CODEWITHSUNDEEP
pandasnumpydataframe
gibbz00
gibbz00

Reputation: 1987

Pandas cumsum from the second element of each group

I have a dataframe that looks like the following. The rightmost column is my desired column:

Group   Value   Target_CumSum   
1        3         0     
1        2         2  
1        5         7
1        4         11
2        1         0
2        5         5
2        9         14
2        3         17

How do I perform the cumsum() from the second element of each group as opposed to the very first one?

df = pd.DataFrame({'Group': [1,1,1,1,2,2,2,2], 'Value': [3,2,5,4,1,5,9,3], 'Target_CumSum': [0,2,7,11,0,5,14,17]})
#df['MyCumSum']= df.groupby(['Group'])['Value'].cumsum()

Upvotes: 1

Views: 177

Answers (4)

Scott Boston
Scott Boston

Reputation: 153460

Just wanted to offer another solution:

df['Value'].where(df['Group'].duplicated(), 0).groupby(df.Group).cumsum()

Output:

0     0
1     2
2     7
3    11
4     0
5     5
6    14
7    17
Name: Value, dtype: int64

Upvotes: 1

Asish M.
Asish M.

Reputation: 2647

In [87]: df.groupby(['Group']).apply(lambda x: x['Value'].shift(-1).cumsum().shift().fillna(0))
Out[87]:
Group
1      0     0.0
       1     2.0
       2     7.0
       3    11.0
2      4     0.0
       5     5.0
       6    14.0
       7    17.0
Name: Value, dtype: float64

Upvotes: 0

BENY
BENY

Reputation: 323306

IIUC

g=df.groupby('Group').Value
g.cumsum()-g.transform('first')
Out[597]: 
0     0
1     2
2     7
3    11
4     0
5     5
6    14
7    17
Name: Value, dtype: int64

Upvotes: 3

epattaro
epattaro

Reputation: 2438

i dont think theres a built in function for that. so you would have to make a custom function and apply it. hope it helps.

def custom_cumsum (X):

    X[1:] =  np.cumsum(X[1:])
    X.iloc[0] = 0

    return X

df['cumsum'] = df.groupby('Group')['Value'].apply(custom_cumsum)

Upvotes: 1

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