当我只有一个已知的Y坐标方程,即P = a * b(其中a& b定义的值为0.8,150)和x坐标完全未知时,如何在曲线绘图上得到一个点没有连接x和y的方程(例如：y = mx b; #i没有这种方程).因此,现在目标是,如果我将’Y坐标’值设置为120并且需要通过从未知’x-coordiante’值获取距离或路径来绘制曲线上的点.

我尝试了如下代码

import matplotlib.pyplot as plt

import numpy as np

from scipy.interpolate import InterpolatedUnivariateSpline

# given values

y = np.array([0, 38.39, 71.41, 99.66, 123.67, 143.88, 160.61, 174.03, 184.16, 190.8, 193.52])

x = np.array([0, 0.37, 0.74, 1.11, 1.48, 1.85, 2.22, 2.59, 2.96, 3.33, 3.7])

x_val = np.linspace(0,7) #limts on x-axis

a = 0.8

b = 150

y_val = np.multiply(a, b)

yinterp = np.interp(x_val, x, y)

plt.plot(x, y, '-')

plt.plot(x_val, yinterp, 'o')

#here i need to plot a exact point w.r.t to y_val

#and also need to show the distance with a line from the selected x and y coordinates

plt.plot(x_val,y_val, '--')

plt.show()

解决方法:

你想要的是找到数组的根(s)或零(s).这个问题的答案显示了如何做到这一点：How to get values from a graph?

将解决方案应用于此案例如下：

import matplotlib.pyplot as plt

import numpy as np

# given values

y = np.array([0, 38.39, 71.41, 99.66, 123.67, 143.88, 160.61, 174.03, 184.16, 190.8, 193.52])

x = np.array([0, 0.37, 0.74, 1.11, 1.48, 1.85, 2.22, 2.59, 2.96, 3.33, 3.7])

x_val = np.linspace(0,7)

plt.plot(x, y, '-')

def find_roots(x,y):

s = np.abs(np.diff(np.sign(y))).astype(bool)

return x[:-1][s] + np.diff(x)[s]/(np.abs(y[1:][s]/y[:-1][s])+1)

a = 0.8

b = 150

y_val = np.multiply(a, b)

roots = find_roots(x, y-y_val)

plt.plot(roots[0],y_val, marker="o")

plt.plot([roots[0],roots[0],0],[0,y_val,y_val], "--")

plt.xlim(0,None)

plt.ylim(0,None)

plt.show()

如果数组单调递增,您当然也可以简单地插值：

import matplotlib.pyplot as plt

import numpy as np

# given values

y = np.array([0, 38.39, 71.41, 99.66, 123.67, 143.88, 160.61, 174.03, 184.16, 190.8, 193.52])

x = np.array([0, 0.37, 0.74, 1.11, 1.48, 1.85, 2.22, 2.59, 2.96, 3.33, 3.7])

x_val = np.linspace(0,7)

plt.plot(x, y, '-')

a = 0.8

b = 150

y_val = np.multiply(a, b)

root = np.interp(y_val,y,x)

plt.plot(root,y_val, marker="o")

plt.plot([root,root,0],[0,y_val,y_val], "--")

plt.xlim(0,None)

plt.ylim(0,None)

plt.show()

标签：python,numpy,scipy,matplotlib