类型错误与执行（）

Question

我想制作一个程序，根据用户输入绘制某些行。

#import turtle program 
#called it cassy cause I was trying to come up with a fun name for an arrow 
that they call 'classic'. I tried ok. 
import turtle 
cassy = turtle.Turtle() 
cassy.shape('classic') 

#set's the window shape, so I can give myself more room 
#for reference, putting startx and y as none centers it 
turtle.setup(width=1200, height=600, startx=None, starty=None) 



#used to find the height of what the screen size is so I could figure out 
what it was set at. 
"""print(turtle.window_width()) 
print(turtle.window_height())""" 


#gives our wonderful circle a home (where it starts), to the right and 
middle of the box 
#old code (turtle.setworldcoordinates(-20, -20, 0, 20)) 
cassy.hideturtle() 
cassy.penup() 
cassy.goto(580, 0) 

#sets the turtle to be facing left 
cassy.left(180) 
cassy.showturtle() 
cassy.pendown() 

#as always, I gotta have that sweet sweet starting message. 
print("Welcome to the line generator!") 
print("To start, type in a letter") 


#using a dictionary function so I can assign each character a "Line" 
line = { 
    "a" : "cassy.left(20)", 
    "b" : "cassy.forward(20)", 
    "c" : "cassy.right(45)", 
    "d" : "cassy.right(70)",  
    "e" : "cassy.circle(10)", 
    "f" : "cassy.left(15)", 
    "g" : "cassy.forward(15)", 
    "h" : "cassy.left(60)", 
    "i" : "cassy.forward(10)", 
    "j" : "cassy.forward(55)", 
    "k" : "cassy.circle(50, 100)", 
    "l" : "cassy.right(80)", 
    "m" : "cassy.forward(35)", 
    "n" : "cassy.left(75)", 
    "o" : "cassy.forward(45)", 
    "p" : "cassy.right(5)", 
    "q" : "cassy.forward(100)", 
    "r" : "cassy.right(65)", 
    "s" : "cassy.forward(125)", 
    "t" : "cassy.left(25)", 
    "u" : "cassy.forward(150)", 
    "v" : "cassy.right(95)", 
    "w" : "cassy.forward(120)", 
    "x" : "cassy.circle(25, 175)", 
    "y" : "cassy.circle(5, 190)", 
    "z" : "cassy.forward(175)", 
    "0" : "cassy.right(20)", 
    "1" : "cassy.forward(210)", 
    "2" : "cassy.left(120)", 
    "3" : "cassy.forward(115)", 
    "4" : "cassy.forward(15)", 
    "5" : "cassy.circle(30, 150)", 
    "6" : "cassy.left(70)", 
    "7" : "cassy.right(40)", 
    "8" : "cassy.backward(30)", 
    "9" : "cassy.backward(130)" 
} 

#let's the user type in letters/numbers (and only those (unless I want to   
mix things up; we'll see)) 
while True: 
    draw = input("Type anything to draw! ") 

    #if statement to make sure that they only type in numbers or letters 
    if(draw.isalnum() == True): 

     #I can only assume that this executes the code inside the quotes 
     #but I haven't looked too deeply into what exec() does 
     exec(line.get(draw)) 

     #this seperates each character drawn by a space, for funsies (yes I know it's a weird thing to say) 
     cassy.penup() 
     cassy.forward(15) 
     cassy.pendown() 

    #else statement in case they type in other charaters that aren't numbers or letters 
    else: 
     print("Please only use letters or numbers.") 

好了，所以我知道这是一个很多的阅读，但我真正的问题是这行代码

exec(line.get(draw)) 

到目前为止，当我在1个字母同时键入程序才有效。但如果我尝试并键入多个字母，比如那句“你好”它给了我这个错误

Traceback (most recent call last): 
    File "C:/Users/73059/Final Project/Final Project funstuffs.py", line 87, in <module> 
    exec(line.get(draw)) 
TypeError: exec() arg 1 must be a string, bytes or code object 

我真的不知道了很多关于函数的exec（），我已经一直在试图看看它的东西，这是一个项目，我真的没有太多时间。

如果有任何问题，或者你不明白我想完成什么，请告诉我。我有点绝望。

感谢

Answer 1

你的问题是不是真的与exec电话，但与价值你传递作为参数。当您执行line.get(draw)时，可以在line字典中查找变量draw的值。如果该值不存在（因为draw是多字符字符串，则不会），则会得到None。运行exec(None)会为您提供TypeError。

