我使用以下代码绘制图形上的交叉点,然后目视检查交叉点以返回代码并遮蔽可行性区域。
是否有更好的方法来找到可行区域而不是简单地绘制线条并从图形中读取交叉点?
# Filling a polygon locating the corner points
# Then let Matplotlib fill within these points
x= [0.0, 0.0, 6.67,5.0]
y= [0.0, 4.0, .67, 0.0]
fill(x,y)
show()
完整代码示例:
x= arange(-3,10.1,0.1)
y= arange(-3,10.1,0.1)
y1= 0.4*x-2.0
y2= 4.0-0.5*x
xlim(-3,10)
ylim(-3,10)
hlines(0,-3,10,color='k')
vlines(0,-3,10,color='k')
grid(True)
xlabel('x-axis')
ylabel('y-axis')
title ('Shaded Area Shows the Feasible Region')
plot(x,y1,color='b')
plot(x,y2,color='r')
legend(['2x-5y=10','x+2y=8'])
x= [0.0, 0.0, 6.67,5.0]
y= [0.0, 4.0, .67, 0.0]
fill(x,y)
show()
答案 0 :(得分:1)
如果您只想绘制可行性区域,您可以这样做:
x= arange(-3,10.1,0.1)
y= arange(-3,10.1,0.1)
y1= 0.4*x-2.0
y2= 4.0-0.5*x
xlim(-3,10)
ylim(-3,10)
hlines(0,-3,10,color='k')
vlines(0,-3,10,color='k')
grid(True)
xlabel('x-axis')
ylabel('y-axis')
title ('Shaded Area Shows the Feasible Region')
plot(x,y1,color='b')
plot(x,y2,color='r')
legend(['2x-5y=10','x+2y=8'])
bottom = np.maximum(y1, 0)
fill_between(x, bottom, y2, where=(x>0) & (y2>y1))