Question

想象一下，我有一个像这样的张量作为输入：

[[1,1,1,1,1,1,0,0,0,0,0,0,0],
[[1,1,1,1,1,1,0,0,0,0,0,0,0],
[[1,1,1,1,1,1,0,0,0,0,0,0,0],
[[1,1,1,1,1,1,0,0,0,0,0,0,0],
[[1,1,1,1,1,1,0,0,0,0,0,0,0],
[[1,1,1,1,1,1,0,0,0,0,0,0,0],
[[1,1,1,1,1,1,0,0,0,0,0,0,0]
[[1,1,1,1,1,1,0,0,0,0,0,0,0]]

我想输出这个：

[[1,1,1,1,1,1,0,0,0,0,0,0,0],
[[0,1,1,1,1,1,1,0,0,0,0,0,0],
[[0,0,1,1,1,1,1,1,0,0,0,0,0],
[[0,0,0,1,1,1,1,1,1,0,0,0,0],
[[0,0,0,0,1,1,1,1,1,1,0,0,0],
[[0,0,0,0,0,1,1,1,1,1,1,0,0],
[[0,0,0,0,0,0,1,1,1,1,1,1,0]
[[0,0,0,0,0,0,0,1,1,1,1,1,1]]

以任何方式仅使用tf操作（转换为numpy，执行操作，并切换回tf张量是禁止的，因为我正在使用tf优化器进行梯度下降）

Answer 1

这是一个零填充的解决方案 - 用随机填充替换它很简单。

import numpy as np
import tensorflow as tf

x = np.zeros((8, 13), dtype=np.float32)
x[:, :6] = 1
x = tf.constant(x)

s0 = tf.shape(x)[0]
# add an extra zero column on the right
x2 = tf.concat([x, tf.zeros((s0, 1), dtype=x.dtype)], axis=-1)
# flatten the result and remove as many elements as we just added
x2 = tf.reshape(x2, [-1])[:-s0]
# reshape the result to the original shape
x2 = tf.reshape(x2, [s0, -1])

sess = tf.InteractiveSession()
print(x.eval())
print(x2.eval())

Answer 2

感谢user1735003

我发现的另一种方法：

def roll(tensor):
    T=tf.expand_dims(tf.zeros([tensor.shape[1]),0)
        for i in range(tensor.shape[0]):
            row=tf.manip.roll(tensor[i,], shift=i, axis=0)
            T = tf.concat([T,row],axis=1)
    return T

请问如何用tensorflow做这个基本的事情？

2 个答案: