[SPARK-12468] getParamMap in Pyspark ML API returns empty dictionary in example for Documentation

• Type: Bug
• Status: Closed
• Priority: Minor
• Resolution: Duplicate
• Affects Version/s: 1.5.2
• Component/s: PySpark
• Assignee: Unassigned
• Reporter: Zachary Brown
• Created: 21/Dec/15 20:42
• Updated: 20/Apr/16 23:24
• Resolved: 20/Apr/16 23:24

Description

The `extractParamMap()` method for a model that has been fit returns an empty dictionary, e.g. (from the [Pyspark ML API Documentation](http://spark.apache.org/docs/latest/ml-guide.html#example-estimator-transformer-and-param)):

```python
from pyspark.mllib.linalg import Vectors
from pyspark.ml.classification import LogisticRegression
from pyspark.ml.param import Param, Params

1. Prepare training data from a list of (label, features) tuples.
   training = sqlContext.createDataFrame([
   (1.0, Vectors.dense([0.0, 1.1, 0.1])),
   (0.0, Vectors.dense([2.0, 1.0, -1.0])),
   (0.0, Vectors.dense([2.0, 1.3, 1.0])),
   (1.0, Vectors.dense([0.0, 1.2, -0.5]))], ["label", "features"])

1. Create a LogisticRegression instance. This instance is an Estimator.
   lr = LogisticRegression(maxIter=10, regParam=0.01)
2. Print out the parameters, documentation, and any default values.
   print "LogisticRegression parameters:\n" + lr.explainParams() + "\n"

1. Learn a LogisticRegression model. This uses the parameters stored in lr.
   model1 = lr.fit(training)

1. Since model1 is a Model (i.e., a transformer produced by an Estimator),
2. we can view the parameters it used during fit().
3. This prints the parameter (name: value) pairs, where names are unique IDs for this
4. LogisticRegression instance.
   print "Model 1 was fit using parameters: "
   print model1.extractParamMap()
   ```

Issue Links

• duplicates: SPARK-10931 PySpark ML Models should contain Param values
• Is contained by: SPARK-14771 Python ML Param and UID issues
• links to: [Github] Pull Request #10419 (ZacharySBrown)