Create Spark DataFrame. Can not infer schema for type: <type 'float'>

Question

Could someone help me solve this problem I have with Spark DataFrame?

When I do myFloatRDD.toDF() I get an error:

TypeError: Can not infer schema for type: type 'float'

I don't understand why...

Example:

myFloatRdd = sc.parallelize([1.0,2.0,3.0]) df = myFloatRdd.toDF() 

Thanks

Answer 1

SparkSession.createDataFrame, which is used under the hood, requires an RDD / list of Row/tuple/list/dict* or pandas.DataFrame, unless schema with DataType is provided. Try to convert float to tuple like this:

myFloatRdd.map(lambda x: (x, )).toDF() 

or even better:

from pyspark.sql import Row  row = Row("val") # Or some other column name myFloatRdd.map(row).toDF() 

To create a DataFrame from a list of scalars you'll have to use SparkSession.createDataFrame directly and provide a schema***:

from pyspark.sql.types import FloatType  df = spark.createDataFrame([1.0, 2.0, 3.0], FloatType())  df.show()  ## +-----+ ## |value| ## +-----+ ## |  1.0| ## |  2.0| ## |  3.0| ## +-----+ 

but for a simple range it would be better to use SparkSession.range:

from pyspark.sql.functions import col  spark.range(1, 4).select(col("id").cast("double")) 

---

* No longer supported.

** Spark SQL also provides a limited support for schema inference on Python objects exposing __dict__.

*** Supported only in Spark 2.0 or later.