symmetric_difference Output as Two Separate Lists












0















Let's say we have two lists of values, whereby each list only contains unique values unto itself. There will never be duplicate values in a single list.



L1 | L2

-------

a  |  a

b  |  d

c  |  e

d  |  g

e  |  h

f  |  i

   |  j


We can get the differences of these lists using set(L1).symmetric_difference(L2), but unfortunately that lumps the results together in a single list. For example, the output of list(set(L1).symmetric_difference(L2)) is ['c', 'b', 'h', 'i', 'j', 'f', 'g'].



Is there a way to obtain two separate lists of output from list(set(L1).symmetric_difference(L2)) like ['c', 'b', 'f',] and ['h', 'i', 'j', 'g'] instead?



Or is there a way to obtain two separate lists as output while comparing the two sets/lists against each other only once?






arrays python-3.x list symmetric-difference







share|improve this question





























    0















    Let's say we have two lists of values, whereby each list only contains unique values unto itself. There will never be duplicate values in a single list.



    L1 | L2
    
-------
    
a  |  a
    
b  |  d
    
c  |  e
    
d  |  g
    
e  |  h
    
f  |  i
    
   |  j


    We can get the differences of these lists using set(L1).symmetric_difference(L2), but unfortunately that lumps the results together in a single list. For example, the output of list(set(L1).symmetric_difference(L2)) is ['c', 'b', 'h', 'i', 'j', 'f', 'g'].



    Is there a way to obtain two separate lists of output from list(set(L1).symmetric_difference(L2)) like ['c', 'b', 'f',] and ['h', 'i', 'j', 'g'] instead?



    Or is there a way to obtain two separate lists as output while comparing the two sets/lists against each other only once?






    arrays python-3.x list symmetric-difference







    share|improve this question



























      0












      0








      0








      Let's say we have two lists of values, whereby each list only contains unique values unto itself. There will never be duplicate values in a single list.



      L1 | L2
      
-------
      
a  |  a
      
b  |  d
      
c  |  e
      
d  |  g
      
e  |  h
      
f  |  i
      
   |  j


      We can get the differences of these lists using set(L1).symmetric_difference(L2), but unfortunately that lumps the results together in a single list. For example, the output of list(set(L1).symmetric_difference(L2)) is ['c', 'b', 'h', 'i', 'j', 'f', 'g'].



      Is there a way to obtain two separate lists of output from list(set(L1).symmetric_difference(L2)) like ['c', 'b', 'f',] and ['h', 'i', 'j', 'g'] instead?



      Or is there a way to obtain two separate lists as output while comparing the two sets/lists against each other only once?






      arrays python-3.x list symmetric-difference







      share|improve this question
















      Let's say we have two lists of values, whereby each list only contains unique values unto itself. There will never be duplicate values in a single list.



      L1 | L2
      
-------
      
a  |  a
      
b  |  d
      
c  |  e
      
d  |  g
      
e  |  h
      
f  |  i
      
   |  j


      We can get the differences of these lists using set(L1).symmetric_difference(L2), but unfortunately that lumps the results together in a single list. For example, the output of list(set(L1).symmetric_difference(L2)) is ['c', 'b', 'h', 'i', 'j', 'f', 'g'].



      Is there a way to obtain two separate lists of output from list(set(L1).symmetric_difference(L2)) like ['c', 'b', 'f',] and ['h', 'i', 'j', 'g'] instead?



      Or is there a way to obtain two separate lists as output while comparing the two sets/lists against each other only once?






      arrays python-3.x list symmetric-difference




      arrays python-3.x list symmetric-difference






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Nov 15 '18 at 0:05







      Anthony

















      asked Nov 14 '18 at 23:31









      AnthonyAnthony

      1,375527




      1,375527
























          1 Answer
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          1














          You can simply do the following:



          dif_1_from_2 = list(set(L1) - set(L2))
          
dif_2_from_1  = list(set(L2) - set(L1))


          And you can create a function to do that like this:



          def get_symmetric_difference(L1, L2):
          
    return list(set(L1)-set(L2)), list(set(L2)-set(L1))


          and then you call it like this:



          print(get_symmetric_difference(L1, L2))


          Hope this helps.






          share|improve this answer


























          • ok, thx, but is there no way to get the two lists of output while comparing the sets only once??

