python - 转换矩阵中的多个 ngram,概率不加到 1
问题描述
我正在尝试找到一种方法,使用 python 和 numpy 为给定文本使用 unigrams、bigrams 和 trigrams 制作转换矩阵。每行的概率应该等于 1。我先用二元组做了这个,效果很好:
distinct_words = list(word_dict.keys())
dwc = len(distinct_words)
matrix = np.zeros((dwc, dwc), dtype=np.float)
for i in range(len(distinct_words)):
word = distinct_words[i]
first_word_idx = i
total = 0
for bigram, count in ngrams.items():
word_1, word_2 = bigram.split(" ")
if word_1 == word:
total += count
for bigram, count in ngrams.items():
word_1, word_2 = bigram.split(" ")
if word_1 == word:
second_word_idx = index_dict[word_2]
matrix[first_word_idx,second_word_idx] = count / total
但现在我想添加 unigrams 和 trigrams 并加权它们的概率(trigrams * .6、bigrams * .2、unigrams *.2)。我不认为我的 python 非常简洁,这是一个问题,但我也不知道如何使用多个 n-gram(和权重,虽然老实说权重是次要的),所以我仍然可以获得所有概率从任何给定的行加起来为一个。
distinct_words = list(word_dict.keys())
dwc = len(distinct_words)
matrix = np.zeros((dwc, dwc), dtype=np.float)
for i in range(len(distinct_words)):
word = distinct_words[i]
first_word_index = i
bi_total = 0
tri_total=0
tri_prob = 0
bi_prob = 0
uni_prob = word_dict[word] / len(distinct_words)
if i < len(distinct_words)-1:
for trigram, count in trigrams.items():
word_1, word_2, word_3 = trigram.split()
if word_1 + word_2 == word + distinct_words[i+1]:
tri_total += count
for trigram, count in trigrams.items():
word_1, word_2, word_3 = trigram.split()
if word_1 + word_2 == word + distinct_words[i+1]:
second_word_index = index_dict[word_2]
tri_prob = count/bigrams[word_1 + " " + word_2]
for bigram, count in bigrams.items():
word_1, word_2 = bigram.split(" ")
if word_1 == word:
bi_total += count
for bigram, count in bigrams.items():
word_1, word_2 = bigram.split(" ")
if word_1 == word:
second_word_index = index_dict[word_2]
bi_prob = count / bi_total
matrix[first_word_index,second_word_index] = (tri_prob * .4) + (bi_prob * .2) + (word_dict[word]/len(word_dict) *.2)
我正在阅读本讲座以了解如何设置我的概率矩阵,这似乎很有意义,但我不确定我哪里出错了。
如果它有帮助,我的 n_grams 就是来自这个——它只是产生一个 n_gram 的字典作为一个字符串和它的计数。
def get_ngram(words, n):
word_dict = {}
for i, word in enumerate(words):
if i > (n-2):
n_gram = []
for num in range(n):
index = i - num
n_gram.append(words[index])
if len(n_gram) > 1:
formatted_gram = ""
for word in reversed(n_gram):
formatted_gram += word + " "
else:
formatted_gram = n_gram[0]
stripped = formatted_gram.strip() if formatted_gram else n_gram[0]
if stripped in word_dict:
word_dict[stripped] += 1
else:
word_dict[stripped] = 1
return word_dict
解决方案
我已经实现了一个用于计算 unigrams、bigrams 和 trigrams 的示例。您可以zip
轻松地用于连接项目。此外,Counter
用于计数项目并defaultdict
用于项目的概率。defaultdict
当键未在集合中映射时很重要,返回零。否则,您必须添加 if 子句以避免None
.
from collections import Counter, defaultdict
def calculate_grams(items_list):
# count items in list
counts = Counter()
for item in items_list:
counts[item] += 1
# calculate probabilities, defaultdict returns 0 if not found
prob = defaultdict(float)
for item, count in counts.most_common():
prob[item] = count / len(items_list)
return prob
def calculate_bigrams(words):
# tuple first and second items
return calculate_grams(list(zip(words, words[1:])))
def calculate_trigrams(words):
# tuple first, second and third items
return calculate_grams(list(zip(words, words[1:], words[2:])))
dataset = ['a', 'b', 'b', 'c', 'a', 'a', 'a', 'b', 'e', 'e', 'c']
# create dictionary
dictionary = set(dataset)
print("Dictionary", dictionary)
unigrams = calculate_grams(dataset)
print("Unigrams", unigrams)
bigrams = calculate_bigrams(dataset)
print("Bigrams", bigrams)
trigrams = calculate_trigrams(dataset)
print("Trigrams", trigrams)
# Testing
test_words = ['a', 'b']
print("Testing", test_words)
for c in dictionary:
# calculate each probabilities
unigram_prob = unigrams[c]
bigram_prob = bigrams[(test_words[-1], c)]
trigram_prob = bigrams[(test_words[-2], test_words[-1], c)]
# calculate total probability
prob = .2 * unigram_prob + .2 * bigram_prob + .4 * trigram_prob
print(c, prob)
输出:
Unigrams defaultdict(<class 'float'>, {'a': 0.36363636363636365, 'b': 0.2727272727272727, 'c': 0.18181818181818182, 'e': 0.18181818181818182})
Bigrams defaultdict(<class 'float'>, {('a', 'b'): 0.2, ('a', 'a'): 0.2, ('b', 'b'): 0.1, ('b', 'c'): 0.1, ('c', 'a'): 0.1, ('b', 'e'): 0.1, ('e', 'e'): 0.1, ('e', 'c'): 0.1})
Trigrams defaultdict(<class 'float'>, {('a', 'b', 'b'): 0.1111111111111111, ('b', 'b', 'c'): 0.1111111111111111, ('b', 'c', 'a'): 0.1111111111111111, ('c', 'a', 'a'): 0.1111111111111111, ('a', 'a', 'a'): 0.1111111111111111, ('a', 'a', 'b'): 0.1111111111111111, ('a', 'b', 'e'): 0.1111111111111111, ('b', 'e', 'e'): 0.1111111111111111, ('e', 'e', 'c'): 0.1111111111111111})
Testing ['a', 'b']
e 0.05636363636363637
b 0.07454545454545455
c 0.05636363636363637
a 0.07272727272727274
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