python - 如何将 4 个子张量交错分配给更大的张量？

Question

我想通过 4 个具有交错索引的子张量来获得更大的张量。

这是我的问题：

形状为 [batch x 2m x 2n x 1] 的四个输入；

输出形状为 [batch x 4m x 4n x 1]；

index_y_0_1 = [0,1,4,5,8,9...]，index_y_2_3 = [2,3,6,7,10,11...]

index_x_0_1 = [0,1,4,5,8,9...]，index_x_2_3 = [2,3,6,7,10,11...]

输出[index_y_0_1,index_x_0_1] = 输入0

输出[index_y_0_1,index_x_2_3] = 输入1

输出[index_y_2_3 ,index_x_0_1] = 输入2

输出[index_y_2_3 ,index_x_2_3 ] = Input3

这是我对python代码的问题：

import numpy as np

UpperLeft = np.ones((3,2,4,1))
UpperRight = np.ones((3,2,4,1))*2
BottonLeft = np.ones((3,2,4,1))*3
BottonRight = np.ones((3,2,4,1))*4

output = np.zeros((UpperLeft.shape[0], UpperLeft.shape[1]*2, UpperLeft.shape[2]*2, 1))

assert(output.shape[1]%4 == 0)
assert(output.shape[2]%4 == 0)

# UpperLeft Assignment
start_y = 0
start_x = 0
output[:,(start_y + 0)::4, (start_x + 0)::4, :] = UpperLeft[:,0::2, 0::2, :]
output[:,(start_y + 0)::4, (start_x + 1)::4, :] = UpperLeft[:,0::2, 1::2, :]
output[:,(start_y + 1)::4, (start_x + 0)::4, :] = UpperLeft[:,1::2, 0::2, :]
output[:,(start_y + 1)::4, (start_x + 1)::4, :] = UpperLeft[:,1::2, 1::2, :]

# UpperRight Assignment
start_y = 0
start_x = 2
output[:,(start_y + 0)::4, (start_x + 0)::4, :] = UpperRight[:,0::2, 0::2, :]
output[:,(start_y + 0)::4, (start_x + 1)::4, :] = UpperRight[:,0::2, 1::2, :]
output[:,(start_y + 1)::4, (start_x + 0)::4, :] = UpperRight[:,1::2, 0::2, :]
output[:,(start_y + 1)::4, (start_x + 1)::4, :] = UpperRight[:,1::2, 1::2, :]

# BottonLeft Assignment
start_y = 2
start_x = 0
output[:,(start_y + 0)::4, (start_x + 0)::4, :] = BottonLeft[:,0::2, 0::2, :]
output[:,(start_y + 0)::4, (start_x + 1)::4, :] = BottonLeft[:,0::2, 1::2, :]
output[:,(start_y + 1)::4, (start_x + 0)::4, :] = BottonLeft[:,1::2, 0::2, :]
output[:,(start_y + 1)::4, (start_x + 1)::4, :] = BottonLeft[:,1::2, 1::2, :]

# BottonRight Assignment
start_y = 2
start_x = 2
output[:,(start_y + 0)::4, (start_x + 0)::4, :] = BottonRight[:,0::2, 0::2, :]
output[:,(start_y + 0)::4, (start_x + 1)::4, :] = BottonRight[:,0::2, 1::2, :]
output[:,(start_y + 1)::4, (start_x + 0)::4, :] = BottonRight[:,1::2, 0::2, :]
output[:,(start_y + 1)::4, (start_x + 1)::4, :] = BottonRight[:,1::2, 1::2, :]

show_out = output[0,:,:,0]

我怎样才能在张量流上做到这一点？谢谢！

标签： pythontensorflow

Answer 1

你可以有一个像这样在一个轴上交错张量的函数。

import tensorflow as tf

# Interlaves tensors across one axis
def interleave(tensors, size=1, axis=-1):
    # Reshape tensors
    tensors_res = []
    for tensor in tensors:
        s = tf.shape(tensor)
        new_s = tf.concat([s[:axis], [s[axis] // size, size], s[axis:][1:]], axis=0)
        tensors_res.append(tf.reshape(tensor, new_s))
    # Concatenate across new dimension
    if isinstance(axis, (tf.Tensor, tf.Variable)):
        selected_axis = tf.cond(tf.less(axis, 0), lambda: axis - 1, lambda: axis)
        concat_axis = tf.cond(tf.less(axis, 0), lambda: axis, lambda: axis + 1)
    else:
        selected_axis = (axis - 1) if axis < 0 else axis
        concat_axis = axis if axis < 0 else (axis + 1)
    tensors_concat = tf.concat(tensors_res, axis=concat_axis)
    # Reshape concatenation
    concat_s = tf.shape(tensors_concat)
    res_s = tf.concat([concat_s[:selected_axis], [-1], concat_s[concat_axis:][1:]], axis=0)
    return tf.reshape(tensors_concat, res_s)

然后，您可以使用它首先在一个维度上交错，然后在另一个维度上交错。

import tensorflow as tf

with tf.Graph().as_default(), tf.Session() as sess:
    # Input data
    UpperLeft = tf.ones((3, 2, 4, 1))
    UpperRight = tf.ones((3, 2, 4, 1)) * 2
    BottomLeft = tf.ones((3, 2, 4, 1)) * 3
    BottomRight = tf.ones((3, 2, 4, 1)) * 4
    # Interleave across axis 2
    Upper = interleave([UpperLeft, UpperRight], size=2, axis=2)
    Bottom = interleave([BottomLeft, BottomRight], size=2, axis=2)
    # Interleave across axis 1
    Result = interleave([Upper, Bottom], size=2, axis=1)
    # Show result
    print(sess.run(Result)[0, :, :, 0])
    # [[1. 1. 2. 2. 1. 1. 2. 2.]
    #  [1. 1. 2. 2. 1. 1. 2. 2.]
    #  [3. 3. 4. 4. 3. 3. 4. 4.]
    #  [3. 3. 4. 4. 3. 3. 4. 4.]]