I wrote this code in Spark ML
import org.apache.spark.ml.classification.LogisticRegression
import org.apache.spark.ml.Pipeline
val lr = new LogisticRegression()
val pipeline = new Pipeline()
.setStages(Array(fooIndexer, fooHotEncoder, assembler, lr))
val model = pipeline.fit(training)
This code takes a long time to run. Is it possible that after running pipeline.fit I save the model on HDFS so that I don't have to run it again and again?
Edit: Also, how to load it back from HDFS when I have to apply transform on the model so that I can make predictions.
Straight from the official documentation - saving:
// Now we can optionally save the fitted pipeline to disk
model.write.overwrite().save("/tmp/spark-logistic-regression-model")
and loading:
// And load it back in during production
val sameModel = PipelineModel.load("/tmp/spark-logistic-regression-model")
Related:
Save ML model for future usage
Related
I would like to concatenate several trained Pipelines to one, which is similar to
"Spark add new fitted stage to a exitsting PipelineModel without fitting again" however the solution as below is for PySpark.
> pipe_model_new = PipelineModel(stages = [pipe_model , pipe_model2])
> final_df = pipe_model_new.transform(df1)
In Apache Spark 2.0 "PipelineModel"'s constructor is marked as private, hence it can not be called outside. While in "Pipeline" class, only "fit" method creates "PipelineModel"
val pipelineModel = new PipelineModel("randomUID", trainedStages)
val df_final_full = pipelineModel.transform(df)
Error:(266, 26) constructor PipelineModel in class PipelineModel cannot be accessed in class Preprocessor
val pipelineModel = new PipelineModel("randomUID", trainedStages)
There is nothing* wrong with using Pipeline and invoking fit method. If a stage is a Transfomer, and PipelineModel is**, fit works like identity.
You can check relevant Python:
if isinstance(stage, Transformer):
transformers.append(stage)
dataset = stage.transform(dataset)
and Scala code:
This means that fitting process will only validate the schema and create a new PipelineModel object.
case t: Transformer =>
t
* The only possible concern is presence of non-lazy Transformers, though, with exception to deprecated OneHotEncoder, Spark core API doesn't provide such.
** In Python:
from pyspark.ml import Transformer, PipelineModel
issubclass(PipelineModel, Transformer)
True
In Scala
import scala.reflect.runtime.universe.typeOf
import org.apache.spark.ml._
typeOf[PipelineModel] <:< typeOf[Transformer]
Boolean = true
I want to train the logistic regression model using Apache Spark in Java. As first step I would like to train the model just once and save the model parameters (intercept and Coefficient). Subsequently use the saved model parameters to score at a later point in time. I am able to save the model in parquet file, using the following code
LogisticRegressionModel trainedLRModel = logReg.fit(data);
trainedLRModel.write().overwrite().save("mypath");
When I load the model to score, I get the following error.
LogisticRegression lr = new LogisticRegression();
lr.load("//saved_model_path");
Exception in thread "main" java.lang.NoSuchMethodException: org.apache.spark.ml.classification.LogisticRegressionModel.<init>(java.lang.String)
at java.lang.Class.getConstructor0(Class.java:3082)
at java.lang.Class.getConstructor(Class.java:1825)
at org.apache.spark.ml.util.DefaultParamsReader.load(ReadWrite.scala:325)
at org.apache.spark.ml.util.MLReadable$class.load(ReadWrite.scala:215)
at org.apache.spark.ml.classification.LogisticRegression$.load(LogisticRegression.scala:672)
at org.apache.spark.ml.classification.LogisticRegression.load(LogisticRegression.scala)
Is there a way to train and save model and then evaluate(score) later? I am using Spark ML 2.1.0 in Java.
I face the same problem with pyspark 2.1.1, when i change from LogisticRegression to LogisticRegressionModel , everything works well.
LogisticRegression.load("/model/path") # not works
LogisticRegressionModel.load("/model/path") # works well
TL;DR Use LogisticRegressionModel.load.
load(path: String): LogisticRegressionModel Reads an ML instance from the input path, a shortcut of read.load(path).
