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- #!/usr/bin/env/python
- # -*- coding:utf-8 -*-
- # paper3_process: 第三类word试卷模式, 题目和答案分开的情况
- # 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
- # 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
- """
- 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
- """
- from structure.ans_structure import *
- from utils.insert_keywords import get_con
- from utils.item_resplit import resplit
- from utils.washutil import table_label_cleal
- from structure.stems_structure import stems_structure_byno
- from utils.item_type_line import get_item_head_info
- from utils.topic_no import judge_item_no_type, get_right_no
- from utils.stem_ans_split import stem_ans_split
- from collections import Counter
- from pprint import pprint
- def items_ans_reform(items_list, ans_list, subject):
- """
- 第三种word试卷格式, 题目和答案分开的情况
- 答案也有几种类型:带题型?
- :param sent_list:
- :param split_point:
- :return:
- """
- con1 = list(filter(lambda x: x.strip() != "", items_list)) # 题目
- anss1 = list(filter(lambda x: x.strip() != "", ans_list)) # 答案,list中的每个元素为一行
- if re.match(".+?省.+?试[卷题]", con1[-1]):
- con1 = con1[:-1]
- if re.match(".+?省.+?试[卷题]|.*?答题?[卷卡页]", anss1[0]):
- anss1 = anss1[1:]
- #--------------答案页也包含题目的情况----------但可能题目不存在-----------------------
- ans_n = re.findall("【答案】", "\n".join(anss1))
- 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
- print("答案页中有相同个数的答案和解析")
- item_res = split_by_keywords(anss1)
- if type(item_res) != str:
- # 还要判断题目是否为空
- if len([i["item_id"] for i in item_res[0] if len(i["stem"].strip()) < 5]) < 2:
- return item_res
- # ----------------- 【解析 题目】----------------------------
- print('---------------解析 题目-------------------')
- ress = stems_structure_byno(con1, subject)
- if type(ress) == str:
- return ress
- else:
- item_res, all_type, item_type_classify, item_no_type, \
- item_type_num, new_item_no, item_groups = ress # 全题目(不含解析)的结构化
- # 将空题目去掉
- new_res = []
- for k, sub_res in enumerate(item_res):
- if sub_res['stem'].strip():
- sub_res['stem'] = del_no(sub_res['stem'])
- new_res.append(sub_res)
- item_res = new_res
- # 先对题目的切分结果进行纠正!!!!!
- item_res = resplit(item_res)
- print("item_type_classify:", item_type_classify)
- print("item_type_num:", item_type_num)
- print('----------解析 答案---------------')
- # -------------解析 答案---------------------------
- # 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
- # 2>>答案中不含题型关键字,只按序号排列
- # 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
- rd1_is_fail = 0
- 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))
- if have_type_line:
- # 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
- anss1_cy = anss1.copy() # 复制一份,保证不能影响后面
- 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 \
- (re.search(r"[\u4e00-\u9fa5]", anss1_cy[0]) is None
- 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):
- del anss1_cy[0]
- # 答案中的题型
- 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)|"
- 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)",
- "\n" + "\n".join(anss1_cy))
- all_type2 = ["".join(a) for a in all_type2]
- # '本大题' 后面处理
- print("答案中的题型:", all_type2)
- ans_str = "\n" + "\n".join(anss1_cy)
- try:
- item_res, rd1_is_fail = anss_structure_with_type(item_res, ans_str, all_type, all_type2, item_type_num, item_type_classify)
- except:
- rd1_is_fail = 1
- # 没有题型行或第一次解析失败
- rd2_is_fail = 0
- 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
- 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')
- anss1 = list(
- map(lambda x: re.sub(r"(\n|^)\s*[一二三四五六七八九十]\s*[、..、::]?\s*(<p>)?"
- r"(\s*.{2,5}题.+?分\s*[.。]?\s*$|.*?[((].+?[得共]\d+分.*?[))].*?$"
- r"|\s*.{2,5}题\s*([((].+?[))])?).*?$|(\n|^)\s*[^\d]{2,5}题(.+?分\s*[))])?\s*$", "", x), anss1))
- # print("anss1:", anss1)
- raw_item_res = item_res
- # try:
- item_res = ans_structure_step1(anss1, item_type_classify, item_res) # 答案整体结构化
- if str(raw_item_res) != str(item_res):
- rd2_is_fail = 1
- # except:
- # rd2_is_fail = 1
- # for i, one_item in enumerate(item_res):
- # item_res[i].update({'key': "", 'parse': ""})
- # return item_res, item_no_type, rd2_is_fail
- for i, one_item in enumerate(item_res):
- if 'key' not in one_item:
- item_res[i]['key'] = ""
- if 'parse' not in one_item:
- item_res[i]['parse'] = ""
- return item_res, item_no_type, rd2_is_fail, item_groups
- def split_by_keywords(con_list):
- """
- 第一种试卷格式:教师用卷,含答案和解析关键字
- 切分思路:
- 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’
- :param con_list:
- :return: 每个切分后的题目组成的dict
- """
- # items_con = "\n" + "\n".join(con_list)
- # judge_item_no_type(items_con)
- # item_no_type = 1
- # all_con = table_label_cleal()
- # item_no = [int(no) for no in re.findall(r'\n+\s*([1-9][0-9]?)\s*[..、、]', all_con)]
- # if len(item_no) <= 2:
- # item_no_type = 2
- # item_no = [int(no) for no in re.findall(r'\n+\s*[((]\s*([1-9][0-9]?)\s*[))]\s*[..、、]?', all_con)]
- # if len(item_no) > 3:
- # 去掉多余空格,作用不大
- con2 = ["【delete】" if (k < len(con_list) - 1 and v.strip() == "" and (
- re.match(r"【(答案|解析)】|(答案|解析)\s*[::]|<imgsrc\d+|\s+", con_list[k + 1].strip()) or
- 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)",
- con_list[k + 1].strip()) is None))
- or (k > 0 and v.strip() == "" and (
- re.match(r"【(答案|解析)】$|(答案|解析)\s*[::]", con_list[k - 1].strip()) or
- 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",
- con_list[k - 1].strip())))
- else v for k, v in enumerate(con_list)]
- con3 = list(filter(lambda x: x != "【delete】", con2))
- while con3 and con3[-1].strip() == "":
- del con3[-1]
- while con3 and con3[0].strip() == "":
- del con3[0]
- con3.append("") # 不然最后一个题就漏掉了
- # 开头没用信息处理
- 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])
- while con3 and (re.search(r"[\u4e00-\u9fa5]", con3[0]) is None
- 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
- and re.match("\s*[1-9]\s*[、..、].+?", con3[0]) is None)):
- del con3[0]
- # ----------------------------------开始结构化---------------------------------------------
- items_con = "\n" + "\n".join(con3)
- # 初步获取题号,题号类型
- items_con, item_no_info, item_no_type = judge_item_no_type(items_con)
- # 1、获取题型行信息、按题型行切分
- con4, title_info_dict, choice_class = get_item_head_info(items_con)
- all_type = title_info_dict["all_type"]
- 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"]
- # 2、据是否有题型行分两步进行
- # 没有做拆图处理
- res = []
- if not all_type:
- print("不存在大题题型行或题型行格式有问题")
- if len(re.findall(r"\n\s*【答案】", items_con)) != len(re.findall(r"\n\s*【解析】", items_con)):
- return "不存在大题题型行或题型行格式有问题"
- else:
- item_no = []
- subcon = re.split(r"((?<=\n)\s*【答案】|(?<=\n)\s*【解析】)\n?", items_con.strip())
- pattern1 = re.compile(r"([1-9]|[1-4][0-9])\s*[..、、].+?")
- if re.match(pattern1, subcon[0].strip()):
- st_id = re.match(pattern1, subcon[0].strip()).group(1)
- if int(st_id) > 1:
- item_no.append(int(st_id))
- else:
- item_no.append(1)
- else:
- item_no.append(1)
- if len(subcon) == 5: # 只有1道题
- dd = dict(zip(["stem", "key", "parse"],
- re.split(r"(?<=\n)\s*【答案】|(?<=\n)\s*【解析】", table_label_cleal(items_con))))
- dd["type"] = ""
- dd["stem"] = re.sub(r"^\d+\s*[..、、]", "", dd["stem"][:5]) + dd["stem"][5:]
- dd["score"] = 0
- dd["errmsgs"] = []
- dd["item_id"] = item_no[0] # 要用实际id 不是索引序号
- res.append(dd)
- else:
- # ------在下一题【解析】在本题【答案】之间找到下一题【stem】的位置--------
- all_item, item_no, errmsg_dict, count = get_con(subcon, item_no_type, item_no, index=0)
- # item_no.extend(local_item_no)
- for idk, one_item in enumerate(all_item):
- if one_item:
- dd = dict(zip(["stem", "key", "parse"],
- re.split(r"(?<=\n)\s*【答案】\n?|(?<=\n)\s*【解析】\n?",
- table_label_cleal(one_item))))
- dd["type"] = ""
- dd["stem"] = re.sub(r"\d+\s*[..、、]", "", dd["stem"][:5]) + dd["stem"][5:]
- dd["score"] = 0
- dd["errmsgs"] = [errmsg_dict[idk]] if idk in errmsg_dict else []
- dd["item_id"] = item_no[idk]
- res.append(dd)
- else:
- if len(all_type) != len(con4):
- print("存在题型行没有换行")
- 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
- else:
- # if "非选择题" in all_type:
- # return "第" + str(all_type.index("非选择题")+1) + "大题的题型不明确"
- index = 0 # 每个大题的第一题的题号索引位置
- for num, one_type in enumerate(con4):
- count = 1
- if len(re.findall(r"\n\s*【答案】", one_type)) == len(re.findall(r"\n\s*【解析】", one_type)):
- subcon = re.split(r"((?<=\n)\s*【答案】|(?<=\n)\s*【解析】)\n?", one_type.strip())
- # index根据第一道题的题号进行纠正
- item_no = []
- pattern1 = re.compile(r"([1-9]|[1-4][0-9])\s*[..、、].+?")
