子都加0,使它们等于最大长度。好的!知道了。我错过了更新一些参数,如iter和epoch,以便进行最终计算。现在它运行良好。谢谢