Tensorflow 将 [X, 10, 10, 1] 矩阵除以 [X, 1]

Question

我正在尝试将 [X, 10, 10, 1] 矩阵除以每个 [10, 10, 1] 矩阵的总和 - 因此将 [X, 10, 10, 1] 除以 [X, 1] 但是使用tf.broadcast_to 给出了一个“不兼容的形状”错误，如下图所示：

t1 = tf.ones([128, 10, 10, 1]) 
t1_sum = tf.reduce_sum(t1, [1, 2])  # Yields a [128, 1]
t1_sum_reshaped = tf.broadcast_to(t1_sum, (t1.shape))
divided = tf.math.divide(t1, t1_sum_reshaped)

使用 tf.broadcast_to 时出现 X=128 错误：形状不兼容：[128,1] 与 [128,10,10,1] [Op:BroadcastTo]

我让部门工作的解决方法如下，但它似乎非常慢（而且很笨）

t1 = tf.ones([128, 10, 10, 1])
t1_sum = tf.reduce_sum(t1, [1, 2])                                                                            # Yields a [128, 1]
t1_sum = tf.stack([t1_sum, t1_sum, t1_sum, t1_sum, t1_sum, t1_sum, t1_sum, t1_sum, t1_sum, t1_sum], axis=-2)  # Yields a [128, 10, 1]
t1_sum = tf.stack([t1_sum, t1_sum, t1_sum, t1_sum, t1_sum, t1_sum, t1_sum, t1_sum, t1_sum, t1_sum], axis=-2)  # Yields a [128, 10, 10, 1]
divided = tf.math.divide(t1, t1_sum)

有一个更好的方法吗？

Answer 1

似乎您可以将 keepdims=True 参数添加到您的代码中，它将按预期工作：

t1 = tf.ones([128, 10, 10, 1])
t1_sum = tf.reduce_sum(t1, [1, 2], keepdims=True) # Yields a [128, 1, 1, 1]
t1_sum_reshaped = tf.broadcast_to(t1_sum, t1.shape)
divided = tf.math.divide(t1, t1_sum_reshaped)

完成此操作后，您可以删除行t1_sum_reshaped = tf.broadcast_to(t1_sum, t1.shape因为它已经具有所需的形状。

t1 = tf.ones([128, 10, 10, 1])
t1_sum = tf.reduce_sum(t1, [1, 2], keepdims=True) # Yields a [128, 1, 1, 1]
divided = tf.math.divide(t1, t1_sum)