How to create column with mean of rows before the date of current row

I need get mean of rows, but with date before the current row date.

I have this code, but is it takes a long time in datasets with 50k rows:

import pandas as pd

data = {
  'id': [1,2,3,4,5],
  'home_goals': [1,0,3,1,2],
  'away_goals': [1,1,2,0,1],
  'home_name': ['a','b','a','b','a'],
  'away_name': ['b','a','b','a','b'],
  'date': ['2020-01-01', '2020-01-02', '2020-01-03', '2020-01-04', '2020-01-05'],
}

df = pd.DataFrame(data=data)
for i, row in df.iterrows():
  rows_before_date = df[df['date'] < row['date']]
  home_matches = rows_before_date[rows_before_date['home_name'] == row['home_name']]
  away_matches = rows_before_date[rows_before_date['away_name'] == row['away_name']]
  if len(home_matches) == 0 or len(away_matches) == 0: continue

  df.loc[i, 'home_mean'] = home_matches['home_goals'].sum() / len(home_matches)
  df.loc[i, 'away_mean'] = away_matches['away_goals'].sum() / len(away_matches)

I wanted to know if it is possible to make a more optimized and readable code.

>Solution :

Assuming prior sorting of the dates, you can use a shifted expanding.mean per group:

df['home_mean'] = (df.groupby('home_name')['home_goals']
                     .apply(lambda s: s.expanding().mean().shift())
                   )
df['away_mean'] = (df.groupby('away_name')['away_goals']
                     .apply(lambda s: s.expanding().mean().shift())
                  )

output:

   id  home_goals  away_goals home_name away_name        date  home_mean  away_mean
0   1           1           1         a         b  2020-01-01        NaN        NaN
1   2           0           1         b         a  2020-01-02        NaN        NaN
2   3           3           2         a         b  2020-01-03        1.0        1.0
3   4           1           0         b         a  2020-01-04        0.0        1.0
4   5           2           1         a         b  2020-01-05        2.0        1.5