让点跟随更大的点
Getting dots to follow bigger dot
所以这是我制作的一个程序,目的是让一些点被一个更大的点吸引,并让那个更大的点长大。我现在面临的问题是这些点没有跟随更大的点,而是似乎远离了它。我让它更接近的方法是将点转换为 (0,0),另一个转换为 [t2.xcor() - t1.xcor() , t2.ycor( )- t1.ycor()] ,然后用勾股定理求C,然后用反余弦求出它需要面对的角度,才能向更大的点移动。
from turtle import *
import sys
from math import *
#grows t1 shape + has it follow cursor
def grow(x, y):
t1.ondrag(None)
t1.goto(x,y)
global big, nig
t1.shapesize(big,nig)
big += .004
nig += .004
t1.ondrag(grow)
follow()
#has create()'d dots follow t1
def follow():
global count
#t1.ondrag(None)
screen.tracer(0,0)
for p in lx:
#print(lx[0:5])
t2.goto(p, ly[count])
t2.dot(4, "white")
if ly[count] != 0:
yb = abs(t2.ycor() - t1.ycor())
xb = abs((t2.xcor() - t1.xcor()))
c = sqrt((xb**2 + yb**2))
#print(y,x,c)
#print(lx)
t2.seth(360 - degrees(acos(yb/c)))
else:
t2.seth(0)
t2.forward(20)
t2.dot(4, "purple")
lx.pop(count)
ly.pop(count)
lx.insert(count, t2.xcor())
ly.insert(count, t2.ycor())
count += 1
#print(lx[0:5])
#screen.update()
screen.tracer(1,10)
count = 0
#t1.ondrag(follow)
#quits program
def quit():
screen.bye()
sys.exit(0)
#create()'s dots with t2
def create():
screen.tracer(0,0)
global nux, nuy, count3
while nuy > -400:
t2.goto(nux, nuy)
if t2.pos() != t1.pos():
t2.dot(4, "purple")
lx.append(t2.xcor())
ly.append(t2.ycor())
nux += 50
count3 += 1
if count3 == 17:
nuy = nuy - 50
nux = -400
count3 = 0
screen.tracer(1, 10)
#variables
count3 = count = 0
big = nig = .02
lx = []
ly = []
nux = -400
nuy = 300
screen = Screen()
screen.screensize(4000,4000)
t2 = Turtle()
t2.ht()
t2.pu()
t2.speed(0)
t2.shape("turtle")
t1 = Turtle()
t1.shape("circle")
t1.penup()
t1.speed(0)
t1.color("purple")
t1.shapesize(.2, .2)
create()
screen.listen()
screen.onkeypress(quit, "Escape")
t1.ondrag(grow)
#t1.ondrag(follow)
#screen.update()
screen.mainloop()
我发现您的代码有两个(类似的)问题。首先,您可以在重新发明 turtle 的 .towards()
方法时扔掉花哨的数学,它可以为您提供所需的角度。其次,您正在重新发明 stamps,与大多数 turtle 元素不同,它可以通过 clearstamp()
从屏幕上干净地清除。此外,您使用的是并行坐标数组,这表明缺少适当的数据结构。我已将其替换为包含位置和邮票元组的单个数组。
我已经调整了你的程序的动态,使点独立行动(在计时器上)而不依赖于光标的移动。 IE。无论光标是否移动,它们都会向光标移动。另外,我让光标仅在点到达并消失时才增长:
from turtle import Turtle, Screen
CURSOR_SIZE = 20
def move(x, y):
""" has it follow cursor """
t1.ondrag(None)
t1.goto(x, y)
screen.update()
t1.ondrag(move)
def grow():
""" grows t1 shape """
global t1_size
t1_size += 0.4
t1.shapesize(t1_size / CURSOR_SIZE)
screen.update()
def follow():
""" has create()'d dots follow t1 """
global circles
new_circles = []
for (x, y), stamp in circles:
t2.clearstamp(stamp)
t2.goto(x, y)
t2.setheading(t2.towards(t1))
t2.forward(2)
if t2.distance(t1) > t1_size // 2:
new_circles.append((t2.position(), t2.stamp()))
else:
grow() # we ate one, make t1 fatter
screen.update()
circles = new_circles
if circles:
screen.ontimer(follow, 50)
def create():
""" create()'s dots with t2 """
count = 0
nux, nuy = -400, 300
while nuy > -400:
t2.goto(nux, nuy)
if t2.distance(t1) > t1_size // 2:
circles.append((t2.position(), t2.stamp()))
nux += 50
count += 1
if count == 17:
nuy -= 50
nux = -400
count = 0
screen.update()
# variables
t1_size = 4
circles = []
screen = Screen()
screen.screensize(900, 900)
t2 = Turtle('circle', visible=False)
t2.shapesize(4 / CURSOR_SIZE)
t2.speed('fastest')
t2.color('purple')
t2.penup()
t1 = Turtle('circle')
t1.shapesize(t1_size / CURSOR_SIZE)
t1.speed('fastest')
t1.color('orange')
t1.penup()
t1.ondrag(move)
screen.tracer(False)
create()
follow()
screen.mainloop()
您应该能够重新编写此代码以执行您想要的任何操作。我强烈建议您花一些时间阅读 Turtle 文档,这样您就不需要重新发明它的许多功能。
