需要幫助來了解此Python Viterbi算法

Question

我正在嘗試將在此Stack Overflow答案中找到的Viterbi算法的Python實現轉換為Ruby。 完整的腳本可以在此問題的底部找到我的評論。

不幸的是，我對Python知之甚少，因此事實證明翻譯比我想要的要困難得多。 盡管如此，我還是取得了一些進展。 現在，唯一使我的大腦完全融化的線是：

prob_k, k = max((probs[j] * word_prob(text[j:i]), j) for j in range(max(0, i - max_word_length), i))

有人可以解釋一下它在做什么嗎？

這是完整的Python腳本：

import re
from itertools import groupby

# text will be a compound word such as 'wickedweather'.
def viterbi_segment(text):
  probs, lasts = [1.0], [0]

  # Iterate over the letters in the compound.
  # eg. [w, ickedweather], [wi, ckedweather], and so on.
  for i in range(1, len(text) + 1):
    # I've no idea what this line is doing and I can't figure out how to split it up?
    prob_k, k = max((probs[j] * word_prob(text[j:i]), j) for j in range(max(0, i - max_word_length), i))
    # Append values to arrays.
    probs.append(prob_k)
    lasts.append(k)

  words = []
  i = len(text)
  while 0 < i:
    words.append(text[lasts[i]:i])
    i = lasts[i]
  words.reverse()
  return words, probs[-1]

# Calc the probability of a word based on occurrences in the dictionary.
def word_prob(word):
  # dictionary.get(key) will return the value for the specified key.
  # In this case, thats the number of occurances of thw word in the
  # dictionary. The second argument is a default value to return if
  # the word is not found.
  return dictionary.get(word, 0) / total

# This ensures we ony deal with full words rather than each
# individual letter. Normalize the words basically.
def words(text):
  return re.findall('[a-z]+', text.lower())

# This gets us a hash where the keys are words and the values are the
# number of ocurrances in the dictionary.
dictionary = dict((w, len(list(ws)))
  # /usr/share/dixt/words is a file of newline delimitated words.
  for w, ws in groupby(sorted(words(open('/usr/share/dict/words').read()))))

# Assign the length of the longest word in the dictionary.
max_word_length = max(map(len, dictionary))

# Assign the total number of words in the dictionary. It's a float
# because we're going to divide by it later on.
total = float(sum(dictionary.values()))

# Run the algo over a file of newline delimited compound words.
compounds = words(open('compounds.txt').read())
for comp in compounds:
  print comp, ": ", viterbi_segment(comp)

Answer 1

您正在查看列表理解 。

擴展版本如下所示：

all_probs = []

for j in range(max(0, i - max_word_length), i):
    all_probs.append((probs[j] * word_prob(text[j:i]), j))

prob_k, k = max(all_probs)

我希望這有助於解釋。 如果不是，請隨時編輯您的問題並指向您不理解的陳述。

Answer 2

這是一個有效的ruby實現，以防其他人對其有所使用。 我翻譯了上面討論的列表理解，我認為這是適當的慣用的不可讀紅寶石級別。

def viterbi(text)
  probabilities = [1.0]
  lasts = [0]

  # Iterate over the letters in the compound.
  # eg. [h ellodarkness],[he llodarkness],...

  (1..(text.length + 1)).each do |i|
    prob_k, k = ([0, i - maximum_word_length].max...i).map { |j| [probabilities[j] * word_probability(text[j...i]), j] }.map { |s| s }.max_by(&:first)
    probabilities << prob_k
    lasts << k
  end

  words = []
  i = text.length
  while i.positive?
    words << text[lasts[i]...i]
    i = lasts[i]
  end
  words.reverse!
  [words, probabilities.last]
end

def word_probability(word)
  word_counts[word].to_f / word_counts_sum.to_f
end

def word_counts_sum
  @word_counts_sum ||= word_counts.values.sum.to_f
end

def maximum_word_length
  @maximum_word_length ||= word_counts.keys.map(&:length).max
end

def word_counts
  return @word_counts if @word_counts 
  @word_counts = {"hello" => 12, "darkness" => 6, "friend" => 79, "my" => 1, "old" => 5}
  @word_counts.default = 0
  @word_counts
end

puts "Best split is %s with probability %.6f" % viterbi("hellodarknessmyoldfriend")

=> Best split is ["hello", "darkness", "my", "old", "friend"] with probability 0.000002

主要的煩惱是python和ruby（打開/關閉間隔）中的范圍定義不同。 該算法非常快。

使用可能性而不是概率可能會比較有利，因為重復的乘法可能導致下溢和/或累積較長單詞的浮點錯誤。