- if re.match(pattern1, subcon[0].strip()):
- st_id = re.match(pattern1, subcon[0].strip()).group(1)
- if num == 0 and int(st_id) != 1:
- index = int(st_id) - 1
- item_no.append(int(st_id))
- else:
- item_no.append(index+1)
- if len(subcon) == 5: # 只有1道题
- dd = dict(zip(["stem", "key", "parse"],
- re.split(r"(?<=\n)\s*【答案】|(?<=\n)\s*【解析】", table_label_cleal(one_type))))
- dd["type"] = all_type[num]
- dd["stem"] = re.sub(r"^\d+\s*[..、、]", "", dd["stem"][:5]) + dd["stem"][5:]
- dd["score"] = each_item_score[num]
- dd["errmsgs"] = []
- dd["item_id"] = item_no[0] # 要用实际id 不是索引序号
- if not dd["score"] and each_item_score2 and str(dd["item_id"]) in each_item_score2.keys():
- dd["score"] = each_item_score2[str(dd["item_id"])]
- if select_type_id and dd["item_id"] in select_type_id:
- dd['is_optional'] = 'true'
- if dd["score"] == 0.0 and title_info_dict["total_score"][num] > 0.0:
- dd["score"] = title_info_dict["total_score"][num]
- res.append(dd)
- else:
- # ------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--------
- all_item, item_no, errmsg_dict, count = get_con(subcon, item_no_type, item_no,
- all_type=all_type, num=num, index=index)
- # item_no.extend(local_item_no)
- for idk, one_item in enumerate(all_item):
- dd = dict(zip(["stem", "key", "parse"],
- re.split(r"(?<=\n)\s*【答案】\n?|(?<=\n)\s*【解析】\n?",
- table_label_cleal(one_item))))
- dd["type"] = all_type[num]
- dd["stem"] = re.sub(r"\d+\s*[..、、]", "", dd["stem"][:5]) + dd["stem"][5:]
- dd["score"] = each_item_score[num]
- dd["errmsgs"] = [errmsg_dict[idk]] if idk in errmsg_dict else []
- dd["item_id"] = item_no[idk] # idk+1+index 为序号
- if choice_class:
- for k, v in choice_class.items():
- if dd["item_id"] in v:
- dd["type"] = k + "选题"
- # elif len(choice_class) == 1:
- # dd["type"] = "多选题" if k == "单" else "单选题"
- if not dd["score"] and each_item_score2 and str(dd["item_id"]) in each_item_score2.keys():
- dd["score"] = each_item_score2[str(dd["item_id"])]
- if select_type_id and dd["item_id"] in select_type_id:
- dd['is_optional'] = 'true'
- res.append(dd)
- # pprint(res)
- else:
- 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"
- index += count
- for i, one_item in enumerate(res):
- if 'key' not in one_item:
- res[i]['key'] = ""
- if 'parse' not in one_item:
- res[i]['parse'] = ""
- return res, item_no_type
- def split_by_topicno(con_list, subject):
- """
- 第二种试卷格式: 不同时或都不含有{答案}和{解析}关键字
- 按题号切分每个题目
- 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
- :param con_list: 所有行文本组成的list
- :return: [{},{}]
- """
- con1 = list(filter(lambda x: x.strip() != "", con_list))
- ress = stems_structure_byno(con1, subject) # 按题号切分后的初步结构化
- if type(ress) == str:
- return ress
- else:
- res, all_type, item_type_classify, item_no_type, item_type_num, new_item_no, item_groups = ress
- # res, all_type, item_type_classify = stems_structure_byno(con1)
- print("item_type_num:", item_type_num)
- # pprint(res)
- # 可能存在有的题目有解析,有的没有
- last_comstem_id = 0
- ans_groups = {}
- no_ans_n = 0
- for k, one_res in enumerate(res):
- if item_groups["is_groups"]:
- if "com_stem" in one_res:
- last_comstem_id = k
- if re.search('\n【(答案|[解分][析答]|详解|点[评睛]|考点|专题)】', one_res["stem"]):
- case = "case1" # 默认有“答案”关键字
- if re.search(r'\n【答案】|答案\s*[::]', one_res["stem"]) is None:
- # 没“答案”关键字
- case = "case0"
- dd1 = stem_ans_split(one_res, case) # 对切分后的每道题再细分
- one_res["stem"] = dd1["stem"]
- del dd1["stem"]
- if not dd1["key"] and not dd1["parse"]:
- no_ans_n += 1
- elif subject == "地理" and no_ans_n == k-last_comstem_id > 0:
- if (k+1 < len(res) and "com_stem" in res[k+1]) or len(re.findall("【详解】", dd1["parse"])) > 1\
- or re.findall(r"(?<=[】\s\n^])\d{1,2}\s*[、..、]", dd1["key"]) > 1:
- # 默认是前后都是题组的情况
- ans_groups["{}-{}".format(last_comstem_id+1, k+1)] = dd1
- dd1 = {"key": "", "parse": ""}
- no_ans_n = 0
- one_res.update(dd1)
- else: # 没有解析的情况
- one_res.update({"key": "", "parse": ""})
- no_ans_n += 1
- one_res["stem"] = del_no(one_res["stem"], item_no_type)
- if 'pic' in one_res:
- one_res["stem"] += "\n" + "\n".join(one_res["pic"])
- del one_res["pic"]
- # 先对题目的切分结果进行纠正!!!!!
