这段代码是简体吗？我应该使用更多功能吗？

Question

所以我正在开发这个程序，它打开一个外部文件，然后运行它以查看它是否包含特定信息。有没有一种方法来简化它，或者它是现在写这个最有效的方式吗？

def printGender(alist): 
    if "Female" in alist: 
     print(alist) 
     print("Female Students") 

def maleInfo(blist): 
    if "2010" in blist: 
     print(blist) 
     print("Students who enrolled in 2010") 

def csc2010(clist): 
    if "CSC" in clist and "2010" in clist and "Female" in clist: 
     print(clist) 
     print("Female students who registered in CSC in 2010") 

def main(): 
    ref = open("file1.txt","r") 

    studentList = ref.readlines() 
    ask = 10 
    while ask != 0: 
    print("1) print all female info") 
    print("2) display all male info from 2010") 
    print("3) display female students who registered for CSC in 2010") 
    ask = int(input("Enter option 1, 2, 3 or 0 to quit: ")) 
    if ask == 1: 
     for i in range(len(studentList)): 
      alist = studentList[i] 
      printGender(alist) 
    elif ask == 2: 
     for i in range(len(studentList)): 
      blist = studentList[i] 
      maleInfo(blist) 
    elif ask == 3: 
     for i in range(len(studentList)): 
      clist = studentList[i] 
      csc2010(clist) 
    elif ask == 0: 
     print("You chose to quit") 
     break 
    else: 
     print("Not a Valid input") 
     continue 

    ref.close() 

main() 

有没有简化此代码的方法，以便我不在主函数中创建三个独立列表。

if ask == 1: 
     for i in range(len(studentList)): 
      alist = studentList[i] 
      printGender(alist) 
    elif ask == 2: 
     for i in range(len(studentList)): 
      blist = studentList[i] 
      maleInfo(blist) 
    elif ask == 3: 
     for i in range(len(studentList)): 
      clist = studentList[i] 
      csc2010(clist) 
    elif ask == 0: 
     print("You chose to quit") 
     break 
    else: 
    ect... 

我很好奇，看看是否有更短的方法来获得相同的结果。也许使用运行那段代码的函数，但我不知道该怎么做。

Answer 1

一些问题需要注意的：

• 构建

  ​​

  是非常讨厌的;如果你确实需要i你应该使用

  for i, student in enumerate(student_list): 
      print_gender(student) 
  

  否则

  for student in student_list: 
      print_gender(student) 
  

• 你的函数命名不好;他们不会做他们所说的事！ printGender列印女学生，printMale列印2010年的学生等。同样，您的变数名称选择不当; alist是不是学生名单，它是单身学生。

• 你似乎有一个每个学生的文本字符串，在猜测像2009, 176915, Jones, Susan, Female, CSC;但你不会试图分离出来的领域。这将导致像2009, 292010, Male, Jill, Female, RCSCA这样的学生在2009年和2010年都会报告为学生的问题（学生编号错误匹配），以及女性和男性（姓氏错误匹配）以及CSC（错误匹配课程名）。您真的需要使用更好的数据格式 - 无论是.csv还是.json还是数据库，任何返回命名字段的数据 - 都可以解决此问题。

• 您的搜索选项是非正交的并且仅限于预先编码的选项;例如，您无法在2007年为所有CSC学生进行搜索，而无需重新编写程序。

修复这些问题会导致你喜欢的东西

import json 

def record_print_format(record): 
    return "{Year:4} {Id:6} {Gender:6} {Firstname:>20} {Lastname:<20} {Course:6}".format(**record) 

def show_records(records, format_fn=record_print_format): 
    for r in records: 
     print(format_fn(r)) 
    num = len(records) 
    print("{} records:".format(num)) 

def filter_records(records, field, value): 
    return [r for r in records if r[field] == value] 

def match_year(records, year): 
    return filter_records(records, "Year", year) 

def match_gender(records, gender): 
    return filter_records(records, "Gender", gender) 

def match_course(records, course): 
    return filter_records(records, "Course", course) 

def main(): 
    with open("student_list.json") as studentfile: 
     all_students = json.load(studentfile) 
     records = all_students 

    while True: 
     print("1: Filter by year") 
     print("2: Filter by gender") 
     print("3: Filter by course") 
     print("8: Show results") 
     print("9: Clear filters") 
     print("0: Exit") 

     option = input("Please pick an option: ").strip() 

     if option == "1": 
      year = input("Which year? ").strip() 
      records = match_year(records, year) 
     elif option == "2": 
      gender = input("Which gender? [Male|Female] ").strip() 
      records = match_gender(records, gender) 
     elif option == "3": 
      course = input("Which course? ").strip() 
      records = match_course(records, course) 
     elif option == "8": 
      show_records(records) 
     elif option == "9": 
      records = all_students 
     elif option == "0": 
      print("Goodbye!") 
      break 
     else: 
      print("I don't recognize that option.") 

if __name__=="__main__": 
    main()