(tensorflow 2.4.1)如何得到一个参差不齐的张量具有确定的最后一维形状？

Question

like this:
'''

a = [1.0, 2.0, 3.0]
b = [[a, a, a, a], [a, a, a], [a, a, a, a, a, a, a], [a, a]]
c = tf.ragged.constant(b, dtype=tf.float32)

''' I got a tensor with shape: [4, None, None], but i expect [4, None, 3],

Answer 1

According to the documentation ( https://www.tensorflow.org/api_docs/python/tf/RaggedTensor#uniform_inner_dimensions_2 ), to get a uniform inner dimension on a ragged tensor, you need to start from a multidimensional tensor of values so TensorFlow knows that dimension is uniform

For your example case, you would want to start by making b a 16x3 matrix and then using one of the "from_row_" methods (eg tf.RaggedTensor.from_row_lengths() ) to partition it into your ragged splits. Eg something like

import numpy as np
import tensorflow as tf

a = np.array([1.0, 2.0, 3.0])
b = np.tile(a, (16,1))
c = tf.RaggedTensor.from_row_lengths(values= b, row_lengths = [4,3,7,2])

should get you the ragged tensor you want with shape [4,None,3] .

Answer 2

I know I'm too late, but for the next person that encounters this issue: You can specify how many ragged dimensions the tensor should have using ragged_rank . You can do as follows

a = [1.0, 2.0, 3.0]
b = [[a, a, a, a], [a, a, a], [a, a, a, a, a, a, a], [a, a]]
c = tf.ragged.constant(b, dtype=tf.float32, ragged_rank=1) #<--here
c.shape

>>> TensorShape([4, None, 3])