Pyspark:根据两个RDD中两列的条件计算两个对应列的总和

Pyspark: Calculating sum of two correspoding columns, based on conditions of two columns in two RDDs

我有两个具有相同列的 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 miduid 计算 frequencies 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: 我也是通过这种方式实现的解决方案(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|
+---+---+---------+

只需要按mid和uid进行分组,并进行求和即可:

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|
# +---+---+---------+

我也是这样解决的(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|
+---+---+---------+