结对编程代码分析

结对编程代码分析

该同学用python语言实现中小学生试卷生成系统。

一、项目结构：

• letters1~letters9依次存放张三1到王五3的过往试卷，create.py为项目的主程序，userData.json 存放用户数据

• userData.json内容如图：
  

  • 每个用户存有三个属性，account是为用户名，password为密码，grade为用户的预设年级。

二、主程序代码分析：

• 参数及主要函数总览：

• login()函数：

  # 用于处理用户登录操作的函数
  def login():
      get: list[str] = []
      flag: bool = False # 标识是否登录成功
      while 1:
          if flag:
              get = input("请输入正确的用户名、密码:").split(' ')#空格分隔
          elif not flag:
              get = input("请输入用户名、密码:").split(' ')
              flag = True
  
          for y in user_lib:
              if get[0] == user_lib[y]['account'] and get[1] == user_lib[y]['password']:
                  return {'uid': y, 'grade': user_lib[y]['grade']}
  

• create_problem()函数：

  # 用于生产单行题目的函数
  def create_problem(grade: str):
      question: str = ''  # 返回值
      length: int = randint(2, 5)  # 可操作数的个数 考虑到实际情况，不可能只有一个操作数。
      # print(f"length:{length}")  # 注释掉
      num_square = randint(1, length)  # 有几个数被赋予特殊符号（开方和平方）
      num_tangle = randint(1, length)  # 有几个数被赋予特殊符号（三角函数）
      # print(f"num:{num_square}")  # 注释掉
      data_lib: list[int] = []  # 给操作数编号，为了保证随机时不出现重复数值，使用数组
      for i in range(0, length):
          data_lib.append(i + 1)
      # print(f"data_lib:{data_lib}")  # 注释掉
      square_map: dict = {}
      tangle_map: dict = {}
      leftbracket: int = 0  # 计数
      leftbracket_flag: bool = False  #
      no_double_bracket: bool = False  # 防止出现(x)的情况
      # 变量定义————————————————————————————————————————————————
  
      if grade == 'primary':  # 还未设计添加括号进去
          for i in range(0, length):
              no_double_bracket = False
              if randint(0, 1) == 1 and length > 2 and i != length - 1:  # 随机生成左括号
                  question += '('
                  leftbracket += 1
                  leftbracket_flag = True
                  no_double_bracket = True
              if i == length - 1:  # 只剩一个数了，不用再加符号了
                  question += f"{randint(1, 100)}"
                  for i in range(0, leftbracket):  # 补全括号部分
                      question += ')'
                  question += '=\n'
                  # if check_string(re_exp, question):# 为了避免等式两端出现括号
                  #     return re.findall(re_exp, question)[0]+'\n'
                  # else:
                  #     return question
                  return question
              question += f"{randint(1, 100)}"
  
              for i in range(0, leftbracket):  # 随机补全右括号
                  if randint(0, 1) == 1 and leftbracket_flag and not no_double_bracket:
                      question += ')'
                      leftbracket -= 1
                      if leftbracket == 0:
                          leftbracket_flag = False
              question += f"{symbol_lib[randint(0, 3)]}"
      # 小学题目————————————————————————————————————————————————
  
      elif grade == 'junior':
          for i in range(0, len(data_lib)):  # 给随机几位操作数生成
              a = randint(0, len(data_lib) - 1)  # 0对应1，1对应2，2对应3,3对应4
              symbol_index = randint(4, 5)
              # print(data_lib[a])
              square_map[data_lib[a]] = symbol_lib[symbol_index]
              data_lib.pop(a)
              num_square -= 1
              if num_square == 0:
                  break
          # print(square_map)
  
          for i in range(0, length):
              no_double_bracket = False
              if randint(0, 1) == 1 and length > 2 and i != length - 1:  # 随机生成左括号
                  question += '('
                  leftbracket += 1
                  leftbracket_flag = True
                  no_double_bracket = True
  
              if i == length - 1:  # 只剩一个数了，不用再加符号了
                  if square_map.__contains__(i + 1):  # dict里面有和i对应的键值，说明需要加平方或者开方
                      question += f"({randint(1, 100)}{square_map[i + 1]})"  # 决定暂时给平方、开方加上括号
                  else:
                      question += f"{randint(1, 100)}"
                  for i in range(0, leftbracket):  # 补全括号部分
                      question += ')'
                  question += '=\n'
                  # if check_string(re_exp, question):  # 为了避免等式两端出现括号
                  #     return re.findall(re_exp, question)[0] + '\n'
                  # else:
                  #     return question
                  return question
  
              r: str = symbol_lib[randint(0, 3)]
              if square_map.__contains__(i + 1):
                  question += f"({randint(1, 100)}{square_map[i + 1]})"  # 决定暂时给平方、开方加上括号
              else:
                  question += f"{randint(1, 100)}"
  
              for i in range(0, leftbracket):  # 随机补全右括号
                  if randint(0, 1) == 1 and leftbracket_flag and not no_double_bracket:
                      question += ')'
                      leftbracket -= 1
                      if leftbracket == 0:
                          leftbracket_flag = False
              question += f"{r}"
      # 初中题目————————————————————————————————————————————————
  
      elif grade == 'senior':
          num_square = randint(0, length)
          for i in range(0, len(data_lib)):  # 给随机几位操作数生成
              a = randint(0, len(data_lib) - 1)
              symbol_index = randint(4, 5)
              # print(data_lib[a])
              square_map[data_lib[a]] = symbol_lib[symbol_index]
              data_lib.pop(a)
              num_square -= 1
              if num_square == 0:
                  break
          # print(square_map)  # 注释掉
  
          data_lib = []  # 为tangle的判定重置data_lib
          for i in range(0, length):
              data_lib.append(i + 1)
          for i in range(0, len(data_lib)):  # 给随机几位操作数生成
              a = randint(0, len(data_lib) - 1)
              symbol_index = randint(6, 8)
              tangle_map[data_lib[a]] = symbol_lib[symbol_index]
              data_lib.pop(a)
              num_tangle -= 1
              if num_tangle == 0:
                  break
          # print(tangle_map)  # 注释掉
  
          for i in range(0, length):
              no_double_bracket = False
              if randint(0, 1) == 1 and length > 2 and i != length - 1:  # 随机生成左括号
                  question += '('
                  leftbracket += 1
                  leftbracket_flag = True
                  no_double_bracket = True
  
              if i == length - 1:  # 只剩一个数了，不用再加符号了
                  if square_map.__contains__(i + 1) and tangle_map.__contains__(i + 1):  # dict里面有和i对应的键值，说明需要加平方或者开方
                      question += f"{tangle_map[i + 1]}({randint(1, 100)}{square_map[i + 1]})"
                  elif square_map.__contains__(i + 1) and not tangle_map.__contains__(i + 1):
                      question += f"({randint(1, 100)}{square_map[i + 1]})"
                  elif not square_map.__contains__(i + 1) and tangle_map.__contains__(i + 1):
                      question += f"{tangle_map[i + 1]}({randint(1, 100)})"
                  else:
                      question += f"{randint(1, 100)}"
                  for i in range(0, leftbracket):  # 补全括号部分
                      question += ')'
                  question += '=\n'
                  return question
  
              r: str = symbol_lib[randint(0, 3)]
              if square_map.__contains__(i + 1) and tangle_map.__contains__(i + 1):  # dict里面有和i对应的键值，说明需要加平方或者开方
                  question += f"{tangle_map[i + 1]}({randint(1, 100)}{square_map[i + 1]})"
              elif square_map.__contains__(i + 1) and not tangle_map.__contains__(i + 1):
                  question += f"({randint(1, 100)}{square_map[i + 1]})"
              elif not square_map.__contains__(i + 1) and tangle_map.__contains__(i + 1):
                  question += f"{tangle_map[i + 1]}({randint(1, 100)})"
              else:
                  question += f"{randint(1, 100)}"
              for i in range(0, leftbracket):  # 随机补全右括号
                  if randint(0, 1) == 1 and leftbracket_flag and not no_double_bracket:
                      question += ')'
                      leftbracket -= 1
                      if leftbracket == 0:
                          leftbracket_flag = False
              question += f"{r}"
      # 高中题目————————————————————————————————————————————————
      return question
  

• change_grade()函数：

  # 用于切换当前用户出题年级的函数
  def change_grade(change: str):
      if change in {'小学', '初中', '高中'}:
          return base[change]
      while 1:
          change = input("请输入小学、初中和高中三个选项中的一个:").split("切换为")[0]
          if change in {'小学', '初中', '高中'}:
              return base[change]
  

• print_letter()函数：

  # 用于输出试卷到txt文件的函数
  def print_letter(uid: str, key: str, number: int):
      file_url = f"./letters{uid}/{datetime.now().strftime('%Y%m%d%H%M%S')}.txt"  
      # 使用datatime类的now()函数获取当前时间
      # %Y%m%d%H%M%S 格式化选择输出
      question: str
      file = open(file_url, 'w+')
      for i in range(0, number):
          while 1:
              question = create_problem(key) # 调用试题生成函数
              if check_in_history_letters(uid, question):
                  break
              else:
                  pass
          question += '\n'
          file.write(question)
      file.close()
      return f"********已生成{file_url}********"
  

