我有两个相同列的RDD:
rdd1: -
+-----------------+ |mid|uid|frequency| +-----------------+ | m1| u1| 1| | m1| u2| 1| | m2| u1| 2| +-----------------+
rdd2: -
+-----------------+ |mid|uid|frequency| +-----------------+ | m1| u1| 10| | m2| u1| 98| | m3| u2| 21| +-----------------+
我想根据frequencies
和mid
计算uid
的总和。结果应该是这样的:
+-----------------+ |mid|uid|frequency| +-----------------+ | m1| u1| 11| | m2| u1| 100| | m3| u2| 21| +-----------------+
提前致谢。
编辑: 我也以这种方式实现了解决方案(使用map-reduce):
from pyspark.sql.functions import col
data1 = [("m1","u1",1),("m1","u2",1),("m2","u1",2)]
data2 = [("m1","u1",10),("m2","u1",98),("m3","u2",21)]
df1 = sqlContext.createDataFrame(data1,['mid','uid','frequency'])
df2 = sqlContext.createDataFrame(data2,['mid','uid','frequency'])
df3 = df1.unionAll(df2)
df4 = df3.map(lambda bbb: ((bbb['mid'], bbb['uid']), int(bbb['frequency'])))\
.reduceByKey(lambda a, b: a+b)
p = df4.map(lambda p: (p[0][0], p[0][1], p[1])).toDF()
p = p.select(col("_1").alias("mid"), \
col("_2").alias("uid"), \
col("_3").alias("frequency"))
p.show()
输出:
+---+---+---------+ |mid|uid|frequency| +---+---+---------+ | m2| u1| 100| | m1| u1| 11| | m1| u2| 1| | m3| u2| 21| +---+---+---------+
答案 0 :(得分:1)
你只需要在mid和uid之间执行一个组并执行求和操作:
data1 = [("m1","u1",1),("m1","u2",1),("m2","u1",2)]
data2 = [("m1","u1",10),("m2","u1",98),("m3","u2",21)]
df1 = sqlContext.createDataFrame(data1,['mid','uid','frequency'])
df2 = sqlContext.createDataFrame(data2,['mid','uid','frequency'])
df3 = df1.unionAll(df2)
df4 = df3.groupBy(df3.mid,df3.uid).sum() \
.withColumnRenamed("sum(frequency)","frequency")
df4.show()
# +---+---+---------+
# |mid|uid|frequency|
# +---+---+---------+
# | m1| u1| 11|
# | m1| u2| 1|
# | m2| u1| 100|
# | m3| u2| 21|
# +---+---+---------+
答案 1 :(得分:0)
我也以这种方式实现了解决方案(使用map-reduce):
from pyspark.sql.functions import col
data1 = [("m1","u1",1),("m1","u2",1),("m2","u1",2)]
data2 = [("m1","u1",10),("m2","u1",98),("m3","u2",21)]
df1 = sqlContext.createDataFrame(data1,['mid','uid','frequency'])
df2 = sqlContext.createDataFrame(data2,['mid','uid','frequency'])
df3 = df1.unionAll(df2)
df4 = df3.map(lambda bbb: ((bbb['mid'], bbb['uid']), int(bbb['frequency'])))\
.reduceByKey(lambda a, b: a+b)
p = df4.map(lambda p: (p[0][0], p[0][1], p[1])).toDF()
p = p.select(col("_1").alias("mid"), \
col("_2").alias("uid"), \
col("_3").alias("frequency"))
p.show()
输出:
+---+---+---------+ |mid|uid|frequency| +---+---+---------+ | m2| u1| 100| | m1| u1| 11| | m1| u2| 1| | m3| u2| 21| +---+---+---------+