qqp*_*lot 3 python tf-idf pyspark
我看过类似的帖子,但没有完整的答案,因此在这里发帖。
我在 Spark 中使用 TF-IDF 来获取文档中具有最大 tf-idf 值的单词。我使用下面的一段代码。
from pyspark.ml.feature import HashingTF, IDF, Tokenizer, CountVectorizer, StopWordsRemover
tokenizer = Tokenizer(inputCol="doc_cln", outputCol="tokens")
remover1 = StopWordsRemover(inputCol="tokens", 
outputCol="stopWordsRemovedTokens")
stopwordList =["word1","word2","word3"]
remover2 = StopWordsRemover(inputCol="stopWordsRemovedTokens", 
outputCol="filtered" ,stopWords=stopwordList)
hashingTF = HashingTF(inputCol="filtered", outputCol="rawFeatures", numFeatures=2000)
idf = IDF(inputCol="rawFeatures", outputCol="features", minDocFreq=5)
from pyspark.ml import Pipeline
pipeline = Pipeline(stages=[tokenizer, remover1, remover2, hashingTF, idf])
model = pipeline.fit(df)
results = model.transform(df)
results.cache()
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我得到的结果是
|[a8g4i9g5y, hwcdn] |(2000,[905,1104],[7.34977707433047,7.076179741760428]) 
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在哪里
filtered: array (nullable = true)
features: vector (nullable = true)
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如何从“特征”中提取数组?理想情况下,我想得到对应于最高 tfidf 的单词,如下所示
|a8g4i9g5y|7.34977707433047
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提前致谢!
您的feature列的类型vector来自 package pyspark.ml.linalg。它可以是
根据您拥有的数据(2000,[905,1104],[7.34977707433047,7.076179741760428]),显然它是SparseVector,它可以分解为 3 个主要组成部分:
size: 2000indices: [905,1104]values: [7.34977707433047,7.076179741760428]而你正在寻找的是该values向量的属性。
使用其他“文字”PySpark SQL 类型,例如StringTypeor IntegerType,您可以使用 SQL 函数包 ( docs )访问其属性(和聚合函数)。然而vector不是文字 SQL 类型,访问其属性的唯一方法是通过 UDF,如下所示:
# Important: `vector.values` returns ndarray from numpy.
# PySpark doesn't understand ndarray, therefore you'd want to 
# convert it to normal Python list using `tolist`
def extract_values_from_vector(vector):
    return vector.values.tolist()
# Just a regular UDF
def extract_values_from_vector_udf(col):
    return udf(extract_values_from_vector, ArrayType(DoubleType()))
# And use that UDF to get your values
results.select(extract_values_from_vector_udf('features'), 'features')
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