Refactor payout calculation logic in app.py

- Improved the payout calculation by grouping entries based on tied positions, allowing for more efficient processing and accurate distribution of payouts.
- Replaced the previous row-wise application of the payout function with a grouped approach, enhancing performance and clarity in handling multiple tied entries.

app.py

@@ -445,10 +445,19 @@ if selected_tab == 'Data Load':
             working_df['finish'] = working_df['index']
             working_df = working_df.drop(['sorted', 'index'], axis=1)
             try:
-                working_df['payout'] = working_df.apply(
-                    lambda row: get_payout_for_position(row['finish'], st.session_state['payout_info'], row['dupes']), 
-                    axis=1
-                )
+                # Calculate payouts efficiently by processing each unique tie group once
+                working_df['payout'] = 0  # Initialize payout column
+                
+                # Group by dupes and finish position to process each tie group
+                for (dupes, finish), group in working_df.groupby(['dupes', 'finish']):
+                    if dupes == 1:
+                        # Single entry - no tie
+                        payout = get_payout_for_position(finish, st.session_state['payout_info'], 1)
+                        working_df.loc[group.index, 'payout'] = payout
+                    else:
+                        # Multiple entries tied - calculate split payout once and apply to all
+                        split_payout = get_payout_for_position(finish, st.session_state['payout_info'], dupes)
+                        working_df.loc[group.index, 'payout'] = split_payout
             except:
                 pass
         
@@ -527,10 +536,19 @@ if selected_tab == 'Data Load':
             working_df['finish'] = working_df['index']
             working_df = working_df.drop(['sorted', 'index'], axis=1)
             try:
-                working_df['payout'] = working_df.apply(
-                    lambda row: get_payout_for_position(row['finish'], st.session_state['payout_info'], row['dupes']), 
-                    axis=1
-                )
+                # Calculate payouts efficiently by processing each unique tie group once
+                working_df['payout'] = 0  # Initialize payout column
+                
+                # Group by dupes and finish position to process each tie group
+                for (dupes, finish), group in working_df.groupby(['dupes', 'finish']):
+                    if dupes == 1:
+                        # Single entry - no tie
+                        payout = get_payout_for_position(finish, st.session_state['payout_info'], 1)
+                        working_df.loc[group.index, 'payout'] = payout
+                    else:
+                        # Multiple entries tied - calculate split payout once and apply to all
+                        split_payout = get_payout_for_position(finish, st.session_state['payout_info'], dupes)
+                        working_df.loc[group.index, 'payout'] = split_payout
             except:
                 pass