312 lines
12 KiB
Python
312 lines
12 KiB
Python
#!/usr/bin/env python3
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"""KOCR v4 — K3 记账凭证解析器（基于坐标）"""
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import re
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from datetime import datetime
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class K3VoucherParser:
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"""基于坐标解析 K3 标准记账凭证"""
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def __init__(self, lines):
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self.lines = sorted(lines, key=lambda l: (l[3], l[2]))
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def parse(self):
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result = {
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'company': '',
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'date': None,
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'year': None,
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'period': None,
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'voucher_word': '记',
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'voucher_no': None,
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'attachments': 0,
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'entries': [],
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'total_str': '',
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'total_amount': 0,
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'preparer': '',
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'checker': '',
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'cashier': '',
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'handler': '',
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'poster': '',
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}
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for text, conf, cx, cy, box in self.lines:
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if cy < 480 and '物业' in text:
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result['company'] = text
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elif '业务日期' in text or '日期' in text:
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m = re.search(r'(\d{4}[-\u5e74]\d{1,2}[-\u6708]\d{1,2})', text)
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if m:
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date_str = m.group(1).replace('年', '-').replace('月', '-').replace('日', '')
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try:
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result['date'] = datetime.strptime(date_str, '%Y-%m-%d')
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except:
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pass
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elif '凭证' in text:
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# 兼容各种格式：记-45、记 - 45、记-45（1/3）、45(1/3)
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m = re.search(r'[记凭证]\s*[-—]\s*(\d+)', text)
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if m:
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result['voucher_no'] = int(m.group(1))
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if not result['voucher_no']:
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m2 = re.search(r'(?:记\s*)?[-—]?\s*(\d+)\s*[（(]\d+[/]\d+[）)]', text)
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if m2:
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result['voucher_no'] = int(m2.group(1))
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# 兜底：直接找"凭证号"后面的数字
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if not result['voucher_no']:
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m3 = re.search(r'凭证号\s*[：:]\s*记?\s*[-—]?\s*(\d+)', text)
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if m3:
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result['voucher_no'] = int(m3.group(1))
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elif '附件数' in text:
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m = re.search(r'附件数[：:](\d+)', text)
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if m:
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result['attachments'] = int(m.group(1))
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elif '第' in text and '期' in text:
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m = re.search(r'第(\d+)期', text)
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if m:
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result['period'] = int(m.group(1))
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elif '合计' in text or '壹' in text or '贰' in text:
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result['total_str'] = text
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m = re.search(r'([\d,]+\.[\d]{2})', text)
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if m:
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result['total_amount'] = float(m.group(1).replace(',', ''))
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elif '制单' in text:
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m = re.search(r'制单[：:](.*)', text)
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if m:
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result['preparer'] = m.group(1).strip()
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elif '审核' in text:
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m = re.search(r'审核[：:](.*)', text)
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if m:
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result['checker'] = m.group(1).strip()
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elif '出纳' in text:
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m = re.search(r'出纳[：:](.*)', text)
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if m:
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result['cashier'] = m.group(1).strip()
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elif '经办' in text:
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m = re.search(r'经办[：:](.*)', text)
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if m:
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result['handler'] = m.group(1).strip()
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elif '过账' in text:
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m = re.search(r'过账[：:](.*)', text)
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if m:
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result['poster'] = m.group(1).strip()
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if result['date']:
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result['year'] = result['date'].year
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result['entries'] = self._parse_entries()
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return result
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def _parse_entries(self):
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"""基于坐标解析分录"""
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if not self.lines:
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return []
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lines = sorted(self.lines, key=lambda l: (l[3], l[2]))
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# === 1. 确定数据区域 ===
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y_headers = []
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for text, conf, cx, cy, box in lines:
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if text.strip() in ('借方', '贷方'):
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y_headers.append(cy)
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y_start = min(y_headers) if y_headers else 0
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y_end = 999999
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for text, conf, cx, cy, box in lines:
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if '合计' in text:
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y_end = min(y_end, cy)
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y_data_start = max(0, y_start - 150)
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# === 2. 获取图像宽度 ===
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all_xs = []
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for _, _, _, _, box in lines:
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if box:
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for p in box:
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all_xs.append(p[0])
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img_w = max(all_xs) if all_xs else 3000
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# === 3. 筛选数据区域 ===
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data_lines = [(t, c, x, y, b) for t, c, x, y, b in lines
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if y_data_start <= y <= y_end and c >= 0.4]
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# === 4. 提取科目代码 ===
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codes = []
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seen_codes = set()
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skip_keywords = ['科目', '摘要', '借方', '贷方', '出纳',
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'制单', '审核', '经办', '过账', '附件',
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'Manager', 'HONOR', 'Magic6', '合计',
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'凭证', '号']
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for text, conf, cx, cy, box in data_lines:
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if any(k in text for k in skip_keywords):
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continue
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text_clean = text.strip().replace('：', '').replace(' ', '')
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if text_clean in {'借', '贷', '科', '摘要', 'RMB', ''}:
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continue
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m = re.match(r'(\d[\d.]*)([-－]?\s*.*)', text.strip())
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if m and m.group(1).count('.') >= 1:
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code_part = m.group(1).rstrip('.')
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rest = m.group(2).lstrip('－-—= ').strip()
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if re.match(r'^\d{3,}\.', code_part):
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first_dot = code_part.find('.')
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code_num = int(code_part[:first_dot]) if first_dot > 0 else int(code_part)
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if code_num >= 1000 and code_part not in seen_codes:
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seen_codes.add(code_part)
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codes.append({
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'code': code_part,
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'name': rest,
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'desc': '',
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'debit': 0.0,
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'credit': 0.0,
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'y_pos': cy,
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'x_pos': cx,
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})
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if not codes:
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return self._parse_entries_fallback()
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# === 4.5 摘要提取：从剩余 OCR 文字中找摘要 ===
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code_y_min = min(c['y_pos'] for c in codes)
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desc_candidates = []
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search_lines = self.lines
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for text, conf, cx, cy, box in search_lines:
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if conf < 0.5:
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continue
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is_code = False
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for c in codes:
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if abs(c['y_pos'] - cy) < 20:
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is_code = True
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break
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if is_code:
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continue
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if re.search(r'\d+\.\d{2}', text):
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continue
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if any(k in text for k in skip_keywords + ['合计', '亿', '千', '百', '拾', '万']):
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continue
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if re.match(r'^[\d.\-,]+$', text.strip()):
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continue
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if len(text.strip()) < 3:
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continue
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if cx < 300:
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continue
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if text.strip().startswith('·') or '工商银行' in text:
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continue
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if '支行' in text or re.search(r'\d{15,}', text):
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continue
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if text.strip() in ('未到账', '已到账', '未达', '已达'):
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continue
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if cy < code_y_min - 50:
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continue
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# 去重：跳过 y 距离 < 30 的重复文本
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if any(abs(d['y'] - cy) < 30 for d in desc_candidates):
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continue
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desc_candidates.append({'text': text, 'y': cy, 'x': cx})
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desc_candidates.sort(key=lambda d: d['y'])
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for i, desc in enumerate(desc_candidates):
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if i < len(codes):
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codes[i]['desc'] = desc['text']
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# === 5. 提取金额 ===
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amounts = []
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skip_amount_keywords = ['Manager', 'HONOR', 'Magic6', 'RMB',
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'日', '期', '凭证', '号', '附件',
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'制单', '审核', '出纳', '经办',
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'科目', '摘要']
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for text, conf, cx, cy, box in data_lines:
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if any(k in text for k in skip_amount_keywords):
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continue
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m = re.search(r'([\d,]+\.[\d]{2})', text.replace(',', ''))
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if m:
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val = float(m.group(1).replace(',', ''))
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# 排除年份/月份误识别
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if 1900 < val < 2100:
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continue
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if re.match(r'^\d{3,}\.', text.strip()):
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continue
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if val < 0.5:
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continue
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amounts.append({'val': val, 'x': cx, 'y': cy})
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# === 6. 找到借贷分界线 ===
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amount_xs = [a['x'] for a in amounts]
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if amount_xs:
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amount_xs.sort()
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max_gap = 0
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split_idx = len(amount_xs) // 2
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for i in range(1, len(amount_xs)):
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gap = amount_xs[i] - amount_xs[i-1]
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if gap > max_gap:
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max_gap = gap
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split_idx = i
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bound_x = (amount_xs[split_idx-1] + amount_xs[split_idx]) / 2 if split_idx > 0 else amount_xs[0]
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else:
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bound_x = img_w * 0.5
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# === 7. 交错匹配金额到科目 ===
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debit_amts = sorted([a for a in amounts if a['x'] < bound_x], key=lambda a: a['y'])
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credit_amts = sorted([a for a in amounts if a['x'] >= bound_x], key=lambda a: a['y'])
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interleaved = []
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max_len = max(len(debit_amts), len(credit_amts))
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for i in range(max_len):
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if i < len(debit_amts):
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interleaved.append(('dr', debit_amts[i]))
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if i < len(credit_amts):
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interleaved.append(('cr', credit_amts[i]))
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codes.sort(key=lambda c: c['y_pos'])
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for i, code in enumerate(codes):
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if i < len(interleaved):
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side, amt = interleaved[i]
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if side == 'dr':
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code['debit'] = amt['val']
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else:
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code['credit'] = amt['val']
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# === 8. 构建结果 ===
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entries = []
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for code in codes:
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entries.append({
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'id': len(entries) + 1,
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'description': code['desc'] or '会计凭证',
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'account_code': code['code'],
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'account_name': code['name'] or '未知科目',
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'debit': code['debit'],
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'credit': code['credit'],
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})
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for e in entries:
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e['account_name'] = e['account_name'].replace('--', '-').strip()
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return entries
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def _parse_entries_fallback(self):
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"""兜底：基础科目代码检测"""
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entries = []
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for text, conf, cx, cy, box in self.lines:
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if conf < 0.5:
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continue
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m = re.match(r'(\d{3,}\.[\d.]+)', text.strip())
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if m:
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code = m.group(1).rstrip('.')
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name = text[m.end():].lstrip('－-—= ')
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entries.append({
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'id': len(entries) + 1,
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'description': '',
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'account_code': code,
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'account_name': name or '未知科目',
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'debit': 0.0,
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'credit': 0.0,
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})
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return entries
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