Python2.7测试

Question

我做了这个测试，但是代码看起来真的很长。有人能找到使我的代码更短的方法吗？短得多是最好的，但任何事情都有帮助。我认为主要的问题是问题的重复。

choice=raw_input("What type of quiz do you want to do: maths or science? ") 
topic=open("topic.txt", "a+") 
topic.write(choice + '\n') 
topic.close() 
difficulty=raw_input("What difficulty do you want to play on: easy, medium or hard? ") 
diff=open("difficulty.txt", "a+") 
diff.write(difficulty + '\n') 
diff.close()

score = 0


def str_answercheck(score, user_answer, str_answer):
    if user_answer.upper()==str_answer:
        print("Well done")
        score+=1
    else:
        print ("Wrong, it was "+str_answer)
    raw_input("Press enter to continue")
    return score

def int_answercheck(score, user_answer, int_answer):
    if int(user_answer.upper())==int_answer:
        print("Well done")
        score+=1
    else:
        print ("Wrong, it was "+str(int_answer))
    raw_input("Press enter to continue")
    return score

if choice.lower() == "maths" and difficulty.lower() == "easy": 
    easym=open("mathseasy.txt" , "r") 
    lines = easym.readlines() 
    print lines[0]
    print lines[1]
    print("A. 4"+'\n'+"B. 6") 
    str_answer_one="A"
    int_answer_one=4
    user_answer_one=raw_input("Select your answer. ")
    if user_answer_one.isdigit()==True:
        first_score=int_answercheck(score, user_answer_one, int_answer_one)
    else:
        first_score=str_answercheck(score, user_answer_one, str_answer_one)

    print lines[2]
    print("A. 5"+'\n'+"B. 6") 
    str_answer_two="A"
    int_answer_two=5
    user_answer_two=raw_input("Select your answer. ")
    if user_answer_two.isdigit()==True:
        second_score=int_answercheck(first_score, user_answer_two, int_answer_two)
    else:
        second_score=str_answercheck(first_score, user_answer_two, str_answer_two)

    print lines[3]
    print("A. 15"+'\n'+"B. 20") 
    str_answer_three="B"
    int_answer_three=20
    user_answer_three=raw_input("Select your answer. ")
    if user_answer_three.isdigit()==True:
        third_score=int_answercheck(second_score, user_answer_three, int_answer_three)
    else:
        third_score=str_answercheck(second_score, user_answer_three, str_answer_three)

    print lines[4]
    print("A. 13"+'\n'+"B. 15") 
    str_answer_four="A"
    int_answer_four=13
    user_answer_four=raw_input("Select your answer. ")
    if user_answer_four.isdigit()==True:
        fourth_score=int_answercheck(third_score, user_answer_four, int_answer_four)
    else:
        fourth_score=str_answercheck(third_score, user_answer_four, str_answer_four)

    print lines[5]
    print("A. 100"+'\n'+"B. 110") 
    str_answer_five="B"
    int_answer_five=110
    user_answer_five=raw_input("Select your answer. ")
    if user_answer_five.isdigit()==True:
        fifth_score=int_answercheck(fourth_score, user_answer_five, int_answer_five)
    else:
        fifth_score=str_answercheck(fourth_score, user_answer_five, str_answer_five)




if choice.lower() == "maths" and difficulty.lower() == "medium": 
    mediumm=open("mathsmedium.txt" , "r") 
    lines = mediumm.readlines() 
    print lines[0]
    print lines[1]
    print("A. 30"+'\n'+"B. 35") 
    str_answer_one="A"
    int_answer_one=30
    user_answer_one=raw_input("Select your answer. ")
    if user_answer_one.isdigit()==True:
        first_score=int_answercheck(score, user_answer_one, int_answer_one)
    else:
        first_score=str_answercheck(score, user_answer_one, str_answer_one)

    print lines[2]
    print("A. 100"+'\n'+"B. 110") 
    str_answer_two="B"
    int_answer_two=110
    user_answer_two=raw_input("Select your answer. ")
    if user_answer_two.isdigit()==True:
        second_score=int_answercheck(first_score, user_answer_two, int_answer_two)
    else:
        second_score=str_answercheck(first_score, user_answer_two, str_answer_two)

    print lines[3]
    print("A. 13"+'\n'+"B. 15") 
    str_answer_three="A"
    int_answer_three=13
    user_answer_three=raw_input("Select your answer. ")
    if user_answer_three.isdigit()==True:
        third_score=int_answercheck(second_score, user_answer_three, int_answer_three)
    else:
        third_score=str_answercheck(second_score, user_answer_three, str_answer_three)

    print lines[4]
    print("A. 30"+'\n'+"B. 32") 
    str_answer_four="B"
    int_answer_four=32
    user_answer_four=raw_input("Select your answer. ")
    if user_answer_four.isdigit()==True:
        fourth_score=int_answercheck(third_score, user_answer_four, int_answer_four)
    else:
        fourth_score=str_answercheck(third_score, user_answer_four, str_answer_four)

    print lines[5]
    print("A. 21"+'\n'+"B. 29") 
    str_answer_five="B"
    int_answer_five=29
    user_answer_five=raw_input("Select your answer. ")
    if user_answer_five.isdigit()==True:
        fifth_score=int_answercheck(fourth_score, user_answer_five, int_answer_five)
    else:
        fifth_score=str_answercheck(fourth_score, user_answer_five, str_answer_five)

以下是mathseasy.txt中的数据：

Welcome to the easy maths quiz.
What's 2+2?
What's 11-6?
What's 5*4?
What's 26/2?
What's 11*10?