所以这是我制作的一个程序,目的是让一些点被一个更大的点吸引,并让那个更大的点长大。我现在面临的问题是这些点没有跟随更大的点,而是似乎远离了它。我让它更接近的方法是将点转换为 (0,0),另一个转换为 [t2.xcor() - t1.xcor() , t2.ycor( )- t1.ycor()] ,然后用勾股定理求C,然后用反余弦求出它需要面对的角度,才能向更大的点移动。
from turtle import *
import sys
from math import *
#grows t1 shape + has it follow cursor
def grow(x, y):
t1.ondrag(None)
t1.goto(x,y)
global big, nig
t1.shapesize(big,nig)
big += .004
nig += .004
t1.ondrag(grow)
follow()
#has create()'d dots follow t1
def follow():
global count
#t1.ondrag(None)
screen.tracer(0,0)
for p in lx:
#print(lx[0:5])
t2.goto(p, ly[count])
t2.dot(4, "white")
if ly[count] != 0:
yb = abs(t2.ycor() - t1.ycor())
xb = abs((t2.xcor() - t1.xcor()))
c = sqrt((xb**2 + yb**2))
#print(y,x,c)
#print(lx)
t2.seth(360 - degrees(acos(yb/c)))
else:
t2.seth(0)
t2.forward(20)
t2.dot(4, "purple")
lx.pop(count)
ly.pop(count)
lx.insert(count, t2.xcor())
ly.insert(count, t2.ycor())
count += 1
#print(lx[0:5])
#screen.update()
screen.tracer(1,10)
count = 0
#t1.ondrag(follow)
#quits program
def quit():
screen.bye()
sys.exit(0)
#create()'s dots with t2
def create():
screen.tracer(0,0)
global nux, nuy, count3
while nuy > -400:
t2.goto(nux, nuy)
if t2.pos() != t1.pos():
t2.dot(4, "purple")
lx.append(t2.xcor())
ly.append(t2.ycor())
nux += 50
count3 += 1
if count3 == 17:
nuy = nuy - 50
nux = -400
count3 = 0
screen.tracer(1, 10)
#variables
count3 = count = 0
big = nig = .02
lx = []
ly = []
nux = -400
nuy = 300
screen = Screen()
screen.screensize(4000,4000)
t2 = Turtle()
t2.ht()
t2.pu()
t2.speed(0)
t2.shape("turtle")
t1 = Turtle()
t1.shape("circle")
t1.penup()
t1.speed(0)
t1.color("purple")
t1.shapesize(.2, .2)
create()
screen.listen()
screen.onkeypress(quit, "Escape")
t1.ondrag(grow)
#t1.ondrag(follow)
#screen.update()
screen.mainloop()
我发现您的代码有两个(类似的)问题。首先,您可以在重新发明 turtle 的 .towards()
方法时扔掉花哨的数学,它可以为您提供所需的角度。其次,您正在重新发明 stamps,与大多数 turtle 元素不同,它可以通过 clearstamp()
从屏幕上干净地清除。此外,您使用的是并行坐标数组,这表明缺少适当的数据结构。我已将其替换为包含位置和邮票元组的单个数组。
我已经调整了你的程序的动态,使点独立行动(在计时器上)而不依赖于光标的移动。 IE。无论光标是否移动,它们都会向光标移动。另外,我让光标仅在点到达并消失时才增长:
from turtle import Turtle, Screen
CURSOR_SIZE = 20
def move(x, y):
""" has it follow cursor """
t1.ondrag(None)
t1.goto(x, y)
screen.update()
t1.ondrag(move)
def grow():
""" grows t1 shape """
global t1_size
t1_size += 0.4
t1.shapesize(t1_size / CURSOR_SIZE)
screen.update()
def follow():
""" has create()'d dots follow t1 """
global circles
new_circles = []
for (x, y), stamp in circles:
t2.clearstamp(stamp)
t2.goto(x, y)
t2.setheading(t2.towards(t1))
t2.forward(2)
if t2.distance(t1) > t1_size // 2:
new_circles.append((t2.position(), t2.stamp()))
else:
grow() # we ate one, make t1 fatter
screen.update()
circles = new_circles
if circles:
screen.ontimer(follow, 50)
def create():
""" create()'s dots with t2 """
count = 0
nux, nuy = -400, 300
while nuy > -400:
t2.goto(nux, nuy)
if t2.distance(t1) > t1_size // 2:
circles.append((t2.position(), t2.stamp()))
nux += 50
count += 1
if count == 17:
nuy -= 50
nux = -400
count = 0
screen.update()
# variables
t1_size = 4
circles = []
screen = Screen()
screen.screensize(900, 900)
t2 = Turtle('circle', visible=False)
t2.shapesize(4 / CURSOR_SIZE)
t2.speed('fastest')
t2.color('purple')
t2.penup()
t1 = Turtle('circle')
t1.shapesize(t1_size / CURSOR_SIZE)
t1.speed('fastest')
t1.color('orange')
t1.penup()
t1.ondrag(move)
screen.tracer(False)
create()
follow()
screen.mainloop()
您应该能够重新编写此代码以执行您想要的任何操作。我强烈建议您花一些时间阅读 Turtle 文档,这样您就不需要重新发明它的许多功能。