Block Diagonal Matrices in Tensorflow

Question

Let's say I have a few tensors A_i of varying shape [N_i, N_i]. Is it possible in tensorflow to create a block diagonal matrix with these matrices on the diagonal? The only way I can think of now is completely construct it myself by stacking and adding tf.zeros.

Answer 1

I agree that it would be nice to have a C++ op that does this. In the meantime, here's what I do (getting the static shape information right is a bit fiddly):

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

As an example:

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

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

Prints:

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

Answer 2

For anyone visiting this now - tensorflow now has tf.linalg.LinearOperatorBlockDiag. Following the example from Allen above:

import tensorflow as tf

tfl = tf.linalg

blocks = [tf.constant([[1.0]]), tf.constant([[1.0, 2.0], [3.0, 4.0]])]

linop_blocks = [tfl.LinearOperatorFullMatrix(block) for block in blocks]
linop_block_diagonal = tfl.LinearOperatorBlockDiag(linop_blocks)

>>> print(linop_block_diagonal.to_dense())
tf.Tensor(
[[1. 0. 0.]
 [0. 1. 2.]
 [0. 3. 4.]], shape=(3, 3), dtype=float32)