sum of specific rows pandas dataframe

Question

I have a dataframe and would like to add sums of specific rows into this dataframe. So for example I have

df = pd.DataFrame({'prod':['a','a','a','b','b','b','c','c','c'], 'attribute':['x','y','z','x','y','z','x','y','z'],
                  'number1':[1,2,2,3,4,3,5,1,1], 'number2':[10,2,3,3,1,2,3,1,1], 'number3':[1,4,3,5,7,1,3,0,1]})

How can I add for each prod a, b and c the sum of number 1/2/3 of the attributes y and z as a new row? So it looks like this

    prod    attribute   number1 number2 number3
0   a       x           1       10      1
1   a       y           2       2       4
2   a       z           2       3       3
3   a       sum_yz      4       5       7
4   b       x           3       3       5
5   b       y           4       1       7
6   b       z           3       2       1
7   b       sum_yz      7       3       8
8   c       x           5       3       3
9   c       y           1       1       0
10  c       z           1       1       1
11  c       sum_yz      2       2       1

halfer · Accepted Answer · 2021-04-09 23:27:34Z

You need concat with a condtional groupby.

You can filter the dataframe by using isin and add a new column with assign.

First let's select the target cols to sum.

cols = [col for col in df.columns if 'number' in col]

df1 = pd.concat(
    [
        df,
        df[df["attribute"].isin(["y", "z"])]
        .groupby("prod")[cols]
        .sum()
        .assign(attribute="sum_yz")
        .reset_index(),
    ]
).sort_values("prod")


print(df1)

  prod attribute  number1  number2  number3
0    a         x        1       10        1
1    a         y        2        2        4
2    a         z        2        3        3
0    a    sum_yz        4        5        7
3    b         x        3        3        5
4    b         y        4        1        7
5    b         z        3        2        1
1    b    sum_yz        7        3        8
6    c         x        5        3        3
7    c         y        1        1        0
8    c         z        1        1        1
2    c    sum_yz        2        2        1

Liam Galbraith · Accepted Answer · 2020-12-07 09:31:31Z

2

You could make a separate DataFrane and append it back to the original DataFrame, something like this (this code is untested):

# Filter to the desired attributes
sum_yz = df[df['attribute'].isin(['y', 'z'])]
# Set the new 'attribute' value
sum_yz['attribute'] = 'sum_yz'
# Group by and sum
sum_yz = sum_yz.groupby(['prod', 'attribute']).sum().reset_index()

# Add it the end of the data frame
df = pd.concat([df, sum_yz])

answered Dec 7, 2020 at 9:31

Liam Galbraith

3041 silver badge6 bronze badges

Comments

Niels Henkens · Accepted Answer · 2020-12-07 09:31:54Z

2

You can use df.groupby() and then combine the groupby-outcome with the original df

# Create groupby DataFrame
df_grp = df[df['attribute'].isin(['y', 'z'])].groupby(['prod']).sum()
df_grp.reset_index(inplace=True)
df_grp['attribute'] = 'sum_yz'

# Combine with original dataframe
df = pd.concat([df, df_grp])

answered Dec 7, 2020 at 9:31

Niels Henkens

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Comments

sammywemmy · Accepted Answer · 2020-12-07 09:38:25Z

You can use pandas concat after the groupby :

result = df.groupby(["prod", df.attribute.isin(["y", "z"])]).sum().loc[:, True, :]
result = result.reset_index()
result.insert(1, "attribute", "sum_yz")
pd.concat([df, result]).sort_values("prod", ignore_index=True)

  prod  attribute   number1 number2 number3
0   a      x           1    10      1
1   a      y           2    2       4
2   a      z           2    3       3
3   a     sum_yz       4    5       7
4   b      x           3    3       5
5   b      y           4    1       7
6   b      z           3    2       1
7   b    sum_yz        7    3       8
8   c      x           5    3       3
9   c      y           1    1       0
10  c      z           1    1       1
11  c    sum_yz        2    2       1

jezrael · Accepted Answer · 2020-12-07 09:45:06Z

One idea with dictionaries, but slowier if large DataFrame:

def f(x):
    d =  x[x['attribute'].isin(['y','z'])].sum()
    d1 = {'prod': x.name, 'attribute':'sum_yz'}
    x = x.append({**d, **d1},ignore_index=True)
    return x

df = df.groupby('prod', sort=False).apply(f).reset_index(drop=True)
print (df)
   prod attribute  number1  number2  number3
0     a         x        1       10        1
1     a         y        2        2        4
2     a         z        2        3        3
3     a    sum_yz        4        5        7
4     b         x        3        3        5
5     b         y        4        1        7
6     b         z        3        2        1
7     b    sum_yz        7        3        8
8     c         x        5        3        3
9     c         y        1        1        0
10    c         z        1        1        1
11    c    sum_yz        2        2        1

Or if possible sorting values of product first filter by Series.isin, aggregate sum, add to original with replace NaN by DataFrame.fillna and last sorting by DataFrame.sort_values with ignore_index for default index:

df = (df.append(df[df['attribute'].isin(['y', 'z'])]
                   .groupby('prod', as_index=False)
                   .sum()
         ).fillna({'attribute': 'sum_yz'})
          .sort_values('prod', ignore_index=True))
         
print (df)
   prod attribute  number1  number2  number3
0     a         x        1       10        1
1     a         y        2        2        4
2     a         z        2        3        3
3     a    sum_yz        4        5        7
4     b         x        3        3        5
5     b         y        4        1        7
6     b         z        3        2        1
7     b    sum_yz        7        3        8
8     c         x        5        3        3
9     c         y        1        1        0
10    c         z        1        1        1
11    c    sum_yz        2        2        1

gilf0yle · Accepted Answer · 2020-12-07 10:34:11Z

0

this is simple and works fine

dr=df[df['attribute']!='x'].groupby('prod').sum().reset_index()
dr['attribute']='sum_yz'
result=pd.concat([df,dr]).sort_values('prod')

answered Dec 7, 2020 at 10:34

gilf0yle

1,1026 silver badges13 bronze badges

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