three_parse_structure.py 22 KB

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  1. #!/usr/bin/env/python
  2. # -*- coding:utf-8 -*-
  3. # paper3_process: 第三类word试卷模式, 题目和答案分开的情况
  4. # split2one_item:将所有行文本 按题型分大类,再在每个大类中切分每个题目
  5. # split2one_item_by_topicno:将所有行文本 按题型分大类,再在每个大类中按题号切分每个题目
  6. """
  7. 总共3种方案:1、教师用卷;2、按题号切分;3、划分试题和答案,再按题号切分
  8. """
  9. from structure.ans_structure import *
  10. from utils.insert_keywords import get_con
  11. from utils.item_resplit import resplit
  12. from utils.washutil import table_label_cleal
  13. from structure.stems_structure import stems_structure_byno
  14. from utils.item_type_line import get_item_head_info
  15. from utils.topic_no import judge_item_no_type, get_right_no
  16. from utils.stem_ans_split import stem_ans_split
  17. from collections import Counter
  18. from pprint import pprint
  19. def items_ans_reform(items_list, ans_list, subject):
  20. """
  21. 第三种word试卷格式, 题目和答案分开的情况
  22. 答案也有几种类型:带题型?
  23. :param sent_list:
  24. :param split_point:
  25. :return:
  26. """
  27. con1 = list(filter(lambda x: x.strip() != "", items_list)) # 题目
  28. anss1 = list(filter(lambda x: x.strip() != "", ans_list)) # 答案,list中的每个元素为一行
  29. if re.match(".+?省.+?试[卷题]", con1[-1]):
  30. con1 = con1[:-1]
  31. if re.match(".+?省.+?试[卷题]|.*?答题?[卷卡页]", anss1[0]):
  32. anss1 = anss1[1:]
  33. #--------------答案页也包含题目的情况----------但可能题目不存在-----------------------
  34. ans_n = re.findall("【答案】", "\n".join(anss1))
  35. if subject != "地理" and ans_n and len(ans_n) == len(re.findall("【解析】", "\n".join(anss1)))>2: # 带相同个数的答案和解析
  36. print("答案页中有相同个数的答案和解析")
  37. item_res = split_by_keywords(anss1)
  38. if type(item_res) != str:
  39. # 还要判断题目是否为空
  40. if len([i["item_id"] for i in item_res[0] if len(i["stem"].strip()) < 5]) < 2:
  41. return item_res
  42. # ----------------- 【解析 题目】----------------------------
  43. print('---------------解析 题目-------------------')
  44. ress = stems_structure_byno(con1, subject)
  45. if type(ress) == str:
  46. return ress
  47. else:
  48. item_res, all_type, item_type_classify, item_no_type, \
  49. item_type_num, new_item_no, item_groups = ress # 全题目(不含解析)的结构化
  50. # 将空题目去掉
  51. new_res = []
  52. for k, sub_res in enumerate(item_res):
  53. if sub_res['stem'].strip():
  54. sub_res['stem'] = del_no(sub_res['stem'])
  55. new_res.append(sub_res)
  56. item_res = new_res
  57. # 先对题目的切分结果进行纠正!!!!!
  58. item_res = resplit(item_res)
  59. print("item_type_classify:", item_type_classify)
  60. print("item_type_num:", item_type_num)
  61. print('----------解析 答案---------------')
  62. # -------------解析 答案---------------------------
  63. # 分两种情况:1>>答案中又按题型排列, 如一、选择题 1.答案 2.答案
  64. # 2>>答案中不含题型关键字,只按序号排列
  65. # 3>>答案中不含题型关键字,且题目中也没有,all_type, item_type_classify为空
  66. rd1_is_fail = 0
  67. have_type_line = re.search(r"[一二三四五六七八九十]\s*[、..、]\s*[^必考基础综合中等((\[]{2,5}题", "\n".join(anss1))
  68. if have_type_line:
  69. # 这里的anss1的清洗不应该影响rd2_is_fail中的原始文本!!先不修改看看再说
  70. anss1_cy = anss1.copy() # 复制一份,保证不能影响后面
  71. while re.search(r"<td><p>[A-F]</p></td>|</td><td>[A-F]</td><td>|([A-F]\s*){3,}", anss1_cy[0]) is None and \
  72. (re.search(r"[\u4e00-\u9fa5]", anss1_cy[0]) is None
  73. or re.search(r"[一二三四五六七八九十]\s*[、..、]\s*(<imgsrc.*?/>)?\s*.{2,5}题", anss1_cy[0]) is None):
  74. del anss1_cy[0]
  75. # 答案中的题型
  76. all_type2 = re.findall(r"\n\s*[一二三四五六七八九十]\s*[、..、::]\s*([^必考基础综合中共等::((\[]{2,5}题)|"
  77. r"\n\s*[、..、::]?\s*(单选题|非?选择题|不定选择题|多选题|填空题|计算题|[解简]答题|实验题|作图题|论述题|探究题)",
  78. "\n" + "\n".join(anss1_cy))
  79. all_type2 = ["".join(a) for a in all_type2]
  80. # '本大题' 后面处理
  81. print("答案中的题型:", all_type2)
  82. ans_str = "\n" + "\n".join(anss1_cy)
  83. try:
  84. item_res, rd1_is_fail = anss_structure_with_type(item_res, ans_str, all_type, all_type2, item_type_num, item_type_classify)
  85. except:
  86. rd1_is_fail = 1
  87. # 没有题型行或第一次解析失败
  88. rd2_is_fail = 0
  89. if not have_type_line or rd1_is_fail: # 答案中没有题型行 或题型行名称不规范
  90. print('没有题型行或题目和答案的题型个数不一致或第一次解析失败')
  91. anss1 = list(
  92. map(lambda x: re.sub(r"(\n|^)\s*[一二三四五六七八九十]\s*[、..、::]?\s*(<p>)?"
  93. r"(\s*.{2,5}题.+?分\s*[.。]?\s*$|.*?[((].+?[得共]\d+分.*?[))].*?$"
  94. r"|\s*.{2,5}题\s*([((].+?[))])?).*?$|(\n|^)\s*[^\d]{2,5}题(.+?分\s*[))])?\s*$", "", x), anss1))
  95. # print("anss1:", anss1)
  96. raw_item_res = item_res
  97. # try:
  98. item_res = ans_structure_step1(anss1, item_type_classify, item_res) # 答案整体结构化
  99. if str(raw_item_res) != str(item_res):
  100. rd2_is_fail = 1
  101. # except:
  102. # rd2_is_fail = 1
  103. # for i, one_item in enumerate(item_res):
  104. # item_res[i].update({'key': "", 'parse': ""})
  105. # return item_res, item_no_type, rd2_is_fail
  106. for i, one_item in enumerate(item_res):
  107. if 'key' not in one_item:
  108. item_res[i]['key'] = ""
  109. if 'parse' not in one_item:
  110. item_res[i]['parse'] = ""
  111. return item_res, item_no_type, rd2_is_fail, item_groups
  112. def split_by_keywords(con_list):
  113. """
  114. 第一种试卷格式:教师用卷,含答案和解析关键字
  115. 切分思路:
  116. 1.根据大题型分,再按【答案|解析】初步拆分题目,再在‘解析’和‘答案’间细分‘题干’和‘解析’
  117. :param con_list:
  118. :return: 每个切分后的题目组成的dict
  119. """
  120. # items_con = "\n" + "\n".join(con_list)
  121. # judge_item_no_type(items_con)
  122. # item_no_type = 1
  123. # all_con = table_label_cleal()
  124. # item_no = [int(no) for no in re.findall(r'\n+\s*([1-9][0-9]?)\s*[..、、]', all_con)]
  125. # if len(item_no) <= 2:
  126. # item_no_type = 2
  127. # item_no = [int(no) for no in re.findall(r'\n+\s*[((]\s*([1-9][0-9]?)\s*[))]\s*[..、、]?', all_con)]
  128. # if len(item_no) > 3:
  129. # 去掉多余空格,作用不大
  130. con2 = ["【delete】" if (k < len(con_list) - 1 and v.strip() == "" and (
  131. re.match(r"【(答案|解析)】|(答案|解析)\s*[::]|<imgsrc\d+|\s+", con_list[k + 1].strip()) or
  132. re.match(r"(([1-9]|[1-4][0-9])\s*[..、、]|[一二三四五六七八九十]\s*[、..、]\s*[^必考基础综合中等]{2,4}题)",
  133. con_list[k + 1].strip()) is None))
  134. or (k > 0 and v.strip() == "" and (
  135. re.match(r"【(答案|解析)】$|(答案|解析)\s*[::]", con_list[k - 1].strip()) or
  136. re.match(r"[a-z<>/\s]*?[一二三四五六七八九十]\s*[、..、]\s*[^必考基础综合中等]{2,4}题",
  137. con_list[k - 1].strip())))
  138. else v for k, v in enumerate(con_list)]
  139. con3 = list(filter(lambda x: x != "【delete】", con2))
  140. while con3 and con3[-1].strip() == "":
  141. del con3[-1]
  142. while con3 and con3[0].strip() == "":
  143. del con3[0]
  144. con3.append("") # 不然最后一个题就漏掉了
  145. # 开头没用信息处理
  146. con3[0] = re.sub(r"([一二三四五六七八九十]\s*[、..、]\s*[^必考基础综合中等]{2,4}题)", r"\n\1", con3[0])
  147. while con3 and (re.search(r"[\u4e00-\u9fa5]", con3[0]) is None
  148. or (re.search(r"[一二三四五六七八九十]\s*[、..、]\s*[^必考基础综合中等]{2,4}题", con3[0]) is None
  149. and re.match("\s*[1-9]\s*[、..、].+?", con3[0]) is None)):
  150. del con3[0]
  151. # ----------------------------------开始结构化---------------------------------------------
  152. items_con = "\n" + "\n".join(con3)
  153. # 初步获取题号,题号类型
  154. items_con, item_no_info, item_no_type = judge_item_no_type(items_con)
  155. # 1、获取题型行信息、按题型行切分
  156. con4, title_info_dict, choice_class = get_item_head_info(items_con)
  157. all_type = title_info_dict["all_type"]
  158. select_type_id = title_info_dict["select_type_id"]
  159. each_item_score, each_item_score2 = title_info_dict["each_item_score"], title_info_dict["each_item_score2"]
  160. # 2、据是否有题型行分两步进行
  161. # 没有做拆图处理
  162. res = []
  163. if not all_type:
  164. print("不存在大题题型行或题型行格式有问题")
  165. if len(re.findall(r"\n\s*【答案】", items_con)) != len(re.findall(r"\n\s*【解析】", items_con)):
  166. return "不存在大题题型行或题型行格式有问题"
  167. else:
  168. item_no = []
  169. subcon = re.split(r"((?<=\n)\s*【答案】|(?<=\n)\s*【解析】)\n?", items_con.strip())
  170. pattern1 = re.compile(r"([1-9]|[1-4][0-9])\s*[..、、].+?")
  171. if re.match(pattern1, subcon[0].strip()):
  172. st_id = re.match(pattern1, subcon[0].strip()).group(1)
  173. if int(st_id) > 1:
  174. item_no.append(int(st_id))
  175. else:
  176. item_no.append(1)
  177. else:
  178. item_no.append(1)
  179. if len(subcon) == 5: # 只有1道题
  180. dd = dict(zip(["stem", "key", "parse"],
  181. re.split(r"(?<=\n)\s*【答案】|(?<=\n)\s*【解析】", table_label_cleal(items_con))))
  182. dd["type"] = ""
  183. dd["stem"] = re.sub(r"^\d+\s*[..、、]", "", dd["stem"][:5]) + dd["stem"][5:]
  184. dd["score"] = 0
  185. dd["errmsgs"] = []
  186. dd["item_id"] = item_no[0] # 要用实际id 不是索引序号
  187. res.append(dd)
  188. else:
  189. # ------在下一题【解析】在本题【答案】之间找到下一题【stem】的位置--------
  190. all_item, item_no, errmsg_dict, count = get_con(subcon, item_no_type, item_no, index=0)
  191. # item_no.extend(local_item_no)
  192. for idk, one_item in enumerate(all_item):
  193. if one_item:
  194. dd = dict(zip(["stem", "key", "parse"],
  195. re.split(r"(?<=\n)\s*【答案】\n?|(?<=\n)\s*【解析】\n?",
  196. table_label_cleal(one_item))))
  197. dd["type"] = ""
  198. dd["stem"] = re.sub(r"\d+\s*[..、、]", "", dd["stem"][:5]) + dd["stem"][5:]
  199. dd["score"] = 0
  200. dd["errmsgs"] = [errmsg_dict[idk]] if idk in errmsg_dict else []
  201. dd["item_id"] = item_no[idk]
  202. res.append(dd)
  203. else:
  204. if len(all_type) != len(con4):
  205. print("存在题型行没有换行")
  206. return "存在题型行末尾没有换行,请在所有题型行末尾重新换行" # 放第【2】种方案中进行处理
  207. else:
  208. # if "非选择题" in all_type:
  209. # return "第" + str(all_type.index("非选择题")+1) + "大题的题型不明确"
  210. index = 0 # 每个大题的第一题的题号索引位置
  211. for num, one_type in enumerate(con4):
  212. count = 1
  213. if len(re.findall(r"\n\s*【答案】", one_type)) == len(re.findall(r"\n\s*【解析】", one_type)):
  214. subcon = re.split(r"((?<=\n)\s*【答案】|(?<=\n)\s*【解析】)\n?", one_type.strip())
  215. # index根据第一道题的题号进行纠正
  216. item_no = []
  217. pattern1 = re.compile(r"([1-9]|[1-4][0-9])\s*[..、、].+?")
  218. if re.match(pattern1, subcon[0].strip()):
  219. st_id = re.match(pattern1, subcon[0].strip()).group(1)
  220. if num == 0 and int(st_id) != 1:
  221. index = int(st_id) - 1
  222. item_no.append(int(st_id))
  223. else:
  224. item_no.append(index+1)
  225. if len(subcon) == 5: # 只有1道题
  226. dd = dict(zip(["stem", "key", "parse"],
  227. re.split(r"(?<=\n)\s*【答案】|(?<=\n)\s*【解析】", table_label_cleal(one_type))))
  228. dd["type"] = all_type[num]
  229. dd["stem"] = re.sub(r"^\d+\s*[..、、]", "", dd["stem"][:5]) + dd["stem"][5:]
  230. dd["score"] = each_item_score[num]
  231. dd["errmsgs"] = []
  232. dd["item_id"] = item_no[0] # 要用实际id 不是索引序号
  233. if not dd["score"] and each_item_score2 and str(dd["item_id"]) in each_item_score2.keys():
  234. dd["score"] = each_item_score2[str(dd["item_id"])]
  235. if select_type_id and dd["item_id"] in select_type_id:
  236. dd['is_optional'] = 'true'
  237. if dd["score"] == 0.0 and title_info_dict["total_score"][num] > 0.0:
  238. dd["score"] = title_info_dict["total_score"][num]
  239. res.append(dd)
  240. else:
  241. # ------在下一题【解析】在本题【答案】之间找到下一题【stem】的位置,再按此3个关键字进行 切分--------
  242. all_item, item_no, errmsg_dict, count = get_con(subcon, item_no_type, item_no,
  243. all_type=all_type, num=num, index=index)
  244. # item_no.extend(local_item_no)
  245. for idk, one_item in enumerate(all_item):
  246. dd = dict(zip(["stem", "key", "parse"],
  247. re.split(r"(?<=\n)\s*【答案】\n?|(?<=\n)\s*【解析】\n?",
  248. table_label_cleal(one_item))))
  249. dd["type"] = all_type[num]
  250. dd["stem"] = re.sub(r"\d+\s*[..、、]", "", dd["stem"][:5]) + dd["stem"][5:]
  251. dd["score"] = each_item_score[num]
  252. dd["errmsgs"] = [errmsg_dict[idk]] if idk in errmsg_dict else []
  253. dd["item_id"] = item_no[idk] # idk+1+index 为序号
  254. if choice_class:
  255. for k, v in choice_class.items():
  256. if dd["item_id"] in v:
  257. dd["type"] = k + "选题"
  258. # elif len(choice_class) == 1:
  259. # dd["type"] = "多选题" if k == "单" else "单选题"
  260. if not dd["score"] and each_item_score2 and str(dd["item_id"]) in each_item_score2.keys():
  261. dd["score"] = each_item_score2[str(dd["item_id"])]
  262. if select_type_id and dd["item_id"] in select_type_id:
  263. dd['is_optional'] = 'true'
  264. res.append(dd)
  265. # pprint(res)
  266. else:
  267. return "第" + str(num + 1) + "大题《" + all_type[num] + "》中【答案】或【解析】格式有误或其中某道题中出现多个相同关键字或漏关键字"
  268. index += count
  269. for i, one_item in enumerate(res):
  270. if 'key' not in one_item:
  271. res[i]['key'] = ""
  272. if 'parse' not in one_item:
  273. res[i]['parse'] = ""
  274. return res, item_no_type
  275. def split_by_topicno(con_list, subject):
  276. """
  277. 第二种试卷格式: 不同时或都不含有{答案}和{解析}关键字
  278. 按题号切分每个题目
  279. 将所有行文本 按题型分大类,再在每个大类中切分每个题目
  280. :param con_list: 所有行文本组成的list
  281. :return: [{},{}]
  282. """
  283. con1 = list(filter(lambda x: x.strip() != "", con_list))
  284. ress = stems_structure_byno(con1, subject) # 按题号切分后的初步结构化
  285. if type(ress) == str:
  286. return ress
  287. else:
  288. res, all_type, item_type_classify, item_no_type, item_type_num, new_item_no, item_groups = ress
  289. # res, all_type, item_type_classify = stems_structure_byno(con1)
  290. print("item_type_num:", item_type_num)
  291. # pprint(res)
  292. # 可能存在有的题目有解析,有的没有
  293. last_comstem_id = 0
  294. ans_groups = {}
  295. no_ans_n = 0
  296. for k, one_res in enumerate(res):
  297. if item_groups["is_groups"]:
  298. if "com_stem" in one_res:
  299. last_comstem_id = k
  300. if re.search('\n【(答案|[解分][析答]|详解|点[评睛]|考点|专题)】', one_res["stem"]):
  301. case = "case1" # 默认有“答案”关键字
  302. if re.search(r'\n【答案】|答案\s*[::]', one_res["stem"]) is None:
  303. # 没“答案”关键字
  304. case = "case0"
  305. dd1 = stem_ans_split(one_res, case) # 对切分后的每道题再细分
  306. one_res["stem"] = dd1["stem"]
  307. del dd1["stem"]
  308. if not dd1["key"] and not dd1["parse"]:
  309. no_ans_n += 1
  310. elif subject == "地理" and no_ans_n == k-last_comstem_id > 0:
  311. if (k+1 < len(res) and "com_stem" in res[k+1]) or len(re.findall("【详解】", dd1["parse"])) > 1\
  312. or re.findall(r"(?<=[】\s\n^])\d{1,2}\s*[、..、]", dd1["key"]) > 1:
  313. # 默认是前后都是题组的情况
  314. ans_groups["{}-{}".format(last_comstem_id+1, k+1)] = dd1
  315. dd1 = {"key": "", "parse": ""}
  316. no_ans_n = 0
  317. one_res.update(dd1)
  318. else: # 没有解析的情况
  319. one_res.update({"key": "", "parse": ""})
  320. no_ans_n += 1
  321. one_res["stem"] = del_no(one_res["stem"], item_no_type)
  322. if 'pic' in one_res:
  323. one_res["stem"] += "\n" + "\n".join(one_res["pic"])
  324. del one_res["pic"]
  325. # 先对题目的切分结果进行纠正!!!!!
  326. res = resplit(res)
  327. # 对最后一个题后面带个别答案(无答案页)
  328. if res:
  329. pattern1 = re.search('\n\s*([1-9]|[1-4][0-9])\s*[..、、]\s*(解\s*[::]|【解析|【答案)', res[-1]["stem"])
  330. if pattern1:
  331. breakp = pattern1.start()
  332. ans_str = res[-1]["stem"][breakp:]
  333. ans_no_info = pre_get_item_no(ans_str, item_no_type)
  334. ans_no, ans_no_idx = get_right_no(ans_no_info)
  335. all_ans = [del_no(ans_str[i:j]) for i, j in zip(ans_no_idx, ans_no_idx[1:] + [None])]
  336. res[-1]["stem"] = res[-1]["stem"][:breakp]
  337. res = get_ans_match(res, all_ans, ans_no)
  338. else:
  339. ans_str = res[-1]["stem"] + res[-1]["parse"]
  340. ans_no_info = pre_get_item_no(ans_str, item_no_type)
  341. ans_no, ans_no_idx = get_right_no(ans_no_info)
  342. if len(ans_no) == len(res):
  343. all_ans = [del_no(ans_str[i:j]) for i, j in zip(ans_no_idx, ans_no_idx[1:] + [None])]
  344. res[-1]["stem"] = res[-1]["stem"][:ans_no_idx[0]]
  345. res = get_ans_match(res, all_ans, ans_no)
  346. elif ans_no_idx:
  347. try:
  348. ans_no1, table_ans, st = get_table_ans(res[-1]["stem"][:ans_no_idx[0]], flag=1)
  349. if table_ans and 0 < ans_no[0] - ans_no1[-1] < 3:
  350. all_ans = table_ans
  351. all_ans.extend([del_no(ans_str[i:j]) for i, j in zip(ans_no_idx, ans_no_idx[1:] + [None])])
  352. new_ans_no = ans_no1
  353. new_ans_no.extend(ans_no)
  354. if st >= 0:
  355. res[-1]["stem"] = res[-1]["stem"][:st]
  356. else:
  357. res[-1]["stem"] = res[-1]["stem"][:ans_no_idx[0]]
  358. res = get_ans_match(res, all_ans, new_ans_no)
  359. except:
  360. if len(ans_no)>4:
  361. all_ans = [del_no(ans_str[i:j]) for i, j in zip(ans_no_idx, ans_no_idx[1:] + [None])]
  362. res[-1]["stem"] = res[-1]["stem"][:ans_no_idx[0]]
  363. res = get_ans_match(res, all_ans, ans_no)
  364. # 没有识别出答案切分点的情况,很可能答案里的部分也当成题文进行拆分,所以先判断下是否有相同的id
  365. all_no = [one_res['item_id'] for one_res in res]
  366. if len(list(set(all_no))) - len(all_no) < -2:
  367. Count_no = sorted(dict(Counter(all_no)).items(), key=lambda d: d[1], reverse=True)
  368. if Count_no[0][1] > 1:
  369. split_idx = [i for i, no in enumerate(all_no) if no == Count_no[0][0]][1]
  370. for one_res in res[split_idx:]:
  371. if re.search("[((]\s+[))]|(等于|存在|[是有为])多少|求.*?[??]",
  372. one_res["stem"] + "\n" + one_res["parse"]) is None:
  373. bef_no = [k for k, j in enumerate(res[:split_idx]) if j["item_id"]==one_res["item_id"]]
  374. if bef_no and not res[:split_idx][bef_no[0]]["parse"]:
  375. res[:split_idx][bef_no[0]]["parse"] = one_res["stem"] + "\n" + one_res["parse"]
  376. return res[:split_idx],item_no_type
  377. return res, item_no_type, item_groups, ans_groups