これに基づいてモデルを構築するとしましょう:
$ wget https://gist.githubusercontent.com/alvations/1c1b388456dc3760ffb487ce950712ac/raw/86cdf7de279a2b9bceeb3adb481e42691d12fbba/something.txt
$ lmplz -o 5 < something.txt > something.arpa
困惑の式から ( https://web.stanford.edu/class/cs124/lec/languagemodeling.pdf )
逆対数式の合計を適用して内部変数を取得し、次に n 乗根を取ると、困惑数は異常に小さくなります。
>>> import kenlm
>>> m = kenlm.Model('something.arpa')
# Sentence seen in data.
>>> s = 'The development of a forward-looking and comprehensive European migration policy,'
>>> list(m.full_scores(s))
[(-0.8502398729324341, 2, False), (-3.0185394287109375, 3, False), (-0.3004383146762848, 4, False), (-1.0249041318893433, 5, False), (-0.6545327305793762, 5, False), (-0.29304179549217224, 5, False), (-0.4497605562210083, 5, False), (-0.49850910902023315, 5, False), (-0.3856896460056305, 5, False), (-0.3572353720664978, 5, False), (-1.7523181438446045, 1, False)]
>>> n = len(s.split())
>>> sum_inv_logs = -1 * sum(score for score, _, _ in m.full_scores(s))
>>> math.pow(sum_inv_logs, 1.0/n)
1.2536033936438895
データにない文で再試行:
# Sentence not seen in data.
>>> s = 'The European developement of a forward-looking and comphrensive society is doh.'
>>> sum_inv_logs = -1 * sum(score for score, _, _ in m.full_scores(s))
>>> sum_inv_logs
35.59524390101433
>>> n = len(s.split())
>>> math.pow(sum_inv_logs, 1.0/n)
1.383679905428275
そして、完全にドメインデータが不足している状態で再試行します:
>>> s = """On the evening of 5 May 2017, just before the French Presidential Election on 7 May, it was reported that nine gigabytes of Macron's campaign emails had been anonymously posted to Pastebin, a document-sharing site. In a statement on the same evening, Macron's political movement, En Marche!, said: "The En Marche! Movement has been the victim of a massive and co-ordinated hack this evening which has given rise to the diffusion on social media of various internal information"""
>>> sum_inv_logs = -1 * sum(score for score, _, _ in m.full_scores(s))
>>> sum_inv_logs
282.61719834804535
>>> n = len(list(m.full_scores(s)))
>>> n
79
>>> math.pow(sum_inv_logs, 1.0/n)
1.0740582373271952
文が長いほど困惑度が低いと予想されるが、その差が 1.0 未満で小数の範囲にあるのは奇妙である。
上記は、KenLM で perplexity を計算する正しい方法ですか? そうでない場合は、Python API を使用して KenLM を使用して複雑さを計算する方法を知っている人はいますか?