Python 用tensorflow 2实现梯度惩罚损失

早上好

我正在尝试实现本文所述的1D数据的改进WGAN：

它已在keras contrib github中作为示例实施： 尽管如此，梯度惩罚损失的这种实现不再适用于tf2。K.gradients（）返回[None]

ValueError: in user code:

    /usr/local/lib/python3.6/dist-packages/tensorflow/python/keras/engine/training.py:505 train_function  *
        outputs = self.distribute_strategy.run(
    /usr/local/lib/python3.6/dist-packages/tensorflow/python/distribute/distribute_lib.py:951 run  **
        return self._extended.call_for_each_replica(fn, args=args, kwargs=kwargs)
    /usr/local/lib/python3.6/dist-packages/tensorflow/python/distribute/distribute_lib.py:2290 call_for_each_replica
        return self._call_for_each_replica(fn, args, kwargs)
    /usr/local/lib/python3.6/dist-packages/tensorflow/python/distribute/distribute_lib.py:2649 _call_for_each_replica
        return fn(*args, **kwargs)
    /usr/local/lib/python3.6/dist-packages/tensorflow/python/keras/engine/training.py:467 train_step  **
        y, y_pred, sample_weight, regularization_losses=self.losses)
    /usr/local/lib/python3.6/dist-packages/tensorflow/python/keras/engine/compile_utils.py:204 __call__
        loss_value = loss_obj(y_t, y_p, sample_weight=sw)
    /usr/local/lib/python3.6/dist-packages/tensorflow/python/keras/losses.py:143 __call__
        losses = self.call(y_true, y_pred)
    /usr/local/lib/python3.6/dist-packages/tensorflow/python/keras/losses.py:246 call
        return self.fn(y_true, y_pred, **self._fn_kwargs)
    <ipython-input-7-4f0896d0107b>:104 gradient_penalty_loss
        gradients_sqr = K.square(gradients)
    /usr/local/lib/python3.6/dist-packages/tensorflow/python/keras/backend.py:2189 square
        return math_ops.square(x)
    /usr/local/lib/python3.6/dist-packages/tensorflow/python/ops/gen_math_ops.py:9964 square
        "Square", x=x, name=name)
    /usr/local/lib/python3.6/dist-packages/tensorflow/python/framework/op_def_library.py:488 _apply_op_helper
        (input_name, err))

    ValueError: Tried to convert 'x' to a tensor and failed. Error: None values not supported.

尽管如此，我得到了越来越高的损失函数，梯度惩罚损失被忽略了吗？

---

如果按照更新中的建议执行，tf将忽略丢失功能

对于Tensorflow 2，似乎不可能以旧的方式实现这一点。我最终修改了代码，使其适应这种创建模型的方式。我的建议是什么

使用keras创建gen/disc型号

加入他们，扩展tf.keras.Model类，比如WGAN：

---

我正在看一些非常相似的东西。现在做出任何承诺都为时过早，但我只是通过禁用急切执行（

tensorflow.compat.v1.disable\u eager\u execution（）

）成功地消除了所讨论的错误。如果你确实需要急切的执行，那么我认为你需要gradienttape。我现在还不想写这篇文章作为答案，但我想如果它仍然阻碍了你，那么你可能会喜欢一些可能很快就会奏效的东西，而不是以后会奏效的东西。你好，约西亚！谢谢你的回答，问题似乎正朝着这个方向发展。我设法解决了它，捕捉到了错误，它似乎只在构造计算图时产生了错误。（我会把代码贴在帖子里）。然后它似乎运行良好！我不知道的是，如果在构造图时忽略错误，会使tensorflow忽略损失函数

def gradient_penalty_loss(y_true, y_pred, averaged_samples):
  gradients = K.gradients(y_pred, averaged_samples)[0]
  try:
    gradients_sqr = K.square(gradients)
  except ValueError:
    print("Gradients returned None")
    return 0
  gradients_sqr_sum = K.sum(gradients_sqr, axis=np.arange(1, len(gradients_sqr.shape)))
  gradient_l2_norm = K.sqrt(gradients_sqr_sum)

  gradient_penalty = K.square(1 - gradient_l2_norm)

  return K.mean(gradient_penalty)