word2vec,总和或平均词嵌入?
word2vec, sum or average word embeddings?
我使用 word2vec 将一个小短语(3 到 4 个词)表示为一个独特的向量,方法是添加每个单独的词嵌入或计算词嵌入的平均值。
根据我所做的实验,我总是得到相同的余弦相似度。我怀疑这与 word2vec 生成的词向量在训练后被标准化为单位长度(欧几里德范数)有关?或者我的代码有 BUG,或者我遗漏了什么。
代码如下:
import numpy as np
from nltk import PunktWordTokenizer
from gensim.models import Word2Vec
from numpy.linalg import norm
from scipy.spatial.distance import cosine
def pattern2vector(tokens, word2vec, AVG=False):
pattern_vector = np.zeros(word2vec.layer1_size)
n_words = 0
if len(tokens) > 1:
for t in tokens:
try:
vector = word2vec[t.strip()]
pattern_vector = np.add(pattern_vector,vector)
n_words += 1
except KeyError, e:
continue
if AVG is True:
pattern_vector = np.divide(pattern_vector,n_words)
elif len(tokens) == 1:
try:
pattern_vector = word2vec[tokens[0].strip()]
except KeyError:
pass
return pattern_vector
def main():
print "Loading word2vec model ...\n"
word2vecmodelpath = "/data/word2vec/vectors_200.bin"
word2vec = Word2Vec.load_word2vec_format(word2vecmodelpath, binary=True)
pattern_1 = 'founder and ceo'
pattern_2 = 'co-founder and former chairman'
tokens_1 = PunktWordTokenizer().tokenize(pattern_1)
tokens_2 = PunktWordTokenizer().tokenize(pattern_2)
print "vec1", tokens_1
print "vec2", tokens_2
p1 = pattern2vector(tokens_1, word2vec, False)
p2 = pattern2vector(tokens_2, word2vec, False)
print "\nSUM"
print "dot(vec1,vec2)", np.dot(p1,p2)
print "norm(p1)", norm(p1)
print "norm(p2)", norm(p2)
print "dot((norm)vec1,norm(vec2))", np.dot(norm(p1),norm(p2))
print "cosine(vec1,vec2)", np.divide(np.dot(p1,p2),np.dot(norm(p1),norm(p2)))
print "\n"
print "AVG"
p1 = pattern2vector(tokens_1, word2vec, True)
p2 = pattern2vector(tokens_2, word2vec, True)
print "dot(vec1,vec2)", np.dot(p1,p2)
print "norm(p1)", norm(p1)
print "norm(p2)", norm(p2)
print "dot(norm(vec1),norm(vec2))", np.dot(norm(p1),norm(p2))
print "cosine(vec1,vec2)", np.divide(np.dot(p1,p2),np.dot(norm(p1),norm(p2)))
if __name__ == "__main__":
main()
这是输出:
Loading word2vec model ...
Dimensions 200
vec1 ['founder', 'and', 'ceo']
vec2 ['co-founder', 'and', 'former', 'chairman']
SUM
dot(vec1,vec2) 5.4008677771
norm(p1) 2.19382594282
norm(p2) 2.87226958166
dot((norm)vec1,norm(vec2)) 6.30125952303
cosine(vec1,vec2) 0.857109242583
AVG
dot(vec1,vec2) 0.450072314758
norm(p1) 0.731275314273
norm(p2) 0.718067395416
dot(norm(vec1),norm(vec2)) 0.525104960252
cosine(vec1,vec2) 0.857109242583
我使用此处定义的余弦相似度 Cosine Similarity (Wikipedia)。范数和点积的值确实不同。
谁能解释为什么余弦是一样的?
谢谢,
大卫
余弦测量两个向量之间的角度,不考虑任何一个向量的长度。当你除以短语的长度时,你只是在缩短向量,而不是改变它的 angular 位置。所以你的结果对我来说看起来是正确的。
我使用 word2vec 将一个小短语(3 到 4 个词)表示为一个独特的向量,方法是添加每个单独的词嵌入或计算词嵌入的平均值。
根据我所做的实验,我总是得到相同的余弦相似度。我怀疑这与 word2vec 生成的词向量在训练后被标准化为单位长度(欧几里德范数)有关?或者我的代码有 BUG,或者我遗漏了什么。
代码如下:
import numpy as np
from nltk import PunktWordTokenizer
from gensim.models import Word2Vec
from numpy.linalg import norm
from scipy.spatial.distance import cosine
def pattern2vector(tokens, word2vec, AVG=False):
pattern_vector = np.zeros(word2vec.layer1_size)
n_words = 0
if len(tokens) > 1:
for t in tokens:
try:
vector = word2vec[t.strip()]
pattern_vector = np.add(pattern_vector,vector)
n_words += 1
except KeyError, e:
continue
if AVG is True:
pattern_vector = np.divide(pattern_vector,n_words)
elif len(tokens) == 1:
try:
pattern_vector = word2vec[tokens[0].strip()]
except KeyError:
pass
return pattern_vector
def main():
print "Loading word2vec model ...\n"
word2vecmodelpath = "/data/word2vec/vectors_200.bin"
word2vec = Word2Vec.load_word2vec_format(word2vecmodelpath, binary=True)
pattern_1 = 'founder and ceo'
pattern_2 = 'co-founder and former chairman'
tokens_1 = PunktWordTokenizer().tokenize(pattern_1)
tokens_2 = PunktWordTokenizer().tokenize(pattern_2)
print "vec1", tokens_1
print "vec2", tokens_2
p1 = pattern2vector(tokens_1, word2vec, False)
p2 = pattern2vector(tokens_2, word2vec, False)
print "\nSUM"
print "dot(vec1,vec2)", np.dot(p1,p2)
print "norm(p1)", norm(p1)
print "norm(p2)", norm(p2)
print "dot((norm)vec1,norm(vec2))", np.dot(norm(p1),norm(p2))
print "cosine(vec1,vec2)", np.divide(np.dot(p1,p2),np.dot(norm(p1),norm(p2)))
print "\n"
print "AVG"
p1 = pattern2vector(tokens_1, word2vec, True)
p2 = pattern2vector(tokens_2, word2vec, True)
print "dot(vec1,vec2)", np.dot(p1,p2)
print "norm(p1)", norm(p1)
print "norm(p2)", norm(p2)
print "dot(norm(vec1),norm(vec2))", np.dot(norm(p1),norm(p2))
print "cosine(vec1,vec2)", np.divide(np.dot(p1,p2),np.dot(norm(p1),norm(p2)))
if __name__ == "__main__":
main()
这是输出:
Loading word2vec model ...
Dimensions 200
vec1 ['founder', 'and', 'ceo']
vec2 ['co-founder', 'and', 'former', 'chairman']
SUM
dot(vec1,vec2) 5.4008677771
norm(p1) 2.19382594282
norm(p2) 2.87226958166
dot((norm)vec1,norm(vec2)) 6.30125952303
cosine(vec1,vec2) 0.857109242583
AVG
dot(vec1,vec2) 0.450072314758
norm(p1) 0.731275314273
norm(p2) 0.718067395416
dot(norm(vec1),norm(vec2)) 0.525104960252
cosine(vec1,vec2) 0.857109242583
我使用此处定义的余弦相似度 Cosine Similarity (Wikipedia)。范数和点积的值确实不同。
谁能解释为什么余弦是一样的?
谢谢, 大卫
余弦测量两个向量之间的角度,不考虑任何一个向量的长度。当你除以短语的长度时,你只是在缩短向量,而不是改变它的 angular 位置。所以你的结果对我来说看起来是正确的。