Question

我有一个线性方程和二次方程的系统，我可以用numpy和scipy进行设置，这样我就可以得到一个图形解决方案。考虑示例代码：

#!/usr/bin/env python
# Python 2.7.1+

import numpy as np #
import matplotlib.pyplot as plt #

# d is a constant;
d=3
# h is variable; depends on x, which is also variable

# linear function:
# condition for h: d-2x=8h; returns h
def hcond(x):
  return (d-2*x)/8.0

# quadratic function:  
# condition for h: h^2+x^2=d*x ; returns h
def hquad(x):
  return np.sqrt(d*x-x**2)

# x indices data
xi = np.arange(0,3,0.01)

# function values in respect to x indices data
hc = hcond(xi)
hq = hquad(xi)

fig = plt.figure() 
sp = fig.add_subplot(111)

myplot = sp.plot(xi,hc)
myplot2 = sp.plot(xi,hq)

plt.show()

该代码以此图表结果：

很明显，两个函数相交，因此有一个解决方案。

如何在保持大多数功能定义完整的同时自动解决什么是解决方案（交叉点）？

Answer 1

事实证明，可以使用scipy.optimize.fsolve来解决这个问题，只需要注意OP中的函数是以y=f(x)格式定义的;虽然fsolve将需要f(x)-y=0格式。这是固定代码：

#!/usr/bin/env python
# Python 2.7.1+

import numpy as np #
import matplotlib.pyplot as plt #
import scipy
import scipy.optimize

# d is a constant;
d=3
# h is variable; depends on x, which is also variable

# linear function:
# condition for h: d-2x=8h; returns h
def hcond(x):
  return (d-2*x)/8.0

# quadratic function:
# condition for h: h^2+x^2=d*x ; returns h
def hquad(x):
  return np.sqrt(d*x-x**2)

# for optimize.fsolve;
# note, here the functions must be equal to 0;
# we defined h=(d-2x)/8 and h=sqrt(d*x-x^2);
# now we just rewrite in form (d-2x)/16-h=0 and sqrt(d*x-x^2)-h=0;
# thus, below x[0] is (guess for) x, and x[1] is (guess for) h!
def twofuncs(x):
  y = [ hcond(x[0])-x[1], hquad(x[0])-x[1] ]
  return y

# x indices data
xi = np.arange(0,3,0.01)

# function values in respect to x indices data
hc = hcond(xi)
hq = hquad(xi)

fig = plt.figure()
sp = fig.add_subplot(111)

myplot = sp.plot(xi,hc)
myplot2 = sp.plot(xi,hq)


# start from x=0 as guess for both functions
xsolv = scipy.optimize.fsolve(twofuncs, [0, 0])
print(xsolv)
print("xsolv: {0}\n".format(xsolv))

# plot solution with red marker 'o'
myplot3 = sp.plot(xsolv[0],xsolv[1],'ro')


plt.show()

exit

...结果如下：

xsolv: [ 0.04478625  0.36380344]

......或者，在情节图像上：

参考文献：

在numpy / matplotlib中以图形和数字方式求解线性二次方程组？

1 个答案: