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- #!/usr/bin/env/python
- # -*- coding:utf-8 -*-
- import re, os
- import configs
- from utils.washutil import table_label_cleal
- import numpy as np
- from PIL import Image
- def option2block(option_con, item_no_type):
- """
- 选择题选项切分
- 对于选项切分部分,最好也像题号一样先自我切分纠错,但这样老师如果手误打错了字母,可能就解析出错!!!!!
- :return:
- """
- def del_table(ss):
- ss = re.sub(r"</?t[dr]>|</?tbody>|</?table>|</?div>|</?p>", "", ss.replace("<td><p>", " "))
- return ss
- # print('***********',option_con)
- if '<table><tbody><tr>' in option_con and \
- len(re.findall('<tr><td><p>(A\s*[..、、::].+?|\(A\)\s*[..、、]?.+?)</tr>', option_con.strip())) == 1:
- st_opt = re.search('<table><tbody><tr><td><p>(A\s*[..、、::].+?|\(A\)\s*[..、、]?.+?)</tr>',
- option_con.strip()).start()
- option_con = option_con.strip()[0:st_opt] + '\n' + del_table(option_con.strip()[st_opt:])
- # print("option_con:", option_con)
- option_con = re.sub(r"</table>\n*\s*(<p>)?\s*(A\s*[、、..::]|\(A\)\s*[、、..]?)(.+?)", r"</table>【【A、】】\3",
- option_con, flags=re.S)
- if re.search("\n\s*C", option_con) is None and re.search("\n\s*c", option_con):
- option_con = re.sub("\n\s*c", "\nC", option_con)
- # option_con = re.sub(r"(\n\s*(<img\s*src=\".*?\"\s(width|height|eq-code|data-latex|ocr-latex)=.*?[\"/]>\s*)+?\s*)(A[、、..::].+?)", r"\1\n\3", option_con.strip())
- option_con = re.sub(r"(\n\s*(<img\s*src=((?![/>]).)*?/>\s*)+?\s*)(A[、、..::].+?)", r"\1\n\4", option_con.strip())
- con = re.sub(r"\n\s*([A-H])\s*[、、..::](.+?)", r"\n【【\1、】】\2", option_con.strip()) # 行首的A、不能考虑,故得用strip
- if item_no_type == 1 and len(re.findall(r'【【[A-H]\s*[..、、]】】', con)) <= 2 and \
- len(re.findall(r'\([A-H]\)', con)) > 2: # 针对题干是第一种类型,选项是第二种类型的情况
- item_no_type = 2
- if item_no_type == 2:
- con = re.sub(r"\n\s*\(([A-Hc])\)\s*[、、..]?(.+?)", r"\n【【\1、】】\2", option_con)
- con = con.replace("</table>【【", "</table>\n【【")
- # print(11111,option_con)
- if item_no_type == 1:
- if len(re.findall(r'【【[A-H]\s*[..、、]】】', con)) <= 3:
- while re.search(r"\n\s*[A-H]\s*<img\s*src=.+?", con.replace(" ", "")): # 2020/7/15
- con = re.sub(r"\n\s*([A-H])\s*(<img\s*src=.+?)", "\n" + r"【【\1、】】\2", con)
- while re.search(r"(\n\s*<img\s*src=.+?)([A-H][..、、])(.+?)", con.replace(" ", "")):
- con = re.sub(r"(\n\s*<img\s*src=.+?)(?<!【)([A-H]\s*[..、、::])(.+?)", r"\1" + "\n" + r"【【\2】】\3", con)
- while re.search(r"(\n【【[A-H][..、、]】】.+?)(?<!【)([A-Hc][..、、])\n+(.+?)(?<!【)([A-H][..、、])(.+?)",
- con.replace(" ", ""), re.S):
- con = re.sub(r"(\n\s*【【[A-H]\s*[..、、]】】.+?)(?<!【)([A-H]\s*[..、、::])\s*\n+(.+?)"
- r"(?<!【)([A-H]\s*[..、、::])(.+?)", r"\1【【\2】】\3【【\4】】\5", con, flags=re.S)
- while re.search(r"(\n【【[A-H][..、、]】】.+?)(?<!【)([A-H][..、、])\n+(.+?)", con.replace(" ", ""), re.S):
- con = re.sub(r"(\n\s*【【[A-H]\s*[..、、]】】.+?)(?<!【)([A-H]\s*[..、、::])\s*\n+(.+?)",
- r"\1【【\2】】\3", con, flags=re.S)
- while re.search(r"(\n【【[A-H][..、、]】】.+?)(?<!【)([A-H][..、、])(.+?)", con.replace(" ", "")):
- con = re.sub(r"(\n\s*【【[A-H]\s*[..、、]】】.+?)(?<!【)([A-H]\s*[..、、::])(.+?)", r"\1【【\2】】\3", con)
- while re.search(r"(\n【【[A-H][..、、]】】[^【]+?/>\s+)(?<!【)([B-H][..、、])(.+?)", con.replace(" ", ""), re.S):
- con = re.sub(r"(\n\s*【【[A-H]\s*[..、、]】】[^【]+?/>\s+)(?<!【)([B-H]\s*[..、、::])\s*(.+?)",
- r"\1【【\2】】\3", con, flags=re.S) # 选项子母前面是图片 9/8
- if item_no_type == 2:
- if len(re.findall(r'【【[A-H][..、、]】】', con)) <= 3:
- while re.search(r"\n\s*\([A-H]\)\s*<imgsrc=.+?", con.replace(" ", "")): # 2020/7/15
- con = re.sub(r"\n\s*\(([A-H])\)\s*(<img src=.+?)", "\n" + r"【【\1、】】\2", con)
- while re.search(r"(\n\s*<imgsrc=.+?)(\([A-H]\)[..、、]?)(.+?)", con.replace(" ", "")):
- con = re.sub(r"(\n\s*<img src=.+?)\(([A-H])\)\s*[..、、]?(.+?)", r"\1" + "\n" + r"【【\2、】】\3", con)
- while re.search(r"(\n【【[A-H]、】】.+?)\(([A-H])\)[..、、]?\n+(.+?)\(([A-H])\)[..、、]?(.+?)",
- con.replace(" ", ""), re.S):
- con = re.sub(r"(\n\s*【【[A-H]、】】.+?)\(([A-H])\)\s*[..、、]?\s*\n+(.+?)"
- r"\(([A-H])\)\s*[..、、]?(.+?)", r"\1【【\2、】】\3【【\4、】】\5", con, flags=re.S)
- while re.search(r"(\n【【[A-H]、】】.+?)\(([A-H])\)[..、、]?\n+(.+?)", con.replace(" ", ""),re.S):
- con = re.sub(r"(\n\s*【【[A-H]、】】.+?)\(([A-H])\)\s*[..、、]?\s*\n+(.+?)",
- r"\1【【\2、】】\3", con, flags=re.S)
- while re.search(r"(\n【【[A-H]、】】.+?)\(([A-H])\)[..、、]?(.+?)", con.replace(" ", "")):
- con = re.sub(r"(\n\s*【【[A-H]、】】.+?)\(([A-H])\)\s*[..、、]?(.+?)", r"\1【【\2、】】\3", con)
- con_list = re.split(r"【【[A-H]\s*[..、、]】】", con)
- if len(con_list) > 1:
- stem_opt = table_label_cleal(con_list[0])
- con_list = list(map(del_table, con_list[1:]))
- con_list.insert(0, stem_opt) # 题干中的表格不需要清洗
- return con_list, con
- recur_n = 1 # 递归次数
- def option_structure(one_item, con, ans, item_no_type, is_danti=0, is_slave=0):
- """
- 选择题选项拆分结构化
- 还需要判断一下 选项个数与题型的对应!!!!
