Question

我有三个张量，张量流A, B and C，A和B都是(m, n, r)形状，C是形状(m, n, 1)的二元张量1}}。

我想根据C的值从A或B中选择元素。显而易见的工具是tf.select，但是没有广播语义，所以我需要首先将C明确地广播到与A和B相同的形状。

这是我第一次尝试如何做到这一点，但它并不像我将张量（tf.shape(A)[2]）混合到形状列表中。

import tensorflow as tf
A = tf.random_normal([20, 100, 10])
B = tf.random_normal([20, 100, 10])
C = tf.random_normal([20, 100, 1])
C = tf.greater_equal(C, tf.zeros_like(C))

C = tf.tile(C, [1,1,tf.shape(A)[2]])
D = tf.select(C, A, B)

这里有什么正确的方法？

Answer 1

编辑：在0.12rc0之后的所有TensorFlow版本中，问题中的代码都可以直接使用。 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()已在TensorFlow 1.0中重命名为tf.stack()。

您的解决方案非常接近工作。您应该替换以下行：

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无法隐式地将张量和Python文字列表转换为张量。tf.pack()获取张量列表，因此它将转换每个元素它的输入（1，1和tf.shape(C)[2]）到张量。由于每个元素都是一个标量，因此结果将是一个向量。）

Answer 2

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)

Answer 3

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)

Answer 4

在最新的Tensorflow版本（2.0）中，您可以按以下方式使用tf.broadcast_to：

import tensorflow as tf

A = tf.random_normal([20, 100, 10])
B = tf.random_normal([20, 100, 10])
C = tf.random_normal([20, 100, 1])
C = tf.greater_equal(C, tf.zeros_like(C))
C = tf.broadcast_to(C, A.shape)

D = tf.where(C,A,B)

如何明确地广播张量以匹配张量流中的另一个形状？

4 个答案: