张量流中张量的快速高效交织

我的尝试就是这样。如果我假设给定的形状,那么这将产生所需的输出。

A = tf.constant([[1, 2, 3], [4, 5, 6], [7, 8, 9]])
B = tf.constant([[10, 11, 12], [13, 14, 15], [16, 17, 18]])
C = tf.constant([[19, 20, 21], [22, 23, 24], [25, 26, 27]])

with tf.Session() as sess :

    sess.run(tf.global_variables_initializer())

    ODD = tf.concat( [A[0::2],B[0::2],C[0::2]], axis=0)

    EVENANDODD = tf.concat([ODD[0::2],
                           tf.concat([A[1::2], B[1::2], C[1::2]], axis=0)], axis=0)

    FINAL = tf.concat([EVENANDODD,
                      ODD[1::2]], axis=0)


    print( sess.run(FINAL) )

[19 20 21] [ 4 5 6] [13 14 15] [ 7 8 9]

注意:我无法解决backprop和性能点。

import numpy as np

a = np.array([[1, 2, 3], [4, 5, 6], [7, 8, 9]])
b = np.array([[10, 11, 12], [13, 14, 15], [16, 17, 18]])
c = np.array([[19, 20, 21], [22, 23, 24], [25, 26, 27]])

abc = np.concatenate((a, b, c), axis=0)
abc = abc.reshape((3, 3, 3))
abc = abc.transpose((1, 0, 2))
abc = abc.reshape((9, 3))  #  this gives your objective

tf.gather 是一种选择。

import tensorflow as tf

a = tf.constant([[1, 2, 3], [4, 5, 6], [7, 8, 9]])
b = tf.constant([[10, 11, 12], [13, 14, 15], [16, 17, 18]])
c = tf.constant([[19, 20, 21], [22, 23, 24], [25, 26, 27]])

abc = tf.concat((a, b, c), axis=0)
abc = tf.gather(abc, [0, 5, 2, 1, 7, 8, 3, 5, 6, 4], axis=0)
sess = tf.InteractiveSession()
print(sess.run(abc))

如果没有弄错,在tf.聚集,backprop可以用a,b,c的元素(仅为1或0)进行,但不能用索引(在本例中, [0, 5, 2, 1, 7, 8, 3, 5, 6, 4] ).