Is there a way to drop duplicated rows based on an unhashable column?how to impute missing values on numpy...
Trouble with Impersonal Passive Voice usage
Can polymorphing monsters spam their ability to effectively give themselves a massive health pool?
Number of FLOP (Floating Point Operations) for exponentiation
Am I a Rude Number?
Why do members of Congress in committee hearings ask witnesses the same question multiple times?
What makes the Forgotten Realms "forgotten"?
integral inequality of length of curve
Word or phrase for showing great skill at something without formal training in it
Dilemma of explaining to interviewer that he is the reason for declining second interview
I am on the US no-fly list. What can I do in order to be allowed on flights which go through US airspace?
How is the Incom shipyard still in business?
How do creatures spend Hit Dice after a short rest (if they can do so)?
What is better: yes / no radio, or simple checkbox?
How can I improve my fireworks photography?
Strange Sign on Lab Door
Explain the objections to these measures against human trafficking
How to prove teleportation does not violate non-cloning theorem?
Everyone is beautiful
What to do if authors don't respond to my serious concerns about their paper?
How do we succintly describe a boxing match?
Why do neural networks need so many training examples to perform?
Why don't I see the difference between two different files in insert mode in vim?
Reference on complex cobordism
If I delete my router's history can my ISP still provide it to my parents?
Is there a way to drop duplicated rows based on an unhashable column?
how to impute missing values on numpy array created by train_test_split from pandas.DataFrame?How do I convert a pandas dataframe to a 1d array?How to change a cell in Pandas dataframe with respective frequency of the cell in respective columnPopulate column based on previous row with a twistCounting the occurrence of each string in a pandas dataframe columnCreate new data frames from existing data frame based on unique column valuesHow to statistically prove that a column in a dataframe is not neededShould I use pandas get_dummies and create additional columns or use my own encoding code that keeps 1 column?How can I merge 2+ DataFrame objects without duplicating column names?operations on multiple entries in one column based on conditions meet from multiple column entries
$begingroup$
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 1 hour 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 1 hour 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 1 hour 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 1 hour ago
n1k31t4
6,1362319
asked 5 hours ago
Fabrice BOUCHARELFabrice BOUCHAREL
585
edited 1 hour ago
n1k31t4
6,1362319
asked 5 hours ago
Fabrice BOUCHARELFabrice BOUCHAREL
585
edited 1 hour ago
n1k31t4
6,1362319
edited 1 hour ago
n1k31t4
6,1362319
edited 1 hour 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 |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "557"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: false,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: null,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fdatascience.stackexchange.com%2fquestions%2f46541%2fis-there-a-way-to-drop-duplicated-rows-based-on-an-unhashable-column%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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 |
Thanks for contributing an answer to Data Science Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fdatascience.stackexchange.com%2fquestions%2f46541%2fis-there-a-way-to-drop-duplicated-rows-based-on-an-unhashable-column%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
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