from nltk.metrics.distance import edit_distance

str1 = 'bad'
str2 = 'dad'
print(edit_distance(str1, str2))

import nltk

#这里展示2元语法
text1 = 'Chief Executive Officer'

#bigram考虑匹配开头和结束，所有使用pad_right和pad_left
ceo_bigrams = nltk.bigrams(text1.split(),pad_right=True,pad_left=True)

print(list(ceo_bigrams))
[(None, 'Chief'), ('Chief', 'Executive'), 
('Executive', 'Officer'), ('Officer', None)]

import nltk

#这里展示2元语法
def bigram_distance(text1, text2):
  #bigram考虑匹配开头和结束，所以使用pad_right和pad_left
  text1_bigrams = nltk.bigrams(text1.split(),pad_right=True,pad_left=True)
  
  text2_bigrams = nltk.bigrams(text2.split(), pad_right=True, pad_left=True)
  
  #交集的长度
  distance = len(set(text1_bigrams).intersection(set(text2_bigrams)))
  
  return distance


text1 = 'Chief Executive Officer is manager'

text2 = 'Chief Technology Officer is technology manager'

print(bigram_distance(text1, text2)) #相似度为3

from nltk.metrics.distance import jaccard_distance

#这里我们以单个的字符代表文本
set1 = set(['a','b','c','d','a'])
set2 = set(['a','b','e','g','a'])

print(jaccard_distance(set1, set2))

from nltk.metrics.distance import jaccard_distance,masi_distance

#这里我们以单个的字符代表文本
set1 = set(['a','b','c','d','a'])
set2 = set(['a','b','e','g','a'])

print(jaccard_distance(set1, set2))
print(masi_distance(set1, set2))

word_space = [w1,w2,w3,w4]

text1 = 'w1 w2 w1 w4 w1'
text2 = 'w1 w3 w2'

#按照word_space位置，计算每个位置词语出现的次数

text1_vector = [3,1,0,1]
text2_vector = [1,1,1,0]

from nltk.cluster.util import cosine_distance

text1_vector = [3,1,0,1]
text2_vector = [1,1,1,0]

print(cosine_distance(text1_vector,text2_vector))