CODEWITHSUNDEEP
python-3.xpandasdataframepandas-groupby
harry04
harry04

Reputation: 962

Pandas Groupby to get average of each group

Suppose I have a df -

Player       Challenge      Description
James          ABC              Desc1
Bob            ABC              Desc1
Bob            XYZ              Desc X
Bob            ABX101           Desc4
Alex           XYZ              Desc X           
Mark           ABC123           Desc 123 
Jessica        ABC123           Desc 123
Lynn           XYZ              Desc X
Bob            ABX101           Desc4
Alex           ABX101           Desc 4
Mark           ABC              Desc 1
Lynn           ABC              Desc 1
Mark           POQ              Desc 3
Mark           XYZ              Desc X
Mark           ABC              Desc 1

I can group these by Player and challenge using groupby -

df.groupby(by=['Player', 'Challenge'])

but how can I get something like a count of the challenges for each player (possibly in the next column) and then average the challenges per player?

Upvotes: 0

Views: 47

Answers (2)

NesteruS
NesteruS

Reputation: 64

You can try to use pivot:

df.pivot(index='foo', columns='bar')

Upvotes: 1

ansev
ansev

Reputation: 30940

Use:

count_challenge=df.groupby('Player').Challenge.count()
print(count_challenge)

Player
Alex       2
Bob        4
James      1
Jessica    1
Lynn       2
Mark       5
Name: Challenge, dtype: int64

If you don't want count duplicates:

count_challenge=df.drop_duplicates(['Challenge','Player']).groupby('Player').Challenge.count()
print(count_challenge)
Player
Alex       2
Bob        3
James      1
Jessica    1
Lynn       2
Mark       4
Name: Challenge, dtype: int64

Then you can calculate the mean:

count_challenge.mean()

if you want how many challenges of each type for each player

count_differents_challenge=df.groupby('Player').Challenge.value_counts()
print(count_differents_challenge)

Player   Challenge
Alex     ABX101       1
         XYZ          1
Bob      ABX101       2
         ABC          1
         XYZ          1
James    ABC          1
Jessica  ABC123       1
Lynn     ABC          1
         XYZ          1
Mark     ABC          2
         ABC123       1
         POQ          1
         XYZ          1
Name: Challenge, dtype: int64

Upvotes: 2

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