How to permutate tranposition in tensorflow?

Question

From the 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.

But it's still a little unclear to me how should I be slicing the input tensor. E.g. from the docs too:

tf.transpose(x, perm=[0, 2, 1]) ==> [[[1  4]                                       [2  5]                                       [3  6]]                                       [[7 10]                                       [8 11]                                       [9 12]]] 

Why is it that perm=[0,2,1] produces a 1x3x2 tensor?

After some trial and error:

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 

[out]:

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]]]) 

And if I transpose it:

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

I get a 4x3x2 to a 2x3x4 and that sounds logical.

[out]:

[[[ 1 13]   [ 5 17]   [ 9 21]]   [[ 2 14]   [ 6 18]   [10 22]]   [[ 3 15]   [ 7 19]   [11 23]]   [[ 4 16]   [ 8 20]   [12 24]]] 

But when I use the perm parameter the output, I'm not sure what I'm really getting:

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

[out]:

[[[ 1  5  9]   [ 2  6 10]   [ 3  7 11]   [ 4  8 12]]   [[13 17 21]   [14 18 22]   [15 19 23]   [16 20 24]]] 

Why does perm=[0,2,1] returns a 2x4x3 matrix from a 2x3x4 ?

Trying it again with perm=[1,0,2]:

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

[out]:

[[[ 1  2  3  4]   [13 14 15 16]]   [[ 5  6  7  8]   [17 18 19 20]]   [[ 9 10 11 12]   [21 22 23 24]]] 

Why does perm=[1,0,2] return a 3x2x4 from a 2x3x4?

Does it mean that the perm parameter is taking my np.shape and transposing the tensor based on the elements based on my array shape?

I.e. :

_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])) 

[out]:

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

Answer 1

I think perm is permuting the dimensions. For example perm=[0,2,1] is short for dim_0 -> dim_0, dim_1 -> dim_2, dim_2 -> dim_1. So for a 2D tensor, perm=[1,0] is just matrix transpose. Does this answer your question?