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修复因明日方舟新增“中坚寻访”导致抽卡模拟不可用的问题 #1418

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merged 1 commit into from
May 16, 2023
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修复因明日方舟新增“中坚寻访”导致抽卡模拟不可用的问题 #1418

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29 changes: 20 additions & 9 deletions plugins/draw_card/handles/prts_handle.py
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Original file line number Diff line number Diff line change
Expand Up @@ -27,6 +27,7 @@
class Operator(BaseData):
recruit_only: bool # 公招限定
event_only: bool # 活动获得干员
core_only:bool #中坚干员
# special_only: bool # 升变/异格干员


Expand Down Expand Up @@ -54,7 +55,7 @@ def get_card(self, add: float) -> Operator:
all_operators = [
x
for x in self.ALL_OPERATOR
if x.star == star and not any([x.limited, x.recruit_only, x.event_only])
if x.star == star and not any([x.limited, x.recruit_only, x.event_only,x.core_only])
]
acquire_operator = None

Expand Down Expand Up @@ -151,11 +152,16 @@ def _init_data(self):
Operator(
name=value["名称"],
star=int(value["星级"]),
limited="干员寻访" not in value["获取途径"],
limited="标准寻访" not in value["获取途径"] and "中坚寻访" not in value["获取途径"],
recruit_only=True
if "干员寻访" not in value["获取途径"] and "公开招募" in value["获取途径"]
if "标准寻访" not in value["获取途径"] and "中坚寻访" not in value["获取途径"] and "公开招募" in value["获取途径"]
else False,
event_only=True
if "活动获取" in value["获取途径"]
else False,
core_only=True
if "标准寻访" not in value["获取途径"] and "中坚寻访" in value["获取途径"]
else False,
event_only=True if "活动获取" in value["获取途径"] else False,
)
for key, value in self.load_data().items()
if "阿米娅" not in key
Expand Down Expand Up @@ -271,22 +277,27 @@ async def update_up_char(self):
for char in chars:
star = char.split("（")[0].count("★")
name = re.split(r"[：（]", char)[1] if "★（" not in char else re.split("）：", char)[1] # 有的括号在前面有的在后面
dual_up = False
if "\\" in name:
names = name.split("\\")
dual_up = True
elif "/" in name:
names = name.split("/")
dual_up = True
else:
names = [name] # 既有用/分割的，又有用\分割的

names = [name.replace("[限定]", "").strip() for name in names]
zoom = 1
if "权值" in char:
match = re.search(r"（在.*?以.*?(\d+).*?倍权值.*?）", char)
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Contributor Author

@RenYuan2004 RenYuan2004 May 13, 2023 •
edited
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原代码此处错误抓取了“5”（倍）并将up权值设置为了5.0，实际概率经计算后约为0.03（截止至2023.5.14，为5/52*2%）

zoom = 0.03
else:
match = re.search(r"（占.*?的.*?(\d+).*?%）", char)
zoom = 1
if match:
zoom = float(match.group(1))
zoom = zoom / 100 if zoom > 10 else zoom
if dual_up == True:
zoom = float(match.group(1))/2
else:
zoom = float(match.group(1))
zoom = zoom / 100 if zoom > 1 else zoom
for name in names:
up_chars.append(
UpChar(name=name, star=star, limited=False, zoom=zoom)
Expand Down