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我正在使用以下代码在图表上绘制交点,然后目视检查交点以返回代码并为可行性区域着色。

有没有比简单地绘制线并从图中读取交点更好的方法来找到可行区域?

# 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()
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1 回答 1

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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))

在此处输入图像描述

于 2015-01-26T02:55:40.667 回答