Mixing
Python pandas remove duplicate rows that have a column value “NaN”
0
The need to rows that have NaN values in them but are also duplicates. For example this table:
A B C
0 foo 2 3
1 foo nan nan
2 foo 1 4
3 bar nan nan
4 foo nan nan
Should become this:
A B C
0 foo 2 3
2 foo 1 4
3 bar nan nan
How can i do that?
python pandas duplicates
asked Nov 21 '18 at 13:26
Lame FanelloLame Fanello
7310
add a comment |
0
The need to rows that have NaN values in them but are also duplicates. For example this table:
A B C
0 foo 2 3
1 foo nan nan
2 foo 1 4
3 bar nan nan
4 foo nan nan
Should become this:
A B C
0 foo 2 3
2 foo 1 4
3 bar nan nan
How can i do that?
python pandas duplicates
asked Nov 21 '18 at 13:26
Lame FanelloLame Fanello
7310
drop_duplicates doesnt drop different column values and drop_na drops every column with nan values. I need to drop every duplicate of a specific column where the row has a NaN value.
– Lame Fanello
Nov 21 '18 at 13:31
add a comment |
0
0
0
The need to rows that have NaN values in them but are also duplicates. For example this table:
A B C
0 foo 2 3
1 foo nan nan
2 foo 1 4
3 bar nan nan
4 foo nan nan
Should become this:
A B C
0 foo 2 3
2 foo 1 4
3 bar nan nan
How can i do that?
python pandas duplicates
asked Nov 21 '18 at 13:26
Lame FanelloLame Fanello
7310
The need to rows that have NaN values in them but are also duplicates. For example this table:
A B C
0 foo 2 3
1 foo nan nan
2 foo 1 4
3 bar nan nan
4 foo nan nan
Should become this:
A B C
0 foo 2 3
2 foo 1 4
3 bar nan nan
How can i do that?
python pandas duplicates
python pandas duplicates
asked Nov 21 '18 at 13:26
Lame FanelloLame Fanello
7310
asked Nov 21 '18 at 13:26
Lame FanelloLame Fanello
7310
asked Nov 21 '18 at 13:26
Lame FanelloLame Fanello
7310
asked Nov 21 '18 at 13:26
Lame FanelloLame Fanello
7310
asked Nov 21 '18 at 13:26
Lame FanelloLame Fanello
7310
7310
drop_duplicates doesnt drop different column values and drop_na drops every column with nan values. I need to drop every duplicate of a specific column where the row has a NaN value.
– Lame Fanello
Nov 21 '18 at 13:31
add a comment |
drop_duplicates doesnt drop different column values and drop_na drops every column with nan values. I need to drop every duplicate of a specific column where the row has a NaN value.
– Lame Fanello
Nov 21 '18 at 13:31
drop_duplicates doesnt drop different column values and drop_na drops every column with nan values. I need to drop every duplicate of a specific column where the row has a NaN value.
– Lame Fanello
Nov 21 '18 at 13:31
drop_duplicates doesnt drop different column values and drop_na drops every column with nan values. I need to drop every duplicate of a specific column where the row has a NaN value.
– Lame Fanello
Nov 21 '18 at 13:31
add a comment |
2 Answers
2
active
oldest
votes
2
Use boolean indexing:
df = df[~df['A'].duplicated(keep=False) | df[['B','C']].notnull().any(axis=1)]
print (df)
A B C
0 foo 2.0 3.0
2 foo 1.0 4.0
3 bar NaN NaN
Explanation:
Test column A for not duplicates - duplicated with ~ for invert boolean mask:
print (~df['A'].duplicated(keep=False))
0 False
1 False
2 False
3 True
4 False
Name: A, dtype: bool
Check non missing values in B,C columns:
print (df[['B','C']].notnull())
B C
0 True True
1 False False
2 True True
3 False False
4 False False
And then at least one True per row with DataFrame.any:
print (df[['B','C']].notnull().any(axis=1))
0 True
1 False
2 True
3 False
4 False
dtype: bool
Chain together by | for bitwise OR:
print (~df['A'].duplicated(keep=False) | df[['B','C']].notnull().any(axis=1))
0 True
1 False
2 True
3 True
4 False
dtype: bool
answered Nov 21 '18 at 13:31
jezraeljezrael
335k25281357
1
To the downvoter: what's wrong with this answer?
