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Groupby() in pandas in Python

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0

I have a dataset with the following columns:

Country, Year, Population, Suicide case, Country GDP

Problem: I Want to calculate (Suicide case / Population )*100 for each country

My Approach :

import pandas as pd
fileName = pd.read_csv("File Path")
pd.groupby("Country")

How should I extend my code for the calculation above?

python pandas

share|improve this question

edited Mar 27 at 9:03

Sreekiran

1,14011 gold badge44 silver badges2121 bronze badges

asked Mar 27 at 8:26

Soma AnchalSoma Anchal

2255 bronze badges

add a comment | 

0

I have a dataset with the following columns:

Country, Year, Population, Suicide case, Country GDP

Problem: I Want to calculate (Suicide case / Population )*100 for each country

My Approach :

import pandas as pd
fileName = pd.read_csv("File Path")
pd.groupby("Country")

How should I extend my code for the calculation above?

python pandas

share|improve this question

edited Mar 27 at 9:03

Sreekiran

1,14011 gold badge44 silver badges2121 bronze badges

asked Mar 27 at 8:26

Soma AnchalSoma Anchal

2255 bronze badges

add a comment | 

I have a dataset with the following columns:

Country, Year, Population, Suicide case, Country GDP

Problem: I Want to calculate (Suicide case / Population )*100 for each country

My Approach :

import pandas as pd
fileName = pd.read_csv("File Path")
pd.groupby("Country")

How should I extend my code for the calculation above?

python pandas

share|improve this question

edited Mar 27 at 9:03

Sreekiran

1,14011 gold badge44 silver badges2121 bronze badges

asked Mar 27 at 8:26

Soma AnchalSoma Anchal

2255 bronze badges

import pandas as pd
fileName = pd.read_csv("File Path")
pd.groupby("Country")

share|improve this question

edited Mar 27 at 9:03

Sreekiran

1,14011 gold badge44 silver badges2121 bronze badges

asked Mar 27 at 8:26

Soma AnchalSoma Anchal

2255 bronze badges

share|improve this question

edited Mar 27 at 9:03

Sreekiran

1,14011 gold badge44 silver badges2121 bronze badges

asked Mar 27 at 8:26

Soma AnchalSoma Anchal

2255 bronze badges

edited Mar 27 at 9:03

Sreekiran

1,14011 gold badge44 silver badges2121 bronze badges

edited Mar 27 at 9:03

Sreekiran

1,14011 gold badge44 silver badges2121 bronze badges

asked Mar 27 at 8:26

Soma AnchalSoma Anchal

2255 bronze badges

asked Mar 27 at 8:26

Soma AnchalSoma Anchal

2255 bronze badges

3 Answers
3

active

oldest

votes

2

Here you have with an example. May be it could be better, but this should work for you.

import pandas as pd
df = pd.DataFrame("Country":["France", "UK", "France", "UK"], 
 "Population":[1, 2, 3, 4],
 "Suicide case":[5, 3, 6, 2])
df_grouped = df.groupby("Country").sum()
(df_grouped["Suicide case"]/df_grouped["Population"])*100

share|improve this answer

answered Mar 27 at 8:58

AngeloAngelo

1811313 bronze badges

add a comment | 

2

Also a more concise one is:

df.groupby('Country').apply(lambda x: x['Suicide case'].sum()/
 float(x['Population'].sum())*100)

share|improve this answer

answered Mar 27 at 9:52

LoochieLoochie

1,03633 silver badges1111 bronze badges

add a comment | 

1

If I understood your question correctly then you can try below code to get your desired result:

fileName = fileName.groupby(['Year','Country']).sum()
fileName['New_var'] = (fileName['Suicide case']/ fileName['Population'])*100

you also need to the year in the group otherwise year-wise will also get aggregate.

share|improve this answer

answered Mar 27 at 9:32

Ghanshyam SavaliyaGhanshyam Savaliya

1631111 bronze badges

add a comment | 

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2

Here you have with an example. May be it could be better, but this should work for you.

import pandas as pd
df = pd.DataFrame("Country":["France", "UK", "France", "UK"], 
 "Population":[1, 2, 3, 4],
 "Suicide case":[5, 3, 6, 2])
df_grouped = df.groupby("Country").sum()
(df_grouped["Suicide case"]/df_grouped["Population"])*100

share|improve this answer

answered Mar 27 at 8:58

AngeloAngelo

1811313 bronze badges

add a comment | 

2

Here you have with an example. May be it could be better, but this should work for you.

