Pandas dataframe，如何按单列分组并将总和应用于多列并添加新的总和列？

Question

This should be an easy one, but somehow I couldn't find a solution that works.

I have a pandas dataframe which looks like this:

Slno   Date         col2   col3   col4   col5  col6
0     01/02/20      2      1      2      5     d
1     03/02/20      5      1      2      4     g
2     04/02/20      5      1      2      5     h
3     05/02/20      4      1      2      6     e
4     08/02/20      8      1      2      5     g
5     05/02/20      8      1      2      8     r

**I want to group by Date and get the sum() of col2, col3, col4, col5 row wise as new column Total

Here is what I tried:

df_new[Total] = df.groupby(['Date', sort=False])["col2", "col3", col4", "col5].sum(axis = 1)

It gives ValueError: Wrong number of items passed 4, placement implies 1

also I have tried

         df = (df.groupby(['Date'])
         .agg(Total=('ConfirmedIndianNational', 'ConfirmedForeignNational', 'Cured', 'Deaths', 'sum'))
         .reset_index())

It gives TypeError: aggregate() missing 1 required positional argument: 'arg'

I am new python, searched all the possible solution but of new use.

Answer 1

You can set Date as index then take sum of the columns on axis=1, then groupby level=0 and transform sum

df['Total'] = (df.set_index('Date')[["col2", "col3","col4", "col5"]].sum(1)
           .groupby(level=0).transform('sum').to_numpy())

---

print(df)

   Slno      Date  col2  col3  col4  col5 col6  Total
0     0  01/02/20     2     1     2     5    d     10
1     1  03/02/20     5     1     2     4    g     12
2     2  04/02/20     5     1     2     5    h     13
3     3  05/02/20     4     1     2     6    e     32 # this is duplicated per group
4     4  08/02/20     8     1     2     5    g     16
5     5  05/02/20     8     1     2     8    r     32 # this is duplicated per group

Answer 2

It is not clear is you have multiple values per date. If that is the case, your group by should aggregate that column by what ever function you want and the perform the sum of the rows with the aggregation.

For example you want the max value for each column per date:

max_df = df.groupby('Date')(['Date'])["col2", "col3", "col4", "col5"].max()

then:

max_df.loc[:,'sum_cols'] = max_df[["col2", "col3", "col4", "col5"]].sum(axis = 1)

If you only have one row per date, you can do:

df.loc[:,'sum_cols'] = df[["col2", "col3", "col4", "col5"]].sum(axis = 1)

Answer 3

Why don't you do a simple:

df[Total] = df.groupby(['Date').col1.sum()+df.groupby(['Date').col2.sum()+df.groupby(['Date').col3.sum()+df.groupby(['Date').col4.sum()

It should be ok.