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1

1

Sample DataFrame:

pd.DataFrame('Name':['John','Peter','John','John','Donald'],
 'City':['Boston','Japan','Boston','Dallas','Japan'],
 'Age':[23,31,21,21,22])

What i want is to get list of indices of all the rows which has same 'Name' and 'City' but different age, using pandas.

In this case : it should return [0,2]

pandas dataframe

share|improve this question

asked Mar 23 at 15:32

Naushad ShukoorNaushad Shukoor

216

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  What should happen when there is a 6th row John Boston 23? Do you want indices 0,2 and 5 then?
  
  – ALollz
  Mar 23 at 20:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Okay...i hate to break it now, but i'm removing all the duplicates(all values including Age) beforehand. So, the above case would'nt happen at all.
  
  – Naushad Shukoor
  Mar 25 at 10:46
  

  
  
  
  

add a comment | 

1

1

Sample DataFrame:

pd.DataFrame('Name':['John','Peter','John','John','Donald'],
 'City':['Boston','Japan','Boston','Dallas','Japan'],
 'Age':[23,31,21,21,22])

What i want is to get list of indices of all the rows which has same 'Name' and 'City' but different age, using pandas.

In this case : it should return [0,2]

pandas dataframe

share|improve this question

asked Mar 23 at 15:32

Naushad ShukoorNaushad Shukoor

216

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  What should happen when there is a 6th row John Boston 23? Do you want indices 0,2 and 5 then?
  
  – ALollz
  Mar 23 at 20:43
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  Okay...i hate to break it now, but i'm removing all the duplicates(all values including Age) beforehand. So, the above case would'nt happen at all.
  
  – Naushad Shukoor
  Mar 25 at 10:46
  

  
  
  
  

add a comment | 

Sample DataFrame:

pd.DataFrame('Name':['John','Peter','John','John','Donald'],
 'City':['Boston','Japan','Boston','Dallas','Japan'],
 'Age':[23,31,21,21,22])

What i want is to get list of indices of all the rows which has same 'Name' and 'City' but different age, using pandas.

In this case : it should return [0,2]

pandas dataframe

share|improve this question

asked Mar 23 at 15:32

Naushad ShukoorNaushad Shukoor

216

pd.DataFrame('Name':['John','Peter','John','John','Donald'],
 'City':['Boston','Japan','Boston','Dallas','Japan'],
 'Age':[23,31,21,21,22])

share|improve this question

asked Mar 23 at 15:32

Naushad ShukoorNaushad Shukoor

216

share|improve this question

asked Mar 23 at 15:32

Naushad ShukoorNaushad Shukoor

216

asked Mar 23 at 15:32

Naushad ShukoorNaushad Shukoor

216

asked Mar 23 at 15:32

Naushad ShukoorNaushad Shukoor

216

3 Answers
3

active

oldest

votes

3

Try this below:

df[df.duplicated(['Name','City'],keep=False)&~df.duplicated(keep=False)]

 Name City Age
0 John Boston 23
2 John Boston 21

EDIT: The scenario that @ALollz had pointed out can be acheived using:

df = pd.DataFrame('Name':['John','Peter','John','John','Donald', 'John'],
 'City':['Boston','Japan','Boston','Dallas','Japan', 'Boston'],
 'Age':[23,31,21,21,22, 23])
df[df.duplicated(['Name','City'],keep=False)].drop_duplicates()

Output:

 Name City Age
0 John Boston 23
2 John Boston 21

share|improve this answer

edited Mar 24 at 6:04

answered Mar 23 at 15:35

anky_91anky_91

13.1k3922

add a comment | 

1

I want is to get list of indices of all the rows which has same 'Name' and 'City' but different age

I think this is a bit ambiguous, because what if a Name-City group has a combination of entries with the same age and some that differ? Depending upon your desired output groupby + transform + nunique to filter may be required.

Sample Data:

Note, the edge case I added here, where John Boston 23 is duplicated:

import pandas as pd
df = pd.DataFrame('Name':['John','Peter','John','John','Donald', 'John'],
 'City':['Boston','Japan','Boston','Dallas','Japan', 'Boston'],
 'Age':[23,31,21,21,22, 23])

# Name City Age
#0 John Boston 23
#1 Peter Japan 31
#2 John Boston 21
#3 John Dallas 21
#4 Donald Japan 22
#5 John Boston 23

Code:

df[df.groupby(['Name', 'City']).Age.transform(pd.Series.nunique).gt(1)]

# Name City Age
#0 John Boston 23
#2 John Boston 21
#5 John Boston 23

---

With other solutions, the exact duplication may lead to an unwanted output:

df[df.duplicated(['Name','City'],keep=False)&~df.duplicated(keep=False)]
# Name City Age
#2 John Boston 21

share|improve this answer

edited Mar 23 at 20:42

answered Mar 23 at 20:35

ALollzALollz

18.4k51840

add a comment | 

0

Another method could be by using groupby():

df[df.groupby(['Name', 'City']).transform(len)['Age']>1]

or may be in two steps as using duplicated():

df =df.set_index('Age')
df[df.duplicated(['Name', 'City'], keep = False)].reset_index()

share|improve this answer

edited Mar 23 at 16:27

answered Mar 23 at 16:18

LoochieLoochie

984311

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  this doesn't give the desired results. also, i'm skeptical on how the groupby would fare on >~300 columns
  
  – Naushad Shukoor
  Mar 23 at 17:18
  

  
  
  
  

add a comment | 

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3

Try this below:

df[df.duplicated(['Name','City'],keep=False)&~df.duplicated(keep=False)]

 Name City Age
0 John Boston 23
2 John Boston 21

EDIT: The scenario that @ALollz had pointed out can be acheived using:

df = pd.DataFrame('Name':['John','Peter','John','John','Donald', 'John'],
 'City':['Boston','Japan','Boston','Dallas','Japan', 'Boston'],
 'Age':[23,31,21,21,22, 23])
df[df.duplicated(['Name','City'],keep=False)].drop_duplicates()

Output:

 Name City Age
0 John Boston 23
2 John Boston 21

share|improve this answer

edited Mar 24 at 6:04

answered Mar 23 at 15:35

anky_91anky_91

13.1k3922

add a comment | 

3

Try this below:

df[df.duplicated(['Name','City'],keep=False)&~df.duplicated(keep=False)]

 Name City Age
0 John Boston 23
2 John Boston 21

EDIT: The scenario that @ALollz had pointed out can be acheived using:

df = pd.DataFrame('Name':['John','Peter','John','John','Donald', 'John'],
 'City':['Boston','Japan','Boston','Dallas','Japan', 'Boston'],
 'Age':[23,31,21,21,22, 23])
df[df.duplicated(['Name','City'],keep=False)].drop_duplicates()

Output:

 Name City Age
0 John Boston 23
2 John Boston 21

share|improve this answer

edited Mar 24 at 6:04

answered Mar 23 at 15:35

anky_91anky_91

13.1k3922

add a comment | 

Try this below:

df[df.duplicated(['Name','City'],keep=False)&~df.duplicated(keep=False)]

 Name City Age
0 John Boston 23
2 John Boston 21

EDIT: The scenario that @ALollz had pointed out can be acheived using:

df = pd.DataFrame('Name':['John','Peter','John','John','Donald', 'John'],
 'City':['Boston','Japan','Boston','Dallas','Japan', 'Boston'],
 'Age':[23,31,21,21,22, 23])
df[df.duplicated(['Name','City'],keep=False)].drop_duplicates()

Output:

 Name City Age
0 John Boston 23
2 John Boston 21

share|improve this answer

edited Mar 24 at 6:04

answered Mar 23 at 15:35

anky_91anky_91

13.1k3922

df[df.duplicated(['Name','City'],keep=False)&~df.duplicated(keep=False)]

 Name City Age
0 John Boston 23
2 John Boston 21

df = pd.DataFrame('Name':['John','Peter','John','John','Donald', 'John'],
 'City':['Boston','Japan','Boston','Dallas','Japan', 'Boston'],
 'Age':[23,31,21,21,22, 23])
df[df.duplicated(['Name','City'],keep=False)].drop_duplicates()

 Name City Age
0 John Boston 23
2 John Boston 21

share|improve this answer

edited Mar 24 at 6:04

answered Mar 23 at 15:35

anky_91anky_91

13.1k3922

answered Mar 23 at 15:35

anky_91anky_91

13.1k3922

answered Mar 23 at 15:35

anky_91anky_91

13.1k3922

1

I want is to get list of indices of all the rows which has same 'Name' and 'City' but different age

I think this is a bit ambiguous, because what if a Name-City group has a combination of entries with the same age and some that differ? Depending upon your desired output groupby + transform + nunique to filter may be required.

Sample Data:

Note, the edge case I added here, where John Boston 23 is duplicated:

import pandas as pd
df = pd.DataFrame('Name':['John','Peter','John','John','Donald', 'John'],
 'City':['Boston','Japan','Boston','Dallas','Japan', 'Boston'],
 'Age':[23,31,21,21,22, 23])

# Name City Age
#0 John Boston 23
#1 Peter Japan 31
#2 John Boston 21
#3 John Dallas 21
#4 Donald Japan 22
#5 John Boston 23

Code:

df[df.groupby(['Name', 'City']).Age.transform(pd.Series.nunique).gt(1)]

# Name City Age
#0 John Boston 23
#2 John Boston 21
#5 John Boston 23

---

With other solutions, the exact duplication may lead to an unwanted output:

df[df.duplicated(['Name','City'],keep=False)&~df.duplicated(keep=False)]
# Name City Age
#2 John Boston 21

share|improve this answer

edited Mar 23 at 20:42

answered Mar 23 at 20:35

ALollzALollz

18.4k51840

add a comment | 

1

I want is to get list of indices of all the rows which has same 'Name' and 'City' but different age

I think this is a bit ambiguous, because what if a Name-City group has a combination of entries with the same age and some that differ? Depending upon your desired output groupby + transform + nunique to filter may be required.

Sample Data:

Note, the edge case I added here, where John Boston 23 is duplicated:

import pandas as pd
df = pd.DataFrame('Name':['John','Peter','John','John','Donald', 'John'],
 'City':['Boston','Japan','Boston','Dallas','Japan', 'Boston'],
 'Age':[23,31,21,21,22, 23])

# Name City Age
#0 John Boston 23
#1 Peter Japan 31
#2 John Boston 21
#3 John Dallas 21
#4 Donald Japan 22
#5 John Boston 23

Code:

df[df.groupby(['Name', 'City']).Age.transform(pd.Series.nunique).gt(1)]

# Name City Age
#0 John Boston 23
#2 John Boston 21
#5 John Boston 23

---

With other solutions, the exact duplication may lead to an unwanted output:

df[df.duplicated(['Name','City'],keep=False)&~df.duplicated(keep=False)]
# Name City Age
#2 John Boston 21

share|improve this answer

edited Mar 23 at 20:42

answered Mar 23 at 20:35

ALollzALollz

18.4k51840

add a comment | 

I want is to get list of indices of all the rows which has same 'Name' and 'City' but different age

I think this is a bit ambiguous, because what if a Name-City group has a combination of entries with the same age and some that differ? Depending upon your desired output groupby + transform + nunique to filter may be required.

Sample Data:

Note, the edge case I added here, where John Boston 23 is duplicated:

import pandas as pd
df = pd.DataFrame('Name':['John','Peter','John','John','Donald', 'John'],
 'City':['Boston','Japan','Boston','Dallas','Japan', 'Boston'],
 'Age':[23,31,21,21,22, 23])

# Name City Age
#0 John Boston 23
#1 Peter Japan 31
#2 John Boston 21
#3 John Dallas 21
#4 Donald Japan 22
#5 John Boston 23

Code:

df[df.groupby(['Name', 'City']).Age.transform(pd.Series.nunique).gt(1)]

# Name City Age
#0 John Boston 23
#2 John Boston 21
#5 John Boston 23

---

With other solutions, the exact duplication may lead to an unwanted output:

df[df.duplicated(['Name','City'],keep=False)&~df.duplicated(keep=False)]
# Name City Age
#2 John Boston 21

share|improve this answer

edited Mar 23 at 20:42

answered Mar 23 at 20:35

ALollzALollz

18.4k51840

import pandas as pd
df = pd.DataFrame('Name':['John','Peter','John','John','Donald', 'John'],
 'City':['Boston','Japan','Boston','Dallas','Japan', 'Boston'],
 'Age':[23,31,21,21,22, 23])

# Name City Age
#0 John Boston 23
#1 Peter Japan 31
#2 John Boston 21
#3 John Dallas 21
#4 Donald Japan 22
#5 John Boston 23

df[df.groupby(['Name', 'City']).Age.transform(pd.Series.nunique).gt(1)]

# Name City Age
#0 John Boston 23
#2 John Boston 21
#5 John Boston 23

df[df.duplicated(['Name','City'],keep=False)&~df.duplicated(keep=False)]
# Name City Age
#2 John Boston 21

share|improve this answer

edited Mar 23 at 20:42

answered Mar 23 at 20:35

ALollzALollz

18.4k51840

answered Mar 23 at 20:35

ALollzALollz

18.4k51840

answered Mar 23 at 20:35

ALollzALollz

18.4k51840

0

Another method could be by using groupby():

df[df.groupby(['Name', 'City']).transform(len)['Age']>1]

or may be in two steps as using duplicated():

df =df.set_index('Age')
df[df.duplicated(['Name', 'City'], keep = False)].reset_index()

share|improve this answer

edited Mar 23 at 16:27

answered Mar 23 at 16:18

LoochieLoochie

984311

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  this doesn't give the desired results. also, i'm skeptical on how the groupby would fare on >~300 columns
  
  – Naushad Shukoor
  Mar 23 at 17:18
  

  
  
  
  

add a comment | 

0

Another method could be by using groupby():

df[df.groupby(['Name', 'City']).transform(len)['Age']>1]

or may be in two steps as using duplicated():

df =df.set_index('Age')
df[df.duplicated(['Name', 'City'], keep = False)].reset_index()

share|improve this answer

edited Mar 23 at 16:27

answered Mar 23 at 16:18

LoochieLoochie

984311

• 
  
  
  

  
  

  
  
  
  
  
  

  
  
  this doesn't give the desired results. also, i'm skeptical on how the groupby would fare on >~300 columns
  
  – Naushad Shukoor
  Mar 23 at 17:18
  

  
  
  
  

add a comment | 

Another method could be by using groupby():

df[df.groupby(['Name', 'City']).transform(len)['Age']>1]

or may be in two steps as using duplicated():

df =df.set_index('Age')
df[df.duplicated(['Name', 'City'], keep = False)].reset_index()

share|improve this answer

edited Mar 23 at 16:27

answered Mar 23 at 16:18

LoochieLoochie

984311

df[df.groupby(['Name', 'City']).transform(len)['Age']>1]

df =df.set_index('Age')
df[df.duplicated(['Name', 'City'], keep = False)].reset_index()

share|improve this answer

edited Mar 23 at 16:27

answered Mar 23 at 16:18

LoochieLoochie

984311

answered Mar 23 at 16:18

LoochieLoochie

984311

answered Mar 23 at 16:18

LoochieLoochie

984311