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

Question

I have a dataset containing workers with their demographic information like age gender,address etc and their work locations. I created an RDD from the dataset and converted

Answer 1

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

Answer 2

About this issue, due to the fact that I'm working in a project with pyspark where I have to use cosine similarity, I have to say that the code of @MaFF is correct, indeed, I hesitated when I see his code, due to the fact he was using the dot product of the vectors' L2 Norm, and the theroy says: Mathematically, it is the ratio of the dot product of the vectors and the product of the magnitude of the two vectors.

And here is my code adapted with the same results, so I came to the conclusion that SKLearn caculates tfidf in a different way, so if you try to replay this excersice using sklearn, you will get a different result.

d = [{'id': '1', 'office': 'Delhi, Mumbai, Gandhinagar'}, {'id': '2', 'office': 'Delhi, Mandi'}, {'id': '3', 'office': 'Hyderbad, Jaipur'}]
df_fussion = spark.createDataFrame(d)
df_fussion = df_fussion.withColumn('office', F.split('office', ', '))


from pyspark.ml.feature import HashingTF, IDF
hashingTF = HashingTF(inputCol="office", outputCol="tf")
tf = hashingTF.transform(df_fussion)

idf = IDF(inputCol="tf", outputCol="feature").fit(tf)
data = idf.transform(tf)   

@udf
def sim_cos(v1,v2):
    try:
        p = 2
        return float(v1.dot(v2))/float(v1.norm(p)*v2.norm(p))
    except:
        return 0

result = data.alias("i").join(data.alias("j"), F.col("i.ID") < F.col("j.ID"))\
    .select(
        F.col("i.ID").alias("i"),
        F.col("j.ID").alias("j"),
        sim_cos("i.feature", "j.feature").alias("sim_cosine"))\
    .sort("i", "j")
result.show()

I also want to share with you some simply test that I did with simply vectors where the results are corrects:

Kind regards,