Python3 turtleを使用して衝突をテストするために移動位置を検出する

Question

import turtle 
import time 
import random 

# randomizing the ball's Y scale 
pong_y = (random.random()) * 350 
# setting screan size 
SIZE_X = 1280 
SIZE_Y = 760 
turtle.setup(SIZE_X, SIZE_Y) 

# the speed/grid 
SQUARE_SIZE = 20 

# refresh every 100 miliseconds 
TIME_STEP = 100 

TIME_STEP_BALL = 10 

# cloning turtles 
player1 = turtle.clone() 
player2 = turtle.clone() 
ball = turtle.clone() 

ball.ht() 
ball.penup() 
ball.fillcolor("red") 
ball.shape("circle") 
turtle.ht() 

# setting starting players position 
player1.penup() 
player1.shape("square") 
player1.ht() 
player1.goto(SIZE_X/2 - SQUARE_SIZE - 2, 0) 
player1.st() 
player2.penup() 
player2.shape("square") 
player2.ht() 
player2.goto(-SIZE_X/2 + SQUARE_SIZE + 2, 0) 
player2.st() 
ball.goto(- SIZE_X/2 + SQUARE_SIZE, 0) 
ball.st() 
# defining ARROW keys 
UP_ARROW = "Up" 
DOWN_ARROW = "Down" 

# giving values 

UP = 0 
DOWN = 1 
# starting p1's direction == up 
p1direction = UP 
# starting p2's direction == down 
p2direction = DOWN 


# up function 
def p1up(): 
    global p1direction 
    p1direction = UP 


# down functionw 
def p1down(): 
    global p1direction 
    p1direction = DOWN 


# p2 up function 
def p2up(): 
    global p2direction 
    p2direction = UP 


# p2 down function 
def p2down(): 
    global p2direction 
    p2direction = DOWN 


# detecting key press of p1 
turtle.onkeypress(p1up, "Up") 
turtle.onkeypress(p1down, "Down") 
# detecting keypress of p2 
turtle.onkeypress(p2up, "w") 
turtle.onkeypress(p2down, "s") 

turtle.listen() 


# making the moovment of p1 
def move_p1(): 
    global SQUARE_SIZE, TIME_STEP, p2_x, p2_y 
    # defining p1's x and y 
    player1_pos = player1.pos() 
    p1_x = player1_pos[0] 
    p1_y = player1_pos[1] 
    if p1direction == UP: 
     player1.goto(p1_x, p1_y + SQUARE_SIZE) 
    if p1direction == DOWN: 
     player1.goto(p1_x, p1_y - SQUARE_SIZE) 

    turtle.ontimer(move_p1, TIME_STEP) 


# p2 function 
def move_p2(): 
    global SQUARE_SIZE, TIME_STEP, p1_x, p1_y, player1_pos 
    # defining p2's x and y 
    player2_pos = player2.pos() 
    p2_x = player2_pos[0] 
    p2_y = player2_pos[1] 

    # choosing which position to go 
    if p2direction == UP: 
     player2.goto(p2_x, p2_y + SQUARE_SIZE) 
    if p2direction == DOWN: 
     player2.goto(p2_x, p2_y - SQUARE_SIZE) 

    turtle.ontimer(move_p2, TIME_STEP) 


def move_ball(): 
    global pong_y 
    ball_pos = ball.pos() 
    ball_x = ball_pos[0] 
    ball_y = ball_pos[1] 
    if ball_pos[0] < -SIZE_X/2: 
     print("hey") 

    if -80 < ball.pos()[0] - player1.pos()[0] < 80 and -80 < ball.pos()[1] - player1.pos()[1] < 80: 
     ball.goto(ball_x - SIZE_X, pong_y) 
    if -80 < player2.pos()[0] - ball.pos()[0] < 80 and -80 < player2.pos()[1] - ball.pos()[1] < 80: 
     print("suc") 
     ball.goto(ball_x + SIZE_X, pong_y) 


move_p2() 
move_p1() 
move_ball() 
turtle.mainloop() 

私のコードでは、2番目の衝突、プレイヤー1（右のプレーヤー）との衝突は機能しません。プレイヤー（if -80 < ball.pos()[0] - player1.pos()[0] < 80 and -80 < ball.pos()[1] - player1.pos()[1] < 80、およびif -80 < player2.pos()[0] - ball.pos()[0] < 80 and -80 < player2.pos()[1] - ball.pos()[1] < 80)の両方に同じif条件。（ところで、それはmove_ball()機能である）私を助けてください！

Answer 1

はあなたのコードを見ている、あなたは戻ってglobal文をおよそ読みする必要があるとする場合それは必要ではありません。あなたは、カメが何を提供しなければならないかを、以下のものの代わりに利用するべきです：

if -80 < ball.pos()[0] - player1.pos()[0] < 80 and -80 < ball.pos()[1] - player1.pos()[1] < 80: 

我々は亀distance()メソッドを使用することができます

if ball.distance(player1) < 80: 

を、同様に亀の見出しを活用します。実際に再生可能なコードの簡略化は次のとおりです。

from random import randint, choice 
from turtle import Turtle, Screen 

# setting screen size 
SIZE_X, SIZE_Y = 1280, 760 
screen = Screen() 
screen.setup(SIZE_X, SIZE_Y) 

# the speed/grid 
SQUARE_SIZE = 20 

# randomizing the ball's Y scale 
pong_y = randint(SQUARE_SIZE - SIZE_Y/2, SIZE_Y/2 - SQUARE_SIZE) 

# refresh every 100 milliseconds 
TIME_STEP = 100 

# create turtles and set player starting positions 
player1 = Turtle('square', visible=False) 
player1.speed('fastest') 
player1.penup() 
player1.setx(SIZE_X/2 - SQUARE_SIZE * 2) 
player1.setheading(choice([-90, 90])) 
player1.showturtle() 

player2 = Turtle('square', visible=False) 
player2.speed('fastest') 
player2.penup() 
player2.setx(SQUARE_SIZE * 2 - SIZE_X/2) 
player2.setheading(choice([-90, 90])) 
player2.showturtle() 

ball = Turtle('circle', visible=False) 
ball.speed('fastest') 
ball.penup() 
ball.fillcolor('red') 
ball.goto(0, pong_y) 
ball.setheading(choice([180, 0])) 
ball.showturtle() 

# giving values 
UP, DOWN = 0, 1 

# up & down functions 
def p1up(): 
    screen.onkeypress(None, 'Up') 
    player1.setheading(90) 
    screen.onkeypress(p1up, 'Up') 

def p1down(): 
    screen.onkeypress(None, 'Down') 
    player1.setheading(-90) 
    screen.onkeypress(p1down, 'Down') 

# p2 up & down functions 
def p2up(): 
    screen.onkeypress(None, 'w') 
    player2.setheading(90) 
    screen.onkeypress(p2up, 'w') 

def p2down(): 
    screen.onkeypress(None, 's') 
    player2.setheading(-90) 
    screen.onkeypress(p2down, 's') 

# detect key press of p1 
screen.onkeypress(p1up, 'Up') 
screen.onkeypress(p1down, 'Down') 

# detect keypress of p2 
screen.onkeypress(p2up, 'w') 
screen.onkeypress(p2down, 's') 

screen.listen() 

# move objects 
def move_p1(): 
    player1.forward(SQUARE_SIZE) 
    screen.ontimer(move_p1, TIME_STEP) 

def move_p2(): 
    player2.forward(SQUARE_SIZE) 
    screen.ontimer(move_p2, TIME_STEP) 

def move_ball(): 
    global pong_y 

    ball_x = ball.xcor() 

    if ball_x < -SIZE_X/2 or ball_x > SIZE_X/2: # out of bounds, start over 
     ball.hideturtle() 
     pong_y = randint(SQUARE_SIZE - SIZE_Y/2, SIZE_Y/2 - SQUARE_SIZE) 
     ball.goto(0, pong_y) 
     ball.setheading(choice([180, 0])) 
     ball.showturtle() 

    elif ball.heading() == 0 and ball.distance(player1) < 80 or ball.heading() == 180 and ball.distance(player2) < 80: 
     ball.setheading(180 - ball.heading()) # player hit ball, reverse direction 

    ball.forward(SQUARE_SIZE) 

    screen.ontimer(move_ball, TIME_STEP) 

move_p2() 
move_p1() 
move_ball() 

screen.mainloop() 

これはまだ実際のゲームである必要があります。