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[英]Does the gensim `Word2Vec()` constructor make a completely independent model?
[英]How to remove a word completely from a Word2Vec model in gensim?
给出一个模型,例如
from gensim.models.word2vec import Word2Vec
documents = ["Human machine interface for lab abc computer applications",
"A survey of user opinion of computer system response time",
"The EPS user interface management system",
"System and human system engineering testing of EPS",
"Relation of user perceived response time to error measurement",
"The generation of random binary unordered trees",
"The intersection graph of paths in trees",
"Graph minors IV Widths of trees and well quasi ordering",
"Graph minors A survey"]
texts = [d.lower().split() for d in documents]
w2v_model = Word2Vec(texts, size=5, window=5, min_count=1, workers=10)
可以从w2v词汇表中删除单词,例如
# Originally, it's there.
>>> print(w2v_model['graph'])
[-0.00401433 0.08862179 0.08601206 0.05281207 -0.00673626]
>>> print(w2v_model.wv.vocab['graph'])
Vocab(count:3, index:5, sample_int:750148289)
# Find most similar words.
>>> print(w2v_model.most_similar('graph'))
[('binary', 0.6781558990478516), ('a', 0.6284914612770081), ('unordered', 0.5971308350563049), ('perceived', 0.5612867474555969), ('iv', 0.5470727682113647), ('error', 0.5346164703369141), ('machine', 0.480206698179245), ('quasi', 0.256790429353714), ('relation', 0.2496253103017807), ('trees', 0.2276223599910736)]
# We can delete it from the dictionary
>>> del w2v_model.wv.vocab['graph']
>>> print(w2v_model['graph'])
KeyError: "word 'graph' not in vocabulary"
但是当我们在删除graph
后对其他单词进行相似性时,我们会看到单词graph
弹出,例如
>>> w2v_model.most_similar('binary')
[('unordered', 0.8710334300994873), ('ordering', 0.8463168144226074), ('perceived', 0.7764195203781128), ('error', 0.7316686511039734), ('graph', 0.6781558990478516), ('generation', 0.5770125389099121), ('computer', 0.40017056465148926), ('a', 0.2762695848941803), ('testing', 0.26335978507995605), ('trees', 0.1948457509279251)]
如何从gensim中的Word2Vec模型中完全删除单词?
要回答@ vumaasha的评论:
你能否提供一些关于你想删除一个单词的细节
让我说我的语料库中所有单词中的单词世界,以学习所有单词之间的密集关系。
但是当我想生成类似的单词时,它应该只来自域特定单词的子集。
可以从.most_similar()
生成足够多的内容然后过滤单词,但是让我们说特定域的空间很小,我可能正在寻找排名第1000的最相似的单词,效率很低。
如果从单词向量中完全删除单词会更好,那么.most_similar()
单词将不会返回特定域之外的单词。
我写了一个函数,它从KeyedVectors中删除不在预定义单词列表中的单词。
def restrict_w2v(w2v, restricted_word_set):
new_vectors = []
new_vocab = {}
new_index2entity = []
new_vectors_norm = []
for i in range(len(w2v.vocab)):
word = w2v.index2entity[i]
vec = w2v.vectors[i]
vocab = w2v.vocab[word]
vec_norm = w2v.vectors_norm[i]
if word in restricted_word_set:
vocab.index = len(new_index2entity)
new_index2entity.append(word)
new_vocab[word] = vocab
new_vectors.append(vec)
new_vectors_norm.append(vec_norm)
w2v.vocab = new_vocab
w2v.vectors = new_vectors
w2v.index2entity = new_index2entity
w2v.index2word = new_index2entity
w2v.vectors_norm = new_vectors_norm
它会重写与基于Word2VecKeyedVectors的单词相关的所有变量。
用法:
w2v = KeyedVectors.load_word2vec_format("GoogleNews-vectors-negative300.bin.gz", binary=True)
w2v.most_similar("beer")
[( '啤酒',0.8409687876701355),
( '啤酒',0.7733745574951172),
('啤酒',0.71753990650177),
('drink',0.668931245803833),
('lagers',0.6570086479187012),
('Yuengling_Lager',0.655455470085144),
('microbrew',0.6534324884414673),
('Brooklyn_Lager',0.6501551866531372),
('suds',0.6497018337249756),
('brewed_beer',0.6490240097045898)]
restricted_word_set = {"beer", "wine", "computer", "python", "bash", "lagers"}
restrict_w2v(w2v, restricted_word_set)
w2v.most_similar("beer")
[('lagers',0.6570085287094116),
('wine',0.6217695474624634),
('bash',0.20583480596542358),
('computer',0.06677375733852386),
