tensorflow - Tensorflow 中的 block 对角矩阵

Question

假设我有几个不同形状的张量 A_i [N_i, N_i]。在 tensorflow 中是否可以在对角线上使用这些矩阵创建块对角矩阵？我现在能想到的唯一方法就是通过堆叠和添加 tf.zeros 完全自己构建它。

Answer 1

我同意有一个 C++ op 来做这件事会很好。与此同时，这就是我所做的(正确获取静态形状信息有点繁琐):

import tensorflow as tf

def block_diagonal(matrices, dtype=tf.float32):
  r"""Constructs block-diagonal matrices from a list of batched 2D tensors.

  Args:
    matrices: A list of Tensors with shape [..., N_i, M_i] (i.e. a list of
      matrices with the same batch dimension).
    dtype: Data type to use. The Tensors in `matrices` must match this dtype.
  Returns:
    A matrix with the input matrices stacked along its main diagonal, having
    shape [..., \sum_i N_i, \sum_i M_i].

  """
  matrices = [tf.convert_to_tensor(matrix, dtype=dtype) for matrix in matrices]
  blocked_rows = tf.Dimension(0)
  blocked_cols = tf.Dimension(0)
  batch_shape = tf.TensorShape(None)
  for matrix in matrices:
    full_matrix_shape = matrix.get_shape().with_rank_at_least(2)
    batch_shape = batch_shape.merge_with(full_matrix_shape[:-2])
    blocked_rows += full_matrix_shape[-2]
    blocked_cols += full_matrix_shape[-1]
  ret_columns_list = []
  for matrix in matrices:
    matrix_shape = tf.shape(matrix)
    ret_columns_list.append(matrix_shape[-1])
  ret_columns = tf.add_n(ret_columns_list)
  row_blocks = []
  current_column = 0
  for matrix in matrices:
    matrix_shape = tf.shape(matrix)
    row_before_length = current_column
    current_column += matrix_shape[-1]
    row_after_length = ret_columns - current_column
    row_blocks.append(tf.pad(
        tensor=matrix,
        paddings=tf.concat(
            [tf.zeros([tf.rank(matrix) - 1, 2], dtype=tf.int32),
             [(row_before_length, row_after_length)]],
            axis=0)))
  blocked = tf.concat(row_blocks, -2)
  blocked.set_shape(batch_shape.concatenate((blocked_rows, blocked_cols)))
  return blocked

举个例子:

blocked_tensor = block_diagonal(
    [tf.constant([[1.]]),
     tf.constant([[1., 2.], [3., 4.]])])

with tf.Session():
  print(blocked_tensor.eval())

打印:

[[ 1.  0.  0.]
 [ 0.  1.  2.]
 [ 0.  3.  4.]]