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csj
2026-09-17 11:12:57 +08:00
parent 45b6f693d6
commit ced5db5473
11 changed files with 550 additions and 3 deletions
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native-engine/match_explanation.py
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from domain import common, date_distance, merchant_similarity
def check(title, points, maximum, detail, issue=''):
return dict(title=title, points=points, maximum=maximum, detail=detail,
issue=issue if points < maximum else '')
def values(items, field):
return '、'.join(dict.fromkeys(value for item in items for value in item['ocr'][field])) or '未识别'
def missing_fields(invoices, payments, field):
missing = []
if not any(item['ocr'][field] for item in invoices):
missing.append('发票')
if not any(item['ocr'][field] for item in payments):
missing.append('付款截图')
return '和'.join(missing)
def date_check(invoices, payments, maximum=15, tiered=True):
distance = date_distance(invoices, payments)
missing = missing_fields(invoices, payments, 'dates')
detail = f"发票日期：{values(invoices, 'dates')}；付款日期：{values(payments, 'dates')}。"
if missing:
return check('日期接近', 0, maximum, detail + f'{missing}未识别到日期，无法比较；请打开原图核对。', '日期未识别')
partial = 8 if maximum == 15 else 5
points = (maximum if distance <= 3 else partial if distance <= 7 else 0) if tiered else (maximum if distance <= 7 else 0)
rule = f'相差 0–3 天得 {maximum} 分，4–7 天得 {partial} 分，超过 7 天不得分。' if tiered else f'相差不超过 7 天得 {maximum} 分，否则不得分。'
return check('日期接近', points, maximum, detail + f'最近相差 {distance} 天。{rule}开票和付款时间可能不同，请核对实际业务日期。',
f'日期相差 {distance} 天')
def single_checks(match, hundred_point=False):
invoice, payment = match['invoices'][0], match['payments'][0]
invoices, payments = [invoice], [payment]
shared = common(invoice, payment, 'amounts')
amount_max, merchant_max, date_max, order_max, unique_max = (40, 20, 10, 15, 15) if hundred_point else (55, 25, 15, 20, 20)
checks = [check('共同金额', amount_max if shared else 0, amount_max,
'识别到共同金额：' + '、'.join('¥' + amount for amount in shared) +
'。此项比较 OCR 提取的金额，不保证它就是价税合计或实付总额，请以原图为准。', '未识别到共同金额')]
similarity = merchant_similarity(invoice, payment)
merchant_points = int(merchant_max * similarity + .5) if similarity >= .35 else 0
missing = missing_fields(invoices, payments, 'merchants')
detail = f"发票商户：{values(invoices, 'merchants')}；付款商户：{values(payments, 'merchants')}。"
if missing:
detail += f'{missing}未识别到商户，无法比较。请核对原图中的销售方和收款方。'
issue = '商户未识别'
else:
detail += f'最高名称相似度 {int(similarity * 100 + .5)}%；达到 35% 才按相似度 × {merchant_max} 分四舍五入计分。名称简称、收款平台或 OCR 误识别可能影响结果，请核对原图。'
issue = '商户未充分印证'
checks.append(check('商户相似', merchant_points, merchant_max, detail, issue))
checks.append(date_check(invoices, payments, date_max))
orders = common(invoice, payment, 'orderNumbers')
missing = missing_fields(invoices, payments, 'orderNumbers')
detail = f"发票单号：{values(invoices, 'orderNumbers')}；付款单号：{values(payments, 'orderNumbers')}。"
if orders:
detail += '识别到相同交易单号。'
elif missing:
detail += f'{missing}未识别到交易单号，无法交叉验证；不表示单据错误。'
else:
detail += '未找到相同交易单号，可能是不同平台的编号，请人工核对。'
checks.append(check('交易单号', order_max if orders else 0, order_max, detail, '单号未识别' if missing else '单号不一致'))
unique_reason = next((reason for reason in match['reasons'] if '在本批材料中唯一' in reason), '')
checks.append(check('批次金额唯一', unique_max if unique_reason else 0, unique_max,
unique_reason + '，获得唯一性加分。' if unique_reason else
'自动匹配时，同一金额还出现在其他发票或付款材料中，未获得唯一性加分；请核对商户、日期或单号，避免同金额串单。',
'同金额存在其他材料'))
return checks
def explain_match(match):
if match.get('matchType') != 'auto':
return None
invoices, payments = match['invoices'], match['payments']
reasons = match.get('reasons', [])
title = '单张发票与付款截图'
single_pair = False
if any(reason.startswith('同一行程多人发票合计一致') for reason in reasons):
title = '同一行程多人发票'
checks = [
check('行程及合计金额', 95, 95, '同一行程的多张发票合计与铁路付款材料一致，存在付款凭证，最近日期相差不超过 14 天。此规则基础分为 95 分。'),
check('补充凭证', 5 if len(payments) >= 2 else 0, 5,
'已归入至少两份相关付款／订单材料。' if len(payments) >= 2 else
'目前仅有一份相关付款材料，缺少第二份订单／付款材料的交叉印证，因此未获得额外 5 分。', '缺少补充凭证')
]
elif any(reason.startswith('多张发票合计一致') for reason in reasons):
title = '多张发票合并付款'
same_route = '车次、路线及行程日期一致' in reasons
checks = [
check('唯一合计组合', 70, 70, '找到唯一的多张发票组合，其合计与付款金额一致，基础得 70 分。'),
check('补充凭证', 10 if len(payments) >= 2 else 0, 10,
'至少两份相关付款／订单材料相互印证。' if len(payments) >= 2 else
'只有一份付款材料，缺少其他相关订单／付款材料印证，未获得 10 分。', '缺少补充凭证'),
date_check(invoices, payments, 10, tiered=False),
check('同一行程', 10 if same_route else 0, 10,
'已识别为出行材料，提取的车次、路线及行程日期相同。' if same_route else
'未满足“所有发票均识别为出行材料且路线、时间、车次相同”的条件，未获得 10 分；非出行类报销不适用此加分项，并不代表配对错误。',
'未获得同一行程加分')
]
elif any(reason.startswith('多笔付款合计一致') for reason in reasons):
title = '一张发票分多笔付款'
checks = [
check('唯一分笔付款组合', 95, 95, '多笔付款合计与发票中提取的某一金额一致，且组合唯一；原匹配规则固定为 95 分。'),
check('规则保留分', 0, 5, '该规则没有逐笔验证商户、日期和单号，固定保留 5 分供人工复核；不是少匹配了 5% 的金额或材料。', '分笔付款需人工复核')
]
elif len(invoices) == 1 and len(payments) == 1 and any(reason.startswith('金额一致 ¥') for reason in reasons):
single_pair = True
checks = single_checks(match)
else:
return dict(version=2, title='历史匹配记录', rawScore=match['score'], checks=[],
note='这条记录没有足够的原始评分依据，无法准确还原各项得分；保留原评分，不推测扣分原因。可查看下方原匹配依据并人工核对。')
raw_score = sum(item['points'] for item in checks)
if min(100, raw_score) != match['score']:
return dict(version=2, title=title, rawScore=match['score'], checks=[],
note='当前识别信息与历史评分不一致，无法准确还原各项得分；保留原评分及配对结果，请结合下方原匹配依据核对。')
if single_pair:
checks = single_checks(match, hundred_point=True)
return dict(version=2, title=title, rawScore=sum(item['points'] for item in checks), checks=checks,
note='每项满分相加恰好为 100 分，实际得分相加就是下方百分比；各项未得分相加就是距满分的差额。这是证据评分，不是金额或材料的匹配比例，也不是匹配成功概率；即使满分也请核对原始单据。')