时髦虽然循环/时髦的语法错误Python 2.7

Question

我试过几件事情来解决这个循环，它只是不会工作。至于现在它给了我一个语法错误，突出显示第一个打印语句中的yes后面的引号......我没有看到任何错误吗？

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

编辑：你的问题的解决方案是由森美阿鲁斯给出的 - 我提出的观点仅仅是良好的Python的做法。

打印出他们与各种类型的字符串（在你的情况，字符串，然后INT，其次是字符串中的首选，Python的方式，是使用上的字符串操作.format(*args)功能。在这种情况下，你可以使用

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

如果你要打印出多个参数，第一个是由{0}引用，接下来通过“{1}”，等等。

虽然它也适用于使用C-esque %运营商格式化蜇（例如You have answered yes %d times " % Ycount，这我不是首选。

尝试使用大括号语法。在大型项目中，它将显着提高代码速度（计算并打印一个字符串，而不是打印三个字符串），而且通常对Python更具惯用性。

Answer 2

我已经在python 2下运行了你的代码，它按预期工作。

在python3但是，也有让你运行它需要做一些改变：

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

和raw_input更名为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