Python 3.x Kmeans唯一的单词taggs

我想从K-Means聚类中获得一个唯一标记的列表。我有以下代码：

def cluster_tagging(variable_a_taggear):

document = result[variable_a_taggear]
vectorizer = TfidfVectorizer(ngram_range=(1, 5))

X = vectorizer.fit_transform(document)

true_k = 180
puntos2= true_k

if model_setting == 'MiniBatchKMeans':
    #model = MiniBatchKMeans(n_clusters=true_k, init='k-means++', max_iter=1000, n_init=1)
    pass
elif model_setting == 'KMeans':
    model = KMeans(n_clusters=true_k, init='k-means++', max_iter=10000000, n_init=1)


model.fit(X)

order_centroids = model.cluster_centers_.argsort()[:, ::-1]
terms = vectorizer.get_feature_names()
#print(terms[:8])

cluster_ = []
key_ = []
ID = []

cluster_col = 'Cluster_%s'%(variable_a_taggear)
keywords_col = 'Keywords_%s'%(variable_a_taggear)
word_cloud = pd.DataFrame(columns=[cluster_col, keywords_col])

for i in range(puntos2):
    print('Cluster %s:' % (i))
    cluster_.append(i)
    key_1 = []
    key_1 = list(set(key_1))
    key_.append(key_1)

    for ind in order_centroids[i, :8]:
        print('%s' % terms[ind])
        terms_ = terms[ind]
        key_1.append(terms_)

print('first key_', key_)
info = {cluster_col:cluster_,keywords_col:key_}

word_cloud = pd.DataFrame(info)
word_cloud.head()


#print('Prediction')

predicted = model.predict(vectorizer.transform(document))
lst2 = result['Ticket ID']
predictions = pd.DataFrame(list(zip(predicted, lst2)), columns =[cluster_col, 'Ticket ID'])

#predictions = pd.DataFrame(predicted,result['Ticket ID'])
predictions.columns = [cluster_col, 'Ticket ID']
#print(predictions)

resultado = pd.merge(predictions, word_cloud, left_on=cluster_col, right_on=cluster_col, how='inner')
print(resultado.head())
return resultado

正如你通过n-gram观察到的那样，我获得了作为不同n-gram的一部分的重复单词。例如，对于一个集群，我有以下标记：

[[fecha-iniciar'，iniciar'，modificar-fecha-iniciar-cc'，proceder-modificar-fecha-iniciar'，proceder-modificar-fecha-inciar-cc'，fecha-iniciar-cc'，fecha']

如何获得每个集群的唯一单词列表

谢谢

问题：如何获得每个集群的唯一单词列表

您可以使用分隔句子中的单词和

numpy.unique

在数组中获取唯一值

import numpy as np
from nltk.tokenize import word_tokenize

cluster_tags  = ['fecha iniciar', 'iniciar', ..., 'fecha']
one_string = ' '.join(cluster_tags)
np.unique(word_tokenize(one_string))

如果您确定所有单词总是用一个干净的空格分隔，您可以简单地拆分它们

np.unique(' '.join(cluster_tags).split())

奖金提示： 如果你愿意，你可以计算每个单词的频率

# See answer by Max Malysh: https://stackoverflow.com/questions/952914/how-to-make-a-flat-list-out-of-list-of-lists
from collections import Counter
from pandas.core.common import flatten

tokenized = [word_tokenize(text) for text in cluster_tags]
Counter(flatten(tokenized))

问题：如何获得每个集群的唯一单词列表

您可以使用分隔句子中的单词和

numpy.unique

在数组中获取唯一值

import numpy as np
from nltk.tokenize import word_tokenize

cluster_tags  = ['fecha iniciar', 'iniciar', ..., 'fecha']
one_string = ' '.join(cluster_tags)
np.unique(word_tokenize(one_string))

如果您确定所有单词总是用一个干净的空格分隔，您可以简单地拆分它们

np.unique(' '.join(cluster_tags).split())

奖金提示： 如果你愿意，你可以计算每个单词的频率

# See answer by Max Malysh: https://stackoverflow.com/questions/952914/how-to-make-a-flat-list-out-of-list-of-lists
from collections import Counter
from pandas.core.common import flatten

tokenized = [word_tokenize(text) for text in cluster_tags]
Counter(flatten(tokenized))