Is there a way to drop duplicated rows based on an unhashable column?












1












$begingroup$


i have a pandas dataframe df with one column z filled with set values



i want to drop duplicated rows where 2 rows are considered duplicated version of one another when they have same column z values ( which are sets ).



import pandas as pd

lnks = [ ( 'a' , 'b' , { 'a' , 'b' } ) , ( 'b' , 'c' , { 'b' , 'c' } ) , ( 'b' , 'a' , { 'a' , 'b' } ) ]
lbls = [ 'x' , 'y' , 'z' ]
df = pd.DataFrame.from_records( lnks , columns = lbls )


Trying to drop duplicated rows based on column z values :



df.drop_duplicates( subset = 'z' , keep='first')


And i get the error message :



TypeError: unhashable type: 'set'


Is there a way to drop duplicated rows based on a unhashable typed column ?










share|improve this question











$endgroup$












  • $begingroup$
    I assume it is a typo - but there isn't actually a duplicate in row z anyway because one b also has a space: 'b '.
    $endgroup$
    – n1k31t4
    4 hours ago










  • $begingroup$
    right. I've made a correction. thx.
    $endgroup$
    – Fabrice BOUCHAREL
    4 hours ago
















1












$begingroup$


i have a pandas dataframe df with one column z filled with set values



i want to drop duplicated rows where 2 rows are considered duplicated version of one another when they have same column z values ( which are sets ).



import pandas as pd

lnks = [ ( 'a' , 'b' , { 'a' , 'b' } ) , ( 'b' , 'c' , { 'b' , 'c' } ) , ( 'b' , 'a' , { 'a' , 'b' } ) ]
lbls = [ 'x' , 'y' , 'z' ]
df = pd.DataFrame.from_records( lnks , columns = lbls )


Trying to drop duplicated rows based on column z values :



df.drop_duplicates( subset = 'z' , keep='first')


And i get the error message :



TypeError: unhashable type: 'set'


Is there a way to drop duplicated rows based on a unhashable typed column ?










share|improve this question











$endgroup$












  • $begingroup$
    I assume it is a typo - but there isn't actually a duplicate in row z anyway because one b also has a space: 'b '.
    $endgroup$
    – n1k31t4
    4 hours ago










  • $begingroup$
    right. I've made a correction. thx.
    $endgroup$
    – Fabrice BOUCHAREL
    4 hours ago














1












1








1





$begingroup$


i have a pandas dataframe df with one column z filled with set values



i want to drop duplicated rows where 2 rows are considered duplicated version of one another when they have same column z values ( which are sets ).



import pandas as pd

lnks = [ ( 'a' , 'b' , { 'a' , 'b' } ) , ( 'b' , 'c' , { 'b' , 'c' } ) , ( 'b' , 'a' , { 'a' , 'b' } ) ]
lbls = [ 'x' , 'y' , 'z' ]
df = pd.DataFrame.from_records( lnks , columns = lbls )


Trying to drop duplicated rows based on column z values :



df.drop_duplicates( subset = 'z' , keep='first')


And i get the error message :



TypeError: unhashable type: 'set'


Is there a way to drop duplicated rows based on a unhashable typed column ?










share|improve this question











$endgroup$




i have a pandas dataframe df with one column z filled with set values



i want to drop duplicated rows where 2 rows are considered duplicated version of one another when they have same column z values ( which are sets ).



import pandas as pd

lnks = [ ( 'a' , 'b' , { 'a' , 'b' } ) , ( 'b' , 'c' , { 'b' , 'c' } ) , ( 'b' , 'a' , { 'a' , 'b' } ) ]
lbls = [ 'x' , 'y' , 'z' ]
df = pd.DataFrame.from_records( lnks , columns = lbls )


Trying to drop duplicated rows based on column z values :



df.drop_duplicates( subset = 'z' , keep='first')


And i get the error message :



TypeError: unhashable type: 'set'


Is there a way to drop duplicated rows based on a unhashable typed column ?







python pandas dataframe






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 1 hour ago









n1k31t4

6,1362319




6,1362319










asked 5 hours ago









Fabrice BOUCHARELFabrice BOUCHAREL

585




585












  • $begingroup$
    I assume it is a typo - but there isn't actually a duplicate in row z anyway because one b also has a space: 'b '.
    $endgroup$
    – n1k31t4
    4 hours ago