As a matter of fact, as of Spark 2.0.0, the recommended approach to use Spark MLlib, incl. LogisticRegression estimator, is using the brand new and shiny Pipeline API.
import org.apache.spark.ml.classification._
val lr = new LogisticRegression()
import org.apache.spark.ml.feature._
val tok = new Tokenizer().setInputCol("body")
val hashTF = new HashingTF().setInputCol(tok.getOutputCol).setOutputCol("features")
import org.apache.spark.ml._
val pipeline = new Pipeline().setStages(Array(tok, hashTF, lr))
// training dataset
val emails = Seq(("hello world", 1)).toDF("body", "label")
val model = pipeline.fit(emails)
model.write.overwrite.save("mypath")
val loadedModel = PipelineModel.load("mypath")
I am a new beginner in Apache Spark. I trained a LogisticRegression model using crossValidation. For instance:
val cv = new CrossValidator()
.setEstimator(pipeline)
.setEvaluator(new BinaryClassificationEvaluator)
.setEstimatorParamMaps(paramGrid)
.setNumFolds(5)
val cvModel = cv.fit(data)
I was able to train and test my model without any error. Then I saved the model and the pipeline using:
cvModel.save("/path-to-my-model/spark-log-reg-transfer-model")
pipeline.save("/path-to-my-pipeline/spark-log-reg-transfer-pipeline")
Up till this stage, the operations worked perfect. Then later on, I tried to load my model back for prediction on new data points, then the following error occured:
val sameModel = PipelineModel.load("/path-to-my-model/spark-log-reg-transfer-model")
java.lang.IllegalArgumentException: requirement failed: Error loading metadata: Expected class name org.apache.spark.ml.PipelineModel but found class name org.apache.spark.ml.tuning.CrossValidatorModel
Any idea what I may have done wrong? Thanks.
You are trying to load CrossValidator with PipelineModel object.
You should use correct loaders...
val crossValidator = CrossValidator.load("/path-to-my-model/spark-log-reg-transfer-model")
val sameModel = PipelineModel.load("/path-to-my-pipeline/spark-log-reg-transfer-pipeline")
To load a Cross Validator it should be
val crossValidator = CrossValidator.load("/path-to-my-model/spark-log-reg-transfer-model")
To load a Cross Validator Model use
(Note: A Cross Validator becomes a Cross Validator model when you call fit() on CrossValidator)
val crossValidatorModel = CrossValidatorModel.load("/path-to-my-model/spark-log-reg-transfer-model")
Since you are trying to load a model, CrossValidatorModel.load would be the correct one.
With Spark MLLib, I'd build a model (like RandomForest), and then it was possible to eval it outside of Spark by loading the model and using predict on it passing a vector of features.
It seems like with Spark ML, predict is now called transform and only acts on a DataFrame.
Is there any way to build a DataFrame outside of Spark since it seems like one needs a SparkContext to build a DataFrame?
Am I missing something?
Re: Is there any way to build a DataFrame outside of Spark?
It is not possible. DataFrames live inside SQLContext with it living in SparkContext. Perhaps you could work it around somehow, but the whole story is that the connection between DataFrames and SparkContext is by design.
Here is my solution to use spark models outside of spark context (using PMML):
You create model with a pipeline like this:
SparkConf sparkConf = new SparkConf();
SparkSession session = SparkSession.builder().enableHiveSupport().config(sparkConf).getOrCreate();
String tableName = "schema.table";
Properties dbProperties = new Properties();
dbProperties.setProperty("user",vKey);
dbProperties.setProperty("password",password);
dbProperties.setProperty("AuthMech","3");
dbProperties.setProperty("source","jdbc");
dbProperties.setProperty("driver","com.cloudera.impala.jdbc41.Driver");
String tableName = "schema.table";
String simpleUrl = "jdbc:impala://host:21050/schema"
Dataset<Row> data = session.read().jdbc(simpleUrl ,tableName,dbProperties);
String[] inputCols = {"column1"};
StringIndexer indexer = new StringIndexer().setInputCol("column1").setOutputCol("indexed_column1");
StringIndexerModel alphabet = indexer.fit(data);
data = alphabet.transform(data);
VectorAssembler assembler = new VectorAssembler().setInputCols(inputCols).setOutputCol("features");
Predictor p = new GBTRegressor();
p.set("maxIter",20);
p.set("maxDepth",2);
p.set("maxBins",204);
p.setLabelCol("faktor");
PipelineStage[] stages = {indexer,assembler, p};
Pipeline pipeline = new Pipeline();
pipeline.setStages(stages);
PipelineModel pmodel = pipeline.fit(data);
PMML pmml = ConverterUtil.toPMML(data.schema(),pmodel);
FileOutputStream fos = new FileOutputStream("model.pmml");
JAXBUtil.marshalPMML(pmml,new StreamResult(fos));
Using PPML for predictions (locally, without spark context, which can be applied to a Map of arguments and not on a DataFrame):
PMML pmml = org.jpmml.model.PMMLUtil.unmarshal(new FileInputStream(pmmlFile));
ModelEvaluatorFactory modelEvaluatorFactory = ModelEvaluatorFactory.newInstance();
MiningModelEvaluator evaluator = (MiningModelEvaluator) modelEvaluatorFactory.newModelEvaluator(pmml);
inputFieldMap = new HashMap<String, Field>();
Map<FieldName,String> args = new HashMap<FieldName, String>();
Field curField = evaluator.getInputFields().get(0);
args.put(curField.getName(), "1.0");
Map<FieldName, ?> result = evaluator.evaluate(args);
Spent days on this problem too. It's not straightforward. My third suggestion involves code I have written specifically for this purpose.
