Question

我为几个频率绘制了一些曲线Imax与Ciclo de servicio。

为此，我为自变量Fs和Vo

进行了两次嵌套扫描

虽然方程式稍微复杂一点，但我基本上是这样做的：

for Fs in np.linspace(70e3,200e3,4):
    Vo_sweep = np.linspace(0,1,1000)
     delta_I = [I_max(Vo) for Vo in Vo_sweep]
     plt.plot(Vo_sweep,delta_I)

我特别感兴趣的是“ciclo de servicio＆＃39;是0.71。

如何在X轴上显示值0.71以及在每条曲线上截取蓝线的Y值

Answer 1

如果您想使用任意数据（无论您的函数和垂直线可能是什么）绘制拦截标记，我建议您计算交点（它们存在的位置）。几个月前，我给出了一个可能对你有用的答案。您可以查看here。

为您的问题调整类似的解决方案：

import numpy as np
import matplotlib.pyplot as plt

def line_intersection(line1, line2):
    xdiff = (line1[0][0] - line1[1][0], line2[0][0] - line2[1][0])
    ydiff = (line1[0][1] - line1[1][1], line2[0][1] - line2[1][1]) #Typo was here

    def det(a, b):
        return a[0] * b[1] - a[1] * b[0]

    div = det(xdiff, ydiff)
    if div == 0:
        return None

    d = (det(*line1), det(*line2))
    x = det(d, xdiff) / div
    y = det(d, ydiff) / div
    return x, y

def near(a, b, rtol=1e-5, atol=1e-8):
    return abs(a - b) < (atol + rtol * abs(b))
def crosses(line1, line2):
    """
    Return True if line segment line1 intersects line segment line2 and 
    line1 and line2 are not parallel.
    """
    (x1,y1), (x2,y2) = line1
    (u1,v1), (u2,v2) = line2
    (a,b), (c,d) = (x2-x1, u1-u2), (y2-y1, v1-v2)
    e, f = u1-x1, v1-y1
    denom = float(a*d - b*c)
    if near(denom, 0):
        # parallel
        return False
    else:
        t = (e*d - b*f)/denom
        s = (a*f - e*c)/denom
        # When 0<=t<=1 and 0<=s<=1 the point of intersection occurs within the
        # line segments
        return 0<=t<=1 and 0<=s<=1


for Fs in np.linspace(70e3,200e3,4):
    Vo_sweep = np.linspace(0,1,1000)
    delta_I = [i*Fs*np.log((i+1.1)/10) for i in range(len(Vo_sweep))] #[I_max(Vo) for Vo in Vo_sweep]
    plt.plot(Vo_sweep,delta_I)

plt.vlines(0.71,min(delta_I),max(delta_I))

for Fs in np.linspace(70e3,200e3,4):
    x = np.linspace(0,1,1000)
    y = [i*Fs*np.log((i+1.1)/10) for i in range(len(Vo_sweep))]
    for i in range(1,len(delta_I)):
        p1 = np.array([x[i-1],y[i-1]],dtype='float')
        p2 = np.array([x[i],y[i]],dtype='float')
        k1 = np.array([0.71,min(delta_I)],dtype='float')
        k2 = np.array([0.71,max(delta_I)],dtype='float')
        if crosses((p2,p1),(k1,k2)):
            seg = line_intersection((p2,p1),(k1,k2))
            plt.scatter(seg[0],seg[1],c='red',s=90)
            print(seg)

plt.ylim(min(delta_I),max(delta_I))
plt.xlim(0,1)

plt.show()

，结果如下：

如果你感兴趣的话，这个特殊的配方也会打印拦截坐标：

(0.70999999999999985, 211670610.06954533)
(0.70999999999999985, 342704797.25546497)
(0.70999999999999985, 473738984.44138455)
(0.70999999999999985, 604773171.62730408)

由于您没有提供数据，我必须制作快速合成数据，但它应该像I_max()定义中的delta_I函数替换一样简单。

显示拦截值，matplotlib

1 个答案: