CODEWITHSUNDEEP
pythonapache-sparkdataframepysparkapache-spark-sql
Breach
Breach

Reputation: 1328

Create Spark DataFrame. Can not infer schema for type

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

Upvotes: 97

Views: 247110

Answers (4)

Mohammad Esmaili
Mohammad Esmaili

Reputation: 7

from pyspark.sql import Row
myFloatRdd.map(lambda x: Row(x)).toDF()

Upvotes: -2

AnyData
AnyData

Reputation: 21

Inferring the Schema Using Reflection
from pyspark.sql import Row
# spark - sparkSession
sc = spark.sparkContext

# Load a text file and convert each line to a Row.
orders = sc.textFile("/practicedata/orders")
#Split on delimiters
parts = orders.map(lambda l: l.split(","))
#Convert to Row
orders_struct = parts.map(lambda p: Row(order_id=int(p[0]), order_date=p[1], customer_id=p[2], order_status=p[3]))
for i in orders_struct.take(5): print(i)
#convert the RDD to DataFrame

orders_df = spark.createDataFrame(orders_struct)
Programmatically Specifying the Schema
from pyspark.sql import Row
# spark - sparkSession
sc = spark.sparkContext

# Load a text file and convert each line to a Row.
orders = sc.textFile("/practicedata/orders")
#Split on delimiters
parts = orders.map(lambda l: l.split(","))
#Convert to tuple
orders_struct = parts.map(lambda p: (p[0], p[1], p[2], p[3].strip()))

#convert the RDD to DataFrame

orders_df = spark.createDataFrame(orders_struct)

# The schema is encoded in a string.
schemaString = "order_id order_date customer_id status"

fields = [StructField(field_name, StringType(), True) for field_name in schemaString.split()]
schema = Struct

ordersDf = spark.createDataFrame(orders_struct, schema)

Type(fields)

Upvotes: 1

lei
lei

Reputation: 99

from pyspark.sql.types import IntegerType, Row

mylist = [1, 2, 3, 4, None ]
l = map(lambda x : Row(x), mylist)
# notice the parens after the type name
df=spark.createDataFrame(l,["id"])
df.where(df.id.isNull() == False).show()

Basiclly, you need to init your int into Row(), then we can use the schema

Upvotes: 0

zero323
zero323

Reputation: 330413

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.

Upvotes: 152

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