Calculating the cosine similarity between all the rows of a dataframe in pyspark

自古美人都是妖i 提交于 2019-11-28 19:01:42

You can use the mllib package to compute the L2 norm of the TF-IDF of every row. Then multiply the table with itself to get the cosine similarity as the dot product of two by two L2norms:

1. RDD

rdd = sc.parallelize([[1, "Delhi, Mumbai, Gandhinagar"],[2, " Delhi, Mandi"], [3, "Hyderbad, Jaipur"]])
  • Compute TF-IDF:

    documents = rdd.map(lambda l: l[1].replace(" ", "").split(","))
    
    from pyspark.mllib.feature import HashingTF, IDF
    hashingTF = HashingTF()
    tf = hashingTF.transform(documents)
    

You can specify the number of features in HashingTF to make the feature matrix smaller (fewer columns).

    tf.cache()
    idf = IDF().fit(tf)
    tfidf = idf.transform(tf)
  • Compute L2norm:

    from pyspark.mllib.feature import Normalizer
    labels = rdd.map(lambda l: l[0])
    features = tfidf
    
    normalizer = Normalizer()
    data = labels.zip(normalizer.transform(features))
    
  • Compute cosine similarity by multiplying the matrix with itself:

    from pyspark.mllib.linalg.distributed import IndexedRowMatrix
    mat = IndexedRowMatrix(data).toBlockMatrix()
    dot = mat.multiply(mat.transpose())
    dot.toLocalMatrix().toArray()
    
        array([[ 0.        ,  0.        ,  0.        ,  0.        ],
               [ 0.        ,  1.        ,  0.10794634,  0.        ],
               [ 0.        ,  0.10794634,  1.        ,  0.        ],
               [ 0.        ,  0.        ,  0.        ,  1.        ]])
    

    OR: Using a Cartesian product and the function dot on numpy arrays:

    data.cartesian(data)\
        .map(lambda l: ((l[0][0], l[1][0]), l[0][1].dot(l[1][1])))\
        .sortByKey()\
        .collect()
    
        [((1, 1), 1.0),
         ((1, 2), 0.10794633570596117),
         ((1, 3), 0.0),
         ((2, 1), 0.10794633570596117),
         ((2, 2), 1.0),
         ((2, 3), 0.0),
         ((3, 1), 0.0),
         ((3, 2), 0.0),
         ((3, 3), 1.0)]
    

2. DataFrame

Since you seem to be using dataframes, you can use the spark mlpackage instead:

import pyspark.sql.functions as psf
df = rdd.toDF(["ID", "Office_Loc"])\
    .withColumn("Office_Loc", psf.split(psf.regexp_replace("Office_Loc", " ", ""), ','))
  • Compute TF-IDF:

    from pyspark.ml.feature import HashingTF, IDF
    hashingTF = HashingTF(inputCol="Office_Loc", outputCol="tf")
    tf = hashingTF.transform(df)
    
    idf = IDF(inputCol="tf", outputCol="feature").fit(tf)
    tfidf = idf.transform(tf)
    
  • Compute L2 norm:

    from pyspark.ml.feature import Normalizer
    normalizer = Normalizer(inputCol="feature", outputCol="norm")
    data = normalizer.transform(tfidf)
    
  • Compute matrix product:

    from pyspark.mllib.linalg.distributed import IndexedRow, IndexedRowMatrix
    mat = IndexedRowMatrix(
        data.select("ID", "norm")\
            .rdd.map(lambda row: IndexedRow(row.ID, row.norm.toArray()))).toBlockMatrix()
    dot = mat.multiply(mat.transpose())
    dot.toLocalMatrix().toArray()
    

    OR: using a join and a UDF for function dot:

    dot_udf = psf.udf(lambda x,y: float(x.dot(y)), DoubleType())
    data.alias("i").join(data.alias("j"), psf.col("i.ID") < psf.col("j.ID"))\
        .select(
            psf.col("i.ID").alias("i"), 
            psf.col("j.ID").alias("j"), 
            dot_udf("i.norm", "j.norm").alias("dot"))\
        .sort("i", "j")\
        .show()
    
        +---+---+-------------------+
        |  i|  j|                dot|
        +---+---+-------------------+
        |  1|  2|0.10794633570596117|
        |  1|  3|                0.0|
        |  2|  3|                0.0|
        +---+---+-------------------+
    

This tutorial lists different methods to multiply large scale matrices: https://labs.yodas.com/large-scale-matrix-multiplication-with-pyspark-or-how-to-match-two-large-datasets-of-company-1be4b1b2871e

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