Question

根据我收到的建议，我决定重做这个问题，这是我给第一年uni，python编码的一个任务问题。我的代码中有错误，无法确定修复它们的位置。 BUG 1即使笔已启动，程序运行时乌龟也会开始绘图。 BUG 2未定义的键，例如's，7，tab'触发space_bar函数

着色书

在此任务中，您将创建一个儿童着色游戏给定的图片可以通过追踪形状来着色然后填写它。控制如下。

箭头键 - 向左，向右移动“画笔”（乌龟光标）或减少固定的少量。

'z' - 撤消最后一步。

'r'，'g'，'b' - 将画笔颜色更改为红色，绿色或蓝色，分别。（如果是，您可以定义更多颜色你喜欢，但我们至少期望这三个。）

空格键 - 切换绘画模式。在“移动”模式下，这是初始模式，“刷子”（乌龟）四处移动屏幕没有绘图。在“油漆”模式下当它移动时，刷子留下一条彩色线条。最重要的是，当模式从“油漆”改变时为了“移动”，画笔描绘的区域是充满了色彩。

from turtle import *
from functools import partial 
bgpic("Colour_A_Turkey.gif") # change this to change the picture

#control the accuracy/speed of the drawing 
step_size =8
pensize(4)
penup()

# whenever spacebar is pressed the current state  and next state switch values 
current_state = penup
next_state = pendown
def space_bar():
    global current_state, next_state
    next_state()
    current_state, next_state = next_state, current_state
    #if the current stat is penup fill in with current colour
    if current_state == penup:
        end_fill()
    else:
        begin_fill()
onkey(space_bar, " ")

# undo do a mistake function
def mistake():
    undo()
onkey(mistake, "z")

#using partial function to store the following functions
#so they can be called  as arguments from a dictionary
#movement
strait = partial(fd, step_size)
reverse = partial(bk, step_size)
turn_rt = partial(rt, step_size)
turn_lf = partial(lt, step_size)

#colour
brow = partial(color, "brown")
gree = partial(color, "green")
yell = partial(color, "yellow")
oran = partial(color, "orange")
purp = partial(color, "purple")
red = partial(color, "red")
blue = partial(color, "blue")


#create a dictionary to store all the keys and there abilities 
key_action = {"b" : blue, "r" : red, "p" : purp, "o" : oran,\
          "y" : yell, "g" : gree, "w" : brow, "Right" : turn_rt , "Up" : strait,\
          "Down" : reverse, "Left" : turn_lf, "z" : undo()}

#when a key in then above dictionary
#is pressed it's function  is activated
for pressed_key, activated_key in key_action.items():
    onkey(activated_key, pressed_key)  

#make turtle look for key strokes with predefined function
listen()
#finish
done()

Answer 1

请记住，在Python中，一切都是对象，我们的意思是。函数也是对象。我们可以将函数存储在字典中，事实上这正是你在这里想要做的。

turn_lf = lt(step_size )

这里的关键问题是你要存储“一个用lt作为参数调用step_size的函数”，但是你刚刚调用了lt step_size立即作为参数，并存储返回值。

可以说，获得所需内容的最简单方法是使用functools.partial来“绑定”step_size参数。

from functools import partial
turn_lf = partial(lt, step_size) # `lt` is not called yet.

# Now, later on, we can do
turn_lf() # and it's just as if we did `lt(step_size)`.
# So now we can store `turn_lf` in a dict, look it up and call it later, etc.

# Similarly for all the other functions you want to make.

（另一个问题是你没有与此保持一致;如果你想要在一个字典中输入所有内容，那么你也需要指出color函数的绑定。'brown'毕竟，这只是一个字符串。幸运的是，这与其他函数一样简单：我们只是创建partial(color, 'brown')。

至于"z" : delete，嗯 - 我们没有任何参数可以绑定到undo。因此，虽然我们可以遵循模式并编写partial(undo)（注意，没有更多的参数，因为我们没有绑定任何东西），只需编写undo就更有意义了。直接

顺便说一句，我们可以简化这个：

for key in key_action:
    pressed_key = key
    activated_key = key_action[key]

对此：

for pressed_key, activated_key in key_action.items():

Answer 2

为什么要定义所有功能？我们可以直接使用：打破这个词：

key_action = {"b" : "blue", "r" : "red", "p" : "purple", "o" : "orange", "y" : "yellow","g" : "green", "w" : "brown","Right" : rt , "Left": lt, "Up" : fd, "Down" : bk,"z" : undo}


no_input = ["z"]

one_input = ["Right", "Left", "Up", "Down"]

for key in key_action:
    activated_key = key_action[key]
    if key in no_input:
       activated_key()
    elif key in one_input():
        activated_key(step_size)
    else:
        onkey(color(activated_key), key)

Answer 3

你的词典方法是正确的思维方式。但是，您的功能将无法正常运行。使用字典比你想象的简单。首先，您必须对 STRING 分别设置 FUNCTIONS 并正确处理它们。因为每个人的表现都不同：

# Get rid of the top four lines, they will not work
# All actions mapping
key_action = {"b" : "blue", "r" : "red", "p" : "purple", "o" : "orange",
              "y" : "yellow", "g" : "green", "w" : "brown",
              "Right" : rt , "Left": lt, "Up" : fd, "Down" : bk,
              "z" : undo}
# Note down which are special actions
# Functions with no input
foo = ["z"]
# Function with one input
bar = ["Right", "Left", "Up", "Down"]

# Assuming the input key is get here, I don't use turtle don't know how
# the input is read
key = listen()  # PSEUDOCODE!

# Handle the input
if key in foo:
     foo[key]() # Execute no input function
elif key in bar:
     foo[key](step_size)
else:
     onkey(key_action[key], key)

使用onkey（）和词典用乌龟（python）绘制的bug

3 个答案: