如何使用 KenLM 計算困惑度？

Question

假設我們在此基礎上構建了一個模型：

$ wget https://gist.githubusercontent.com/alvations/1c1b388456dc3760ffb487ce950712ac/raw/86cdf7de279a2b9bceeb3adb481e42691d12fbba/something.txt
$ lmplz -o 5 < something.txt > something.arpa

從困惑公式（ https://web.stanford.edu/class/cs124/lec/languagemodeling.pdf ）

應用逆對數公式之和得到內部變量，然后取第n個根，困惑數異常小：

>>> 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

用數據中未找到的句子再試一次：

# 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

並再次嘗試完全域外數據：

>>> 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

雖然，期望較長的句子具有較低的困惑度，但奇怪的是差異小於1.0並且在小數范圍內。

以上是使用 KenLM 計算困惑度的正確方法嗎？ 如果沒有，有沒有人知道如何通過 Python API 計算 KenLM 的困惑？

Answer 1

見https://github.com/kpu/kenlm/blob/master/python/kenlm.pyx#L182

import kenlm

model=kenlm.Model("something.arpa") 
per=model.perplexity("your text sentance")

print(per)

Answer 2

困惑度公式為：

但這是原始概率，所以在代碼中：

 import numpy as np
 import kenlm
 m = kenlm.Model('something.arpa')
 # Because the score is in log base 10, so:
 product_inv_prob = np.prod([math.pow(10.0, score) for score, _, _ in m.full_scores(s)])
 n = len(list(m.full_scores(s)))
 perplexity = math.pow(product_inv_prob, 1.0/n)

或者直接使用 log (base 10) prob：

 sum_inv_logprob = -1 * sum(score for score, _, _ in m.full_scores(s))
 n = len(list(m.full_scores(s)))
 perplexity = math.pow(10.0, sum_inv_logs / n)

來源： https : //www.mail-archive.com/moses-support@mit.edu/msg15341.html

Answer 3

只想評論阿爾瓦斯的回答

sum_inv_logprob = sum(score for score, _, _ in m.full_scores(s))

實際上應該是：

sum_inv_logprob = -1.0 * sum(score for score, _, _ in m.full_scores(s))

Answer 4

你可以簡單地使用

import numpy as np
import kenlm
m = kenlm.Model('something.arpa')
ppl = m.perplexity('something')