beam search 간단한 예 실현 및 설명

from math import log
from numpy import array
from numpy import argmax

# beam search
def beam_search_decoder(data, k):
    sequences = [[list(), 1.0]]
    # walk over each step in sequence
    for row in data:
        all_candidates = list()
        # expand each current candidate
        for i in range(len(sequences)):
            seq, score = sequences[i]
            for j in range(len(row)):
                candidate = [seq + [j], score * -log(row[j])]
                all_candidates.append(candidate)
        # order all candidates by score
        ordered = sorted(all_candidates, key=lambda tup :tup[1])
        # select k best
        sequences = ordered[:k]
    return sequences

def greedy_decoder(data):
    # index for largest probability each row
    return [argmax(s) for s in data]

# define a sequence of 10 words over a vocab of 5 words
data = [[0.1, 0.2, 0.3, 0.4, 0.5],
        [0.5, 0.4, 0.3, 0.2, 0.1],
        [0.1, 0.2, 0.3, 0.4, 0.5],
        [0.5, 0.4, 0.3, 0.2, 0.1],
        [0.1, 0.2, 0.3, 0.4, 0.5],
        [0.5, 0.4, 0.3, 0.2, 0.1],
        [0.1, 0.2, 0.3, 0.4, 0.5],
        [0.5, 0.4, 0.3, 0.2, 0.1],
        [0.1, 0.2, 0.3, 0.4, 0.5],
        [0.5, 0.4, 0.3, 0.2, 0.1]]
data = array(data)
# decode sequence
result = beam_search_decoder(data, 3)
# print result
for seq in result:
    print(seq)

매번 순환 sequences 값

[[[4], 0.6931471805599453], [[3], 0.916290731874155], [[2], 1.2039728043259361]]

[[[4, 0], 0.4804530139182014], [[4, 1], 0.6351243373717793], [[3, 0], 0.6351243373717793]]

[[[4, 0, 4], 0.33302465198892944], [[4, 0, 3], 0.4402346437542523], [[4, 1, 4], 0.4402346437542523]]

최종 print 결과

[[4, 0, 4, 0, 4, 0, 4, 0, 4, 0], 0.025600863289563108]
[[4, 0, 4, 0, 4, 0, 4, 0, 4, 1], 0.03384250043584397]
[[4, 0, 4, 0, 4, 0, 4, 0, 3, 0], 0.03384250043584397]