问题描述
我这里有这段代码,但我不明白如何简化代码,有没有人知道如何简化这样的代码
if ball.ycor() > 340:
ball.sety(340)
ball.ycoor *= -1
if ball.ycor() < -340:
ball.sety(-340)
ball.ycoor *= -1
if ball.xcor() > 490:
ball.goto(0,0)
ball.xcoor *= -1
scoreboard_a += 1
score.clear()
score.write("{} {}".format(scoreboard_a,scoreboard_b),font=("Arial",104,"normal"))
if ball.xcor() < -490:
ball.goto(0,0)
ball.xcoor *= -1
scoreboard_b += 1
score.clear()
score.write("{} {}".format(scoreboard_a,"normal"))
if (ball.xcor() > 440 and ball.xcor() < 450) and ball.ycor() < block_b.ycor() + 40 and ball.ycor() > block_b.ycor() - 40:
ball.setx(440)
ball.xcoor *= -1
if (ball.xcor() < -440 and ball.xcor() > -450) and ball.ycor() > block_a.ycor() - 40 and ball.ycor() < block_a.ycor() + 40:
ball.setx(-440)
ball.xcoor *= -1
解决方法
您应该将互斥的可能性与 if-else
语句结合起来。我的第一个倾向是评估 ball.ycor()
和 ball.xcor()
一次,假设连续调用不会返回不同的值:
x_coor = ball.xcor()
y_coor = ball.ycor()
但我注意到表达式 ball.ycoor *= -1
,它表明这可能会影响后续调用 ball.ycor()
的返回值。所以我决定不尝试优化这些调用:
if ball.ycor() > 340:
ball.sety(340)
ball.ycoor *= -1
elif ball.ycor() < -340:
ball.sety(-340)
ball.ycoor *= -1
if ball.xcor() > 490:
ball.goto(0,0)
ball.xcoor *= -1
scoreboard_a += 1
score.clear()
score.write("{} {}".format(scoreboard_a,scoreboard_b),font=("Arial",104,"normal"))
elif ball.xcor() < -490:
ball.goto(0,0)
ball.xcoor *= -1
scoreboard_b += 1
score.clear()
score.write("{} {}".format(scoreboard_a,"normal"))
elif (-450 < ball.xcor() < -440) and ball.ycor() > block_a.ycor() - 40 and ball.ycor() < block_a.ycor() + 40:
ball.setx(-440)
ball.xcoor *= -1
elif (440 < ball.xcor() < 450) and ball.ycor() < block_b.ycor() + 40 and ball.ycor() > block_b.ycor() - 40:
ball.setx(440)
ball.xcoor *= -1
注意 if x > y and x < z:
可以改写为 if y < x < z:
您可以将一个值夹在多个比较运算符之间。您可以使用 f
字符串将值直接传递到花括号中。
if ball.ycor() > 340:
ball.sety(340)
ball.ycoor *= -1
if ball.ycor() < -340:
ball.sety(-340)
ball.ycoor *= -1
if ball.xcor() > 490:
ball.goto(0,0)
ball.xcoor *= -1
scoreboard_a += 1
score.clear()
score.write(f"{scoreboard_a} {scoreboard_b}","normal"))
if ball.xcor() < -490:
ball.goto(0,0)
ball.xcoor *= -1
scoreboard_b += 1
score.clear()
score.write("{scoreboard_a} {scoreboard_b}","normal"))
if 450 > ball.xcor() > 440 and block_b.ycor() - 40 < ball.ycor() < block_b.ycor() + 40:
ball.setx(440)
ball.xcoor *= -1
if -450 < ball.xcor() < -440 and block_a.ycor() + 40 > ball.ycor() > block_a.ycor() - 40:
ball.setx(-440)
ball.xcoor *= -1
此外,您可能希望将 font
值存储到一个变量中,以便您可以通过使用该变量来使用它,例如 font = ("Arial","normal")
,然后
score.write(f"{scoreboard_a} {scoreboard_b}",font=font)
此外,如果您想让您的代码更高效,请仅将 if
语句用于第一条语句,其余语句使用 elif
。这样,python一旦找到满足条件的语句就不需要费心去检查其他语句了。