I am trying to implement k means clustering using apache spark ml version in 2.0.2. After finding the cluster center , facing the issue in how do identity the data belongs to which cluster point. Please help me.. Thanks in advance. Please find my code :
val tokenFrameprocess = TokenizerProcessor.process("value", "tokenized")
val stopwordRemover = StopWordsProcessor.process("tokenized", "stopwords")
val stemmingProcess = StemmingProcessor.process("value", "stemmed")
val HashingProcess = HashingTFProcessor.process("stopwords" ,"features")
val pipeline = new Pipeline().setStages(Array(tokenFrameprocess,stopwordRemover,stemmingProcess,HashingProcess))
val finalFrameProcess = pipeline.fit(df)
val finalFramedata = finalFrameProcess.transform(df);
val kmeans = new KMeans().setK(4).setSeed(1L)
val model = kmeans.fit(finalFramedata)
val WSSSE = model.computeCost(finalFramedata)
// Shows the result.
println("Cluster Centers: ")
model.clusterCenters.foreach(println)
You have to set the feature column which will be used to predict on while instantiating the Kmeans like following example:
val kmeans = new KMeans().setK(4).setSeed(1L).setFeaturesCol("features").setPredictionCol("prediction")
Once you call the .fit() on the kmeans, It returns a Transformer. In case of your example, variable "model" is a transformer. You can call the .transform() To get the desired prediction on the given data. For example below lines will give you the dataframe with prediction column.
val model = kmeans.fit(finalFramedata)
val transformedDF = model.transform(finalFramedata)
transformedDF.show(false)
Prediction column denotes clusters points. If you give k value as 3 and the prediction column will have values like 0,1,2.
Related
I'd like to use FP-Growth from Spark MLlib in Spark 2.1.
My data has only two columns item_group and item.
I have tried the following but it does not work:
sc = SparkSession.builder.appName("Assoziationsanalyse").getOrCreate()
hiveCtx = SQLContext(sc)
input = hiveCtx.sql("""select * from bosch.input_view""").
groupBy("item_group").
agg(collect_list("item")).
alias("items").
rdd.
map(lambda x : x.items)
model = FPGrowth.train(input, minSupport=0.2, numPartitions=10)
I have solved the problem now using an other approch which I have found in a discussion here.
data=hiveCtx.sql("""select * from bosch.input_view""")
datardd=data.rdd.map(lambda x (x[0],x[1])).groupByKey().mapValues(list).values()
model = FPGrowth.train(datardd, minSupport=0.1, numPartitions=10)
I have this code:
val userIndexer: StringIndexer = new StringIndexer()
.setInputCol("userKey")
.setOutputCol("user")
val alsRatings = userIndexerModel.transform(ratings)
val matrixFactorizationModel = ALS.trainImplicit(alsRatings.rdd, rank = 10, iterations = 10)
val rec = matrixFactorizationModel.recommendProductsForUsers(20)
This gives me back recommendations with user ids. I want to have my user key strings back. What is the more efficient way to do it? Thanks.
PD: I certainly cannot understand why ALS library developers don't accept string labels. It's extremely painful and expensive to deal with conversions (string to int and then int to string) from the outside. Hope there is an issue or something in their backlog.
I generally run the StringIndexer collect the labels in the Driver. And
parallelize the labels with an index. And instead of calling Transform using the StringIndexer. I join the DataFrames to get the same result as a StringIndexer.
val swidConverter = new StringIndexer()
.setInputCol("id")
.setOutputCol("idIndex").fit(df)
val idDf = spark.sparkContext.parallelize(
swidConverter.labels.zipWithIndex
).toDF("id", "idIndex").repartition(PARTITION_SIZE) // set the partition size depending on your data size.
// Joining the idDf(DataFrame) with the actual Data.
val indexedDF = df.join(idDf,idDf.col("id")===df.col("id")).select("idIndex","product_id","rating")
val als = new ALS()
.setMaxIter(5)
.setRegParam(0.01)
.setUserCol("idIndex")
.setItemCol("product_id")
.setRatingCol("rating")
val model = als.fit(indexedDF)
val resultRaw = model.recommendForAllUsers(4)
// Joining the idDf(DataFrame) with the Result to get the original ID from the indexed Id.
val resultDf = resultRaw.join(idDf,resultRaw.col("idIndex")===idDf.col("idIndex")).select("id","recommendations")
Firstly, I use spark 1.6.0. I want to use L1 penalty in pyspark.ml.regression.LinearRegressionModel for features selection.
But I can not get the detailed coefficients when calling the function:
lr = LogisticRegression(elasticNetParam=1.0, regParam=0.01,maxIter=100,fitIntercept=False,standardization=False)
model = lr.fit(df_one_hot_train)
print model.coefficients.toArray().astype(float).tolist()
I only get sparse list like:
[0,0,0,0,0,..,-0.0871650387514,..,]
While when I use sklearn.linear_model.LogisticRegression model, I can get the detailed list without zero value in coef_ list like:
[0.03098372361467529,-0.13709075166114365,-0.15069548597557908,-0.017968044053830862]
With the better performance in spark, I could finished my work faster. I just want to use L1 penalty for feature selection.
I think I should use more detailed values of coefficients for my feature selection work just as sklearn does, how can I solve my problem?
Below is a working code snip in Spark 2.1.
The key to extract values is :
stages(4).asInstanceOf[LinearRegressionModel]
Spark 1.6 may have something similar.
val holIndIndexer = new StringIndexer().setInputCol("holInd").setOutputCol("holIndIndexer")
val holIndEncoder = new OneHotEncoder().setInputCol("holIndIndexer").setOutputCol("holIndVec")
val time_intervaLEncoder = new OneHotEncoder().setInputCol("time_interval").setOutputCol("time_intervaLVec")
val assemblerL1 = (new VectorAssembler()
.setInputCols(Array("time_intervaLVec", "holIndVec", "length")).setOutputCol("features") )
val lrL1 = new LinearRegression().setFeaturesCol("features").setLabelCol("travel_time")
val pipelineL1 = new Pipeline().setStages(Array(holIndIndexer,holIndEncoder,time_intervaLEncoder,assemblerL1, lrL1))
val modelL1 = pipelineL1.fit(dfTimeMlFull)
val l1Coeff =modelL1.stages(4).asInstanceOf[LinearRegressionModel].coefficients
println(l1Coeff)
Based on the Spark documentation for 1.4 (https://spark.apache.org/docs/1.4.0/mllib-feature-extraction.html) I'm writing a TF-IDF example for converting text documents to vectors of values. The example given shows how this can be done but the input is a RDD of tokens with no keys. This means that my output RDD no longer contains an index or key to refer back to the original document. The example is this:
documents = sc.textFile("...").map(lambda line: line.split(" "))
hashingTF = HashingTF()
tf = hashingTF.transform(documents)
I would like to do something like this:
documents = sc.textFile("...").map(lambda line: (UNIQUE_LINE_KEY, line.split(" ")))
hashingTF = HashingTF()
tf = hashingTF.transform(documents)
and have the resulting tf variable contain the UNIQUE_LINE_KEY value somewhere. Am I just missing something obvious? From the examples it appears there is no good way to link the document RDD with the tf RDD.
I also encountered the same issue. In the example from the docs they encourage you to apply the transformations directly on the RDD.
However, you can apply the transformations on the vectors themselves and this way you can keep the keys whichever way you choose.
val input = sc.textFile("...")
val documents = input.map(doc => doc -> doc.split(" ").toSeq)
val hashingTF = new HashingTF()
val tf = documents.mapValues(hashingTF.transform(_))
tf.cache()
val idf = new IDF().fit(tf.values)
val tfidf = tf.mapValues(idf.transform(_))
Note that this code will yield RDD[(String, Vector)] instead of RDD[Vector]
If you use a version of Spark from after commit 85b96372cf0fd055f89fc639f45c1f2cb02a378f (this includes the 1.4), and use the ml API HashingTF (requires DataFrame input instead of plain RDDs), the original columns in its output. Hope that helps!
Let's say I have a DataFrame (that I read in from a csv on HDFS) and I want to train some algorithms on it via MLlib. How do I convert the rows into LabeledPoints or otherwise utilize MLlib on this dataset?
Assuming you're using Scala:
Let's say your obtain the DataFrame as follows:
val results : DataFrame = sqlContext.sql(...)
Step 1: call results.printSchema() -- this will show you not only the columns in the DataFrame and (this is important) their order, but also what Spark SQL thinks are their types. Once you see this output things get a lot less mysterious.
Step 2: Get an RDD[Row] out of the DataFrame:
val rows: RDD[Row] = results.rdd
Step 3: Now it's just a matter of pulling whatever fields interest you out of the individual rows. For this you need to know the 0-based position of each field and it's type, and luckily you obtained all that in Step 1 above. For example,
let's say you did a SELECT x, y, z, w FROM ... and printing the schema yielded
root
|-- x double (nullable = ...)
|-- y string (nullable = ...)
|-- z integer (nullable = ...)
|-- w binary (nullable = ...)
And let's say all you wanted to use x and z. You can pull them out into an RDD[(Double, Integer)] as follows:
rows.map(row => {
// x has position 0 and type double
// z has position 2 and type integer
(row.getDouble(0), row.getInt(2))
})
From here you just use Core Spark to create the relevant MLlib objects. Things could get a little more complicated if your SQL returns columns of array type, in which case you'll have to call getList(...) for that column.
