How to get word details from TF Vector RDD in Spark ML Lib?
I have created Term Frequency using HashingTF
in Spark. I have got the term frequencies using tf.transform
for each word.
But the results are showing in this format.
[<hashIndexofHashBucketofWord1>,<hashIndexofHashBucketofWord2> ...]
,[termFrequencyofWord1, termFrequencyOfWord2 ....]
eg:
(1048576,[105,3116],[1.0,2.0])
I am able to get the index in hash bucket, using tf.indexOf("word")
.
But, how can I get the word using the index?
Solution 1:
Well, you can't. Since hashing is non-injective there is no inverse function. In other words infinite number of tokens can map to a single bucket so it is impossible to tell which one is actually there.
If you're using a large hash and number of unique tokens is relatively low then you can try to create a lookup table from bucket to possible tokens from your dataset. It is one-to-many mapping but if above conditions are met number of conflicts should be relatively low.
If you need a reversible transformation you can use combine Tokenizer
and StringIndexer
and build a sparse feature vector manually.
See also: What hashing function does Spark use for HashingTF and how do I duplicate it?
Edit:
In Spark 1.5+ (PySpark 1.6+) you can use CountVectorizer
which applies reversible transformation and stores vocabulary.
Python:
from pyspark.ml.feature import CountVectorizer
df = sc.parallelize([
(1, ["foo", "bar"]), (2, ["foo", "foobar", "baz"])
]).toDF(["id", "tokens"])
vectorizer = CountVectorizer(inputCol="tokens", outputCol="features").fit(df)
vectorizer.vocabulary
## ('foo', 'baz', 'bar', 'foobar')
Scala:
import org.apache.spark.ml.feature.{CountVectorizer, CountVectorizerModel}
val df = sc.parallelize(Seq(
(1, Seq("foo", "bar")), (2, Seq("foo", "foobar", "baz"))
)).toDF("id", "tokens")
val model: CountVectorizerModel = new CountVectorizer()
.setInputCol("tokens")
.setOutputCol("features")
.fit(df)
model.vocabulary
// Array[String] = Array(foo, baz, bar, foobar)
where element at the 0th position corresponds to index 0, element at the 1st position to index 1 and so on.