Python 制作括号程序？我卡住了

这是Python。我正在尝试编写一个程序，要求用户在不使用全局变量的情况下输入字符串。如果字符串只有并排的括号，那么它是偶数。如果它有字母、数字或括号隔开，那么它是不均匀的。例如，and和是偶数，而and和pie和非偶数。下面是我到目前为止写的内容。我的程序一直在无限次地打印“输入字符串”，我现在被这个问题困住了

selection = 0
def recursion():
#The line below keeps on repeating infinitely.
    myString = str(input("Enter your string: "))
    if not myString.replace('()', ''):
        print("The string is even.")
    else:
        print("The string is not even.")

while selection != 3:
    selection = int(input("Select an option: "))

    #Ignore this.
    if selection == 1:
        print("Hi")

    #This won't work.
    elif selection == 2:
        def recursion():
            recursion()

---

这将输出正确的偶数/非偶数答案

 selection = 0
 def recursion():
    myString = str(raw_input("Enter your string: "))  #Use raw_input or ( ) will be () 
    paren = 0
    for char in myString:
        if char == "(":
            paren += 1
        elif char == ")":
            paren -= 1
        else:
            print "Not Even"
            return

        if paren < 0:
            print "Not even"
            return
    if paren != 0:
        print "Not even"
        return
    print "Even"
    return

while selection != 3:
    selection = int(input("Select an option: "))

    #Ignore this.
    if selection == 1:
        print("Hi")

    #This won't work.
    elif selection == 2:
        recursion()    #This isn't a recursive function - naming it as so is...

---

除非您使用的是Python3，否则您应该使用原始输入而不是输入，因为输入以与输入最匹配的任何类型返回结果，而原始输入总是返回字符串。在Python3中，输入总是返回一个字符串。还有，为什么要重新定义递归？从elif语句中调用它。例如：

selection = 0
def recursion():
#The line below keeps on repeating infinitely.
    myString = raw_input("Enter your string: ") # No need to convert to string.
    if not myString.replace('()', ''):
        print("The string is even.")
    else:
        print("The string is not even.")

while selection != 3:
    selection = int(raw_input("Select an option: "))

    #Ignore this.
    if selection == 1:
        print("Hi")

    #This won't work.
    elif selection == 2:
        recursion() # Just call it, as your program already
                    # recurs because of the if statement.

---

什么是def recursion:elif选择项下的递归==2。我的程序一直无限次地打印“输入字符串”。有趣的是，当我运行它时，它会一直打印选择一个选项：无限直到我输入3。这是您最新的代码吗？我不知道你怎么能达到输入字符串的提示。根据Kevin和我在运行你的程序时得到的结果，在我看来，如果你在elic selection==2:下取出def递归，然后放入递归，一切都会正常。剩下的就是修复递归中的if语句，以完成您希望它完成的任务