• read_history_letters()函数：

  # 用于读取老师出题历史记录的函数
  def read_history_letters(uid: str):
      dir_url = f'letters{uid}'
      question_list: list[str] = []
      files: list[str] = []
      for root, dirs, files in os.walk(dir_url, topdown=False):
          # root 根目录、dirs 子目录、files 子目录里的文件
          # print(files)  # 注释掉
          pass
      for file in files:  # 遍历目录里的txt文件
          f = open(f'./{dir_url}/{file}', 'r')
          for line in f:  # 遍历文件的每一行
              if line != '\n':
                  question_list.append(line.split('\n')[0])  # 去除换行转义字符
      question_dict[uid] = question_list
      # print(question_dict)  # 注释掉
  

• check_in_history_letters()函数：

  # 用于检查题目是否已经出现过的函数
  def check_in_history_letters(uid: str, question: str):
      if question_dict[uid].__contains__(question):
          return False
      else:
          return True
  

• main函数

  # 程序运行的主函数
  if __name__ == '__main__':
      user = login()
      key = user.get('grade')
      uid = user.get('uid')
      # print(datetime.now().strftime('%Y%m%d%H%M%S'))#注释掉
      print(f"当前选择为{base[key]}出题")
      read_history_letters(uid)  # 预先读取当前用户的题库
      while 1:
          change: str = ''
          number: int = 0
          error: bool = False
          print("********---********")
          get = input(f"准备生成{base[key]}数学题目，请输入生成题目数量(输入-1将退出当前用户，重新登录):")
  
          if len(get.split("切换为")) == 2:
              change = get.split("切换为")[1]
              if change == '':
                  error = True
          else:
              try:
                  number = int(get)
              except ValueError:
                  error = True
          if number <= 30 and number >= 10:  # 10-30 正常输出
              print(print_letter(uid, key, number))
          elif get == '-1':  # -1重新登录(-1 的判断优先级必须高于 >30 or ＜10 的判断)
              user = login()
              uid = user.get('uid')
              key = user.get('grade')
              read_history_letters(uid)  # 同时更新key和uid
          elif change in {'小学', '初中', '高中'} and key != base[change]:  # 切换年级
              key = change_grade(change)
              print(f"当前选择为{base[key]}出题")
          elif error:
              print("请输入正确的指令，切换年级请输入\"切换为XX\"")
              pass
          elif number > 30 or number < 10:  # 超出题数范围，重新输入
              while 1:
                  get = input(f"准备生成{base[key]}数学题目，请在允许范围内输入生成题目数量(10-30之间):")
                  try:
                      number = int(get)
                      if number <= 30 and number >= 10 or number == -1:
                          break;
                  except ValueError:
                      pass
              if number != -1:
                  print(print_letter(uid, key, number))
          else:
              print("请输入正确的指令，切换年级请输入\"切换为XX\"")
              pass
  

三、代码优缺点分析：

• 优点：

  • 分了多个函数实现功能，每个函数的作用明确，整体的结构清晰

  • 可拓展性强

    代码结构清晰，对各数据结构、函数进行了明确细致地划分，有利于后期功能拓展与项目扩大。

  • 文件路径使用相对路径，清晰明确。

• 缺点：

  • 功能实现不完全：不能在三角函数、平方和开方的计算中掺杂复杂表达式（例如：tan(10+20)、(1+9)^2等等类似题目)，无法生成

  • 括号的使用会出现冗余现象，指：

    (21-(11/(96+18)))=
    

    这种情况，计算式最外层出现的无意义括号。

四、生成结果展示：

• 小学

  33/73-45+72=
  
  44/81=
  
  42-59=
  
  (1/30)-72*(26-40)=
  
  (44/67)+76*(14-47)=
  
  (21-(11/(96+18)))=
  
  65/(85/91*11-2)=
  
  32+(91-61)=
  
  26-(88/15)=
  
  89/13=
  

• 初中

  (85-22*((65^0.5)/28))=
  
  51/15*(77^0.5)=
  
  (81^0.5)-(25^2)=
  
  ((51^0.5)*29*((96^0.5)+((87^2)*31)))=
  
  65-25-(36^2)=
  
  ((90^0.5)-(32/((75^0.5)-(66+(65^2)))))=
  
  (20/(67^2))-51=
  
  (35^2)-((65^0.5)-(60^2))/(68^0.5)+(73^2)=
  
  (59^0.5)+2=
  
  (54^0.5)*(75^2)=
  

• 高中

  (tan(78^0.5)/tan(45^2))+(50^2)=
  
  tan(93^0.5)/(sin(27^2)-sin(37^2))=
  
  (88^2)/78/(tan(99^0.5)*(23+83))=
  
  tan(64^0.5)-(41^2)=
  
  (tan(68^0.5)/cos(96))/sin(34)=
  
  (sin(87^2)+cos(94^0.5)*(cos(35^2)+tan(30^0.5))+sin(2^0.5))=
  
  tan(20^0.5)*cos(23^2)=
  
  (92-((46^0.5)-(16+((92^0.5)-cos(91^2)))))=
  
  (99^2)-sin(100^2)=
  
  (29^2)-cos(3^0.5)/cos(22^0.5)=