index-1 on itterrow makes new row at the end

Question:

Based on this table

pernr plans mnth jum_mnth
123 000 1 NaN
123 001 3 NaN
123 001 6 NaN
789 002 10 NaN
789 003 2 NaN
789 003 2 NaN
789 002 2 NaN

I want to set ‘jum_mnth’ from ‘mnth’. ‘jum_mnth’ have value if:

  1. its last row from same plans
  2. last row from same pernr

so i tried:

for index, row in que.iterrows():
if row['pernr'] != nipp:
    que_cop.at[index-1, 'jum_mnth'] = mon
    nipp = row['pernr']
    plan = row['plans']
    mon = row['mnth']
else:
    if row['plans'] == plan:
        mon = mon + row['mnth']
    else:
        que_cop.at[index-1, 'jum_mnth'] = mon
        print(str(nipp),plan,str(mon))
        plan = row['plans']
        mon = row['mnth']
if index == que_cop.index[-2]:
            que_cop.at[index, 'jum_mnth'] = mon

but it resulting new row ( index -1) at the last like this:

pernr plans mnth jum_mnth
123 000 1 1.0
123 001 3 NaN
123 001 6 9.0
789 002 10 10.0
789 003 2 NaN
789 003 2 4.0
789 002 2 NaN
NaN NaN NaN 0.0

and the last row didnt have jum_mnth (it should have jum_mnth)
expected:

pernr plans mnth jum_mnth
123 000 1 1
123 001 3 NaN
123 001 6 9
789 002 10 10
789 003 2 NaN
789 003 2 4
789 002 2 2

so what happened?
any help i would appreciate it.

Asked By: nabilahnran

||
Source

Answers:

You can use:

grp = (df[['pernr', 'plans']].ne(df[['pernr', 'plans']].shift())
        .any(axis=1).cumsum()
       )

g = df.groupby(grp)['mnth']

df['jum_mnth'] = g.transform('sum').where(g.cumcount(ascending=False).eq(0))

Output:

   pernr plans  mnth  jum_mnth
0    123   000     1       1.0
1    123   001     3       NaN
2    123   001     6       9.0
3    789   002    10      10.0
4    789   003     2       NaN
5    789   003     2       4.0
6    789   002     2       2.0
Answered By: mozway