            – Anthony
            Nov 14 '18 at 23:46











          • Is my updated answer gives you what you asked in your comment? Please let me know.

            – TeeKea
            Nov 14 '18 at 23:48











          • no, unfortunately that's not what i meant. what i'm trying to get at it is that dif_1_from_2 compares the sets once, then dif_2_from_1 compares the sets for a 2nd time. is there no way to compare L1 and L2 once, but get two different lists as the output? does that make sense?

            – Anthony
            Nov 14 '18 at 23:50











          • Not sure, actually. I believe you would really need two operations to do that at the end.

            – TeeKea
            Nov 15 '18 at 0:21











          • ok, interesting. thanks anyways!!

            – Anthony
            Nov 15 '18 at 0:50











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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1














          You can simply do the following:



          dif_1_from_2 = list(set(L1) - set(L2))
          
dif_2_from_1  = list(set(L2) - set(L1))


          And you can create a function to do that like this:



          def get_symmetric_difference(L1, L2):
          
    return list(set(L1)-set(L2)), list(set(L2)-set(L1))


          and then you call it like this:



          print(get_symmetric_difference(L1, L2))


          Hope this helps.






          share|improve this answer


























          • ok, thx, but is there no way to get the two lists of output while comparing the sets only once??

            – Anthony
            Nov 14 '18 at 23:46











          • Is my updated answer gives you what you asked in your comment? Please let me know.

            – TeeKea
            Nov 14 '18 at 23:48











          • no, unfortunately that's not what i meant. what i'm trying to get at it is that dif_1_from_2 compares the sets once, then dif_2_from_1 compares the sets for a 2nd time. is there no way to compare L1 and L2 once, but get two different lists as the output? does that make sense?

            – Anthony
            Nov 14 '18 at 23:50











          • Not sure, actually. I believe you would really need two operations to do that at the end.

            – TeeKea
            Nov 15 '18 at 0:21











          • ok, interesting. thanks anyways!!

            – Anthony
            Nov 15 '18 at 0:50
















          1














          You can simply do the following:



          dif_1_from_2 = list(set(L1) - set(L2))
          
dif_2_from_1  = list(set(L2) - set(L1))


          And you can create a function to do that like this:



          def get_symmetric_difference(L1, L2):
          
    return list(set(L1)-set(L2)), list(set(L2)-set(L1))


          and then you call it like this:



          print(get_symmetric_difference(L1, L2))


          Hope this helps.






          share|improve this answer


























          • ok, thx, but is there no way to get the two lists of output while comparing the sets only once??

            – Anthony
            Nov 14 '18 at 23:46











          • Is my updated answer gives you what you asked in your comment? Please let me know.

            – TeeKea
            Nov 14 '18 at 23:48











          • no, unfortunately that's not what i meant. what i'm trying to get at it is that dif_1_from_2 compares the sets once, then dif_2_from_1 compares the sets for a 2nd time. is there no way to compare L1 and L2 once, but get two different lists as the output? does that make sense?

            – Anthony
            Nov 14 '18 at 23:50











          • Not sure, actually. I believe you would really need two operations to do that at the end.

            – TeeKea
            Nov 15 '18 at 0:21











          • ok, interesting. thanks anyways!!