- res = resplit(res)
- # 对最后一个题后面带个别答案(无答案页)
- if res:
- pattern1 = re.search('\n\s*([1-9]|[1-4][0-9])\s*[..、、]\s*(解\s*[::]|【解析|【答案)', res[-1]["stem"])
- if pattern1:
- breakp = pattern1.start()
- ans_str = res[-1]["stem"][breakp:]
- ans_no_info = pre_get_item_no(ans_str, item_no_type)
- ans_no, ans_no_idx = get_right_no(ans_no_info)
- all_ans = [del_no(ans_str[i:j]) for i, j in zip(ans_no_idx, ans_no_idx[1:] + [None])]
- res[-1]["stem"] = res[-1]["stem"][:breakp]
- res = get_ans_match(res, all_ans, ans_no)
- else:
- ans_str = res[-1]["stem"] + res[-1]["parse"]
- ans_no_info = pre_get_item_no(ans_str, item_no_type)
- ans_no, ans_no_idx = get_right_no(ans_no_info)
- if len(ans_no) == len(res):
- all_ans = [del_no(ans_str[i:j]) for i, j in zip(ans_no_idx, ans_no_idx[1:] + [None])]
- res[-1]["stem"] = res[-1]["stem"][:ans_no_idx[0]]
- res = get_ans_match(res, all_ans, ans_no)
- elif ans_no_idx:
- try:
- ans_no1, table_ans, st = get_table_ans(res[-1]["stem"][:ans_no_idx[0]], flag=1)
- if table_ans and 0 < ans_no[0] - ans_no1[-1] < 3:
- all_ans = table_ans
- all_ans.extend([del_no(ans_str[i:j]) for i, j in zip(ans_no_idx, ans_no_idx[1:] + [None])])
- new_ans_no = ans_no1
- new_ans_no.extend(ans_no)
- if st >= 0:
- res[-1]["stem"] = res[-1]["stem"][:st]
- else:
- res[-1]["stem"] = res[-1]["stem"][:ans_no_idx[0]]
- res = get_ans_match(res, all_ans, new_ans_no)
- except:
- if len(ans_no)>4:
- all_ans = [del_no(ans_str[i:j]) for i, j in zip(ans_no_idx, ans_no_idx[1:] + [None])]
- res[-1]["stem"] = res[-1]["stem"][:ans_no_idx[0]]
- res = get_ans_match(res, all_ans, ans_no)
- # 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
- all_no = [one_res['item_id'] for one_res in res]
- if len(list(set(all_no))) - len(all_no) < -2:
- Count_no = sorted(dict(Counter(all_no)).items(), key=lambda d: d[1], reverse=True)
- if Count_no[0][1] > 1:
- split_idx = [i for i, no in enumerate(all_no) if no == Count_no[0][0]][1]
- for one_res in res[split_idx:]:
- if re.search("[((]\s+[))]|(等于|存在|[是有为])多少|求.*?[??]",
- one_res["stem"] + "\n" + one_res["parse"]) is None:
- bef_no = [k for k, j in enumerate(res[:split_idx]) if j["item_id"]==one_res["item_id"]]
- if bef_no and not res[:split_idx][bef_no[0]]["parse"]:
- res[:split_idx][bef_no[0]]["parse"] = one_res["stem"] + "\n" + one_res["parse"]
- return res[:split_idx],item_no_type
- return res, item_no_type, item_groups, ans_groups
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