  • $begingroup$
    right. I've made a correction. thx.
    $endgroup$
    – Fabrice BOUCHAREL
    4 hours ago


















  • $begingroup$
    I assume it is a typo - but there isn't actually a duplicate in row z anyway because one b also has a space: 'b '.
    $endgroup$
    – n1k31t4
    4 hours ago










  • $begingroup$
    right. I've made a correction. thx.
    $endgroup$
    – Fabrice BOUCHAREL
    4 hours ago
















$begingroup$
I assume it is a typo - but there isn't actually a duplicate in row z anyway because one b also has a space: 'b '.
$endgroup$
– n1k31t4
4 hours ago




$begingroup$
I assume it is a typo - but there isn't actually a duplicate in row z anyway because one b also has a space: 'b '.
$endgroup$
– n1k31t4
4 hours ago












$begingroup$
right. I've made a correction. thx.
$endgroup$
– Fabrice BOUCHAREL
4 hours ago




$begingroup$
right. I've made a correction. thx.
$endgroup$
– Fabrice BOUCHAREL
4 hours ago










1 Answer
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oldest

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2












$begingroup$

It is true that a set is not hashable (it cannot be used as a key in a hashmap a.k.a a dictionary). So what you can do is to just convert the column to a type that is hashable - I would go for a tuple.



I made a new column that is just the "z" column you had, converted to tuples. Then you can use the same method you tried to, on the new column:



In [1] : import pandas as pd 
...:
...: lnks = [ ( 'a' , 'b' , { 'a' , 'b' } ) , ( 'b' , 'c' , { 'b' , 'c' } )
...: , ( 'b' , 'a' , { 'a' , 'b' } ) ]
...: lbls = [ 'x' , 'y' , 'z' ]
...: df = pd.DataFrame.from_records( lnks , columns = lbls)

In [2]: df["z_tuple"] = df.z.apply(lambda x: tuple(x))

In [3]: df.drop_duplicates(subset="z_tuple", keep="first")
Out[3]:
x y z z_tuple
0 a b {b, a} (b, a)
1 b c {c, b} (c, b)


The apply method lets you apply a function to each item in a column, and then returns the values as a new column (a Pandas Series object). This lets you assign it back to the original DataFrame as a new column, as I did.



You can also remove the "z_tuple" column then if you no longer want it:



In [4] : df.drop("z_tuple", axis=1, inplace=True)                               

In [5] : df
Out[5] :
x y z
0 a b {b, a}
1 b c {c, b}
2 b a {b, a}





share|improve this answer









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    active

    oldest

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    $begingroup$

    It is true that a set is not hashable (it cannot be used as a key in a hashmap a.k.a a dictionary). So what you can do is to just convert the column to a type that is hashable - I would go for a tuple.



    I made a new column that is just the "z" column you had, converted to tuples. Then you can use the same method you tried to, on the new column:



    In [1] : import pandas as pd 
    ...:
    ...: lnks = [ ( 'a' , 'b' , { 'a' , 'b' } ) , ( 'b' , 'c' , { 'b' , 'c' } )
    ...: , ( 'b' , 'a' , { 'a' , 'b' } ) ]
    ...: lbls = [ 'x' , 'y' , 'z' ]
    ...: df = pd.DataFrame.from_records( lnks , columns = lbls)

    In [2]: df["z_tuple"] = df.z.apply(lambda x: tuple(x))

    In [3]: df.drop_duplicates(subset="z_tuple", keep="first")
    Out[3]:
    x y z z_tuple
    0 a b {b, a} (b, a)
    1 b c {c, b} (c, b)


    The apply method lets you apply a function to each item in a column, and then returns the values as a new column (a Pandas Series object). This lets you assign it back to the original DataFrame as a new column, as I did.



    You can also remove the "z_tuple" column then if you no longer want it:



    In [4] : df.drop("z_tuple", axis=1, inplace=True)                               

    In [5] : df
    Out[5] :
    x y z
    0 a b {b, a}
    1 b c {c, b}
    2 b a {b, a}





    share|improve this answer









    $endgroup$


















      2












      $begingroup$

      It is true that a set is not hashable (it cannot be used as a key in a hashmap a.k.a a dictionary). So what you can do is to just convert the column to a type that is hashable - I would go for a tuple.



      I made a new column that is just the "z" column you had, converted to tuples. Then you can use the same method you tried to, on the new column:



      In [1] : import pandas as pd 
      ...:
      ...: lnks = [ ( 'a' , 'b' , { 'a' , 'b' } ) , ( 'b' , 'c' , { 'b' , 'c' } )
      ...: , ( 'b' , 'a' , { 'a' , 'b' } ) ]
      ...: lbls = [ 'x' , 'y' , 'z' ]
      ...: df = pd.DataFrame.from_records( lnks , columns = lbls)

      In [2]: df["z_tuple"] = df.z.apply(lambda x: tuple(x))

      In [3]: df.drop_duplicates(subset="z_tuple", keep="first")
      Out[3]:
      x y z z_tuple
      0 a b {b, a} (b, a)
      1 b c {c, b} (c, b)


      The apply method lets you apply a function to each item in a column, and then returns the values as a new column (a Pandas Series object). This lets you assign it back to the original DataFrame as a new column, as I did.



      You can also remove the "z_tuple" column then if you no longer want it:



      In [4] : df.drop("z_tuple", axis=1, inplace=True)                               

      In [5] : df
      Out[5] :
      x y z
      0 a b {b, a}
      1 b c {c, b}
      2 b a {b, a}





      share|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        It is true that a set is not hashable (it cannot be used as a key in a hashmap a.k.a a dictionary). So what you can do is to just convert the column to a type that is hashable - I would go for a tuple.



        I made a new column that is just the "z" column you had, converted to tuples. Then you can use the same method you tried to, on the new column:



        In [1] : import pandas as pd 
        ...:
        ...: lnks = [ ( 'a' , 'b' , { 'a' , 'b' } ) , ( 'b' , 'c' , { 'b' , 'c' } )
        ...: , ( 'b' , 'a' , { 'a' , 'b' } ) ]
        ...: lbls = [ 'x' , 'y' , 'z' ]
        ...: df = pd.DataFrame.from_records( lnks , columns = lbls)

        In [2]: df["z_tuple"] = df.z.apply(lambda x: tuple(x))

        In [3]: df.drop_duplicates(subset="z_tuple", keep="first")
        Out[3]:
        x y z z_tuple
        0 a b {b, a} (b, a)
        1 b c {c, b} (c, b)


        The apply method lets you apply a function to each item in a column, and then returns the values as a new column (a Pandas Series object). This lets you assign it back to the original DataFrame as a new column, as I did.



        You can also remove the "z_tuple" column then if you no longer want it:



        In [4] : df.drop("z_tuple", axis=1, inplace=True)                               

        In [5] : df
        Out[5] :
        x y z
        0 a b {b, a}
        1 b c {c, b}
        2 b a {b, a}





        share|improve this answer









        $endgroup$



        It is true that a set is not hashable (it cannot be used as a key in a hashmap a.k.a a dictionary). So what you can do is to just convert the column to a type that is hashable - I would go for a tuple.



        I made a new column that is just the "z" column you had, converted to tuples. Then you can use the same method you tried to, on the new column:



        In [1] : import pandas as pd 
        ...:
        ...: lnks = [ ( 'a' , 'b' , { 'a' , 'b' } ) , ( 'b' , 'c' , { 'b' , 'c' } )
        ...: , ( 'b' , 'a' , { 'a' , 'b' } ) ]
        ...: lbls = [ 'x' , 'y' , 'z' ]
        ...: df = pd.DataFrame.from_records( lnks , columns = lbls)

        In [2]: df["z_tuple"] = df.z.apply(lambda x: tuple(x))

        In [3]: df.drop_duplicates(subset="z_tuple", keep="first")
        Out[3]:
        x y z z_tuple
        0 a b {b, a} (b, a)
        1 b c {c, b} (c, b)


        The apply method lets you apply a function to each item in a column, and then returns the values as a new column (a Pandas Series object). This lets you assign it back to the original DataFrame as a new column, as I did.



        You can also remove the "z_tuple" column then if you no longer want it:



        In [4] : df.drop("z_tuple", axis=1, inplace=True)                               

        In [5] : df
        Out[5] :
        x y z
        0 a b {b, a}
        1 b c {c, b}
        2 b a {b, a}






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered 4 hours ago









        n1k31t4n1k31t4

        6,1362319




        6,1362319






























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