问题
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()
Results in:
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'
What am I missing?
回答1:
sqlContext
is missing; it needs to be created. The following code works:
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()
Edit:
In Spark 2.0, the above can be achieved with:
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()
回答2:
u can directly do this
a_df = a.toDF()
type(a_df)
来源:https://stackoverflow.com/questions/47341048/converting-rdd-to-dataframe-attributeerror-rdd-object-has-no-attribute-todf