我试图弄清楚为什么即使在我的图形中的点相等之后,我的一个函数中的while
循环仍在运行,这是我将其设置为停止的时候。我做错了什么吗?我试图切换其他方法以使其正常工作,但是没有运气。
这是一款游戏,当角色到达终端盒时,循环需要中断,但在我明确编码后并没有这样做。这是我拥有的第二个功能:
from graphics import *
def field():
#creating the window
win = GraphWin('The Field',400,400)
win.setBackground('white')
#drawing the grid
boxlist = []
for i in range(0,400,40):
for j in range(0,400,40):
box = Rectangle(Point(i,j),Point(i+40,j+40))
box.setOutline('light gray')
box.draw(win)
boxlist.append(box)
#creating other boxes
startbox = Rectangle(Point(0,0),Point(40,40))
startbox.setFill('lime')
startbox.setOutline('light gray')
startbox.draw(win)
endbox = Rectangle(Point(360,360),Point(400,400))
endbox.setFill('red')
endbox.setOutline('light gray')
endbox.draw(win)
boxlist.append(startbox)
boxlist.append(endbox)
#creating Pete
pete = Rectangle(Point(2,2),Point(38,38))
pete.setFill('gold')
pete.draw(win)
return win,boxlist,pete
def move(win2,boxlist,pete,endbox):
peteloc = pete.getCenter()
#creating loop to move pete
while peteloc != endbox.getCenter():
click = win2.getMouse()
x = click.getX()
y = click.getY()
peteloc = pete.getCenter()
petex = peteloc.getX()
petey = peteloc.getY()
#moving pete
if x>=petex+20 and y<=petey+20 and y>=petey-20:
pete.move(40,0)
elif x<=petex-20 and y<=petey+20 and y>=petey-20:
pete.move(-40,0)
elif y>=petey+20 and x<=petex+20 and x>=petex-20:
pete.move(0,40)
elif y<=petey-20 and x<=petex+20 and x>=petex-20:
pete.move(0,-40)
peteloc = pete.getCenter()
# The main function
def main():
win2,boxlist,pete = field()
endbox = boxlist[len(boxlist)-1]
move(win2,boxlist,pete,endbox)
main()
答案 0 :(得分:1)
我认为这可能是由于浮动精度引起的。我猜pete.getCenter()和endbox.getCenter()类似于[float,float],您应该避免在浮点之间使用!=
,例如1.0000001不等于1。
因此,即使角色到达收尾箱,该位置仍会产生一些浮动偏差。
因此,当错误可接受时,您可以将a != b
更改为abs(a - b) > acceptable_error
。示例代码如下:
# while peteloc != endbox.getCenter():
while abs(peteloc.getX() - endbox.getCenter().getX()) > 0.01 and abs(peteloc.getY() - endbox.getCenter().getY()) > 0.01:
希望对您有帮助。
答案 1 :(得分:1)
Zelle图形Point
对象似乎不会比较相等:
>>> from graphics import *
>>> a = Point(100, 100)
>>> b = Point(100, 100)
>>> a == b
False
>>>
我们必须提取坐标并进行自己的比较。尽管@recnac提供了一个可行的解决方案(+1),但我将建议一个更通用的解决方案。我们将创建一个distance()
方法,该方法对从_BBox
继承的任何对象均有效,其中包括Rectangle
,Oval
,Circle
和Line
:
def distance(bbox1, bbox2):
c1 = bbox1.getCenter()
c2 = bbox2.getCenter()
return ((c2.getX() - c1.getX()) ** 2 + (c2.getY() - c1.getY()) ** 2) ** 0.5
我们现在可以测量水平,垂直和对角线上物体之间的距离。由于您的框一次移动20个像素,因此我们可以假设,如果它们彼此相差1个像素,则它们位于同一位置。您的代码已改写为使用distance()
方法和其他调整:
from graphics import *
def field(win):
# drawing the grid
boxlist = []
for i in range(0, 400, 40):
for j in range(0, 400, 40):
box = Rectangle(Point(i, j), Point(i + 40, j + 40))
box.setOutline('light gray')
box.draw(win)
boxlist.append(box)
# creating other boxes
startbox = Rectangle(Point(0, 0), Point(40, 40))
startbox.setFill('lime')
startbox.setOutline('light gray')
startbox.draw(win)
boxlist.append(startbox)
endbox = Rectangle(Point(360, 360), Point(400, 400))
endbox.setFill('red')
endbox.setOutline('light gray')
endbox.draw(win)
boxlist.append(endbox)
# creating Pete
pete = Rectangle(Point(2, 2), Point(38, 38))
pete.setFill('gold')
pete.draw(win)
return boxlist, pete
def distance(bbox1, bbox2):
c1 = bbox1.getCenter()
c2 = bbox2.getCenter()
return ((c2.getX() - c1.getX()) ** 2 + (c2.getY() - c1.getY()) ** 2) ** 0.5
def move(win, pete, endbox):
# creating loop to move pete
while distance(pete, endbox) > 1:
click = win.getMouse()
x, y = click.getX(), click.getY()
peteloc = pete.getCenter()
petex, petey = peteloc.getX(), peteloc.getY()
# moving pete
if x >= petex + 20 and petey - 20 <= y <= petey + 20:
pete.move(40, 0)
elif x <= petex - 20 and petey - 20 <= y <= petey + 20:
pete.move(-40, 0)
elif y >= petey + 20 and petex - 20 <= x <= petex + 20:
pete.move(0, 40)
elif y <= petey - 20 and petex - 20 <= x <= petex + 20:
pete.move(0, -40)
# The main function
def main():
# creating the window
win = GraphWin('The Field', 400, 400)
win.setBackground('white')
boxlist, pete = field(win)
endbox = boxlist[-1]
move(win, pete, endbox)
main()