Python 迭代一个列表,为每个项目分配一个变量并返回它
我的一个项目有点困难。我试图为列表项分配一个变量,调用该项,然后无限期地重复该过程。我在海龟体内做这件事 代码的目的是绘制一个彩色圆圈。目前,我已经设置好,它可以从列表中随机选择一种颜色。我宁愿它从头到尾地浏览列表,并在列表中反复绘制下一种颜色Python 迭代一个列表,为每个项目分配一个变量并返回它,python,loops,range,turtle-graphics,Python,Loops,Range,Turtle Graphics,我的一个项目有点困难。我试图为列表项分配一个变量,调用该项,然后无限期地重复该过程。我在海龟体内做这件事 代码的目的是绘制一个彩色圆圈。目前,我已经设置好,它可以从列表中随机选择一种颜色。我宁愿它从头到尾地浏览列表,并在列表中反复绘制下一种颜色 import turtle as t import random as r # list of shades of blue colourBlue = ['midnight blue', 'navy', 'cornflower blue', 'dark
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,我已经建立了一种在列表中选择项目的方法,但我不确定应该如何将其分配给变量,在
t.color()
函数中调用它,并在整个列表中重复该过程。我想您只想将一个参数传递给circle
:
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]
def colourPick():
for c in colourBlue:
circle(c)
然后使用该参数而不是r.choice(colorblue)
我宁愿它从头到尾反复地浏览清单
画下一种颜色
如果您想按顺序处理颜色列表,但又不想受列表本身的约束,我建议您使用itertools.cycle()
。它将允许您在不考虑实际颜色数量的情况下,根据需要反复查看颜色列表:
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()
如果你只想浏览一下颜色列表,那也很简单。只需将颜色列表本身用作迭代目标:
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()
在朋友的帮助下,我设法想出了一个解决办法
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()
太棒了,非常感谢。我试着用谷歌搜索解决方案,但从未遇到itertools或cycle函数。您可以使用模运算符(
%
)而不是if
语句,方法是执行currentColour=(currentColour+1)%len(colourBlue)
,使currentColour
保持在正确的范围内。