以下是mathsmedium.txt中的数据：

Welcome to the medium maths quiz.
What's 5*6?
What's 79+31?
What's 26/2?
What's 4*8?
What's 50-21

Answer 1

下面是一些缩短代码的方法：

1. 您的str_和int_answercheck基本上是相同的函数，除了一个布尔表达式。与传递和比较这些值不同，为什么不编写一个使用布尔值并执行以下工作的底层函数: def check_answer(was_correct)：如果was_correct: print("Well to“) score+=1 score+=1(”错误，它是"+str_answer) raw_input(“按下回车继续”)返回分数str_answercheck(user_answer，str_answer)：返回check_answer(user_answer.upper() == str_answer.upper()) def int_answercheck(user_answer，int_answer)：返回check_answer(int(user_answer) == int_answer)
2. 您编写了两个应答检查函数，但始终使用if语句来确定要调用哪个函数。将该逻辑放入一个单一的应答检查函数中: def应答检查(user_answer，str_answer，int_answer)：if user_answer.isdigit()：返回int_answercheck(user_answer，int_answer)，否则:返回str_answercheck(user_answer，str_answer)
3. 如果答案检查返回true，则执行加法，而不是更新全局得分。这并不能缩短您的代码，但它是良好的编码风格(在不需要时避免使用全局变量)：如果回答检查(user_answer，str_answer，int_answer)：得分+= 1
4. 您正在从文件中打印行，然后从程序源代码中打印允许的答案。这意味着测试文件和源代码是同步的-它们不是相互独立的。既然源代码是常量的，为什么不把可能的答案放在测试文件中:欢迎参加简单的数学测试。2+2是什么？什么是11-6？如果您做了这些更改，您可以转换以下代码：0打印行1 print (“A.30”+‘\n’+“B.35 ") str_answer_one="A”int_answer_one=30 user_answer_one=raw_input(“选择您的答案”)。如果user_answer_one.isdigit()==True: first_score=int_answercheck(得分，user_answer_one，int_answer_one)，其他:first_score=str_answercheck(得分，user_answer_one，( str_answer_one)变成这样的代码:标题=行0打印头打印行1#问句打印行2#答案1打印行3.#答案2 user_answer =raw_input(“选择您的答案：") str_answer = "A”int_answer = 30如果回答检查(user_answer，str_answer，int_answer)：分数+= 1
5. 如果你有一种从输入文件中读取正确答案的方法就好了！哦，等等！欢迎来到简单的数学测验。2+2是什么？答: A. 4.什么是11-6？答:通过在文件中提供答案，你可以使你的代码更通用。(注:有一个明显的循环解决方案。但是，我在代码中没有看到任何循环，所以我假设您在学习中还没有达到这个阶段。)def问题(名词，行)：I= num *4+1打印行我#问句打印行I+1#答案1打印行I+2#答案2 user_answer =raw_input(“选择答案：")答案=行I+3.split()#”答案：“，"5”str_answer，int_answer =应答1，int(答案2)返回答案检查(user_answer，str_answer，int_answer)：如果答案正确，这将允许您问每个问题并返回真假: score =0如果问题(0，行)：得分+= 1如果问题(1，行)：得分+= 1如果问题(2，行)：分数+= 1(当您到达那里时，循环版本应该是显而易见的)。
6. 您正在处理不同程度的困难，作为单独的代码块。这是必要的，因为您的代码内置了答案。但有了问卷文件中的答案，你就不用再这么做了。相反，只需打开文件并在一个地方处理这些问题:问句= subject.lower() + difficulty.lower() + ".txt“，打开(问题)为幼稚:line= infile.readlines()打印行0评分=0如果问题(0，行)：score += 1如果问题(1，行)：得分+= 1如果问题(2，行)：分数+= 1

Answer 2

做答案检查的一个较短的方法是将两者结合在一起：

def answercheck(score, user_answer, str_answer, int_answer):
    if user_answer.upper() == str_answer or user_answer.upper() == int_answer:
        print("Well done")
        score+=1
    else:
        print ("Wrong, it was "+str(answer))
    raw_input("Press enter to continue")
    return score

注意:这意味着代码的其余部分必须进行编辑，才能始终发送到answercheck。