Question

我对我的一个项目有些困难。我试图将一个变量分配给一个列表项，调用该项，然后无限期地重复该过程。我在Turtle中做这件事。

代码的目的是绘制一个彩色圆圈。目前，我设置它从列表中随机选择一种颜色。我宁愿它从头到尾遍历列表并重复绘制列表中的下一个颜色。

import turtle as t
import random as r

# list of shades of blue
colourBlue = ['midnight blue', 'navy', 'cornflower blue', 'dark slate blue', 
'slate blue', 'medium slate blue', 'light slate blue', 'medium blue', 'royal 
blue', 'blue', 'dodger blue', 'deep sky blue']


# Call a colour from the list and draw a circle of said colour
def circle():
    t.pendown()
    t.begin_fill()        
    t.color(r.choice(colourBlue))
    t.circle(10)
    t.end_fill()
    t.penup()

# Defines a function that loops through ColourBlue list

def colourPick():
    colourBlueLen = len(colourBlue)
    for i in range(11, colourBlueLen):
        i = colourBlue[0]

到目前为止，我已经建立了一种方法来选择列表中的项目，但我不确定如何将其分配给变量，在t.color()函数中调用它并在整个过程中重复该过程清单。

Answer 1

我宁愿它从头到尾反复浏览列表绘制下一个颜色

如果您想按顺序处理颜色列表，但又不想受列表本身约束，我建议itertools.cycle()。它允许您在不考虑实际颜色数量的情况下，一次又一次地遍历颜色列表，而不考虑实际颜色数量：

from itertools import cycle
from turtle import Turtle, Screen

# list of shades of blue
BLUE_SHADES = cycle(['midnight blue', 'navy', 'cornflower blue', 'dark slate blue', \
    'slate blue', 'medium slate blue', 'light slate blue', 'medium blue', 'royal blue', \
    'blue', 'dodger blue', 'deep sky blue'])

# Call a colour from the list and draw a circle of said colour
def circle(turtle):
    turtle.color(next(BLUE_SHADES))
    turtle.begin_fill()
    turtle.circle(50)
    turtle.end_fill()

screen = Screen()

yertle = Turtle(visible=False)
yertle.speed('fastest')

for _ in range(120):
    circle(yertle)
    yertle.right(3)

screen.exitonclick()

如果您只想通过颜色列表工作一次，那也很容易。只需使用颜色列表本身作为迭代目标：

from turtle import Turtle, Screen

# list of shades of blue
BLUE_SHADES = ['midnight blue', 'navy', 'cornflower blue', 'dark slate blue', \
    'slate blue', 'medium slate blue', 'light slate blue', 'medium blue', 'royal blue', \
    'blue', 'dodger blue', 'deep sky blue']

# Call a colour from the list and draw a circle of said colour
def circle(turtle, color):
    turtle.color(color)
    turtle.begin_fill()
    turtle.circle(50)
    turtle.end_fill()

screen = Screen()

yertle = Turtle(visible=False)
yertle.speed('fastest')

for shade in BLUE_SHADES:
    circle(yertle, shade)
    yertle.right(360 / len(BLUE_SHADES))

screen.exitonclick()

Answer 2

我认为您只想将参数传递给circle：

def colourPick():
    for c in colourBlue:
        circle(c)

然后使用该参数代替r.choice(colourBlue)。

Answer 3

我设法在朋友的帮助下找到解决方案。

colourBlue = ['midnight blue', 'navy', 'cornflower blue', 'dark slate blue', 
'slate blue', 'medium slate blue', 'light slate blue', 'medium blue', 'royal 
blue', 'blue', 'dodger blue', 'deep sky blue']

currentColour = 0

# Establishes a function that calls a each colour from the list
def circle():
    t.pendown()
    t.begin_fill()        
    t.color(colourPick())
    t.circle(10)
    t.end_fill()
    t.penup()

def colourPick():
    global currentColour
    colourBlueLen = len(colourBlue)

    # If the last colour in the list has been used, reset it back to 0
    if currentColour == colourBlueLen:
        currentColour = 0

    colour = colourBlue[currentColour]

    # Increment currentColour values
    currentColour += 1

    return colour

circle()

迭代一个列表，为每个项目分配一个变量并返回它

3 个答案: