#!/usr/bin/env/python # -*- coding:utf-8 -*- import re from utils.item_type_line import get_item_head_info from utils.topic_no import judge_item_no_type, get_right_no, pre_get_item_no def stems_structure_byno(stem_con): """ 按题号进行切分; 针对无解析的试卷中所有题目的拆分; :return:{"stem": , "item_id": , "errmsgs": [],"type":,} """ try: while stem_con and ((re.search(r"[\u4e00-\u9fa5]", stem_con[0]) is None) or (re.search(r"[一二三四]\s*[、..、]\s*[^必考基础综合中等]{2,4}题", stem_con[0]) is None and re.search(r"(^|\n)\s*[1-9][0-9]?\s*[..、、]((?!(答题卡|涂黑|本卷|2B铅笔|签字笔|密封线)).)*?$", stem_con[0]) is None)): del stem_con[0] except: return "本份试卷开头格式有问题,请按试卷格式来!" stem_str = "\n" + "\n".join(stem_con) # 题号格式有条件清洗 # def sub1(ss): # if int(ss.group(5)) - int(ss.group(2)) in [1, 2]: # return ss.group(1)+"\n"+ss.group(5)+"、"+ss.group(6) # else: # return ss.group(0) # stem_str = re.sub(r"(\n\s*([1-4][0-9]|[1-9])\s*[..、、]((?!\n\s*([1-4][0-9]|[1-9])\s*[..、、]).)+?)\n+\s*([1-4][0-9]|[1-9])\s*(?![..、、])([^\s\d]+)", sub1, stem_str, flags=re.S) # 也可以用下面方法,但比较啰嗦 while re.search(r"(\n\s*([1-9][0-9]|[1-9])\s*[..、、]((?!\n\s*([1-9][0-9]|[1-9])\s*[..、、]).)+?)" r"\n+\s*([1-9][0-9]|[1-9])\s*(?![..、、])([^\s\d]+)", stem_str, re.S): wrong_id_info = re.search(r"(\n\s*([1-9][0-9]|[1-9])\s*[..、、]((?!\n\s*([1-9][0-9]|[1-9])\s*[..、、]).)+?)" r"\n+\s*([1-9][0-9]|[1-9])\s*(?![..、、])([^\s\d]+)", stem_str, re.S) stem_str = stem_str.replace(wrong_id_info.group(0), wrong_id_info.group(1)+"\n"+wrong_id_info.group(5)+"、"+wrong_id_info.group(6)) # print(stem_str) # 1、获取题型行信息、按题型行切分 con11, title_info_dict, choice_class = get_item_head_info(stem_str) all_type = title_info_dict["all_type"] title_type_num = title_info_dict["title_type_num"] select_type_id = title_info_dict["select_type_id"] each_item_score, each_item_score2 = title_info_dict["each_item_score"], title_info_dict["each_item_score2"] # ------------------------------------------------------- res = [] item_type_classify = {} # 记录每类题型中含有的题目个数,含合并的情况 item_type_num = [] # 题型不合并 pic_no = {} # 记录每个题的图 new_item_no = [] # 将纠错后的题号再记录一份 if not all_type: print("不存在大题题型行或题型行格式或名称有问题") # 初步获取题号,题号类型 stem_str, item_no_info, item_no_type = judge_item_no_type(stem_str) # 获取正确题号的位置,进行切分 new_item_no, items_no_idx = get_right_no(item_no_info) one_item_split = [stem_str[i:j] for i, j in zip(items_no_idx, items_no_idx[1:] + [None])] dd = {} for n, one_item in enumerate(one_item_split): dd["stem"] = one_item dd["item_id"] = new_item_no[n] # 题目本身的题号 dd["type"] = "" # 先题型不备注,根据答案再看 dd["errmsgs"] = [] dd["score"] = 0 # if select_type_id and dd["item_id"] in select_type_id: # dd['is_optional'] = 'true' res.append(dd) dd = {} # 先不做拆图处理了 else: # print(all_type, len(con11)) if len(all_type) == len(con11)-1: # 第一部分的题型行掉了 all_type.insert(0, "") each_item_score.insert(0, 0) # 按题型行拿的分数也会掉,先补上默认的0 elif len(all_type) != len(con11): print("第二种试卷格式:存在题型行没有换行") # 可能造成题目和题型行在同一行 # return "存在题型行末尾没有换行或题型行中题型不明确" # else: # # print(all_type) # if "非选择题" in all_type: # error_info1 = "第" + str(all_type.index("非选择题")+1) + "大题的题型不明确" # if any([True for one_type in all_type if re.search("必考基础综合中等面列下各", one_type)]): # error_info1 = "存在题型行中题型不明确" # --------------------------------------------------------------------- # 思路:>>>>先纠正题号,再拆分题目;等切分好后再纠正(删减添加操作)比较费时 # 按题号切分,可再加些细节!!!! 1>>题号不要求连续 # 初步判断题号类型 stem_str, item_no_info, item_no_type = judge_item_no_type(stem_str) # >>>>切分题目 for num, one_type in enumerate(con11): # 初步获取题号 item_no_info = pre_get_item_no("\n"+one_type, item_no_type) # 获取正确题号的位置,进行切分 if res: items_no_temp, items_no_idx = get_right_no(item_no_info, have_type=1, last_id=res[-1]["item_id"]) else: items_no_temp, items_no_idx = get_right_no(item_no_info) is_from_0 = 1 if items_no_temp and items_no_idx[0] != 0 and items_no_temp[0] > 1: # 以防出现题号漏了的情况 items_no_idx.insert(0, 0) is_from_0 = 0 one_item_split = [("\n" + one_type)[i:j] for i, j in zip(items_no_idx, items_no_idx[1:] + [None])] # ------------------每个大题的第一题前面的图可能有漏的情况------------------------ may_oimt_pic = [] if not is_from_0: may_oimt_pic_info = re.search("\n((\s*).)+?/>)*?)\s*\n?$", one_item_split[0]) if may_oimt_pic_info: if len(one_item_split) > 1 and re.search("如[上下左右]?图", one_item_split[1]): may_oimt_pic.extend(re.findall("", may_oimt_pic_info.group(1))) # print("may_oimt_pic:", may_oimt_pic) # ------------------------------------------------------------------ if len(all_type) == len(con11): if not is_from_0: if all_type[num] and all_type[num].replace("题", "") not in ['选择', '单选', '多选', '不定选择']: one_item_split = one_item_split[1:] else: # 针对选择题第1题或前面几题题号漏了的情况 one_item_split1 = one_item_split[1:] # ------针对分错的细节继续拆分,如上一题选项行与下一题题干没有换行----------------- new_one_item_split = [] pattern_1 = re.compile( r"([CDE]\s*[..、、].+?)(? 0.0: dd["score"] = title_info_dict["total_score"][num] res.append(dd) # 多图在一起的情况进行拆分 # if len(re.findall(r"第?\(?([1-9]|[1-4][0-9])\)?\s*题图", one_item)) > 1 and \ # re.search(r"", pic_info.group(1)) pic_w = re.findall("([1-9]|[1-9][0-9])[))]?\s*(?=题)", pic_info.group(2)) if len(pic_list) >= len(pic_w): pic_no = {int(p): pic_list[len(pic_list) - len(pic_w) + k] for k, p in enumerate(pic_w)} dd["stem"] = dd["stem"].replace(pic_info.group(2), "") for k, pic in enumerate(pic_list[::-1]): if k < len(pic_w): dd["stem"] = dd["stem"].replace(pic, "") # ----------------------------------------------------- dd = {} if pic_no: for i in list(pic_no.keys()): res[i-1]["stem"] = res[i-1]["stem"].strip() + "\n" + pic_no[i] + "\n" + "第" + str(i) + "题图" # ---------------------------------------------------------------------- # 可能出现选择题类的选做题 # --------最后判断一下题量是否正确----------------------------------------- # 针对拆分后题量特别多的情况,将拆分后题量与已知题量一样的题目保留 new_res = [] right_type = [] # 记录与已知题目个数一样的分块{第几个分块,题型} if all_type and sum(list(item_type_classify.values())) > 40: # 很可能存在大片不是题号的题号 title_type_num_all = sum([t[1] for t in title_type_num]) # 题型行给出的题目个数 if title_type_num_all > 0: for idx, type_num in enumerate(item_type_num): if type_num[1] == title_type_num[idx][1] > 0: # 同一题型是否题量一致 right_type.append((idx, type_num[0])) # elif title_type_num[idx][1] == 0 and type_num[1] == 1: # title_type_num[idx][1] = 1 # right_type.append((idx, type_num[0])) elif title_type_num[idx][1] == 0 and type_num[1] >= 1: # 2023.6.26 title_type_num[idx][1] = type_num[1] right_type.append((idx, type_num[0])) print("right_type::item_type_num", right_type, item_type_num) if right_type: for rtype in right_type: for idx, item in enumerate(res): r_st = 0 if rtype[0]==0 else sum([t[1] for t in item_type_num[:rtype[0]]]) r_ed = r_st + item_type_num[rtype[0]][1] if item["type"] == rtype[1] and r_st<=idx