如何明确地广播张量以匹配张量流中的另一个形状？

3 回答

• 12

  EDIT: 在0.12rc0之后的所有版本的TensorFlow中，问题中的代码直接起作用 . TensorFlow会自动将张量和Python数字叠加到张量参数中 . 使用 tf.pack() 的以下解决方案仅在0.12rc0之前的版本中需要 . 请注意， tf.pack() 在TensorFlow 1.0中重命名为tf.stack() .

  ---

  您的解决方案非常接近工作 . 您应该替换该行：

  C = tf.tile(C, [1,1,tf.shape(C)[2]])
  

  ......以下内容：

  C = tf.tile(C, tf.pack([1, 1, tf.shape(A)[2]]))
  

  （问题的原因是TensorFlow不会隐式地将张量和Python文字列表转换为张量.tf.pack()采用张量列表，因此它将转换其输入中的每个元素（ 1 ， 1 和 tf.shape(C)[2] ）由于每个元素都是一个标量，因此结果将是一个向量 . ）

  回复于 2024-05-02T14:48:47+08:00
• 3

  这是一个肮脏的黑客：

  import tensorflow as tf
  
  def broadcast(tensor, shape):
      return tensor + tf.zeros(shape, dtype=tensor.dtype)
  
  A = tf.random_normal([20, 100, 10])
  B = tf.random_normal([20, 100, 10])
  C = tf.random_normal([20, 100, 1])
  
  C = broadcast(C, A.shape)
  D = tf.select(C, A, B)
  

  回复于 2024-05-02T14:48:47+08:00
• 0

  import tensorflow as tf
  
  def broadcast(tensor, shape):
       """Broadcasts ``x`` to have shape ``shape``.
                                                                     |
       Uses ``tf.Assert`` statements to ensure that the broadcast is
       valid.
  
       First calculates the number of missing dimensions in 
       ``tf.shape(x)`` and left-pads the shape of ``x`` with that many 
       ones. Then identifies the dimensions of ``x`` that require
       tiling and tiles those dimensions appropriately.
  
       Args:
           x (tf.Tensor): The tensor to broadcast.
           shape (Union[tf.TensorShape, tf.Tensor, Sequence[int]]): 
               The shape to broadcast to.
  
       Returns:
           tf.Tensor: ``x``, reshaped and tiled to have shape ``shape``.
  
       """
       with tf.name_scope('broadcast') as scope:
           shape_x = tf.shape(x)
           rank_x = tf.shape(shape0)[0]
           shape_t = tf.convert_to_tensor(shape, preferred_dtype=tf.int32)
           rank_t = tf.shape(shape1)[0]
  
           with tf.control_dependencies([tf.Assert(
               rank_t >= rank_x,
               ['len(shape) must be >= tf.rank(x)', shape_x, shape_t],
               summarize=255
           )]):
               missing_dims = tf.ones(tf.stack([rank_t - rank_x], 0), tf.int32)
  
           shape_x_ = tf.concat([missing_dims, shape_x], 0)
           should_tile = tf.equal(shape_x_, 1)
  
           with tf.control_dependencies([tf.Assert(
               tf.reduce_all(tf.logical_or(tf.equal(shape_x_, shape_t), should_tile),
               ['cannot broadcast shapes', shape_x, shape_t],
               summarize=255
           )]):
               multiples = tf.where(should_tile, shape_t, tf.ones_like(shape_t))
               out = tf.tile(tf.reshape(x, shape_x_), multiples, name=scope)
  
           try:
               out.set_shape(shape)
           except:
               pass
  
           return out
  
  A = tf.random_normal([20, 100, 10])
  B = tf.random_normal([20, 100, 10])
  C = tf.random_normal([20, 100, 1])
  
  C = broadcast(C, A.shape)
  D = tf.select(C, A, B)
  

  回复于 2024-05-02T14:48:47+08:00