Pyspark convert a standard list to data frame

Question

The case is really simple, I need to convert a python list into data frame with following code

from pyspark.sql.types import StructType
from pyspark.sql.types import StructField
from pyspark.sql.types import StringType, IntegerType

schema = StructType([StructField("value", IntegerType(), True)])
my_list = [1, 2, 3, 4]
rdd = sc.parallelize(my_list)
df = sqlContext.createDataFrame(rdd, schema)

df.show()

it failed with following error:

    raise TypeError("StructType can not accept object %r in type %s" % (obj, type(obj)))
TypeError: StructType can not accept object 1 in type <class 'int'>

Asked By: seiya

||

Source

Answer 1

Please see the below code:

    from pyspark.sql import Row
    li=[1,2,3,4]
    rdd1 = sc.parallelize(li)
    row_rdd = rdd1.map(lambda x: Row(x))
    df=sqlContext.createDataFrame(row_rdd,['numbers']).show()

df

+-------+
|numbers|
+-------+
|      1|
|      2|
|      3|
|      4|
+-------+

Answered By: user15051990

Answer 2

This solution is also an approach that uses less code, avoids serialization to RDD and is likely easier to understand:

from pyspark.sql.types import IntegerType

# notice the variable name (more below)
mylist = [1, 2, 3, 4]

# notice the parens after the type name
spark.createDataFrame(mylist, IntegerType()).show()

NOTE: About naming your variable list: the term list is a Python builtin function and as such, it is strongly recommended that we avoid using builtin names as the name/label for our variables because we end up overwriting things like the list() function. When prototyping something fast and dirty, a number of folks use something like: mylist.

Answered By: E. Ducateme

Pyspark convert a standard list to data frame

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