Is there a way to drop duplicated rows based on an unhashable column?how to impute missing values on numpy...
Does the "particle exchange" operator have any validity?
Trouble with Impersonal Passive Voice usage
Unwarranted claim of higher degree of accuracy in zircon geochronology
Is there a better way to make this?
Why did Jodrell Bank assist the Soviet Union to collect data from their spacecraft in the mid 1960's?
Why did the villain in the first Men in Black movie care about Earth's Cockroaches?
Closed form for these polynomials?
Using only 1s, make 29 with the minimum number of digits
A starship is travelling at 0.9c and collides with a small rock. Will it leave a clean hole through, or will more happen?
The vanishing of sum of coefficients: symmetric polynomials
How should I handle players who ignore the session zero agreement?
Eww, those bytes are gross
How to deal with an incendiary email that was recalled
Number of FLOP (Floating Point Operations) for exponentiation
Program that converts a number to a letter of the alphabet
Does Windows 10's telemetry include sending *.doc files if Word crashed?
Am I a Rude Number?
Can pricing be copyrighted?
Can a person refuse a presidential pardon?
How can I improve my fireworks photography?
Do authors have to be politically correct in article-writing?
Tikzing a circled star
It took me a lot of time to make this, pls like. (YouTube Comments #1)
Strange Sign on Lab Door
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