RDD with (key, (key2, value))












-1















I have an RDD in pyspark of the form (key, other things), where "other things" is a list of fields. I would like to get another RDD that uses a second key from the list of fields. For example, if my initial RDD is:



(User1, 1990 4 2 green...)

(User1, 1990 2 2 green...)

(User2, 1994 3 8 blue...)

(User1, 1987 3 4 blue...)



I would like to get (User1, [(1990, x), (1987, y)]),(User2, (1994 z))



where x, y, z would be an aggregation on the other fields, eg x is the count of how may rows I have with User1 and 1990 (two in this case), and I get a list with one tuple per year.



I am looking at the key value functions from:
https://www.oreilly.com/library/view/learning-spark/9781449359034/ch04.html



But don't seem to find anything that will give and aggregation twice: once for user and one for year. My initial attempt was with combineByKey() but I get stuck in getting a list from the values.



Any help would be appreciated!






pyspark rdd







share|improve this question



























    -1















    I have an RDD in pyspark of the form (key, other things), where "other things" is a list of fields. I would like to get another RDD that uses a second key from the list of fields. For example, if my initial RDD is:



    (User1, 1990 4 2 green...)

    (User1, 1990 2 2 green...)

    (User2, 1994 3 8 blue...)

    (User1, 1987 3 4 blue...)



    I would like to get (User1, [(1990, x), (1987, y)]),(User2, (1994 z))



    where x, y, z would be an aggregation on the other fields, eg x is the count of how may rows I have with User1 and 1990 (two in this case), and I get a list with one tuple per year.



    I am looking at the key value functions from:
    https://www.oreilly.com/library/view/learning-spark/9781449359034/ch04.html



    But don't seem to find anything that will give and aggregation twice: once for user and one for year. My initial attempt was with combineByKey() but I get stuck in getting a list from the values.



    Any help would be appreciated!






    pyspark rdd







    share|improve this question

























      -1












      -1








      -1








      I have an RDD in pyspark of the form (key, other things), where "other things" is a list of fields. I would like to get another RDD that uses a second key from the list of fields. For example, if my initial RDD is:



      (User1, 1990 4 2 green...)

      (User1, 1990 2 2 green...)

      (User2, 1994 3 8 blue...)

      (User1, 1987 3 4 blue...)



      I would like to get (User1, [(1990, x), (1987, y)]),(User2, (1994 z))



      where x, y, z would be an aggregation on the other fields, eg x is the count of how may rows I have with User1 and 1990 (two in this case), and I get a list with one tuple per year.



      I am looking at the key value functions from:
      https://www.oreilly.com/library/view/learning-spark/9781449359034/ch04.html



      But don't seem to find anything that will give and aggregation twice: once for user and one for year. My initial attempt was with combineByKey() but I get stuck in getting a list from the values.



      Any help would be appreciated!






      pyspark rdd







      share|improve this question














      I have an RDD in pyspark of the form (key, other things), where "other things" is a list of fields. I would like to get another RDD that uses a second key from the list of fields. For example, if my initial RDD is:



      (User1, 1990 4 2 green...)

      (User1, 1990 2 2 green...)

      (User2, 1994 3 8 blue...)

      (User1, 1987 3 4 blue...)



      I would like to get (User1, [(1990, x), (1987, y)]),(User2, (1994 z))



      where x, y, z would be an aggregation on the other fields, eg x is the count of how may rows I have with User1 and 1990 (two in this case), and I get a list with one tuple per year.



      I am looking at the key value functions from:
      https://www.oreilly.com/library/view/learning-spark/9781449359034/ch04.html



      But don't seem to find anything that will give and aggregation twice: once for user and one for year. My initial attempt was with combineByKey() but I get stuck in getting a list from the values.



