Note
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Reconciling two feeds
This example shows resolve_odds. It pairs the odds of one feed to the matches of another when the two feeds name their clubs differently.
# Author: Georgios Douzas <gdouzas@icloud.com>
# Licence: MIT
import matplotlib.pyplot as plt
import pandas as pd
from sportsbet.sources import resolve_odds
The problem
This is the most dangerous step in the library. One feed says Man United. Another says Manchester United. If a
name fails to match, that match has no odds. A missing odd does not look like an error. It looks like a slightly
smaller dataset. That gives you a backtest that is clean, plausible and wrong. A dataloader reconciles the two feeds
for you. This is the machinery underneath.
identity = {'league': 'England', 'division': 1, 'year': 2025}
moment = {'event_status': 'preplay', 'event_time': pd.Timedelta(0)}
def match(home, away, **extra):
"""Build a one-row snapshot for a match."""
fixed = {'date': pd.Timestamp('2025-08-16', tz='UTC'), **identity, **moment}
return {**fixed, 'home_team': home, 'away_team': away, **extra}
stats = pd.DataFrame(
[
match('Man United', 'Arsenal'),
match('Tottenham', 'Everton'),
match('Newcastle', 'Wolves'),
match('Brighton', 'Chelsea'),
],
)
odds = pd.DataFrame(
[
match('Manchester United', 'Arsenal', provider='acme', home_win=1.8),
match('Spurs', 'Everton', provider='acme', home_win=2.1),
match('Newcastle Utd', 'Wolves', provider='acme', home_win=1.9),
match('Brighton', 'Chelsea', provider='acme', home_win=2.4),
],
)
An exact join on the names keeps only Brighton, the one club both feeds spell the same way. It silently drops the
other three:
exact = stats.merge(odds[['home_team', 'away_team']], on=['home_team', 'away_team'])
len(exact)
Out:
1
What it matches on its own
resolve_odds reads past the spelling. Man United and Manchester United are the same club. So are Newcastle
and Newcastle Utd. It pairs them without being told. It cannot place a name like Spurs for Tottenham, so it
drops that match rather than guess. Attaching one club's odds to another would be worse than a missing row.
paired = resolve_odds(stats, odds)
sorted(paired['home_team'])
Out:
['Brighton', 'Man United', 'Newcastle']
That silent drop is the danger. Three of the four matches keep their odds and the fourth does not, and nothing says so. When you know which club a leftover name means, pass it as an alias. Now every match has its odds:
paired = resolve_odds(stats, odds, aliases={'Spurs': 'Tottenham'})
sorted(paired['home_team'])
Out:
['Brighton', 'Man United', 'Newcastle', 'Tottenham']
The odds take the identity of the statistics, so the result writes Manchester United as Man United:
paired[['home_team', 'away_team', 'home_win']]
Out:
home_team away_team home_win
0 Man United Arsenal 1.8
1 Tottenham Everton 2.1
2 Newcastle Wolves 1.9
3 Brighton Chelsea 2.4
A picture of it
The gap between the first bar and the last is the danger. An exact join throws away three quarters of the odds and calls the result a dataset.
counts = {'exact name join': len(exact), 'resolved automatically': 3, 'with one alias': 4}
fig, ax = plt.subplots()
ax.bar(list(counts), list(counts.values()), color=['tab:red', 'tab:blue', 'tab:green'])
ax.set_title('Matches that keep their odds, four in all')
ax.set_ylabel('matches with odds')
Out:
Text(42.722222222222214, 0.5, 'matches with odds')
Total running time of the script: ( 0 minutes 0.130 seconds)
Download Python source code: plot_reconciliation.py