Question

所以我正在开发一个打开外部文件的程序，然后运行它以查看它是否包含特定信息。有没有办法简化它，或者它现在是最有效的方式来写这个？

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()

有没有办法简化这段代码，这样我就不会在main函数中创建三个单独的列表。

    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...

我很想知道是否有更短的方法来获得相同的结果。也许使用运行该部分代码的函数，但我不知道该怎么做。

Answer 1

需要注意的一些问题：

构造

for i in range(len(studentList)):
    alist = studentList[i]
    printGender(alist)

非常讨厌;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)

你的功能名不佳;他们不按照他们的说法去做！ printGender打印女学生，printMale打印2010年的学生等。同样，您的变量名称选择不当; alist 不是学生列表，只有一名学生。
您似乎每个学生都有一个文字字符串，类似于2009, 176915, Jones, Susan, Female, CSC;但你没有尝试将字段分开。这将导致像2009, 292010, Male, Jill, Female, RCSCA这样的学生在2009年和2010年（学生编号的错误匹配）以及女性和男性（姓氏的错误匹配）以及CSC中被报告为学生的恼人问题（课程名称的错误匹配）。你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，还有任何返回命名字段的东西 - 来解决这个问题。
您的搜索选项是非正交的，仅限于预编码选项;例如，你没有办法搜索2007年的所有CSC学生而没有重写你的课程。

修复这些问题会导致类似

的问题

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()

这段代码是简化的吗？我应该使用更多功能吗？

1 个答案: