对于python /编程而言,这是我最大的项目。
我正在编写一个程序,可以为你做SUVAT方程式。 (SUVAT方程用于找到具有恒定速度的物体的位移,开始/结束速度,加速度和行进时间,您可以将它们称为不同的东西。)
我做了这个清单:
variables = ["Displacement", "Start Velocity", "End Velocity", "Acceleration", "Time"]
用于以下while / for循环:
a = 0
while a==0:
for variable in variables:
# choice1 is what the user is looking to calculate
choice1 = raw_input("Welcome to Mattin's SVUVAT Simulator! Choose the value you are trying to find. You can pick from " + str(variables))
# will execute the following code when the for loop reaches an item that matches the raw_input
if choice1 == variable:
print "You chave chosen", choice1
variables.remove(variable) #Removes the chosen variable from the list, so the new list can be used later on
a = 1 # Ends the for loop by making the while loop false
# This part is so that the error message will not show when the raw_input does not match with the 4 items in the list the user has not chosen
else:
if choice1 == "Displacement":
pass
elif choice1 == "Start Velocity":
pass
elif choice1 == "End Velocity":
pass
elif choice1 == "Acceleration":
pass
# This error message will show if the input did not match any item in the list
else:
print "Sorry, I didn't understand that, try again. Make sure your spelling is correct (Case Sensitive), and that you did not inlcude the quotation marks."
希望我在代码中写的评论应该解释我的意图,如果没有,请随时提出任何问题。
问题在于,当我运行代码并输入choice1时,for循环会激活最后一行代码:
else:
print "Sorry, I didn't understand that, try again. Make sure your spelling is correct (Case Sensitive), and that you did not inlcude the quotation marks."
然后提示我再次输入输入,并且会多次执行此操作以获取我正在键入的列表中的项目。
但是,我特意编写了如果我输入的内容与列表中的项目不匹配,for循环当前正在检查,但确实匹配列表中的其他项目之一,那么它应该传递并循环以检查下一个项目。
我可能做了一些愚蠢的事,但我没有看到它,所以请帮我弄清楚我必须做些什么才能得到我想要的结果?我认为这是我错误的语法所以这就是为什么这就是标题。
感谢您的帮助,我很感激。
答案 0 :(得分:2)
除了粘贴代码中的缩进问题外,我还会重写它:
while True:
choice = raw_input('...')
if choice in variables:
print "You chave chosen", choice
# Remove the chosen member from the list
variables = [v for v in variables if v != choice]
# Break out of loop
break
# Print error messages etc.
还要记住,字符串比较区分大小写。 I.e 'Displacement' != 'displacement'
。