在Keras中批量大小的batch_dot。-batch_dotwithvariablebatchsizeinKeras

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Your multiplications should select which axes it uses in the batch dot function.

您的乘法应该选择它在批点函数中使用的轴。

• Axis 0 - the batch dimension, it's your ?
• 轴0 -批次尺寸，它是你的?
• Axis 1 - the dimension you say has length 1
• 轴1 -你说的尺寸长度是1
• Axis 2 - the last dimension, of size 500
• 轴2 -最后一个维度，大小为500

You won't change the batch dimension, so you will use batch_dot always with axes=[1,2]

您不会更改批处理维度，因此您将始终使用batch_dot，并使用坐标轴=[1,2]

But for that to work, you must ajust M to be (?, 500, 500).
For that define M not as (500,500), but as (1,500,500) instead, and repeat it in the first axis for the batch size:

但是要使它起作用，你必须让我成为(?、500、500)。定义M不是为(500,500)，而是为(1,500,500)，并在第一个轴中重复批次大小:

import keras.backend as K

#Being M with shape (1,500,500), we repeat it.   
BatchM = K.repeat_elements(x=M,rep=batch_size,axis=0)
#Not sure if repeating is really necessary, leaving M as (1,500,500) gives the same output shape at the end, but I haven't checked actual numbers for correctness, I believe it's totally ok. 

#Now we can use batch dot properly:
firstMul = K.batch_dot(x[0], BatchM, axes=[1,2]) #will result in (?,500,500)

#we also need to transpose x[1]:
x1T = K.permute_dimensions(x[1],(0,2,1))

#and the second multiplication:
result = K.batch_dot(firstMul, x1T, axes=[1,2])

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I prefer using TensorFlow so I tried to figure it out with TensorFlow in past few days.

我更喜欢使用TensorFlow，所以在过去的几天里我尝试用TensorFlow来解决这个问题。

The first one is much similar to Daniel's solution.

第一个和丹尼尔的解决方案很相似。

x = tf.placeholder('float32',shape=(None,1,3))
M = tf.placeholder('float32',shape=(None,3,3))
tf.matmul(x, M)
# return: 

It needs to feed values to M with fit shapes.

它需要向M提供符合形状的值。

sess = tf.Session()
sess.run(tf.matmul(x,M), feed_dict = {x: [[[1,2,3]]], M: [[[1,2,3],[0,1,0],[0,0,1]]]})
# return : array([[[ 1.,  4.,  6.]]], dtype=float32)

Another way is simple with tf.einsum.

另一种方法是使用tf.einsum。

x = tf.placeholder('float32',shape=(None,1,3))
M = tf.placeholder('float32',shape=(3,3))
tf.einsum('ijk,lm->ikl', x, M)
# return: 

Let's feed some values.

让我们喂一些值。

sess.run(tf.einsum('ijk,kl->ijl', x, M), feed_dict = {x: [[[1,2,3]]], M: [[1,2,3],[0,1,0],[0,0,1]]})
# return: array([[[ 1.,  4.,  6.]]], dtype=float32)

Now M is a 2D tensor and no need to feed batch_size to M.

现在M是一个二维张量，不需要向M输入batch_size。

What's more, now it seems such a question can be solved in TensorFlow with tf.einsum. Does it mean it's a duty for Keras to invoke tf.einsum in some situations? At least I find no where Keras calls tf.einsum. And in my opinion, when batch_dot 3D tensor and 2D tensor Keras behaves weirdly. In Daniel's answer, he pads M to (1,500,500) but in K.batch_dot() M will be adjusted to (500,500,1) automatically. I find tf will adjust it with Broadcasting rules and I'm not sure Keras does the same.

更重要的是，现在似乎这样的问题可以通过tf.einsum的TensorFlow来解决。这是否意味着Keras有责任调用tf ?在某些情况下einsum吗?至少我找不到Keras打电话给tf.einsum的地方。在我看来，当batch_dot 3D张量和2D张量Keras怪异地运行时。在丹尼尔的回答中，他把M(1,500500)写在了(1,500500)上，但在kbatch_dot()中，M将被自动调整为(500,500,1)。我发现tf会根据广播规则进行调整，我不确定Keras也会这样做。