0
我有一个RDD的分区包含元素(熊猫数据框,因为它发生),可以很容易地变成行列表。把它看成是看起来像这样RDD的pyspark行列表DataFrame
rows_list = []
for word in 'quick brown fox'.split():
rows = []
for i,c in enumerate(word):
x = ord(c) + i
row = pyspark.sql.Row(letter=c, number=i, importance=x)
rows.append(row)
rows_list.append(rows)
rdd = sc.parallelize(rows_list)
rdd.take(2)
这给
[[Row(importance=113, letter='q', number=0),
Row(importance=118, letter='u', number=1),
Row(importance=107, letter='i', number=2),
Row(importance=102, letter='c', number=3),
Row(importance=111, letter='k', number=4)],
[Row(importance=98, letter='b', number=0),
Row(importance=115, letter='r', number=1),
Row(importance=113, letter='o', number=2),
Row(importance=122, letter='w', number=3),
Row(importance=114, letter='n', number=4)]]
我希望把它变成一个Spark数据帧。我希望我可以做
rdd.toDF()
,但给人的无用结构
DataFrame[_1: struct<importance:bigint,letter:string,number:bigint>,
_2: struct<importance:bigint,letter:string,number:bigint>,
_3: struct<importance:bigint,letter:string,number:bigint>,
_4: struct<importance:bigint,letter:string,number:bigint>,
_5: struct<importance:bigint,letter:string,number:bigint>]
我真正想要的是一个3列数据框,如该
desired_df = sql_context.createDataFrame(sum(rows_list, []))
让我可执行如下操作:
desired_df.agg(pyspark.sql.functions.sum('number')).take(1)
并得到答案23.
什么是正确的方式去做这件事?