Assuming you're using JAVA (Spark version 1.6.2):
Here is a simple example of JAVA code using DataFrame for machine learning.
It loads a JSON with the following structure,
[{"label":1,"att2":5.037089672359123,"att1":2.4100883023159456}, ... ]
splits the data into training and testing,
train the model using the train data,
apply the model to the test data and
stores the results.
Moreover according to the official documentation the "DataFrame-based API is primary API" for MLlib since the current version 2.0.0. So you can find several examples using DataFrame.
The code:
SparkConf conf = new SparkConf().setAppName("MyApp").setMaster("local[2]");
SparkContext sc = new SparkContext(conf);
String path = "F:\\SparkApp\\test.json";
String outputPath = "F:\\SparkApp\\justTest";
System.setProperty("hadoop.home.dir", "C:\\winutils\\");
SQLContext sqlContext = new org.apache.spark.sql.SQLContext(sc);
DataFrame df = sqlContext.read().json(path);
df.registerTempTable("tmp");
DataFrame newDF = df.sqlContext().sql("SELECT att1, att2, label FROM tmp");
DataFrame dataFixed = newDF.withColumn("label", newDF.col("label").cast("Double"));
VectorAssembler assembler = new VectorAssembler().setInputCols(new String[]{"att1", "att2"}).setOutputCol("features");
StringIndexer indexer = new StringIndexer().setInputCol("label").setOutputCol("labelIndexed");
// Split the data into training and test
DataFrame[] splits = dataFixed.randomSplit(new double[] {0.7, 0.3});
DataFrame trainingData = splits[0];
DataFrame testData = splits[1];
DecisionTreeClassifier dt = new DecisionTreeClassifier().setLabelCol("labelIndexed").setFeaturesCol("features");
Pipeline pipeline = new Pipeline().setStages(new PipelineStage[] {assembler, indexer, dt});
// Train model
PipelineModel model = pipeline.fit(trainingData);
// Make predictions
DataFrame predictions = model.transform(testData);
predictions.rdd().coalesce(1,true,null).saveAsTextFile("justPlay.txt" +System.currentTimeMillis());
RDD based Mllib is on its way to be deprecated, so you should rather use DataFrame based Mllib.
Generally the input to these MLlib apis is a DataFrame containing 2 columns - label and feature. There are various methods to build this DataFrame - low level apis like org.apache.spark.mllib.linalg.{Vector, Vectors}, org.apache.spark.mllib.regression.LabeledPoint, org.apache.spark.mllib.linalg.{Matrix, Matrices} etc. They all take numeric values for feature and label.
Words can be converted to vectors using - org.apache.spark.ml.feature.{Word2Vec, Word2VecModel}. This documentation explains more - https://spark.apache.org/docs/latest/mllib-data-types.html
Once input dataframe with label and feature is created, instantiate the MLlib api and pass in the DataFrame to 'fit' function to get the model and then call 'transform' or 'predict' function on the model to get the results.
Example -
training file looks like -
<numeric label> <a string separated by space>
//Build word vector
val trainingData = spark.read.parquet(<path to training file>)
val sampleDataDf = trainingData
.map { r =>
val s = r.getAs[String]("value").split(" ")
val label = s.head.toDouble
val feature = s.tail
(label, feature)
}.toDF("lable","feature_words")
val word2Vec = new Word2Vec()
.setInputCol("feature_words")
.setOutputCol("feature_vectors")
.setMinCount(0)
.setMaxIter(10)
//build word2Vector model
val model = word2Vec.fit(sampleDataDf)
//convert training text data to vector and labels
val wVectors = model.transform(sampleDataDf)
//train LinearSVM model
val svmAlgorithm = new LinearSVC()
.setFeaturesCol("feature_vectors")
.setMaxIter(100)
.setLabelCol("lable")
.setRegParam(0.01)
.setThreshold(0.5)
.fit(wVectors) //use word vectors created
//Predict new data, same format as training data containing words
val predictionData = spark.read.parquet(<path to prediction file>)
val pDataDf = predictionData
.map { r =>
val s = r.getAs[String]("value").split(" ")
val label = s.head.toDouble
val feature = s.tail
(label, feature)
}.toDF("lable","feature_words")
val pVectors = model.transform(pDataDf)
val predictionlResult = pVectors.map{ r =>
val s = r.getAs[Seq[String]]("feature_words")
val v = r.getAs[Vector]("feature_vectors")
val c = svmAlgorithm.predict(v) // predict using trained SVM
s"$c ${s.mkString(" ")}"
}