Python中的字符串和变量问题

我花了几个小时试图解决一个问题，但没法再进一步了。这是因为缺乏理解为什么一些变量的行为方式与我预期的不同。我想我将问题识别为一个变量被解释为字符串而不是变量

这是我的密码：

# Global variables.
fillers = ["__1__", "__2__", "__3__", "__4__"]
difficulties = ["easy", "medium", "hard"]
easy_text = "The Three Little Pigs is a fable about three __1__ who build three __2__ of different materials. A big bad __3__ blows down the first two pigs' houses, made of straw and sticks respectively, but is unable to destroy the third pig's house, made of __4__."
easy_answers = ["pigs", "houses", "wolf", "bricks"]

所以在这种特殊情况下，我的问题围绕着变量easy_文本和easy_答案的显示

# Function to be called to check if difficulty is available
def difficulty_level(difficulties, user_input):
    for diff in difficulties:
        if diff in user_input:
            return diff
    return None

# User sets a difficulty level and variables for further use are being created accordingly
def diff_validation():
    user_input = raw_input("To do so enter either easy, medium or hard:")
    difficulty = difficulty_level(difficulties, user_input)
    if difficulty != None:
        print "Your difficulty level was sucessfully set to " + difficulty + "!"
        fl_text = difficulty + "_text"
        fl_answers = difficulty + "_answers"
        return fl_text, fl_answers

因此，我在这里根据所选的难度陈述了外语课文和外语答案。在我的例子中，它们变成了简单的文本和简单的答案

    else:
        print "Something went wrong please try again."
        return diff_validation()


# Initialising the game with welcome text and difficulty selection.
print "Hello and welcome to a short little game testing your general knowledge. Dont be intimidated you can choose your own difficulty level."
fl_text, fl_answers = diff_validation()
print fl_answers

现在我不明白我脑子里到底出了什么问题。如果我现在打印fl_答案，我会得到简单的答案，但我想打印列表。所以为了保持它的动态性，我不能只打印简单的答案，它需要了解自己，我想打印简单的答案列表

随着我们在代码中的进一步深入，原因变得越来越清楚：

def play_game(fl_string, fillers):
    replaced = []
    fl_string = fl_string.split()
    for word in fl_string:
        replacement = word_in_pos(word, fillers)
        if replacement != None:
            user_input = raw_input("Type in a: " + replacement + " ")
            answer_nr = 0
            if user_input == fl_answers[answer_nr]:
                word = word.replace(replacement, user_input)
                replaced.append(word)
                answer_nr + 1
            else:
                print "Your answer was wrong please try again."
                print fl_answers[1]
        else:
            replaced.append(word)
    replaced = " ".join(replaced)
    return replaced

print play_game(fl_text, fillers)

我试图打印出FLU答案的列表元素1，以检查该元素是否与用户输入相同。但它确实给了我第二个简单答案。所以我想我确实声明了它应该是对列表的引用，而不是定义的变量

也不是说我希望代码的底线是从fl_文本中得到easy_文本，但它只是不执行它所写的fl_文本。如果我手动输入easy_文本，那么至少它会被执行。程序本身的功能还没有发展到我满意的程度

我希望有人能理解我的问题，帮助我理解我在这里忘记了什么基本原则。

您的函数差异验证返回的是字符串而不是变量。因此，在打印时，应该打印字符串，而不是列表。要从匹配的变量名中获取列表变量，可以使用如下字典：

# Global variables.
fillers = ["__1__", "__2__", "__3__", "__4__"]
difficulties = ["easy", "medium", "hard"]
easy_text = "The Three Little Pigs is a fable about three __1__ who build three __2__ of different materials. A big bad __3__ blows down the first two pigs' houses, made of straw and sticks respectively, but is unable to destroy the third pig's house, made of __4__."
easy_answers = ["pigs", "houses", "wolf", "bricks"]

lists = {"easy_text":easy_text, "easy_answers":easy_answers}

然后检索：

fl_text, fl_answers = diff_validation()
print lists[f1_answers]

---

嘿，谢谢你抽出时间。要访问列表，这很好。但是我怎样才能得到第一个列表元素呢？我尝试打印列表[fl_答案[0]]什么列表？easy_text=三只小猪是一个关于三只用不同材料建造三只小猪的寓言。一个大坏蛋uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu。这个？在这种情况下，如果此答案解决了您的问题，则应打印列表[f1_text]，将问题标记为已解决，以便其他人知道此问题有解决方案。否，请打印fl_答案列表。这个问题还没有解决，但我马上就要解决了。当然，我会在得到最终答案后尽快更新。感谢实际列表[f1\U答案]是列表。因此，要从列表中获取第i个元素，必须编写列表[f1_answers][i]。因此，对于第一个元素，它将是列表[f1_answers][0]