时髦的 While 循环/时髦的语法错误 Python 2.7 [关闭]

Question

我尝试了几种方法来修复此循环，但它无法正常工作。截至目前，它给了我一个语法错误，在第一个打印语句中的 yes 之后突出显示引号......我看不出有什么问题吗？

Ycount = 0
Ncount = 0
Valid = ["Y","y"]
InValid = ["N","n"]
Quit = ["Q","q"]
Uinp = ""

while Uinp != "Q":
    Uinp = raw_input("Did you answer Y or N to the question(enter Q to quit)? ")
    if Uinp in Valid:
        Ycount = Ycount + 1
        print "You have answered yes" ,Ycount, "times"
        print "You have answered no" ,Ncount, "times"
    elif Uinp in InValid:
        Ncount = Ncount + 1
        print "You have answered yes" ,Ycount, "times"
        print "You have answered no" ,Ncount, "times"
    elif Uinp in Quit:
        break

Answer 1

我已经在 python 2 下运行了您的代码，它按预期运行。

然而，在 python3 下，您需要进行一些更改才能运行它：

print "something"不再支持，需要使用 print ("something")

并且raw_input 被重命名为input

Ycount = 0
Ncount = 0
Valid = ["Y","y"]
InValid = ["N","n"]
Quit = ["Q","q"]
Uinp = ""

while Uinp != "Q":
    Uinp = input("Did you answer Y or N to the question(enter Q to quit)? ")
    if Uinp in Valid:
        Ycount = Ycount + 1
        print ("You have answered yes" ,Ycount, "times")
        print ("You have answered no" ,Ncount, "times")
    elif Uinp in InValid:
        Ncount = Ncount + 1
        print ("You have answered yes" ,Ycount, "times")
        print ("You have answered no" ,Ncount, "times")
    elif Uinp in Quit:
        break

Answer 2

编辑：Sammy Arous 给出了您的问题的解决方案——我要说的只是良好的 Python 实践。

打印出具有各种类型的字符串的首选 Pythonic 方法（在您的情况下，字符串，后跟 int，然后是字符串，是使用对字符串进行操作的 .format(*args) 函数。在这种情况下，您可以用

print ("You have answered yes {0} times".format(Ycount))

如果要打印多个参数，第一个由{0} 引用，下一个由'{1}' 引用，依此类推。

虽然使用 C 风格的 % 运算符来格式化字符串也是有效的（例如 You have answered yes %d times " % Ycount，但这不是首选。

尝试使用大括号语法。在大型项目中，它会显着提高您的代码速度（计算并打印一个字符串，而不是打印三个字符串），并且通常更符合 Python 的习惯。