| """ | |
| Author: Awni Hannun | |
| This is an example CTC decoder written in Python. The code is | |
| intended to be a simple example and is not designed to be | |
| especially efficient. | |
| The algorithm is a prefix beam search for a model trained | |
| with the CTC loss function. | |
| For more details checkout either of these references: | |
| https://distill.pub/2017/ctc/#inference | |
| https://arxiv.org/abs/1408.2873 | |
| """ | |
| import numpy as np | |
| import math | |
| import collections | |
| NEG_INF = -float("inf") | |
| def make_new_beam(): | |
| fn = lambda : (NEG_INF, NEG_INF) | |
| return collections.defaultdict(fn) | |
| def logsumexp(*args): | |
| """ | |
| Stable log sum exp. | |
| """ | |
| if all(a == NEG_INF for a in args): | |
| return NEG_INF | |
| a_max = max(args) | |
| lsp = math.log(sum(math.exp(a - a_max) | |
| for a in args)) | |
| return a_max + lsp | |
| def decode(probs, beam_size=100, blank=0): | |
| """ | |
| Performs inference for the given output probabilities. | |
| Arguments: | |
| probs: The output probabilities (e.g. post-softmax) for each | |
| time step. Should be an array of shape (time x output dim). | |
| beam_size (int): Size of the beam to use during inference. | |
| blank (int): Index of the CTC blank label. | |
| Returns the output label sequence and the corresponding negative | |
| log-likelihood estimated by the decoder. | |
| """ | |
| T, S = probs.shape | |
| probs = np.log(probs) | |
| # Elements in the beam are (prefix, (p_blank, p_no_blank)) | |
| # Initialize the beam with the empty sequence, a probability of | |
| # 1 for ending in blank and zero for ending in non-blank | |
| # (in log space). | |
| beam = [(tuple(), (0.0, NEG_INF))] | |
| for t in range(T): # Loop over time | |
| # A default dictionary to store the next step candidates. | |
| next_beam = make_new_beam() | |
| for s in range(S): # Loop over vocab | |
| p = probs[t, s] | |
| # The variables p_b and p_nb are respectively the | |
| # probabilities for the prefix given that it ends in a | |
| # blank and does not end in a blank at this time step. | |
| for prefix, (p_b, p_nb) in beam: # Loop over beam | |
| # If we propose a blank the prefix doesn't change. | |
| # Only the probability of ending in blank gets updated. | |
| if s == blank: | |
| n_p_b, n_p_nb = next_beam[prefix] | |
| n_p_b = logsumexp(n_p_b, p_b + p, p_nb + p) | |
| next_beam[prefix] = (n_p_b, n_p_nb) | |
| continue | |
| # Extend the prefix by the new character s and add it to | |
| # the beam. Only the probability of not ending in blank | |
| # gets updated. | |
| end_t = prefix[-1] if prefix else None | |
| n_prefix = prefix + (s,) | |
| n_p_b, n_p_nb = next_beam[n_prefix] | |
| if s != end_t: | |
| n_p_nb = logsumexp(n_p_nb, p_b + p, p_nb + p) | |
| else: | |
| # We don't include the previous probability of not ending | |
| # in blank (p_nb) if s is repeated at the end. The CTC | |
| # algorithm merges characters not separated by a blank. | |
| n_p_nb = logsumexp(n_p_nb, p_b + p) | |
| # *NB* this would be a good place to include an LM score. | |
| next_beam[n_prefix] = (n_p_b, n_p_nb) | |
| # If s is repeated at the end we also update the unchanged | |
| # prefix. This is the merging case. | |
| if s == end_t: | |
| n_p_b, n_p_nb = next_beam[prefix] | |
| n_p_nb = logsumexp(n_p_nb, p_nb + p) | |
| next_beam[prefix] = (n_p_b, n_p_nb) | |
| # Sort and trim the beam before moving on to the | |
| # next time-step. | |
| beam = sorted(next_beam.items(), | |
| key=lambda x : logsumexp(*x[1]), | |
| reverse=True) | |
| beam = beam[:beam_size] | |
| best = beam[0] | |
| return best[0], -logsumexp(*best[1]) | |
| if __name__ == "__main__": | |
| np.random.seed(3) | |
| time = 50 | |
| output_dim = 20 | |
| probs = np.random.rand(time, output_dim) | |
| probs = probs / np.sum(probs, axis=1, keepdims=True) | |
| labels, score = decode(probs) | |
| print("Score {:.3f}".format(score)) |
You could add either. Like you said for a word level LM you'd only add in the score when the proposed character is a space.
And it's a great observation that every time you add in an LM score the probability goes down so the search will inherently favor prefixes with fewer words hence you need to include a word bonus term. I discuss this in more detail in the distill article https://distill.pub/2017/ctc/#inference.
Thanks for your work,I read the paper(1408.2873),I think you made a mistake on algorithm 1 at page 5.
if c = ℓ end then
p nb (ℓ + ;x 1:t ) ← p(c;xt )pb (ℓ;x 1:t−1 )
p nb (ℓ ;x 1:t ) ← p(c;xt )pb (ℓ;x 1:t−1 )
is the
p nb (ℓ ;x 1:t ) ← p(c;xt )pb (ℓ;x 1:t−1 )
should be
p nb (ℓ ;x 1:t ) ← p(c;xt )pnb (ℓ;x 1:t−1 )
or maybe I misunderstand your algorithm. Thanks
I had a similar issue with the algo. I agree with your suggested correction, @sigpro
I also agree @sigpro. You can also see it reflected in line 102 of this implementation of the algorithm (p_nb rather than p_b):
n_p_nb = logsumexp(n_p_nb, p_nb + p)
agree @sigpro
@YuanEric88, that is handled at the implementation level - beam and next_beam are dictionaries with keys as prefix strings.
Thanks for your work. I recently read your paper (1408.2873) and had two Qs in it.
- Same as what @sigpro mentioned above
Wondering if the counter part to the below section from paper is considered in the logic?
if ℓ + not in Aprev then
....
end if
by counterpart I mean :
if ℓ - not in Aprev then
pnb(ℓ ; x1:t) = p( ℓ _lastElement ; xt) * (pb( ℓ - ; x1:t−1) + pnb( ℓ - ; x1:t−1))
end if
Where,
l_lastElement refer to the last alphabet of l
ℓ - refer to one alphabet less in " ℓ "
for ex:
ℓ : "BAG"
ℓ - : "BA"
ℓ _lastElement : "G"
License for this code in case anyone needs it.
MIT License
Copyright (c) 2022 Awni Hannun
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SOFTWARE.
Would you add character level or word level bi-gram LM in line 96?
For adding word level, I can imagine that I need to check if current (s,) is a space label and then times the prob P(current word | last word), but wouldn't this approach makes the prob of the beam go smaller than the beams that does not add bi-gram?