将RDD转换为DataFrame：AttributeError：'RDD'对象没有'toDF'属性。

Question

from pyspark import SparkContext, SparkConf
from pyspark.sql import SQLContext

conf = SparkConf().setAppName("myApp").setMaster("local")
sc = SparkContext(conf=conf)

a = sc.parallelize([[1, "a"], [2, "b"], [3, "c"], [4, "d"], [5, "e"]]).toDF(['ind', "state"])

a.show()

结果为：

Traceback (most recent call last):
  File "/Users/ktemlyakov/messing_around/SparkStuff/mock_maersk_data.py", line 7, in <module>
    a = sc.parallelize([[1, "a"], [2, "b"], [3, "c"], [4, "d"], [5, "e"]]).toDF(['ind', "state"])
AttributeError: 'RDD' object has no attribute 'toDF'

我漏掉了什么？

- Akavall

Answer 1

sqlContext 没有被创建，需要创建它。以下代码可行：

from pyspark import SparkContext, SparkConf
from pyspark.sql import SQLContext
from pyspark import sql

conf = SparkConf().setAppName("myFirstApp").setMaster("local")
sc = SparkContext(conf=conf)
sqlContext = sql.SQLContext(sc)

a = sc.parallelize([[1, "a"], [2, "b"], [3, "c"], [4, "d"], [5, "e"]]).toDF(['ind', "state"])

a.show()

编辑：

在Spark 2.0中，以上内容可通过以下方式实现：

from pyspark import SparkConf
from pyspark.sql import SparkSession

spark = SparkSession.builder.master("local").config(conf=SparkConf()).getOrCreate()

a = spark.createDataFrame([[1, "a"], [2, "b"], [3, "c"], [4, "d"], [5, "e"]], ['ind', "state"])
a.show()

- Akavall

Answer 2

你可以直接这样做

a_df = a.toDF()
type(a_df)

- pranav