tensorflow中两个同级张量之间的广播

Question

我有两个具有形状的张量x和s：

> x.shape
TensorShape([Dimension(None), Dimension(3), Dimension(5), Dimension(5)])
> s.shape
TensorShape([Dimension(None), Dimension(12), Dimension(5), Dimension(5)])

我想通过维数x广播s和s之间的点积，如下所示：

> x_s.shape
TensorShape([Dimension(None), Dimension(4), Dimension(5), Dimension(5)])

哪里

x_s[i, 0, k, l] = sum([x[i, j, k, l] * s[i, j, k, l] for j in range (3)])
x_s[i, 1, k, l] = sum([x[i, j-3, k, l] * s[i, j, k, l] for j in range (3, 6)])
x_s[i, 2, k, l] = sum([x[i, j-6, k, l] * s[i, j, k, l] for j in range (6, 9)])
x_s[i, 3, k, l] = sum([x[i, j-9, k, l] * s[i, j, k, l] for j in range (9, 12)])

我有这样的实施：

s_t = tf.transpose(s, [0, 2, 3, 1]) # [None, 5, 5, 12]
x_t = tf.transpose(x, [0, 2, 3, 1]) # [None, 5, 5, 3]
x_t = tf.tile(x_t, [1, 1, 1, 4]) # [None, 5, 5, 12]

x_s = x_t * s_t # [None, 5, 5, 12]
x_s = tf.reshape(x_s, [tf.shape(x_s)[0], 5, 5, 4, 3]) # [None, 5, 5, 4, 3]
x_s = tf.reduce_sum(x_s, axis=-1) # [None, 5, 5, 4]
x_s = tf.transpose(x_s, [0, 3, 1, 2]) # [None, 4, 5, 5]

我知道这在内存中是不有效的，因为tile。此外，reshape's、transpose's element-wise和reduce_sum的操作也会对大张量的性能造成损害。还有其他方法可以让它变得更干净吗？

Answer 1

你有证据证明reshape很贵吗？以下是使用整形和维度广播：

x_s = tf.reduce_sum(tf.reshape(s, (-1, 4, 3, 5, 5)) *
                    tf.expand_dims(x, axis=1), axis=2)

Answer 2

只是一些建议，也许不会比你的更快。首先将s与tf.split拆分为四个张量，然后使用tf.tensordot获得最终结果，如下所示

splits = tf.split(s, [3] * 4, axis=1)
splits = map(lambda split: tf.tensordot(split, x, axes=[[1], [1]]), splits)
x_s = tf.stack(splits, axis=1)