            – Anthony
            Nov 15 '18 at 0:50














          1












          1








          1







          You can simply do the following:



          dif_1_from_2 = list(set(L1) - set(L2))
          
dif_2_from_1  = list(set(L2) - set(L1))


          And you can create a function to do that like this:



          def get_symmetric_difference(L1, L2):
          
    return list(set(L1)-set(L2)), list(set(L2)-set(L1))


          and then you call it like this:



          print(get_symmetric_difference(L1, L2))


          Hope this helps.






          share|improve this answer















          You can simply do the following:



          dif_1_from_2 = list(set(L1) - set(L2))
          
dif_2_from_1  = list(set(L2) - set(L1))


          And you can create a function to do that like this:



          def get_symmetric_difference(L1, L2):
          
    return list(set(L1)-set(L2)), list(set(L2)-set(L1))


          and then you call it like this:



          print(get_symmetric_difference(L1, L2))


          Hope this helps.







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited Nov 14 '18 at 23:46

























          answered Nov 14 '18 at 23:41









          TeeKeaTeeKea

          3,22851731




          3,22851731













          • ok, thx, but is there no way to get the two lists of output while comparing the sets only once??

            – Anthony
            Nov 14 '18 at 23:46











          • Is my updated answer gives you what you asked in your comment? Please let me know.

            – TeeKea
            Nov 14 '18 at 23:48











          • no, unfortunately that's not what i meant. what i'm trying to get at it is that dif_1_from_2 compares the sets once, then dif_2_from_1 compares the sets for a 2nd time. is there no way to compare L1 and L2 once, but get two different lists as the output? does that make sense?

            – Anthony
            Nov 14 '18 at 23:50











          • Not sure, actually. I believe you would really need two operations to do that at the end.

            – TeeKea
            Nov 15 '18 at 0:21











          • ok, interesting. thanks anyways!!

            – Anthony
            Nov 15 '18 at 0:50



















          • ok, thx, but is there no way to get the two lists of output while comparing the sets only once??

            – Anthony
            Nov 14 '18 at 23:46











          • Is my updated answer gives you what you asked in your comment? Please let me know.

            – TeeKea
            Nov 14 '18 at 23:48











          • no, unfortunately that's not what i meant. what i'm trying to get at it is that dif_1_from_2 compares the sets once, then dif_2_from_1 compares the sets for a 2nd time. is there no way to compare L1 and L2 once, but get two different lists as the output? does that make sense?

            – Anthony
            Nov 14 '18 at 23:50











          • Not sure, actually. I believe you would really need two operations to do that at the end.

            – TeeKea
            Nov 15 '18 at 0:21











          • ok, interesting. thanks anyways!!

            – Anthony
            Nov 15 '18 at 0:50

















          ok, thx, but is there no way to get the two lists of output while comparing the sets only once??

          – Anthony
          Nov 14 '18 at 23:46





          ok, thx, but is there no way to get the two lists of output while comparing the sets only once??

          – Anthony
          Nov 14 '18 at 23:46













          Is my updated answer gives you what you asked in your comment? Please let me know.

          – TeeKea
          Nov 14 '18 at 23:48





          Is my updated answer gives you what you asked in your comment? Please let me know.

          – TeeKea
          Nov 14 '18 at 23:48













          no, unfortunately that's not what i meant. what i'm trying to get at it is that dif_1_from_2 compares the sets once, then dif_2_from_1 compares the sets for a 2nd time. is there no way to compare L1 and L2 once, but get two different lists as the output? does that make sense?

          – Anthony
          Nov 14 '18 at 23:50





          no, unfortunately that's not what i meant. what i'm trying to get at it is that dif_1_from_2 compares the sets once, then dif_2_from_1 compares the sets for a 2nd time. is there no way to compare L1 and L2 once, but get two different lists as the output? does that make sense?

          – Anthony
          Nov 14 '18 at 23:50













          Not sure, actually. I believe you would really need two operations to do that at the end.

          – TeeKea
          Nov 15 '18 at 0:21





          Not sure, actually. I believe you would really need two operations to do that at the end.

          – TeeKea
          Nov 15 '18 at 0:21













          ok, interesting. thanks anyways!!

          – Anthony
          Nov 15 '18 at 0:50





          ok, interesting. thanks anyways!!

          – Anthony
          Nov 15 '18 at 0:50




















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