If for some reason you can't use .toDF() method cannot, the solution I propose is this:
data = sqlContext.createDataFrame(sc.parallelize([('a','b','c', 1,4), ('o','u','w', 9,3), ('s','q','a', 8,6), ('l','g','z', 8,3), \
('a','b','c', 9,8), ('s','q','a', 10,10), ('l','g','z', 20,20), ('o','u','w', 77,77)]))
This will create a DF with names "_n" where n is the number of the column. If you want to rename the columns I suggest that you look this post: How to change dataframe column names in pyspark?. But all you need to do is:
data_named = data.selectExpr("_1 as One", "_2 as Two", "_3 as Three", "_4 as Four", "_5 as Five")
Now let's see the DF:
data_named.show()
And this will output:
+---+---+-----+----+----+
|One|Two|Three|Four|Five|
+---+---+-----+----+----+
| a| b| c| 1| 4|
| o| u| w| 9| 3|
| s| q| a| 8| 6|
| l| g| z| 8| 3|
| a| b| c| 9| 8|
| s| q| a| 10| 10|
| l| g| z| 20| 20|
| o| u| w| 77| 77|
+---+---+-----+----+----+
EDIT: Try again, because you should be able to use .toDF() in spark 1.6.1