      Any help would be appreciated!






      pyspark rdd




      pyspark rdd






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Jan 1 at 10:55









      PandaPanda

      31




      31
























          1 Answer
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          0














          You can do the following using groupby:



          # sample rdd
          
l = [("User1", "1990"), 
          
     ("User1", "1990"),
          
     ("User2", "1994"),
          
     ("User1", "1987") ]
          

          
rd = sc.parallelize(l)
          

          
# returns a tuples of count of year
          
def f(l):
          
    dd = {}
          
    for i in l:
          
        if i not in dd:
          
            dd[i] =1
          
        else:
          
            dd[i]+=1
          
    return list(dd.items())
          

          
# using groupby and applying the function on x[1] (which is a list)
          
rd1 = rd.groupByKey().map(lambda x : (x[0], f(x[1]))).collect()
          

          
[('User1', [('1990', 2), ('1987', 1)]), ('User2', [('1994', 1)])]





          share|improve this answer
























          • This works fine, thanks

            – Panda
            Jan 1 at 15:49











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          1 Answer
          1






          active

          oldest

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          active

          oldest

          votes






          active

          oldest

          votes









          0














          You can do the following using groupby:



          # sample rdd
          
l = [("User1", "1990"), 
          
     ("User1", "1990"),
          
     ("User2", "1994"),
          
     ("User1", "1987") ]
          

          
rd = sc.parallelize(l)
          

          
# returns a tuples of count of year
          
def f(l):
          
    dd = {}
          
    for i in l:
          
        if i not in dd:
          
            dd[i] =1
          
        else:
          
            dd[i]+=1
          
    return list(dd.items())
          

          
# using groupby and applying the function on x[1] (which is a list)
          
rd1 = rd.groupByKey().map(lambda x : (x[0], f(x[1]))).collect()
          

          
[('User1', [('1990', 2), ('1987', 1)]), ('User2', [('1994', 1)])]





          share|improve this answer
























          • This works fine, thanks

            – Panda
            Jan 1 at 15:49
















          0














          You can do the following using groupby:



          # sample rdd
          
l = [("User1", "1990"), 
          
     ("User1", "1990"),
          
     ("User2", "1994"),
          
     ("User1", "1987") ]
          

          
rd = sc.parallelize(l)
          

          
# returns a tuples of count of year
          
def f(l):
          
    dd = {}
          
    for i in l:
          
        if i not in dd:
          
            dd[i] =1
          
        else:
          
            dd[i]+=1
          
    return list(dd.items())
          

          
# using groupby and applying the function on x[1] (which is a list)
          
rd1 = rd.groupByKey().map(lambda x : (x[0], f(x[1]))).collect()
          

          
[('User1', [('1990', 2), ('1987', 1)]), ('User2', [('1994', 1)])]





          share|improve this answer
























          • This works fine, thanks

            – Panda
            Jan 1 at 15:49














          0












          0








          0







          You can do the following using groupby:



          # sample rdd
          
l = [("User1", "1990"), 
          
     ("User1", "1990"),
          
     ("User2", "1994"),
          
     ("User1", "1987") ]
          

          
rd = sc.parallelize(l)
          

          
# returns a tuples of count of year
          
def f(l):
          
    dd = {}
          
    for i in l:
          
        if i not in dd:
          
            dd[i] =1
          
        else:
          
            dd[i]+=1
          
    return list(dd.items())
          

          
# using groupby and applying the function on x[1] (which is a list)
          
rd1 = rd.groupByKey().map(lambda x : (x[0], f(x[1]))).collect()
          

          
[('User1', [('1990', 2), ('1987', 1)]), ('User2', [('1994', 1)])]





          share|improve this answer













          You can do the following using groupby:



          # sample rdd
          
l = [("User1", "1990"), 
          
     ("User1", "1990"),
          
     ("User2", "1994"),
          
     ("User1", "1987") ]
          

          
rd = sc.parallelize(l)
          

          
# returns a tuples of count of year
          
def f(l):
          
    dd = {}
          
    for i in l:
          
        if i not in dd:
          
            dd[i] =1
          
        else:
          
            dd[i]+=1
          
    return list(dd.items())
          

          
# using groupby and applying the function on x[1] (which is a list)
          
rd1 = rd.groupByKey().map(lambda x : (x[0], f(x[1]))).collect()
          

          
[('User1', [('1990', 2), ('1987', 1)]), ('User2', [('1994', 1)])]






          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Jan 1 at 11:45









          YOLOYOLO

          5,5001424




          5,5001424













          • This works fine, thanks

            – Panda
            Jan 1 at 15:49



















          • This works fine, thanks

            – Panda
            Jan 1 at 15:49

















          This works fine, thanks

          – Panda
          Jan 1 at 15:49





          This works fine, thanks

          – Panda
          Jan 1 at 15:49




















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