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pygame怎么计算距离
在Pygame中,计算两点之间的距离可以使用几何公式。给定两点P1(x1, y1)和P2(x2, y2),距离D可以用以下公式计算:
D = √((x2 - x1)² + (y2 - y1)²)
在Python中,你可以使用math库中的sqrt函数来计算平方根。以下是一个计算两点之间距离的Pygame示例代码:
import math
import pygame
point1 = pygame.math.Vector2(100, 100)
point2 = pygame.math.Vector2(200, 200)
distance = point1.distance_to(point2)
print(f"The distance between the two points is: {distance}")
在这个例子中,我们使用了pygame的math.Vector2类,它内置了一个distance_to方法来计算两个点之间的距离。这样可以简化计算过程,并提供更为直观和易于理解的代码。
end_point_x = start_point[0] + ray_length * math.cos(math.radians(rotate_angle))
end_point_y = start_point[1] - ray_length * math.sin(math.radians(rotate_angle))
end_point = (end_point_x, end_point_y)
# 绘制射线
pygame.draw.line(window_surface, (255, 0, 0), start_point, end_point, 2)
# 刷新屏幕
pygame.display.update()
在上面的示例中,我们在事件循环中检测键盘左键是否按下,并根据按键状态更新线条的旋转角度。您可以通过调整rotate_speed参数来改变旋转速度,从而控制角度的变化速度。这样就可以在Pygame中实现按下键盘左键时旋转线条角度的效果。
2024/4/7 19:02:10
end_point_x = start_point[0] + ray_length * math.cos(math.radians(rotate_angle))
end_point_y = start_point[1] - ray_length * math.sin(math.radians(rotate_angle))
2024/4/7 19:02:16
这两行代码是用来计算旋转后的射线终点坐标的。让我解释一下每个部分的作用:
math.cos(math.radians(rotate_angle)): 这部分计算了给定角度rotate_angle的余弦值,其中math.radians()函数将角度从度转换为弧度。余弦值表示终点在x轴上的偏移量。
math.sin(math.radians(rotate_angle)): 这部分计算了给定角度rotate_angle的正弦值,其中math.radians()函数将角度从度转换为弧度。正弦值表示终点在y轴上的偏移量。
start_point[0] + ray_length * …: 这部分将起点的x坐标与旋转后x轴上的偏移量相加,从而得到终点的x坐标。
start_point[1] - ray_length * …: 这部分将起点的y坐标减去旋转后y轴上的偏移量,从而得到终点的y坐标。
通过这些计算,您可以根据起点、长度和角度来确定旋转后的射线终点坐标。这样就能实现在Pygame中根据旋转角度绘制旋转的射线的效果。
import pygame from pygame.locals import * import sys import math pygame.init() width, height = 800, 600 screen = pygame.display.set_mode((width, height)) c=pygame.time.Clock() q = [ [1, 1, 1, 1, 1, 1, 1, 1], [1, 0, 0, 0, 0, 0, 0, 1], [1, 0, 0, 0, 0, 0, 0, 1], [1, 0, 0, 0, 0, 0, 0, 1], [1, 0, 0, 0, 0, 0, 0, 1], [1, 0, 0, 0, 0, 0, 0, 1], [1, 0, 0, 0, 0, 0, 0, 1], [1, 1, 1, 1, 1, 1, 1, 1] ] px=70 py=70 du=0 left=False right=False down=False up=False img = pygame.image.load("1.png") # 替换为您自己的图像文件路径 img=pygame.transform.scale(img,(30,30)) ci=0 sz=[] qsz=[] xiansz=[(0,0)] xuanzhuanjiao=0 class zidan: global xiansz def __init__(self): self.x = self self.y = self self.endx=self self.endy=self self.xiansz=self def chuangjian(self): self.endx = px self.endy = py p=pygame.draw.rect(screen, (40, 140, 40), (self.x, self.y, 10,20), 0) print(self.x,self.y,"d没碰") if(p.collidelistall(qsz)): xiansz.clear() #self.endx, self.endy = self.x, self.y xiansz.append((self.x,self.y)) print(self.x,self.y,"duixiang",xiansz) def player(x,y,zhuan): #pygame.draw.line(screen, (40, 140, 40), (x, y), (74, 713)) #print(du) screen.blit(zhuan,(x,y)) def dl(x,y): pygame.draw.line(screen, (40, 140, 40), (x, y), (x, 700)) while True: screen.fill((255, 255, 255)) #c.tick(7) for event in pygame.event.get(): if event.type == QUIT: pygame.quit() sys.exit() keys = pygame.key.get_pressed() if keys[pygame.K_UP]: print("up") py=py -10 if keys[pygame.K_LEFT]:#zuo px=px-10 if keys[pygame.K_DOWN]:#xia print("up") py = py + 10 if keys[pygame.K_RIGHT]:#you px=px+10 #print(sz) #在这直接写会变成子弹,写成函数qiang(ci)就能画成一次,不会移动的墙 zi = zidan() zi.x = px zi.y = py sz.append(zi) for h in range(len(q)): #print(q[h],"行",h) for g in range(len(q[h])): if q[h][g]==1: #print(q[h][g],"个",g,h) pz=pygame.draw.rect(screen, (140, 240, 40), (h*60, g*60, 60, 60)) qsz.append(pz) player(px,py,img) dl(px,py) for i in (sz): #print(i) i.x=i.x+60 i.chuangjian() #pygame.draw.line(screen, (40, 140, 40), (px, py), (i.endx, i.endy)) #print(i.endx, i.endy, "点") if(i.x>500): sz.remove(i) for i in (sz): xuanzhuanjiao = xuanzhuanjiao + 1 pygame.draw.line(screen, (40, 140, 40), (px, py), (xiansz[0])) p=pygame.math.Vector2(px,py) p1=pygame.math.Vector2(xiansz[0]) juli=p.distance_to(p1) end_point_x = px +int(juli) * math.cos(math.radians(1)) end_point_y = py - int(juli) * math.sin(math.radians(1)) end_point = (end_point_x, end_point_y) pygame.draw.line(screen, (255, 0, 0), (px, py), end_point, 2) end_point_x = px + int(juli) * math.cos(math.radians(2)) end_point_y = py - int(juli) * math.sin(math.radians(2)) end_point = (end_point_x, end_point_y) pygame.draw.line(screen, (255, 0, 0), (px, py), end_point, 2) end_point_x = px + int(juli) * math.cos(math.radians(3)) end_point_y = py - int(juli) * math.sin(math.radians(3)) end_point = (end_point_x, end_point_y) pygame.draw.line(screen, (255, 0, 0), (px, py), end_point, 2) end_point_x = px + int(juli) * math.cos(math.radians(4)) end_point_y = py - int(juli) * math.sin(math.radians(4)) end_point = (end_point_x, end_point_y) pygame.draw.line(screen, (255, 0, 0), (px, py), end_point, 2) print(i.endx, i.endy, "点", (end_point_x, end_point_y), xiansz,xuanzhuanjiao) #if (i.x > 500): #sz.remove(i) #在这直接写会变成子弹,写成函数qiang(ci)就能画成一次,不会移动的墙 pygame.display.flip()
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