在 countvectorizer() 中找不到 get_feature_names

Question

我正在挖掘有关深度学习库的帖子的 Stack Overflow 数据转储。我想识别语料库中的停用词（例如“python”）。我想要获取我的特征名称，以便我可以识别术语频率最高的单词。

我按如下方式创建文档和语料库：

with open("StackOverflow_2018_Data.csv") as csv_file:
    csv_reader = csv.reader(csv_file, delimiter=',')
    line_count = 0
    pytorch_doc = ''
    tensorflow_doc = ''
    cotag_list = []
    keras_doc = ''
    counte = 0
    for row in csv_reader:
        if row[2] == 'tensorflow':
            tensorflow_doc += row[3] + ' '
        if row[2] == 'keras':
            keras_doc += row[3] + ' '
        if row[2] == 'pytorch':
            pytorch_doc += row[3] + ' '

corpus = [pytorch_doc, tensorflow_doc, keras_doc]
vectorizer = CountVectorizer()
x = vectorizer.fit_transform(corpus)
print(x)
x.toarray()
Dict = []
feat = x.get_feature_names()
for i,arr in enumerate(x):
    for x, ele in enumerate(arr):
        if i == 0:
            Dict += ('pytorch', feat[x], ele)
        if i == 1:
            Dict += ('tensorflow', feat[x], ele)
        if i == 2:
            Dict += ('keras', feat[x], ele)

sorted_arr = sorted(Dict, key=lambda tup: tup[2])

但是，我得到：

  File "sklearn_stopwords.py", line 83, in <module>
    main()
  File "sklearn_stopwords.py", line 50, in main
    feat = x.get_feature_names()
  File "/opt/anaconda3/lib/python3.7/site-packages/scipy/sparse/base.py", line 686, in __getattr__
    raise AttributeError(attr + " not found")
AttributeError: get_feature_names not found

Answer 1

get_feature_names是 CountVectorizer 对象中的一个方法。您正在尝试访问 get_feature_names fit_transform 的结果，它是一个 scipy.sparse 矩阵。

你需要使用vectorizer.get_feature_names().

试试这个 MVCE：

from sklearn.feature_extraction.text import CountVectorizer
vectorizer = CountVectorizer()
corpus = ['This is the first document.',
          'This is the second second document.',
          'And the third one.',
          'Is this the first document?']

X = vectorizer.fit_transform(corpus)

features = vectorizer.get_feature_names()

features

输出：

['and', 'document', 'first', 'is', 'one', 'second', 'the', 'third', 'this']