- :return:
- """
- global recur_n
- # print(con)
- # print('----------------------')
- if recur_n>2:
- if 'options' not in one_item and not is_slave:
- one_item["errmsgs"].append("选项格式不正确")
- recur_n = 1
- return one_item
- ans = re.sub("[;;.]+", "", ans)
- ans2 = []
- for a in ans.split("#"):
- if 0<len(a.replace(" ", "")) < 8:
- ans2.append("、".join(re.findall(r"[A-G]", a)))
- one_item["key"] = "; ".join(ans2)
- options_rank = get_options_arrange(one_item["stem"])
- # print("id:", one_item['item_id'])
- # print("options_rank:",options_rank)
- con_list, repl_con = option2block(con, item_no_type)
- # print(len(con_list), con_list)
- # 初筛
- if len(con_list) < 5:
- opt_letter = re.findall(r"【【([A-H])\s*[..、、]】】", repl_con)
- if opt_letter and opt_letter[0] == 'B' and re.search("<img src=.+?/>\s*A\s*[..、、].+?$", con_list[0]):
- re_split = re.sub("(<img src=.+?/>)\s*A\s*[..、、](.+?)$", r"\1【【A、】】\2", con_list[0])
- con_list[0] = re_split.split("【【A、】】")[0]
- con_list.insert(1, re_split.split("【【A、】】")[1])
- if len(con_list) >= 5:
- pattern_1 = re.compile(r"\s([1-9]|1[0-9])[..、、].+?([是为有]|等于)[((]\s*[))]\n", re.S)
- pattern_2 = re.compile(r"\s\(([1-9]|1[0-9])\).+?([是为有]|等于)[((]\s*[))]\n", re.S)
- pattern_3 = re.compile(r"([是为有]|等于)[((]\s*[))]\n", re.S)
- # 第一个错误针对题目中没有答案解析的情况,不然就是选项切分错误
- if not is_danti:
- if (item_no_type == 1 and any([True for op in con_list[1:] if re.search(pattern_1, op)])) or \
- (item_no_type == 2 and any([True for op in con_list[1:] if re.search(pattern_2, op)])):
- one_item["errmsgs"].append("本题选项与下一题题干间没有换行符,请注意重新换行!!!") # 一般只有一题和上一题连在一起
- if 'item_id' in one_item:
- one_item['spliterr_point'] = one_item['item_id']
- return one_item
- elif any([True for op in con_list[1:] if re.search(pattern_3, op)]):
- one_item["errmsgs"].append("本题的下一题的题号有问题,请注意重新输入!!!")
- if 'item_id' in one_item:
- one_item['spliterr_point'] = one_item['item_id']
- # ------------------------------------------------------------------------
- aft_opt = [] # 针对选项后是题目图片的情况
- if "\n" in con_list[-1]:
- ccon = re.split("\n+", con_list[-1])
- while re.match("<img src=", ccon[-1]) and len(ccon) > 1:
- aft_opt.insert(0, ccon[-1])
- ccon = ccon[:-1]
- if aft_opt:
- con_list[0] += "\n" + "\n".join(aft_opt)
- con_list[-1] = "\n".join(ccon)
- # -------------------------------------------------------------------------
- # 选项纠错
- con_list[0] = re.sub(r"\(\d+分\)", "", con_list[0][:9]) + con_list[0][9:]
- opt_letter = re.findall(r"【【([A-H])\s*[..、、]】】", repl_con)
- # print('/////////////////////////',opt_letter)
- if "".join(sorted(opt_letter)) in "ABCDEFGHIJ" or "".join(sorted(opt_letter)) in ["ABCE", "ABDE", "ACDE", "BCDE"]:
- # con_list = pic_transfer(con_list)
- if con_list:
- return dict(one_item, **{"stem": con_list[0],
- "options": [re.sub("(<br/>|\n)\s*$|\s+$", "", i) for i in con_list[1:]],
- "options_rank": options_rank,
- }) # , "options_num": len(con_list[1:])
- else:
- # 初次选项拆分的错误判断
- con_list = option_label_correct(opt_letter, con_list, repl_con)
- # double_l = [key for key, value in dict(Counter(opt_letter)).items() if value > 1]
- if type(con_list) == str:
- one_item["errmsgs"].append(con_list)
- return one_item
- else:
- # con_list = pic_transfer(con_list)
- if con_list:
- return dict(one_item,
- **{"stem": con_list[0],
- "options": [re.sub("(<br/>|\n)\s*$|\s+$", "", i) for i in con_list[1:]],
- "options_rank": options_rank,
- })
- # return dict(one_item, **dict(zip(["stem","A","B","C","D"], con_list)))
- else:
- # 选项可能放在表格中
- is_fail = 0
- con_list2 = re.split(r"\n+", con)
- errmsgs = ""
- if len(con_list2) == 2: # 选项是4个图片组成的情况
- option_array = len(re.findall("(^|\n)<img src=.+?", con_list2[1].strip()))
- if option_array > 2: # 排列情况
- options_rank = 1
- elif option_array > 1:
- options_rank = 3
- else:
- options_rank = 2
- ims = con_list2[1].split("<img src=")
- if len(ims) == 5 and re.search(r"[\u4e00-\u9fa5]", ims[0]) is None:
- con_list2 = [con_list2[0] if k == 0 else "<img src=" + v
- for k, v in enumerate(con_list2[1].split("<img src="))] # 默认将“<img src=”切分后的第一项丢掉了
- # if len(con_list2) == 5:
- con_list2[0] = re.sub(r"\(\d+分\)", "", con_list2[0].replace(" ", "")[:9]) + con_list2[0][9:]
- return dict(one_item, **{"stem": con_list2[0],
- "options": [re.sub("(<br/>|\n)\s*$|\s+$", "", i) for i in con_list2[1:]],
- "options_rank": options_rank,
- })
- else:
- errmsgs = """选项格式不正确,请改为: A.xxxx B.xxx 或 (A)xxxx (B)xxx,全文选项和题号格式要统一。
- 【注意】1>>选项和题干间要换行,选项不要放在表格中;2>>选项【如A.】重新手输;3>>选项图片时用嵌入式;
- 4>>选项太长时,每项之间要换行,上一项的内容不要与下一项在同一行!!"""
- is_fail = 1
- else:
- con_list3 = re.split(r"\n(?=<img)", con)
- if len(con_list3) == 5:
- return dict(one_item, **{"stem": con_list3[0],
- "options": [re.sub("(<br/>|\n)\s*$|\s+$", "", i) for i in con_list3[1:]],
- "options_rank": options_rank,
- })
- else:
- errmsgs = """选项格式不正确,请改为: A.xxxx B.xxx 或 (A)xxxx (B)xxx,全文选项和题号格式要统一。
- 【注意】1>>选项和题干间要换行,选项不要放在表格中;2>>选项【如A.】重新手输;3>>选项图片时用嵌入式;
- 4>>选项太长时,每项之间要换行,上一项的内容不要与下一项在同一行!!"""
- is_fail = 1
- op_con = re.split("[((]\s*[))]", con)[-1]
- stem_con = "".join(re.split("[((]\s*[))]", con)[:-1])+"( )\n"
- if is_fail:
- if "table" in op_con:
- to_clean_con = re.findall('<table>(((?!(</?table>)).)*)</table>', op_con, re.S)
- if len(to_clean_con) == 1:
- op_con = re.sub("</?table>|</?tr>|</?td>", "", op_con)
- one_item = option_structure(one_item, stem_con+op_con, ans, item_no_type)
- else:
- aa = re.findall("[A-E]", op_con)
- if len(aa)==len(set(aa)) == 4:
- recur_n += 1
- op_con = re.sub("(?<!\\\)([A-E])\s*(?![..、、])", r"\1、", op_con)
- one_item = option_structure(one_item, stem_con + op_con, ans, item_no_type, is_slave=is_slave)
- if not is_slave and 'options' not in one_item and "选项格式不正确" not in "".join(one_item["errmsgs"]):
- one_item["errmsgs"].append(errmsgs)
- return one_item
- def get_options_arrange(cont):
- """
- 判断word中选项每行排版个数
- :return:
- """
- options_rank = 1 # 纵向排列
- option_num = 0
- if '<table><tbody><tr>' in cont:
- table_op = re.findall('<tr>.+?>([A-H]\s*[..、、].+?|\([A-Z]\)\s*[..、、]?.+?)</tr>', cont.strip())
- if table_op:
- option_num = len(re.findall('[A-H]\s*[..、、].+?|\([A-Z]\)\s*[..、、]?.+?', table_op[0]))
- if option_num == 2:
- options_rank = 3
- if option_num > 2:
- options_rank = 2
- else:
- option_list = cont.split("\n")
- for op in option_list:
- if re.search("^\s*[A-H]\s*[..、、].+?|^\s*[A-H]\s*<img src=.+?|^\s*\([A-Z]\)\s*[..、、]?.+?", op.strip()):
- option_num += 1
- if option_num == 2:
- options_rank = 3 # 一排2个
- elif option_num < 2:
- options_rank = 2 # 横向排列
- return options_rank
- def new_options_rank(options):
- """
- 按提分宝产品B5纸176*250mm设置选项的排版形式
- 选项的排版形式暂设置3种:1:纵向排列 2:横向排列 3:一排2个
- 中文字符按5号字体,即10.5磅,英文字符按3/4个中文字符算
- :return:
- """
- options_rank = 1 # 纵向排列
- option_len = []
- for opt in options:
- if re.search("\$.*?\$", opt):
- return 0
- pic_len = []
- if "<img " in opt:
- for img in re.findall("<img src=.*?/>", opt):
- w_info1 = re.search(' style=".*?width: (\d+[.\d]*?)\s*([pxtin]*?);.*?"', img)
- w_info2 = re.search(' width="(\d+[.\d]*?)\s*([pxt]*?)"', img)
- if w_info1:
- if w_info1.group(2) == 'pt':
- pic_len.append((25.4/72)*float(w_info1.group(1)))
- elif w_info1.group(2) == 'px':
- pic_len.append((25.4 / 72) * (3/4) * float(w_info1.group(1)))
- elif w_info1.group(2) == 'in':
- pic_len.append(25.4 * float(w_info1.group(1)))
- elif w_info2:
- pic_len.append((25.4 / 72) * (3 / 4) * float(w_info2.group(1)))
- else:
- print("选项中存在图片宽高未知")
- # 主要没有宽高的图片是用户在编辑器新粘贴的图片,保存在本地,通过读取获取宽高
- w_info3 = re.search('<img src=.*?(/[^/]*?/new_image.*?)"', img)
- if w_info3:
- local_p = configs.IMG_FOLDER + w_info3.group(1)
- if os.path.exists(local_p):
- w = Image.open(local_p).size[0]
- pic_len.append((25.4 / 72) * (3 / 4) * float(w))
- else:
- print("选项中存在d的宽高未知图片不存在本地")
- options_rank = 0
- else:
- options_rank = 0
- opt = opt.replace(img, "")
- # 统计字符长度
- char_en_l = len(re.findall(r"[a-z\d,.!?;'\-/:<>=*+$~%()\[\]{}\" ]", opt))
- opt = re.sub(r"[a-z\d,.!?;'\-/:<>=*+$~%()\[\]{}\" ]", "", opt)
- char_zh_l = len(opt)
- char_len = (10.5/72)*25.4*(char_en_l*0.75+char_zh_l)
- option_len.append(sum(pic_len) + char_len)
- # 以最长的选项长度作为参考:<=6个中文字符则排成1行,<=15个中文字符则排成2排,否则都是纵向排列===>此逻辑不对
- if sum(option_len) + (len(options)*2 + (len(options)-1)*4)*(10.5/72)*25.4 < 176-40:
- options_rank = 2
- else:
- option_len = sorted(option_len, reverse=True)
- if option_len[0]+option_len[1] + (2*2+1*4)*(10.5/72)*25.4 <= 176-40:
- options_rank = 3
- return options_rank
- def option_label_correct(opt_letter, con_list, con):
- """
- 选项少切了会报错,所以优先解决多切的错误问题
- 纠正中标签错误的情况:选项字母不连续或重复;
- opt_letter:选项的字母 con_list:选择题拆分了选项的列表
- """
- lable_sign = re.findall(r"【【([A-H][..、、])】】", con.replace(" ", ""))
- con_list2 = con_list.copy()
- for i, j in enumerate(lable_sign): # 将con_list的选项字母加上
- con_list[i + 1] = j + con_list[i + 1]
- # con_list2 = re.split(r"【【[A-H]\s*[..、、]】】", con)
- p1 = 0 # 选项在con_list中的起始位置
- for k, v in enumerate(con_list[1:]):
- if re.search(r"[((]\s*[))]", v): # 选择题末尾一般都有()
- opt_letter[k] = '0'
- p1 = k + 2
- if p1 and p1 < len(con_list[1:]): # '0'不在最后一个位置
- option_list = con_list2[p1:]
- if len(option_list) >= 4:
- new_con_list = ["".join(con_list[:p1])].extend(option_list)
- return new_con_list
- else: # 只考虑ABCD和ACBD两种情况
- label_str = "".join(opt_letter)
- if re.match("A","".join(opt_letter)) is None:
- label_str = re.sub("[^A]A", "0A", "".join(opt_letter), count=1)
- # print(label_str)
- # -------------------------------------------------------------
- # 若选择题中没有(),题干中还是出现了AA的话,需要判断下是否存在错误
- if re.search("AA", label_str):
- label_bcd_idx = [k for k, i in enumerate(label_str) if i != 'A']
- label_a_idx = [k for k, i in enumerate(label_str) if i == 'A']
- length_all = []
- for i1 in label_bcd_idx: # 先将公式替换,作选项长度判断
- l1 = len(re.sub(r"<img\s*src\s*=\s*((?!/>).)+?/>", "<img>", con_list2[i1+1]).replace(" ",""))
- length_all.append(l1)
- aver_length = np.mean(length_all)
- st_a = label_str.index("AA")
- for i2 in label_a_idx:
- l2 = len(re.sub(r"<img\s*src\s*=\s*((?!/>).)+?/>", "<img>", con_list2[i2+1]).replace(" ",""))
- if abs(l2 - aver_length) >= 12:
- if i2 >= st_a:
- st_a = i2+1
- if st_a < len(label_str)-3:
- label_str = "".join(["0" if k < st_a else i for k, i in enumerate(label_str)])
- # -----------------------------------------------------------------
- label_str = re.sub("A[^BC]", "AA", label_str)
- label_str = re.sub("B[^CD]", "BB", label_str)
- label_str = re.sub("C[^BD]", "CC", label_str)
- label_str = re.sub("D[^E]", "DD", label_str)
- # 统计是否有重复的字符,若有,则进行合并,否则保持原来
- new_con_list = [con_list[0]]
- local_w = 0
- while local_w < len(label_str):
- if local_w == len(label_str) - 1 and label_str[local_w] == '0':
- break
- while label_str[local_w] == '0': # 如果‘0’在中间,则‘0’会被去除
- local_w += 1
- double_num = label_str.count(label_str[local_w])
- if double_num >= 2:
- new_con_list.append(con_list2[local_w + 1] + "".join(con_list[2 + local_w:local_w+double_num + 1]))
- else:
- new_con_list.append(con_list2[local_w + 1])
- local_w += double_num
- new_opt_letter = label_str.replace('AA',"A").replace('BB',"B").replace('CC',"C").replace('DD',"D")
- if len(new_con_list) >= 4:
- if "".join(sorted(new_opt_letter)) in "ABCDEFGHIJ" or "".join(sorted(new_opt_letter)) in ["ABCE", "ABDE", "ACDE", "BCDE"]:
- return new_con_list
- return "选项格式不正确,1、请改为: A.xxxx B.xxx,手动输入选项字母及后面的标点符号;" \
- "2.第一个选项A与题干之间要换行,各选项按ABCD排序;3.选项含图片时用嵌入式;"
- def table_option_struc(stem):
- """
- 表格类的选项结构化,在化学科目的选择题中较常见
- :return: 表格仍然作为表格,选项则根据表格中的选项补充,如A、A B、B
- """
- options = []
- may_options = re.findall("<table>(((?!(</?table>)).)*)</table>", stem)
- if may_options:
- options_data = may_options[-1][0]
- data_col = re.findall("<tr><td>(.*?)</td>", options_data) # 第一列
- if re.search("#?A#B#C#D#", re.sub("[..、、,,\s]", "", "#".join(data_col).strip())+"#"):
- options_str = re.sub("[..、、,,\s]", "", "#".join(data_col).strip()+"#")
- if "A#B#C#D#E#F#" not in options_str:
- if "A#B#C#D#E#" in options_str:
- options = ["A", "B", "C", "D", "E"]
- elif "A#B#C#D#" in options_str:
- options = ["A", "B", "C", "D"]
- else:
- data_rows = re.findall("<tr>(.*?)</tr>", options_data)
- data_row = re.findall("<td>(.*?)</td>", data_rows[0]) # 第一行
- if re.search("#?A#B#C#D#", re.sub("[..、、,,\s]", "", "#".join(data_row).strip()) + "#"):
- options_str = re.sub("[..、、,,\s]", "", "#".join(data_row).strip() + "#")
- if "A#B#C#D#E#F#" not in options_str:
- if "A#B#C#D#E#" in options_str:
- options = ["A", "B", "C", "D", "E"]
- elif "A#B#C#D#" in options_str:
- options = ["A", "B", "C", "D"]
- return options
- if __name__ == '__main__':
- stem ="""
- 下列物质与危险化学品标志的对应关系不正确的是<br/><table><tr><td>A</td><td>B</td><td>C</td><td>D</td></tr><tr><td>汽油</td><td>天然气</td><td>浓硫酸</td><td>氢氧化钠</td></tr><tr><td><img src="files/image2.png" width="125px" height="116px" /></td><td><img src="files/image3.png" width="117px" height="117px" /></td><td><img src="files/image4.png" width="118px" height="119px" /></td><td><img src="files/image5.png" width="122px" height="118px" /></td></tr></table>
- """
- # print(table_option_struc(stem))
- one_item = {
- 'errmsgs': [],
- 'key': 'C',
- 'parse': '【详解】根据题意可知,辐射出的光子能量$\\varepsilon = 3 . 5 2 \\times 1 0 ^ { - 1 9 } '
- 'J$,由光子的能量$\\varepsilon = h v$得<br/>$\\nu = \\frac { \\varepsilon } '
- '{ h } = 5 . 3 1 \\times 1 0 ^ { 1 4 } H z$<br/>故选C。',
- 'stem': '近年来,江西省科学家发明硅衬底氮化镓基系列发光二极管,开创了国际上第三条$L E D$技术路线。某氮化镓基$L E '
- 'D$材料的简化能级如图所示,若能级差为$2.20\\text{eV}$(约$3 . 5 2 \\times 1 0 ^ { - 1 9 '
- '} J$),普朗克常量$h = 6 . 6 3 \\times 1 0 ^ { - 3 4 } J \\cdot '
- 's$,则发光频率约为()<br/><img height="112px" src="/word/media/image5.png" '
- 'width="140px"/><br/>A.$6 . 3 8 \\times 1 0 ^ { 1 4 } H z$B.$5 . 6 7 '
- '\\times 1 0 ^ { 1 4 } H z$C.$5 . 3 1 \\times 1 0 ^ { 1 4 } H z$D.$4 '
- '. 6 7 \\times 1 0 ^ { 1 4 } H z$',
- 'topic_num': 1,
- 'type': '选择题',
- }
- one_item = option_structure(one_item, one_item["stem"], one_item["key"], 1)
- print(one_item)
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