– timgeb
Nov 21 '18 at 14:02
add a comment |
1
Slightly different to jezrael's solution:
>>> df
A B C
0 foo 2.0 3.0
1 foo NaN NaN
2 foo 1.0 4.0
3 bar NaN NaN
4 foo NaN NaN
>>>
>>> df.drop(index=df[df.duplicated(keep=False)].isnull().any(1).index)
A B C
0 foo 2.0 3.0
2 foo 1.0 4.0
3 bar NaN NaN
Steps:
>>> df.duplicated(keep=False)
0 False
1 True
2 False
3 False
4 True
dtype: bool
>>>
>>> df[df.duplicated(keep=False)]
A B C
1 foo NaN NaN
4 foo NaN NaN
>>>
>>> df[df.duplicated(keep=False)].isnull()
A B C
1 False True True
4 False True True
>>>
>>> df[df.duplicated(keep=False)].isnull().any(1).index
Int64Index([1, 4], dtype='int64')
answered Nov 21 '18 at 13:40
timgebtimgeb
50.8k116493
If you downvoted this answer, I would appreciate if you left a comment such that I can improve the answer.
– timgeb
Nov 21 '18 at 14:01
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
StackExchange.using("externalEditor", function () {
StackExchange.using("snippets", function () {
StackExchange.snippets.init();
});
});
}, "code-snippets");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "1"
};
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: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
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
});
}
});
draft saved
draft discarded
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%2fstackoverflow.com%2fquestions%2f53413077%2fpython-pandas-remove-duplicate-rows-that-have-a-column-value-nan%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
2
Use boolean indexing:
df = df[~df['A'].duplicated(keep=False) | df[['B','C']].notnull().any(axis=1)]
print (df)
A B C
0 foo 2.0 3.0
2 foo 1.0 4.0
3 bar NaN NaN
Explanation:
Test column A for not duplicates - duplicated with ~ for invert boolean mask:
print (~df['A'].duplicated(keep=False))
0 False
1 False
2 False
3 True
4 False
Name: A, dtype: bool
Check non missing values in B,C columns:
print (df[['B','C']].notnull())
B C
0 True True
1 False False
2 True True
3 False False
4 False False
And then at least one True per row with DataFrame.any:
print (df[['B','C']].notnull().any(axis=1))
0 True
1 False
2 True
3 False
4 False
dtype: bool
Chain together by | for bitwise OR:
print (~df['A'].duplicated(keep=False) | df[['B','C']].notnull().any(axis=1))
0 True
1 False
2 True
3 True
4 False
dtype: bool
answered Nov 21 '18 at 13:31
jezraeljezrael
335k25281357
1
To the downvoter: what's wrong with this answer?
– timgeb
Nov 21 '18 at 14:02
add a comment |
2
Use boolean indexing:
df = df[~df['A'].duplicated(keep=False) | df[['B','C']].notnull().any(axis=1)]
print (df)
A B C
0 foo 2.0 3.0
2 foo 1.0 4.0
3 bar NaN NaN
Explanation:
Test column A for not duplicates - duplicated with ~ for invert boolean mask:
print (~df['A'].duplicated(keep=False))
0 False
1 False
2 False
3 True
4 False
Name: A, dtype: bool
Check non missing values in B,C columns:
print (df[['B','C']].notnull())
B C
0 True True
1 False False
2 True True
3 False False
4 False False
And then at least one True per row with DataFrame.any:
print (df[['B','C']].notnull().any(axis=1))
0 True
1 False
2 True
3 False
4 False
dtype: bool
Chain together by | for bitwise OR:
print (~df['A'].duplicated(keep=False) | df[['B','C']].notnull().any(axis=1))
0 True
1 False
2 True
3 True
4 False
dtype: bool
answered Nov 21 '18 at 13:31
jezraeljezrael
335k25281357
1
To the downvoter: what's wrong with this answer?
– timgeb
Nov 21 '18 at 14:02
add a comment |
2
2
2
Use boolean indexing:
df = df[~df['A'].duplicated(keep=False) | df[['B','C']].notnull().any(axis=1)]
print (df)
A B C
0 foo 2.0 3.0
2 foo 1.0 4.0
3 bar NaN NaN
Explanation:
Test column A for not duplicates - duplicated with ~ for invert boolean mask:
print (~df['A'].duplicated(keep=False))
0 False
1 False
2 False
3 True
4 False
Name: A, dtype: bool
Check non missing values in B,C columns:
print (df[['B','C']].notnull())
B C
0 True True
1 False False
2 True True
3 False False
4 False False
And then at least one True per row with DataFrame.any:
print (df[['B','C']].notnull().any(axis=1))
0 True
1 False
2 True
3 False
4 False
dtype: bool
Chain together by | for bitwise OR:
print (~df['A'].duplicated(keep=False) | df[['B','C']].notnull().any(axis=1))
0 True
1 False
2 True
3 True
4 False
dtype: bool
answered Nov 21 '18 at 13:31
jezraeljezrael
335k25281357
Use boolean indexing:
df = df[~df['A'].duplicated(keep=False) | df[['B','C']].notnull().any(axis=1)]
print (df)
A B C
0 foo 2.0 3.0
2 foo 1.0 4.0
3 bar NaN NaN
Explanation:
Test column A for not duplicates - duplicated with ~ for invert boolean mask:
print (~df['A'].duplicated(keep=False))
0 False
1 False
2 False
3 True
4 False
Name: A, dtype: bool
Check non missing values in B,C columns:
print (df[['B','C']].notnull())
B C
0 True True
1 False False
2 True True
3 False False
4 False False
And then at least one True per row with DataFrame.any:
print (df[['B','C']].notnull().any(axis=1))
0 True
1 False
2 True
3 False
4 False
dtype: bool
Chain together by | for bitwise OR:
print (~df['A'].duplicated(keep=False) | df[['B','C']].notnull().any(axis=1))
0 True
1 False
2 True
3 True
4 False
dtype: bool
answered Nov 21 '18 at 13:31
jezraeljezrael
335k25281357
edited Nov 21 '18 at 13:37
answered Nov 21 '18 at 13:31
jezraeljezrael
335k25281357
answered Nov 21 '18 at 13:31
jezraeljezrael
335k25281357
answered Nov 21 '18 at 13:31
jezraeljezrael
335k25281357
335k25281357
1
To the downvoter: what's wrong with this answer?
– timgeb
Nov 21 '18 at 14:02
add a comment |
1
To the downvoter: what's wrong with this answer?
– timgeb
Nov 21 '18 at 14:02
1
1
To the downvoter: what's wrong with this answer?
– timgeb
Nov 21 '18 at 14:02
To the downvoter: what's wrong with this answer?
– timgeb
Nov 21 '18 at 14:02
add a comment |
1
Slightly different to jezrael's solution:
>>> df
A B C
0 foo 2.0 3.0
1 foo NaN NaN
2 foo 1.0 4.0
3 bar NaN NaN
4 foo NaN NaN
>>>
>>> df.drop(index=df[df.duplicated(keep=False)].isnull().any(1).index)
A B C
0 foo 2.0 3.0
2 foo 1.0 4.0
3 bar NaN NaN
Steps:
>>> df.duplicated(keep=False)
0 False
1 True
2 False
3 False
4 True
dtype: bool
>>>
>>> df[df.duplicated(keep=False)]
A B C
1 foo NaN NaN
4 foo NaN NaN
>>>
>>> df[df.duplicated(keep=False)].isnull()
A B C
1 False True True
4 False True True
>>>
>>> df[df.duplicated(keep=False)].isnull().any(1).index
Int64Index([1, 4], dtype='int64')
answered Nov 21 '18 at 13:40
timgebtimgeb
50.8k116493
If you downvoted this answer, I would appreciate if you left a comment such that I can improve the answer.
– timgeb
Nov 21 '18 at 14:01
add a comment |
1
Slightly different to jezrael's solution:
>>> df
A B C
0 foo 2.0 3.0
1 foo NaN NaN
2 foo 1.0 4.0
3 bar NaN NaN
4 foo NaN NaN
>>>
>>> df.drop(index=df[df.duplicated(keep=False)].isnull().any(1).index)
A B C
0 foo 2.0 3.0
2 foo 1.0 4.0
3 bar NaN NaN
Steps:
>>> df.duplicated(keep=False)
0 False
1 True
2 False
3 False
4 True
dtype: bool
>>>
>>> df[df.duplicated(keep=False)]
A B C
1 foo NaN NaN
4 foo NaN NaN
>>>
>>> df[df.duplicated(keep=False)].isnull()
A B C
1 False True True
4 False True True
>>>
>>> df[df.duplicated(keep=False)].isnull().any(1).index
Int64Index([1, 4], dtype='int64')
answered Nov 21 '18 at 13:40
timgebtimgeb
50.8k116493
If you downvoted this answer, I would appreciate if you left a comment such that I can improve the answer.
– timgeb
Nov 21 '18 at 14:01
add a comment |
1
1
1
Slightly different to jezrael's solution:
>>> df
A B C
0 foo 2.0 3.0
1 foo NaN NaN
2 foo 1.0 4.0
3 bar NaN NaN
4 foo NaN NaN
>>>
>>> df.drop(index=df[df.duplicated(keep=False)].isnull().any(1).index)
A B C
0 foo 2.0 3.0
2 foo 1.0 4.0
3 bar NaN NaN
Steps:
>>> df.duplicated(keep=False)
0 False
1 True
2 False
3 False
4 True
dtype: bool
>>>
>>> df[df.duplicated(keep=False)]
A B C
1 foo NaN NaN
4 foo NaN NaN
>>>
>>> df[df.duplicated(keep=False)].isnull()
A B C
1 False True True
4 False True True
>>>
>>> df[df.duplicated(keep=False)].isnull().any(1).index
Int64Index([1, 4], dtype='int64')
answered Nov 21 '18 at 13:40
timgebtimgeb
50.8k116493
Slightly different to jezrael's solution:
>>> df
A B C
0 foo 2.0 3.0
1 foo NaN NaN
2 foo 1.0 4.0
3 bar NaN NaN
4 foo NaN NaN
>>>
>>> df.drop(index=df[df.duplicated(keep=False)].isnull().any(1).index)
A B C
0 foo 2.0 3.0
2 foo 1.0 4.0
3 bar NaN NaN
Steps:
>>> df.duplicated(keep=False)
0 False
1 True
2 False
3 False
4 True
dtype: bool
>>>
>>> df[df.duplicated(keep=False)]
A B C
1 foo NaN NaN
4 foo NaN NaN
>>>
>>> df[df.duplicated(keep=False)].isnull()
A B C
1 False True True
4 False True True
>>>
>>> df[df.duplicated(keep=False)].isnull().any(1).index
Int64Index([1, 4], dtype='int64')
answered Nov 21 '18 at 13:40
timgebtimgeb
50.8k116493
answered Nov 21 '18 at 13:40
timgebtimgeb
50.8k116493
answered Nov 21 '18 at 13:40
timgebtimgeb
50.8k116493
answered Nov 21 '18 at 13:40
timgebtimgeb
50.8k116493
50.8k116493
If you downvoted this answer, I would appreciate if you left a comment such that I can improve the answer.
– timgeb
Nov 21 '18 at 14:01
add a comment |
If you downvoted this answer, I would appreciate if you left a comment such that I can improve the answer.
– timgeb
Nov 21 '18 at 14:01
If you downvoted this answer, I would appreciate if you left a comment such that I can improve the answer.
– timgeb
Nov 21 '18 at 14:01
If you downvoted this answer, I would appreciate if you left a comment such that I can improve the answer.
– timgeb
Nov 21 '18 at 14:01
add a comment |
draft saved
draft discarded
Thanks for contributing an answer to Stack Overflow!
- 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.
To learn more, see our tips on writing great answers.
draft saved
draft discarded
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%2fstackoverflow.com%2fquestions%2f53413077%2fpython-pandas-remove-duplicate-rows-that-have-a-column-value-nan%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
drop_duplicates doesnt drop different column values and drop_na drops every column with nan values. I need to drop every duplicate of a specific column where the row has a NaN value.
– Lame Fanello
Nov 21 '18 at 13:31