Removing unwanted characters and editing Column names in pandas
I have pandas df with certain column names. The column names are as below,
u'Kanta/City', u'Aluepaso/Regional Level', u'Akue/District', u'Seotukartakudi/Map code', u'k�/Age', u'2015', u'2016', u'2017', u'2018'.
What I would like to do is, rename the columns in one line of code as below,
'City', 'Regional_Level', 'District', 'Map_Code', 'Age', '2015', '2016', '2017', '2018'.
Is there any efficient way of doing so (with lambda function)?
python pandas lambda multiple-columns rename
edited Nov 22 '18 at 11:05
Mohit Motwani
1,2571422
asked Nov 22 '18 at 10:31
Kripalu SarKripalu Sar
327
add a comment |
I have pandas df with certain column names. The column names are as below,
u'Kanta/City', u'Aluepaso/Regional Level', u'Akue/District', u'Seotukartakudi/Map code', u'k�/Age', u'2015', u'2016', u'2017', u'2018'.
What I would like to do is, rename the columns in one line of code as below,
'City', 'Regional_Level', 'District', 'Map_Code', 'Age', '2015', '2016', '2017', '2018'.
Is there any efficient way of doing so (with lambda function)?
python pandas lambda multiple-columns rename
edited Nov 22 '18 at 11:05
Mohit Motwani
1,2571422
asked Nov 22 '18 at 10:31
Kripalu SarKripalu Sar
327
@ Kripalu Sar what you should do when someone answers your question must learn!
– pygo
Nov 23 '18 at 19:22
add a comment |
I have pandas df with certain column names. The column names are as below,
u'Kanta/City', u'Aluepaso/Regional Level', u'Akue/District', u'Seotukartakudi/Map code', u'k�/Age', u'2015', u'2016', u'2017', u'2018'.
What I would like to do is, rename the columns in one line of code as below,
'City', 'Regional_Level', 'District', 'Map_Code', 'Age', '2015', '2016', '2017', '2018'.
Is there any efficient way of doing so (with lambda function)?
python pandas lambda multiple-columns rename
edited Nov 22 '18 at 11:05
Mohit Motwani
1,2571422
asked Nov 22 '18 at 10:31
Kripalu SarKripalu Sar
327
I have pandas df with certain column names. The column names are as below,
u'Kanta/City', u'Aluepaso/Regional Level', u'Akue/District', u'Seotukartakudi/Map code', u'k�/Age', u'2015', u'2016', u'2017', u'2018'.
What I would like to do is, rename the columns in one line of code as below,
'City', 'Regional_Level', 'District', 'Map_Code', 'Age', '2015', '2016', '2017', '2018'.
Is there any efficient way of doing so (with lambda function)?
python pandas lambda multiple-columns rename
python pandas lambda multiple-columns rename
edited Nov 22 '18 at 11:05
Mohit Motwani
1,2571422
asked Nov 22 '18 at 10:31
Kripalu SarKripalu Sar
327
edited Nov 22 '18 at 11:05
Mohit Motwani
1,2571422
asked Nov 22 '18 at 10:31
Kripalu SarKripalu Sar
327
edited Nov 22 '18 at 11:05
Mohit Motwani
1,2571422
edited Nov 22 '18 at 11:05
Mohit Motwani
1,2571422
edited Nov 22 '18 at 11:05
Mohit Motwani
1,2571422
1,2571422
asked Nov 22 '18 at 10:31
Kripalu SarKripalu Sar
327
asked Nov 22 '18 at 10:31
Kripalu SarKripalu Sar
327
asked Nov 22 '18 at 10:31
Kripalu SarKripalu Sar
327
327
@ Kripalu Sar what you should do when someone answers your question must learn!
– pygo
Nov 23 '18 at 19:22
add a comment |
@ Kripalu Sar what you should do when someone answers your question must learn!
– pygo
Nov 23 '18 at 19:22
@ Kripalu Sar what you should do when someone answers your question must learn!
– pygo
Nov 23 '18 at 19:22
@ Kripalu Sar what you should do when someone answers your question must learn!
– pygo
Nov 23 '18 at 19:22
add a comment |
2 Answers
2
active
oldest
votes
Using lambda:
df.rename(columns=lambda x: x.split('/')[1].replace(' ','_') if '/' in x else x, inplace= True)
df.columns
> Index(['City', 'Regional_Level', 'District', 'Map_code', 'Age', '2015', '2016',
'2017', '2018'],
dtype='object')
answered Nov 22 '18 at 10:35
Mohit MotwaniMohit Motwani
1,2571422
add a comment |
Simplest will be with using replace using regex.
>>> df
Empty DataFrame
Columns: [Kanta/City, Aluepaso/Regional Level, Akue/District, Seotukartakudi/Map code, k�/Age, 2015, 2016, 2017, 2018]
Index:
>>> df.columns.str.replace('.*[\/]', '')
Index(['City', 'Regional Level', 'District', 'Map code', 'Age', '2015', '2016',
'2017', '2018'],
dtype='object')
Regex explanation:
.*matches any character (except for line terminators)
*Quantifier — Matches between zero and unlimited times, as many times as possible, giving back as needed (greedy)
Match a single character present in the list below
[\/]
\matches the characterliterally (case sensitive)
/matches the character/literally (case sensitive)
answered Nov 22 '18 at 11:27
pygopygo
2,5131619
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
Using lambda:
df.rename(columns=lambda x: x.split('/')[1].replace(' ','_') if '/' in x else x, inplace= True)
df.columns
> Index(['City', 'Regional_Level', 'District', 'Map_code', 'Age', '2015', '2016',
'2017', '2018'],
dtype='object')
answered Nov 22 '18 at 10:35
Mohit MotwaniMohit Motwani
1,2571422
add a comment |
Using lambda:
df.rename(columns=lambda x: x.split('/')[1].replace(' ','_') if '/' in x else x, inplace= True)
df.columns
> Index(['City', 'Regional_Level', 'District', 'Map_code', 'Age', '2015', '2016',
'2017', '2018'],
dtype='object')
answered Nov 22 '18 at 10:35
Mohit MotwaniMohit Motwani
1,2571422
add a comment |
Using lambda:
df.rename(columns=lambda x: x.split('/')[1].replace(' ','_') if '/' in x else x, inplace= True)
df.columns
> Index(['City', 'Regional_Level', 'District', 'Map_code', 'Age', '2015', '2016',
'2017', '2018'],
dtype='object')
answered Nov 22 '18 at 10:35
Mohit MotwaniMohit Motwani
1,2571422
Using lambda:
df.rename(columns=lambda x: x.split('/')[1].replace(' ','_') if '/' in x else x, inplace= True)
df.columns
> Index(['City', 'Regional_Level', 'District', 'Map_code', 'Age', '2015', '2016',
'2017', '2018'],
dtype='object')
answered Nov 22 '18 at 10:35
Mohit MotwaniMohit Motwani
1,2571422
answered Nov 22 '18 at 10:35
Mohit MotwaniMohit Motwani
1,2571422
answered Nov 22 '18 at 10:35
Mohit MotwaniMohit Motwani
1,2571422
answered Nov 22 '18 at 10:35
Mohit MotwaniMohit Motwani
1,2571422
1,2571422
add a comment |
add a comment |
Simplest will be with using replace using regex.
>>> df
Empty DataFrame
Columns: [Kanta/City, Aluepaso/Regional Level, Akue/District, Seotukartakudi/Map code, k�/Age, 2015, 2016, 2017, 2018]
Index:
>>> df.columns.str.replace('.*[\/]', '')
Index(['City', 'Regional Level', 'District', 'Map code', 'Age', '2015', '2016',
'2017', '2018'],
dtype='object')
Regex explanation:
.*matches any character (except for line terminators)
*Quantifier — Matches between zero and unlimited times, as many times as possible, giving back as needed (greedy)
Match a single character present in the list below
[\/]
\matches the characterliterally (case sensitive)
/matches the character/literally (case sensitive)
answered Nov 22 '18 at 11:27
pygopygo
2,5131619
add a comment |
Simplest will be with using replace using regex.
>>> df
Empty DataFrame
Columns: [Kanta/City, Aluepaso/Regional Level, Akue/District, Seotukartakudi/Map code, k�/Age, 2015, 2016, 2017, 2018]
Index:
>>> df.columns.str.replace('.*[\/]', '')
Index(['City', 'Regional Level', 'District', 'Map code', 'Age', '2015', '2016',
'2017', '2018'],
dtype='object')
Regex explanation:
.*matches any character (except for line terminators)
*Quantifier — Matches between zero and unlimited times, as many times as possible, giving back as needed (greedy)
Match a single character present in the list below
[\/]
\matches the characterliterally (case sensitive)
/matches the character/literally (case sensitive)
answered Nov 22 '18 at 11:27
pygopygo
2,5131619
add a comment |
Simplest will be with using replace using regex.
>>> df
Empty DataFrame
Columns: [Kanta/City, Aluepaso/Regional Level, Akue/District, Seotukartakudi/Map code, k�/Age, 2015, 2016, 2017, 2018]
Index:
>>> df.columns.str.replace('.*[\/]', '')
Index(['City', 'Regional Level', 'District', 'Map code', 'Age', '2015', '2016',
'2017', '2018'],
dtype='object')
Regex explanation:
.*matches any character (except for line terminators)
*Quantifier — Matches between zero and unlimited times, as many times as possible, giving back as needed (greedy)
Match a single character present in the list below
[\/]
\matches the characterliterally (case sensitive)
/matches the character/literally (case sensitive)
answered Nov 22 '18 at 11:27
pygopygo
2,5131619
Simplest will be with using replace using regex.
>>> df
Empty DataFrame
Columns: [Kanta/City, Aluepaso/Regional Level, Akue/District, Seotukartakudi/Map code, k�/Age, 2015, 2016, 2017, 2018]
Index:
>>> df.columns.str.replace('.*[\/]', '')
Index(['City', 'Regional Level', 'District', 'Map code', 'Age', '2015', '2016',
'2017', '2018'],
dtype='object')
Regex explanation:
.*matches any character (except for line terminators)
*Quantifier — Matches between zero and unlimited times, as many times as possible, giving back as needed (greedy)
Match a single character present in the list below
[\/]
\matches the characterliterally (case sensitive)
/matches the character/literally (case sensitive)
answered Nov 22 '18 at 11:27
pygopygo
2,5131619
edited Nov 22 '18 at 11:32
answered Nov 22 '18 at 11:27
pygopygo
2,5131619
answered Nov 22 '18 at 11:27
pygopygo
2,5131619
answered Nov 22 '18 at 11:27
pygopygo
2,5131619
2,5131619
add a comment |
add a comment |
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@ Kripalu Sar what you should do when someone answers your question must learn!
– pygo
Nov 23 '18 at 19:22