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venture: with several open cards, work the furthest-along one first

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Co-Authored-By: Claude Fable 5.1 <noreply@anthropic.com>
Claude-Session: https://claude.ai/code/session_01UpT7eWzeLhRwuYmsSQnbRb
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claude committed 2026-10-11 14:47:22 -07:00
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commit 66c427eba4
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.github/time/venture.py
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@@ -203,6 +203,8 @@ def tick() -> int:
pitchers = env("TIME_VENTURE_PITCH", "qwen,kimi,grok").split(",") pitchers = env("TIME_VENTURE_PITCH", "qwen,kimi,grok").split(",")
voters = env("TIME_VENTURE_BOARD", "grok,kimi,qwen").split(",") voters = env("TIME_VENTURE_BOARD", "grok,kimi,qwen").split(",")
cards = venture_cards() cards = venture_cards()
order = ["new", "scouted", "pitched", "picked", "built", "launched"] # work the card that is furthest along first
cards.sort(key=lambda c: -order.index(state_of(c)["beat"]) if state_of(c)["beat"] in order else 0)
card = cards[0] if cards else (new_card(today) if env("TIME_VENTURE", "off") == "on" else None) card = cards[0] if cards else (new_card(today) if env("TIME_VENTURE", "off") == "on" else None)
if not card: if not card:
print("venture loop off (TIME_VENTURE=on to start)") print("venture loop off (TIME_VENTURE=on to start)")