Question

有没有办法获得张量（具有批处理尺寸）的乘积行为，类似于批处理尺寸等于1的2D矩阵之间的tf.matmul？

具体来说，我想执行2D矩阵（6,255）和张量（2,255，255,1）（批处理尺寸等于2），其中：

import tensorflow as tf
import numpy as np

im = np.random.rand(255, 255)
A = np.random.rand(6, 255)
B = np.array([im,im]).reshape([-1,255, 255,1])

batch_size = 2
a = tf.placeholder(tf.float64,shape=(6, 255))
b = tf.placeholder(tf.float64,shape=(batch_size,255, 255,1))
out_mat = tf.matmul(a,b) #Didn't work 
with tf.Session() as sess:
    sess.run(out_mat, feed_dict={a: A, b: B})

，结果应为（2，6，255，1）（谢谢@rvinas）形状。

注意：在张量流中，matmul只能处理2D矩阵，而batch_matmul可以只能用（...，n，p）来做（...，m，n），其中A和B中的...相同。

Answer 1

这是使用隐式广播和tf.reduce_sum的一种方法：

class externalClass {
  testData = new dataModel()

  testing(){
    this.testData.data.test1 = "WW"; // i'm able to set value which i shouldn't (data is a getter)
    this.testData.data = {test1:"WW",data:{} // this will trow error as expected
  }
}

以及使用tf.matmul，tf.reshape和tf.transpose的另一种方式：

import tensorflow as tf
import numpy as np

batch_size = 2
dim_1 = 3
dim_2 = 4
dim_3 = 5
dim_4 = 32

im = np.arange(dim_2 * dim_3).reshape(dim_2, dim_3)
A = np.arange(dim_1 * dim_2).reshape(dim_1, dim_2)
B = im
C = np.matmul(A, B)
print('NumPy result (for batch_size=1):\n {}'.format(C))

B = np.repeat(B[None, ..., None], batch_size, axis=0)
B = np.repeat(B, dim_4, axis=3)
print(B.shape)  # B shape=(batch_size, dim_2, dim_3, dim_4)

a = tf.placeholder(tf.float64, shape=(dim_1, dim_2))
b = tf.placeholder(tf.float64, shape=(batch_size, dim_2, dim_3, dim_4))
a_ = a[None, :, :, None, None]  # Shape=(1, dim_1, dim_2, 1, 1)
b_ = b[:, None, :, :, :]  # Shape=(batch_size, 1, dim_2, dim_3, dim_4)
out_mat = tf.reduce_sum(a_ * b_, axis=2)

with tf.Session() as sess:
    c = sess.run(out_mat, feed_dict={a: A, b: B})
    print('TF result (for batch_size={}):\n {}'.format(batch_size, c))
    assert c.shape == (batch_size, dim_1, dim_3, dim_4)

对于您的特定示例，您将设置b_ = tf.transpose(b, [1, 0, 2, 3]) # Shape=(dim_2, batch_size, dim_3, dim_4) b_ = tf.reshape(b_, [dim_2, -1]) # Shape=(dim_2, batch_size * dim_3 * dim_4) matmul = a @ b_ # Shape=(dim_1, batch_size * dim_3 * dim_4) matmul_ = tf.reshape(matmul, [dim_1, batch_size, dim_3, dim_4]) out_mat = tf.transpose(matmul_, [1, 0, 2, 3]) # Shape=(batch_size, dim_1, dim_3, dim_4)，batch_size=2，dim_1=6，dim_2=255和dim_3=255。

有没有办法做张量矩阵乘法？

1 个答案: