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Is there a way to drop duplicated rows based on an unhashable column?
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$begingroup$
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 ?
python pandas dataframe
edited 59 mins ago
n1k31t4
6,1362319
asked 5 hours ago
Fabrice BOUCHARELFabrice BOUCHAREL
585
$endgroup$
add a comment |
$begingroup$
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 ?
python pandas dataframe
edited 59 mins ago
n1k31t4
6,1362319
asked 5 hours ago
Fabrice BOUCHARELFabrice BOUCHAREL
585
$endgroup$
$begingroup$
I assume it is a typo - but there isn't actually a duplicate in row z anyway because onebalso has a space:'b '.
$endgroup$
– n1k31t4
4 hours ago
$begingroup$
right. I've made a correction. thx.
$endgroup$
– Fabrice BOUCHAREL
3 hours ago
add a comment |
$begingroup$
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 ?
python pandas dataframe
edited 59 mins ago
n1k31t4
6,1362319
asked 5 hours ago
Fabrice BOUCHARELFabrice BOUCHAREL
585
$endgroup$
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
python pandas dataframe
edited 59 mins ago
n1k31t4
6,1362319
asked 5 hours ago
Fabrice BOUCHARELFabrice BOUCHAREL
585
edited 59 mins ago
n1k31t4
6,1362319
asked 5 hours ago
Fabrice BOUCHARELFabrice BOUCHAREL
585
edited 59 mins ago
n1k31t4
6,1362319
edited 59 mins ago
n1k31t4
6,1362319
edited 59 mins ago
n1k31t4
6,1362319
6,1362319
asked 5 hours ago
Fabrice BOUCHARELFabrice BOUCHAREL
585
asked 5 hours ago
Fabrice BOUCHARELFabrice BOUCHAREL
585
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 onebalso has a space:'b '.
$endgroup$
– n1k31t4
4 hours ago
$begingroup$
right. I've made a correction. thx.
$endgroup$
– Fabrice BOUCHAREL
3 hours ago
add a comment |
$begingroup$
I assume it is a typo - but there isn't actually a duplicate in row z anyway because onebalso has a space:'b '.
$endgroup$
– n1k31t4
4 hours ago
$begingroup$
right. I've made a correction. thx.
$endgroup$
– Fabrice BOUCHAREL
3 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
3 hours ago
$begingroup$
right. I've made a correction. thx.
$endgroup$
– Fabrice BOUCHAREL
3 hours ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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}
answered 4 hours ago
n1k31t4n1k31t4
6,1362319
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
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votes
active
oldest
votes
$begingroup$
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}
answered 4 hours ago
n1k31t4n1k31t4
6,1362319
$endgroup$
add a comment |
$begingroup$
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}
answered 4 hours ago
n1k31t4n1k31t4
6,1362319
$endgroup$
add a comment |
$begingroup$
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}
answered 4 hours ago
n1k31t4n1k31t4
6,1362319
$endgroup$
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}
answered 4 hours ago
n1k31t4n1k31t4
6,1362319
answered 4 hours ago
n1k31t4n1k31t4
6,1362319
answered 4 hours ago
n1k31t4n1k31t4
6,1362319
answered 4 hours ago
n1k31t4n1k31t4
6,1362319
6,1362319
add a comment |
add a comment |
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$begingroup$
I assume it is a typo - but there isn't actually a duplicate in row z anyway because one
balso has a space:'b '.$endgroup$
– n1k31t4
4 hours ago
$begingroup$
right. I've made a correction. thx.
$endgroup$
– Fabrice BOUCHAREL
3 hours ago