allow plan priority to be specified for plans with identical ouput 🟩

This commit is contained in:
Daniel Roth 2026-02-18 11:20:08 +00:00
parent 508f3f2859
commit b916551921

View file

@ -1,6 +1,8 @@
from collections import defaultdict
from typing import Dict, List, Optional
from sqlalchemy import Tuple
from backend.app.db.functions.recommendations_functions import (
bulk_update_plans,
get_plans_by_portfolio_id,
@ -33,20 +35,43 @@ def process_portfolio(
def choose_cheapest_relevant_plan(
plans: List[Plan], plan_priority_order: Optional[List[int]] = []
plans: List[Plan], plan_priority_order: Optional[List[int]] = None
) -> Plan:
plans_to_consider: List[Plan] = [p for p in plans if p.is_compliant] or plans
plan_priority_order = plan_priority_order or []
def plan_cost(plan: Plan) -> float:
return (
eligible_plans: List[Plan] = [plan for plan in plans if plan.is_compliant] or plans
if not eligible_plans:
raise ValueError("No plans available to choose from.")
for plan in eligible_plans:
if plan.id is None:
# This should never actually happen, but plan.id is optional to cater
# for new plans. We are only working with already persisted plans here
raise ValueError(
f"All plans must have an ID, but found a plan with no ID: {plan}"
)
min_cost: float = min(
(
plan.record.cost_of_works
if plan.record.cost_of_works is not None
else float("inf")
)
for plan in eligible_plans
)
cheapest_plan = min(plans_to_consider, key=plan_cost)
cheapest_plans: List[Plan] = [
plan
for plan in eligible_plans
if (plan.record.cost_of_works or float("inf")) == min_cost
]
return cheapest_plan
for priority_plan_id in plan_priority_order:
for plan in cheapest_plans:
if plan.id == priority_plan_id:
return plan
return cheapest_plans[0]
def _load_plans_for_portfolio(portfolio_id: int) -> List[Plan]: