python - 用列表切片张量 - TensorFlow

Question

有没有办法在 Tensorflow 中完成这种切片方法(使用 numpy 显示的示例)？

z = np.random.random((3,7,7,12))
x = z[...,[0,5]]

这样

x_hat = np.concatenate([z[...,[0]], z[...,[5]]], 3)
assert np.all(x == x_hat)
x.shape # (3, 7, 7, 2)

在Tensorflow中，这个操作

tfz = tf.constant(z)
i = np.array([0,5] dtype=np.int32)
tfx = tfz[...,i]

抛出错误

ValueError: Shapes must be equal rank, but are 0 and 1
From merging shape 0 with other shapes. for 'strided_slice/stack_1' (op: 'Pack') with input shapes: [], [2].

Answer 1

怎么样:

x = tf.stack([tfz[..., i] for i in [0,5]], axis=-1) 

这对我有用:

z = np.random.random((3,7,7,12))
tfz = tf.constant(z)
x = tf.stack([tfz[..., i] for i in [0,5]], axis=-1)

x_hat = np.concatenate([z[...,[0]], z[...,[5]]], 3)

with tf.Session() as sess:
    x_run = sess.run(x)

assert np.all(x_run == x_hat)