我有几种解决此问题的方法。你想要遵循哪些可能取决于你的程序需要什么样的行为。

首先，我建议使用括号索引字典，而不是get方法。这会将您从TypeError（从exec）得到的错误更改为字典中的KeyError。这仍然是一个例外，但它至少可以更准确地指出问题的原因（查找的字符串不在字典中）。

下一个建议是在尝试查找它之前检查多字符字符串，并向用户报告错误。你可以检查len(draw)是否等于1，如果是的话只运行其余的代码。我会将这张支票放在您当前对字母数字字符进行检查的地方。

下面是这两个补丁会是什么样在一起：

while True: 
    draw = input("Type anything to draw! ") 

    if draw.isalnum() and len(draw) == 1: # check length, "== True" and parens were unneeded 
     exec(line[draw]) # use indexing instead of "get" to generate better exception types 

    else: 
     print("Please only enter a single letter or number at a time.") 

另外，如果你想允许更长的输入（与输入的每个字符被另案处理），你可以在你的代码改过来循环而不是试图一次性处理它们。这将是这个样子：

while True: 
    draw = input("Type anything to draw! ") 

    if draw.isalnum(): # no length check needed in this version 

     for character in draw: # loop over characters 
      exec(line[character]) 

      cassy.penup()  # it's not clear if you want these lines inside the inner loop 
      cassy.forward(15) # if not, you can unindent them one level 
      cassy.pendown() 

    else: 
     print("Please only use letters or numbers.") 

虽然这应该工作，使用exec通常是编写代码一个糟糕的方式。更好的方法是将函数（或其他可调用对象）放入字典中，然后调用从索引中返回的值。您可以使用lambda关键字创建匿名功能。 Lambda函数有一些限制，但它们足够满足我们的需求。下面是你可能会看在字典中使用lambda表达式：

line = { 
    "a" : lambda: cassy.left(20), 
    "b" : lambda: cassy.forward(20), 
    "c" : lambda: cassy.right(45), 
    "d" : lambda: cassy.right(70),  
    "e" : lambda: cassy.circle(10), 
    "f" : lambda: cassy.left(15), 
    "g" : lambda: cassy.forward(15), 
    "h" : lambda: cassy.left(60), 
    "i" : lambda: cassy.forward(10), 
    "j" : lambda: cassy.forward(55), 
    "k" : lambda: cassy.circle(50, 100), 
    "l" : lambda: cassy.right(80), 
    "m" : lambda: cassy.forward(35), 
    "n" : lambda: cassy.left(75), 
    "o" : lambda: cassy.forward(45), 
    "p" : lambda: cassy.right(5), 
    "q" : lambda: cassy.forward(100), 
    "r" : lambda: cassy.right(65), 
    "s" : lambda: cassy.forward(125), 
    "t" : lambda: cassy.left(25), 
    "u" : lambda: cassy.forward(150), 
    "v" : lambda: cassy.right(95), 
    "w" : lambda: cassy.forward(120), 
    "x" : lambda: cassy.circle(25, 175), 
    "y" : lambda: cassy.circle(5, 190), 
    "z" : lambda: cassy.forward(175), 
    "0" : lambda: cassy.right(20), 
    "1" : lambda: cassy.forward(210), 
    "2" : lambda: cassy.left(120), 
    "3" : lambda: cassy.forward(115), 
    "4" : lambda: cassy.forward(15), 
    "5" : lambda: cassy.circle(30, 150), 
    "6" : lambda: cassy.left(70), 
    "7" : lambda: cassy.right(40), 
    "8" : lambda: cassy.backward(30), 
    "9" : lambda: cassy.backward(130) 
} 

while True: 
    draw = input("Type anything to draw! ") 

    if(draw.isalnum() == True): 

     for character in draw: # loop over characters 
      line[character]() # call the value looked up in the dict 

      cassy.penup() # it's not clear if you want these lines inside the inner loop 
      cassy.forward(15) # if not, you can unindent them one level 
      cassy.pendown() 

    else: 
     print("Please only use letters or numbers.") 

还有其他几种方法，你可以使可调用，如使用functools.partial乌龟方法及其自变量绑定在一起。