Pandas: Updating the Position of Values in a Row

I have the following DataFrame and its separated by 2 teams P and E and each player and their position are related data so should be kept together P1 & P1_Position.

I'm interested in:

• Viewing specific players from each team P: [('T', '3'), ('K', '2')]
  and E: [('N', '1')]

  I would like to order each row with respect to the players im
  interested in viewing, so that ('T', '3'), ('K', '2') and ('N',
'1') at at the beginning of their teams.

The resulting DataFrame would look something like:

The purpose of the ordering is to have a consistent formatting when attempting to perform a groupby operation E.G All players of interest are at the far left of each entry.

My current thought process is that I will need to use rename(columns={}) across each row using apply, but my first reaction to that is its going to be slow.

def reorderrows(row, players, enemies):
    doSomething(row)

desiredrows = desiredrows.apply(reorderrows, axis=1, players=players, enemies=enemies)

Is there a good way of enforcing that ordering given that each list P: [('T', '3'), ('K', '2')]
and E: [('N', '1')]` could include up to 3 players?

Example DataFrame:

import pandas as pd

df = pd.DataFrame({'P1': ['H', 'K', 'K', 'K'],
                       'P1_Position': [1,2,2,2],
                       'P2': ['K', 'T', 'T', 'T'],
                       'P2_Position': [2,3,3,3],
                       'P3': ['T', 'Y', 'Y', 'Y'],
                       'P3_Position': [3,4,4,4],
                       'E1': ['H', 'A', 'AK', 'AK'],
                       'E1_Position': [3,2,4,4],
                       'E2': ['N', 'H', 'H', 'A'],
                       'E2_Position': [1,3,3,2],
                       'E3': ['S', 'N', 'N', 'N'],
                       'E3_Position': [5,1,1,1],})

First create dictionary for specify P/E players and convert to DataFrame with index column for sorting:

d = {'P': [('T', '3'), ('K', '2')], 'E': [('N', '1')]}

df1 = pd.DataFrame([(k, *x) for k, v in d.items() for x in v],
                   columns=['key','_','Position']).reset_index()
print (df1)
   index key  _ Position
0      0   P  T        3
1      1   P  K        2
2      2   E  N        1

Then reshape DataFrame by split columns names by _ to MultiIndex, rename NaNs to _ and reshape by DataFrame.stack, extract column key by non numeric column from original columns names, convert Position to strings and use left join by helper DataFrame, sorting and reshape back by DataFrame.unstack with count original indices and columns names:

f = lambda x:  '_' if pd.isna(x) else x
df2 = (df.pipe(lambda x: x.set_axis(x.columns.str.split('_', expand=True), axis=1))
          .rename(columns=f)
          .stack(0)
          .reset_index()
          .assign(key=lambda x: x.level_1.str.extract('^(\D+)', expand=False),
                   Position = lambda x: x.Position.astype(str))
          .merge(df1, how='left')
          .sort_values('index')
          .assign(i = lambda x: x.groupby('key').cumcount(),
                   level_0 = lambda x: x['i'] % len(df),
                   level_1 = lambda x: x['key'] + (x['i'] // len(df) + 1).astype(str))
          .set_index(['level_0','level_1'])[['_','Position']].rename(columns={'_':''})
          .unstack()
          .pipe(lambda x: x.set_axis([f'{b}_{a}'.strip('_') for a,b in x.columns], axis=1))
          .reindex(df.columns, axis=1)
          )

print (df2)
        P1 P1_Position P2 P2_Position P3 P3_Position E1 E1_Position E2  \
level_0                                                                  
0        T           3  K           2  H           1  N           1  H   
1        T           3  K           2  Y           4  N           1  S   
2        T           3  K           2  Y           4  N           1  A   
3        T           3  K           2  Y           4  N           1  H   

        E2_Position  E3 E3_Position  
level_0                              
0                 3  AK           4  
1                 5   H           3  
2                 2  AK           4  
3                 3   A           2