Option 1
As other commenters have said, predict(Vector) is now available. However, you need to know how to construct a vector. If you don't, see Option 3.
Option 2
If the goal is to avoid setting up a Spark server (standalone or cluster modes), then its possible to start Spark in local mode. The whole thing will run inside a single JVM.
val spark = SparkSession.builder().config("spark.master", "local[*]").getOrCreate()
// create dataframe from file, or make it up from some data in memory
// use model.transform() to get predictions
But this brings unnecessary dependencies to your prediction module, and it consumes resources in your JVM at runtime. Also, if prediction latency is critical, for example making a prediction within a millisecond as soon as a request comes in, then this option is too slow.
Option 3
MLlib FeatureHasher's output can be used as an input to your learner. The class is good for one hot encoding and also for fixing the size of your feature dimension. You can use it even when all your features are numerical. If you use that in your training, then all you need at prediction time is the hashing logic there. Its implemented as a spark transformer so it's not easy to re-use outside of a spark environment. So I have done the work of pulling out the hashing function to a lib. You apply FeatureHasher and your learner during training as normal. Then here's how you use the slimmed down hasher at prediction time:
// Schema and hash size must stay consistent across training and prediction
val hasher = new FeatureHasherLite(mySchema, myHashSize)
// create sample data-point and hash it
val feature = Map("feature1" -> "value1", "feature2" -> 2.0, "feature3" -> 3, "feature4" -> false)
val featureVector = hasher.hash(feature)
// Make prediction
val prediction = model.predict(featureVector)
You can see details in my github at tilayealemu/sparkmllite. If you'd rather copy my code, take a look at FeatureHasherLite.scala.There are sample codes and unit tests too. Feel free to create an issue if you need help.
I want to tunning my model with grid search and cross validation with spark. In the spark, it must put the base model in a pipeline, the office demo of pipeline use the LogistictRegression as an base model, which can be new as an object. However, the RandomForest model cannot be new by client code, so it seems not be able to use RandomForest in the pipeline api. I don't want to recreate an wheel, so can anybody give some advice?
Thanks
However, the RandomForest model cannot be new by client code, so it seems not be able to use RandomForest in the pipeline api.
Well, that is true but you simply trying to use a wrong class. Instead of mllib.tree.RandomForest you should use ml.classification.RandomForestClassifier. Here is an example based on the one from MLlib docs.
import org.apache.spark.ml.classification.RandomForestClassifier
import org.apache.spark.ml.Pipeline
import org.apache.spark.ml.feature.StringIndexer
import org.apache.spark.mllib.linalg.Vector
import org.apache.spark.mllib.util.MLUtils
import sqlContext.implicits._
case class Record(category: String, features: Vector)
val data = MLUtils.loadLibSVMFile(sc, "data/mllib/sample_libsvm_data.txt")
val splits = data.randomSplit(Array(0.7, 0.3))
val (trainData, testData) = (splits(0), splits(1))
val trainDF = trainData.map(lp => Record(lp.label.toString, lp.features)).toDF
val testDF = testData.map(lp => Record(lp.label.toString, lp.features)).toDF
val indexer = new StringIndexer()
.setInputCol("category")
.setOutputCol("label")
val rf = new RandomForestClassifier()
.setNumTrees(3)
.setFeatureSubsetStrategy("auto")
.setImpurity("gini")
.setMaxDepth(4)
.setMaxBins(32)
val pipeline = new Pipeline()
.setStages(Array(indexer, rf))
val model = pipeline.fit(trainDF)
model.transform(testDF)
There is one thing I couldn't figure out here. As far as I can tell it should be possible to use labels extracted from LabeledPoints directly, but for some reason it doesn't work and pipeline.fit raises IllegalArgumentExcetion:
RandomForestClassifier was given input with invalid label column label, without the number of classes specified.
Hence the ugly trick with StringIndexer. After applying we get required attributes ({"vals":["1.0","0.0"],"type":"nominal","name":"label"}) but some classes in ml seem to work just fine without it.