`uniq` for 2D Anano tensor

I have this code:

def uniq(seq):
  """
  Like Unix tool uniq. Removes repeated entries.
  :param seq: numpy.array. (time,) -> label
  :return: seq
  """
  diffs = np.ones_like(seq)
  diffs[1:] = seq[1:] - seq[:-1]
  idx = diffs.nonzero()
  return seq[idx]

      

Now I want to expand this to support 2D arrays and use Theano. It should be fast on the GPU.

I will get an array with multiple sequences as multiple batches in the format (time, batch) and time_mask

that indirectly indicates the length of each sequence.

My current attempt:

def uniq_with_lengths(seq, time_mask):
  # seq is (time,batch) -> label
  # time_mask is (time,batch) -> 0 or 1
  num_batches = seq.shape[1]
  diffs = T.ones_like(seq)
  diffs = T.set_subtensor(diffs[1:], seq[1:] - seq[:-1])
  time_range = T.arange(seq.shape[0]).dimshuffle([0] + ['x'] * (seq.ndim - 1))
  idx = T.switch(T.neq(diffs, 0) * time_mask, time_range, -1)
  seq_lens = T.sum(T.ge(idx, 0), axis=0)  # (batch,) -> len
  max_seq_len = T.max(seq_lens)

  # I don't know any better way without scan.
  def step(batch_idx, out_seq_b1):
    out_seq = seq[T.ge(idx[:, batch_idx], 0).nonzero(), batch_idx][0]
    return T.concatenate((out_seq, T.zeros((max_seq_len - out_seq.shape[0],), dtype=seq.dtype)))

 out_seqs, _ = theano.scan(
    step,
    sequences=[T.arange(num_batches)],
    outputs_info=[T.zeros((max_seq_len,), dtype=seq.dtype)]
  )
  # out_seqs is (batch,max_seq_len)
  return out_seqs.T, seq_lens

      

How to build out_seqs

directly?

I would do something like out_seqs = seq[idx]

, but I'm not really sure how to express it.

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1 answer


Here's a quick answer that only addresses part of your problem:

def compile_theano_uniq(x):
    diffs = x[1:] - x[:-1]
    diffs = tt.concatenate([tt.ones_like([x[0]], dtype=diffs.dtype), diffs])
    y = diffs.nonzero_values()
    return theano.function(inputs=[x], outputs=y)

theano_uniq = compile_theano_uniq(tt.vector(dtype='int32'))

      

The key nonzero_values()

.

Update: I can't imagine how to do this without using theano.scan

. To be clear and using 0 as a complement, I am assuming that given the input

1 1 2 3 3 4 0
1 2 2 2 3 3 4
1 2 3 4 5 0 0

      



you want the output to be

1 2 3 4 0 0 0
1 2 3 4 0 0 0
1 2 3 4 5 0 0

      

or even

1 2 3 4 0
1 2 3 4 0
1 2 3 4 5

      

You can identify the indices of the items you want to keep without using a scan. Then either the new tensor has to be built from scratch, or the values ​​you want to keep as moved to make the sequences continuous. Neither approach is possible without theano.scan

.

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