How to create dataframe from list in Spark SQL?

问题

Spark version : 2.1

For example, in pyspark, i create a list

test_list = [['Hello', 'world'], ['I', 'am', 'fine']]

then how to create a dataframe form the test_list, where the dataframe's type is like below:

DataFrame[words: array<string>]

回答1:

here is how -

from pyspark.sql.types import *

cSchema = StructType([StructField("WordList", ArrayType(StringType()))])

# notice extra square brackets around each element of list 
test_list = [['Hello', 'world']], [['I', 'am', 'fine']]

df = spark.createDataFrame(test_list,schema=cSchema)

回答2:

i had to work with multiple columns and types - the example below has one string column and one integer column. A slight adjustment to Pushkr's code (above) gives:

from pyspark.sql.types import *

cSchema = StructType([StructField("Words", StringType())\
                      ,StructField("total", IntegerType())])

test_list = [['Hello', 1], ['I am fine', 3]]

df = spark.createDataFrame(test_list,schema=cSchema)

output:

 df.show()
 +---------+-----+
|    Words|total|
+---------+-----+
|    Hello|    1|
|I am fine|    3|
+---------+-----+

回答3:

You should use list of Row objects([Row]) to create data frame.

from pyspark.sql import Row

spark.createDataFrame(list(map(lambda x: Row(words=x), test_list)))

回答4:

   You can create a RDD first from the input and then convert to dataframe from the constructed RDD
   <code>  
     import sqlContext.implicits._
       val testList = Array(Array("Hello", "world"), Array("I", "am", "fine"))
       // CREATE RDD
       val testListRDD = sc.parallelize(testList)
     val flatTestListRDD = testListRDD.flatMap(entry => entry)
     // COnvert RDD to DF 
     val testListDF = flatTestListRDD.toDF
     testListDF.show
    </code>

来源：https://stackoverflow.com/questions/43444925/how-to-create-dataframe-from-list-in-spark-sql

标签

python

apache-spark

pyspark