('python',0.005948573350906372)]
没有直接的方法来做你想要的。 但是,你并没有完全迷失。 most_similar
方法在most_similar
类中WordEmbeddingsKeyedVectors
(检查链接)。 您可以查看此方法并根据需要进行修改。
下面显示的行执行计算相似单词的实际逻辑,您需要将变量limited
与您感兴趣的单词对应的向量。 然后你就完成了
limited = self.vectors_norm if restrict_vocab is None else self.vectors_norm[:restrict_vocab]
dists = dot(limited, mean)
if not topn:
return dists
best = matutils.argsort(dists, topn=topn + len(all_words), reverse=True)
更新:
limited = self.vectors_norm if restrict_vocab is None else self.vectors_norm[:restrict_vocab]
如果你看到这条线,这意味着如果restrict_vocab
使用是它限制前n个词语的词汇,它才有意义,如果你已经整理通过频率的词汇。 如果你没有传递restrict_vocab, self.vectors_norm
是有限的
most_similar方法调用另一个方法init_sims
。 这将初始化[self.vector_norm][4]
的值,如下所示
self.vectors_norm = (self.vectors / sqrt((self.vectors ** 2).sum(-1))[..., newaxis]).astype(REAL)
所以,你可以拾取你感兴趣的单词,准备他们的标准并用它来代替有限的。 这应该工作
请注意,这不会修剪模型本身。 它修剪了相似性查找基于的KeyedVectors
对象。
假设您只想在模型中保留前5000个单词。
wv = w2v_model.wv
words_to_trim = wv.index2word[5000:]
# In op's case
# words_to_trim = ['graph']
ids_to_trim = [wv.vocab[w].index for w in words_to_trim]
for w in words_to_trim:
del wv.vocab[w]
wv.vectors = np.delete(wv.vectors, ids_to_trim, axis=0)
wv.init_sims(replace=True)
for i in sorted(ids_to_trim, reverse=True):
del(wv.index2word[i])
这样做是因为BaseKeyedVectors类包含以下属性:self.vectors,self.vectors_norm,self.vocab,self.vector_size,self.index2word。
这样做的好处是,如果使用save_word2vec_format()
等方法编写KeyedVectors,则文件要小得多。
尝试过并感觉最简单的方法如下:
我的示例代码如下:
line_no = 0 # line0 = header
numEntities=0
targetLines = []
with open(file_entVecs_txt,'r') as fp:
header = fp.readline() # header
while True:
line = fp.readline()
if line == '': #EOF
break
line_no += 1
isLatinFlag = True
for i_l, char in enumerate(line):
if not isLatin(char): # Care about entity that is Latin-only
isLatinFlag = False
break
if char==' ': # reached separator
ent = line[:i_l]
break
if not isLatinFlag:
continue
# Check for numbers in entity
if re.search('\d',ent):
continue
# Check for entities with subheadings '#' (e.g. 'ENTITY/Stereotactic_surgery#History')
if re.match('^ENTITY/.*#',ent):
continue
targetLines.append(line_no)
numEntities += 1
# Update header with new metadata
header_new = re.sub('^\d+',str(numEntities),header,count=1)
# Generate the file
txtWrite('',file_entVecs_SHORT_txt)
txtAppend(header_new,file_entVecs_SHORT_txt)
line_no = 0
ptr = 0
with open(file_entVecs_txt,'r') as fp:
while ptr < len(targetLines):
target_line_no = targetLines[ptr]
while (line_no != target_line_no):
fp.readline()
line_no+=1
line = fp.readline()
line_no+=1
ptr+=1
txtAppend(line,file_entVecs_SHORT_txt)
仅供参考。 失败的尝试我尝试了@ zsozso的方法(使用np.array
建议的np.array
修改),让它隔夜运行至少12小时,它仍然坚持从限制集中获取新词......)。 这可能是因为我有很多实体...但我的文本文件方法在一个小时内就可以工作了。
失败的代码
# [FAILED] Stuck at Building new vocab...
def restrict_w2v(w2v, restricted_word_set):
new_vectors = []
new_vocab = {}
new_index2entity = []
new_vectors_norm = []
print('Building new vocab..')
for i in range(len(w2v.vocab)):
if (i%int(1e6)==0) and (i!=0):
print(f'working on {i}')
word = w2v.index2entity[i]
vec = np.array(w2v.vectors[i])
vocab = w2v.vocab[word]
vec_norm = w2v.vectors_norm[i]
if word in restricted_word_set:
vocab.index = len(new_index2entity)
new_index2entity.append(word)
new_vocab[word] = vocab
new_vectors.append(vec)
new_vectors_norm.append(vec_norm)
print('Assigning new vocab')
w2v.vocab = new_vocab
print('Assigning new vectors')
w2v.vectors = np.array(new_vectors)
print('Assigning new index2entity, index2word')
w2v.index2entity = new_index2entity
w2v.index2word = new_index2entity
print('Assigning new vectors_norm')
w2v.vectors_norm = np.array(new_vectors_norm)
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