我可以在一行中定义一个规则折线函数吗？(1)y=x,y=0

Question

我可以在一行中定义一个线函数吗，因为它是正则的？

图形橙色线：myPoly

条件：

图像在 Y=0 和 Y=X 之间摆动。

直线的斜率为 2 或 -2。

你可以从 (2,0) 开始。

我要

①是否可以在一行中定义以下部分 def myPolyY(Ho,myX): ？

②能分段做还是串联做？

from sympy import *
var('x y')
import matplotlib.pyplot as plt
def myPolyY(Ho,myX):
    myY=0
    xx=[x[0] for x in Ho]
    yy=[x[1] for x in Ho]
    for i in range(len(xx)-1):
        if xx[i] <= myX & myX<= xx[i+1]:
           myY= (yy[i+1]-yy[i])/(xx[i+1]-xx[i])* (myX - xx[i])+yy[i]
    return myY
def myPolyDef(nMax):
    myPoly=[[0,0],[2,0]]
    i=2
    ans = solve([y - x,
                 y - myPoly[i-1][1] - 2 * (x - myPoly[i-1][0])], [x, y])
    myPoly=myPoly+[[ans[x], ans[y]]]
    i=1
    for j in range(nMax):
        i=i+1
        ans=solve([y-0,y-myPoly[i][1]+2*(x-myPoly[i][0])],[x,y])
        myPoly=myPoly+[[ans[x],ans[y]]]
        i=i+1
        ans=solve([y-x,
                   y-myPoly[i  ][1]-2*(x-myPoly[i  ][0])], [x,y])
        myPoly=myPoly+[[ans[x],ans[y]]]
    return myPoly
def myPolyPlot(Ho,myLabel,myLinestyle):
    plt.plot([x[0] for x in Ho],[x[1] for x in Ho], mec='none', ms=4, lw=1, label=myLabel,linestyle=myLinestyle)
    for i in range(len(Ho)):
        if i % 2==0:
           myPos='right'
        else:
            myPos='left'
        plt.text(Ho[i][0], Ho[i][1], '({x}, {y})'.format(x=Ho[i][0], y=Ho[i][1]),
                  fontsize=6, horizontalalignment=myPos)
    return
def main():
    # myN=4
    myN=2
    myX=30
    myPoly=myPolyDef(myN)
    myY=myPolyY(myPoly,30)
    # matplotlib
    myH=max(list(map(lambda x: max(x), myPoly)))
    plt.axes().set_aspect('equal')
    plt.text(myX,myY, '({x}, {y})'.format(x=myX, y=myY),
             fontsize=6, horizontalalignment='right')
    myPolyPlot([[myH,myH],[0,0]],'Y=X','--')
    myPolyPlot(myPoly,'myPoly','-')
    myPolyPlot([[myH,  0],[0,0]],'Y=0','--')
    plt.legend(frameon=False, fontsize=10, numpoints=1, loc='upper left')
    plt.savefig('myPoly.png', dpi=200)
    plt.show()
if __name__ == '__main__':
    main()

(2022-03-03)‖我用的是log.

from sympy import *
import matplotlib.pyplot as plt
var('x y')
def myCal_PolyTopBottom(myT):
    myPolyTB = [[0, 0], [2, 0]]
    for x in range(myT):
        xt = 4*(3**x)
        yt = 4*pow(3, int(log(xt // 2, 3)))
        myPolyTB = myPolyTB + [[xt           , yt ]]
        myPolyTB = myPolyTB + [[xt + yt * 0.5, 0.0]]
    return myPolyTB
def myCal_PolyXYN(myPolyTB,myXmax):
    myPolyXYN = [[0, 0], [2, 0]]
    for x in range(3, int(myXmax)+1):
        y=myCal_PolyXYi(myPolyTB,x)
        myPolyXYN = myPolyXYN + [[float(x), y]]
    return myPolyXYN
def myCal_PolyXYi(Ho,myX):
    xx=[x[0] for x in Ho]
    yy=[x[1] for x in Ho]
    for i in range(len(xx)-1):
        if xx[i] <= myX & myX<= xx[i+1]:
           myY= (yy[i+1]-yy[i])/(xx[i+1]-xx[i])* (myX - xx[i])+yy[i]
    return myY
def myPlot_Poly(Ho,myLabel,myLinestyle):
    plt.plot([x[0] for x in Ho],[x[1] for x in Ho],
             mec='none', ms=4, lw=1, label=myLabel,linestyle=myLinestyle)
    return
def myPlot_Text(Ho,myPos):
    for i in range(len(Ho)):
        plt.text(Ho[i][0], Ho[i][1], '({x}, {y})'.format(x=Ho[i][0], y=Ho[i][1]),
                  fontsize=6, horizontalalignment=myPos)
    return
def main():
    myT=3
    myPolyTB=myCal_PolyTopBottom(myT)
    myH=max(list(map(lambda x: max(x),myPolyTB)))
    myPolyN=myCal_PolyXYN(myPolyTB,myH)
# matplotlib
    plt.axes().set_aspect('equal')
    myPlot_Text(myPolyTB,'right' )
    myPlot_Text(myPolyN ,'left'  )
    myPlot_Poly([[myH,myH],[0,0]],'Y=X'     ,'--')
    myPlot_Poly(  myPolyTB       ,'myPolyTB','-' )
    myPlot_Poly([[myH,  0],[0,0]],'Y=0'     ,'--')
    plt.legend (frameon=False, fontsize=10, numpoints=1, loc='upper left')
    # plt.savefig('myPoly.png', dpi=200)
    plt.show()
if __name__ == '__main__':
    main()

(20220322)OEIS

OEIS:0 2 4 6 12 18

https://oeis.org/search?q=0+2++4++6+12+18+&sort=&language=&go=Search

0 2 4 6 12 18 36

https://oeis.org/search?q=0+2++4++6+12+18+36&sort=&language=&go=Search

对不起......

Answer 1

如果你看第一个三角形，很容易看出方程可以表示为

y = 4 - 2 * abs(4-x)

更一般地，每个三角形都有一个顶值，对应的方程为

y = top - 2 * abs(top-x)

然后，对于每个坐标x，我们要确定对应的top

的值

top = 4 * pow(3, int(log(x//2,3)))

from math import *

myPoly=[[0,0],[2,0]]
for x in range(3, 37):
    val = 4 * pow(3, int(log(x//2,3)))
    y = val - 2 * abs(x - val)
    myPoly = myPoly+[[x,y]]

我可以在一行中定义一个规则折线函数吗？(1)y=x,y=0

Can I define a regular polyline function in one line?(1)y=x,y=0

algorithm

sympy

oeis

(20220322)OEIS