我有一个包含5个文档的文本语料库,每个文档都用/n来分隔,我想为文档中的每个单词提供一个id,并计算其各自的tfidf评分。例如,假设我们有一个名为"corpus.txt“的文本语料库,如下所示:
在计算tfidf时,“跨流文本矢量化scikit python参与稀疏csr”
mylist =list("corpus.text")
vectorizer= CountVectorizer
x_counts = vectorizer_train.fit_transform(mylist)
tfidf_transformer = TfidfTransformer()
x_tfidf = tfidf_transformer.fit_transform(x_counts)输出是
(0,12) 0.1234 #for 1st document
(1,8) 0.3456 #for 2nd document
(1,4) 0.8976
(2,15) 0.6754 #for third document
(2,14) 0.2389
(2,3) 0.7823
(3,11) 0.9897 #for fourth document
(3,13) 0.8213
(3,5) 0.7722
(3,6) 0.2211
(4,7) 0.1100 # for fifth document
(4,10) 0.6690
(4,2) 0.0912
(4,9) 0.2345
(4,1) 0.1234我将此scipy.sparse.csr矩阵转换为列表列表,以删除文档id,并仅保留vocabulary_id及其相应的tfidf评分,使用:
m = x_tfidf.tocoo()
mydata = {k: v for k, v in zip(m.col, m.data)}
key_val_pairs = [str(k) + ":" + str(v) for k, v in mydata.items()] 但问题是,我得到了一个输出,其中vocabulary_id及其相应的tfidf分数是按升序排列的,并且没有任何参考文档。
例如,对于上述给定的语料库,我的当前输出(我已经使用json转储到文本文件中)如下所示:
1:0.1234
2:0.0912
3:0.7823
4:0.8976
5:0.7722
6:0.2211
7:0.1100
8:0.3456
9:0.2345
10:0.6690
11:0.9897
12:0.1234
13:0.8213
14:0.2389
15:0.6754然而,我希望我的文本文件如下所示:
12:0.1234
8:0.3456 4:0.8976
15:0.1234 14:0.2389 3:0.7823
11:0.9897 13:0.8213 5:0.7722 6:0.2211
7:0.1100 10:0.6690 2:0.0912 9:0.2345 1:0.1234知道怎么做吗?
发布于 2016-11-23 05:11:28
我想这就是你需要的。这里,corpus是文档的集合。
from sklearn.feature_extraction.text import TfidfVectorizer
corpus = ["stack over flow stack over flow text vectorization scikit", "stack over flow"]
vectorizer = TfidfVectorizer()
x = vectorizer.fit_transform(corpus) # corpus is a collection of documents
print(vectorizer.vocabulary_) # vocabulary terms and their index
print(x) # tf-idf weights for each terms belong to a particular document这些指纹:
{'vectorization': 5, 'text': 4, 'over': 1, 'flow': 0, 'stack': 3, 'scikit': 2}
(0, 2) 0.33195438857 # first document, word = scikit
(0, 5) 0.33195438857 # word = vectorization
(0, 4) 0.33195438857 # word = text
(0, 0) 0.472376562969 # word = flow
(0, 1) 0.472376562969 # word = over
(0, 3) 0.472376562969 # word = stack
(1, 0) 0.57735026919 # second document
(1, 1) 0.57735026919
(1, 3) 0.57735026919根据这些信息,您可以按照以下方式表示所需的文档:
cx = x.tocoo()
doc_id = -1
for i,j,v in zip(cx.row, cx.col, cx.data):
if doc_id == -1:
print(str(j) + ':' + "{:.4f}".format(v), end=' ')
else:
if doc_id != i:
print()
print(str(j) + ':' + "{:.4f}".format(v), end=' ')
doc_id = i这些指纹:
2:0.3320 5:0.3320 4:0.3320 0:0.4724 1:0.4724 3:0.4724
0:0.5774 1:0.5774 3:0.5774https://stackoverflow.com/questions/40742105
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