如何使用 KenLM 计算困惑度？

Question

假设我们在此基础上建立了一个模型：

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

来自困惑度公式(https://web.stanford.edu/class/cs124/lec/languagemodeling.pdf)

用反对数公式求和得到内变量再取n次方根，困惑数异常小：

>>> 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

用数据中找不到的句子重试：

# 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

并再次尝试使用完全域外的数据：

>>> 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

虽然预计较长的句子具有较低的困惑度，但奇怪的是差异小于1.0并且在小数点范围内。

以上是使用 KenLM 计算困惑度的正确方法吗？如果没有，有谁知道如何通过 Python API?

来计算 KenLM 的困惑度

Answer 1

困惑度公式为：

但这是原始概率，所以在代码中：

 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)

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

Answer 2

见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 3

只想评论一下alvas的回答

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

实际上应该是：

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

Answer 4

你可以简单地使用

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

如何使用 KenLM 计算困惑度？

How to compute perplexity using KenLM?

python

nlp

language-model

kenlm

perplexity