如何在张量流中置换变换位置？

Question

从docs：

Transposes a. Permutes the dimensions according to perm.

The returned tensor's dimension i will correspond to the input dimension perm[i]. If perm is not given, it is set to (n-1...0), where n is the rank of the input tensor. Hence by default, this operation performs a regular matrix transpose on 2-D input Tensors.

但我还是有点不清楚我该如何切片输入张量。E、 g.文件中也提到：

tf.transpose(x, perm=[0, 2, 1]) ==> [[[1  4]
                                      [2  5]
                                      [3  6]]

                                     [[7 10]
                                      [8 11]
                                      [9 12]]]

为什么perm=[0,2,1]会产生1x3x2张量？

经过反复试验：

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

[出局]：

array([[[ 1,  2,  3,  4],
        [ 5,  6,  7,  8],
        [ 9, 10, 11, 12]],

       [[13, 14, 15, 16],
        [17, 18, 19, 20],
        [21, 22, 23, 24]]])

如果我转置它：

fourthreetwo = tf.transpose(twothreefour) 
with tf.Session() as sess:
    init = tf.initialize_all_variables()
    sess.run(init)
    print (fourthreetwo.eval())

我得到一个4x3x2到一个2x3x4，这听起来很合理。

[出局]：

[[[ 1 13]
  [ 5 17]
  [ 9 21]]

 [[ 2 14]
  [ 6 18]
  [10 22]]

 [[ 3 15]
  [ 7 19]
  [11 23]]

 [[ 4 16]
  [ 8 20]
  [12 24]]]

但是，当我使用perm参数输出时，我不确定我真正得到的是什么：

twofourthree = tf.transpose(twothreefour, perm=[0,2,1]) 
with tf.Session() as sess:
    init = tf.initialize_all_variables()
    sess.run(init)
    print (threetwofour.eval())

[出局]：

[[[ 1  5  9]
  [ 2  6 10]
  [ 3  7 11]
  [ 4  8 12]]

 [[13 17 21]
  [14 18 22]
  [15 19 23]
  [16 20 24]]]

为什么perm=[0,2,1]从2x3x4返回2x4x3矩阵？

用perm=[1,0,2]再试一次：

threetwofour = tf.transpose(twothreefour, perm=[1,0,2]) 
with tf.Session() as sess:
    init = tf.initialize_all_variables()
    sess.run(init)
    print (threetwofour.eval())

[出局]：

[[[ 1  2  3  4]
  [13 14 15 16]]

 [[ 5  6  7  8]
  [17 18 19 20]]

 [[ 9 10 11 12]
  [21 22 23 24]]]

为什么perm=[1,0,2]从2x3x4返回3x2x4？

这是否意味着perm参数采用my np.shape并基于基于数组形状的元素转置张量？

即：

_size = (2, 4, 3, 5)
randarray = np.random.randint(5, size=_size)

shape_idx = {i:_s for i, _s in enumerate(_size)}

randarray_t_func = tf.transpose(randarray, perm=[3,0,2,1]) 
with tf.Session() as sess:
    init = tf.initialize_all_variables()
    sess.run(init)
    tranposed_array = randarray_t_func.eval()
    print (tranposed_array.shape)

print (tuple(shape_idx[_s] for _s in [3,0,2,1]))

[出局]：

(5, 2, 3, 4)
(5, 2, 3, 4)

Answer 1

我认为perm正在排列维度。例如，perm=[0,2,1]是dim_0 -> dim_0, dim_1 -> dim_2, dim_2 -> dim_1的缩写。所以对于二维张量，perm=[1,0]只是矩阵转置。这能回答你的问题吗？

Answer 2

A=[2,3,4] matrix, using perm(1,0,2) will get B=[3,2,4].

说明：

Index=(0,1,2)
A    =[2,3,4]
Perm =(1,0,2)
B    =(3,2,4)  --> Perm 1 from Index 1 (3), Perm 0 from Index 0 (2), Perm 2 from Index 2 (4) --> so get (3,2,4)