import pandas as pd
df = pd.DataFrame("Country":["France", "UK", "France", "UK"], 
 "Population":[1, 2, 3, 4],
 "Suicide case":[5, 3, 6, 2])
df_grouped = df.groupby("Country").sum()
(df_grouped["Suicide case"]/df_grouped["Population"])*100

share|improve this answer

answered Mar 27 at 8:58

AngeloAngelo

1811313 bronze badges

add a comment | 

Here you have with an example. May be it could be better, but this should work for you.

import pandas as pd
df = pd.DataFrame("Country":["France", "UK", "France", "UK"], 
 "Population":[1, 2, 3, 4],
 "Suicide case":[5, 3, 6, 2])
df_grouped = df.groupby("Country").sum()
(df_grouped["Suicide case"]/df_grouped["Population"])*100

share|improve this answer

answered Mar 27 at 8:58

AngeloAngelo

1811313 bronze badges

import pandas as pd
df = pd.DataFrame("Country":["France", "UK", "France", "UK"], 
 "Population":[1, 2, 3, 4],
 "Suicide case":[5, 3, 6, 2])
df_grouped = df.groupby("Country").sum()
(df_grouped["Suicide case"]/df_grouped["Population"])*100

share|improve this answer

answered Mar 27 at 8:58

AngeloAngelo

1811313 bronze badges

answered Mar 27 at 8:58

AngeloAngelo

1811313 bronze badges

answered Mar 27 at 8:58

AngeloAngelo

1811313 bronze badges

2

Also a more concise one is:

df.groupby('Country').apply(lambda x: x['Suicide case'].sum()/
 float(x['Population'].sum())*100)

share|improve this answer

answered Mar 27 at 9:52

LoochieLoochie

1,03633 silver badges1111 bronze badges

add a comment | 

2

Also a more concise one is:

df.groupby('Country').apply(lambda x: x['Suicide case'].sum()/
 float(x['Population'].sum())*100)

share|improve this answer

answered Mar 27 at 9:52

LoochieLoochie

1,03633 silver badges1111 bronze badges

add a comment | 

Also a more concise one is:

df.groupby('Country').apply(lambda x: x['Suicide case'].sum()/
 float(x['Population'].sum())*100)

share|improve this answer

answered Mar 27 at 9:52

LoochieLoochie

1,03633 silver badges1111 bronze badges

df.groupby('Country').apply(lambda x: x['Suicide case'].sum()/
 float(x['Population'].sum())*100)

share|improve this answer

answered Mar 27 at 9:52

LoochieLoochie

1,03633 silver badges1111 bronze badges

answered Mar 27 at 9:52

LoochieLoochie

1,03633 silver badges1111 bronze badges

answered Mar 27 at 9:52

LoochieLoochie

1,03633 silver badges1111 bronze badges

1

If I understood your question correctly then you can try below code to get your desired result:

fileName = fileName.groupby(['Year','Country']).sum()
fileName['New_var'] = (fileName['Suicide case']/ fileName['Population'])*100

you also need to the year in the group otherwise year-wise will also get aggregate.

share|improve this answer

answered Mar 27 at 9:32

Ghanshyam SavaliyaGhanshyam Savaliya

1631111 bronze badges

add a comment | 

1

If I understood your question correctly then you can try below code to get your desired result:

fileName = fileName.groupby(['Year','Country']).sum()
fileName['New_var'] = (fileName['Suicide case']/ fileName['Population'])*100

you also need to the year in the group otherwise year-wise will also get aggregate.

share|improve this answer

answered Mar 27 at 9:32

Ghanshyam SavaliyaGhanshyam Savaliya

1631111 bronze badges

add a comment | 

If I understood your question correctly then you can try below code to get your desired result:

fileName = fileName.groupby(['Year','Country']).sum()
fileName['New_var'] = (fileName['Suicide case']/ fileName['Population'])*100

you also need to the year in the group otherwise year-wise will also get aggregate.

share|improve this answer

answered Mar 27 at 9:32

Ghanshyam SavaliyaGhanshyam Savaliya

1631111 bronze badges

fileName = fileName.groupby(['Year','Country']).sum()
fileName['New_var'] = (fileName['Suicide case']/ fileName['Population'])*100

share|improve this answer

answered Mar 27 at 9:32

Ghanshyam SavaliyaGhanshyam Savaliya

1631111 bronze badges

answered Mar 27 at 9:32

Ghanshyam SavaliyaGhanshyam Savaliya

1631111 bronze badges

answered Mar 27 at 9:32

Ghanshyam SavaliyaGhanshyam Savaliya

1631111 bronze badges