下面的代码为我提供了一条最佳拟合线的平坦线，而不是沿着 e^(-x) 模型的一条适合数据的漂亮曲线。谁能告诉我如何修复下面的代码以使其适合我的数据？
import numpy as np
import matplotlib.pyplot as plt
import scipy.optimize
def _eNegX_(p,x):
x0,y0,c,k=p
y = (c * np.exp(-k*(x-x0))) + y0
return y
def _eNegX_residuals(p,x,y):
return y - _eNegX_(p,x)
def Get_eNegX_Coefficients(x,y):
print 'x is: ',x
print 'y is: ',y
# Calculate p_guess for the vectors x,y. Note that p_guess is the
# starting estimate for the minimization.
p_guess=(np.median(x),np.min(y),np.max(y),.01)
# Calls the leastsq() function, which calls the residuals function with an initial
# guess for the parameters and with the x and y vectors. Note that the residuals
# function also calls the _eNegX_ function. This will return the parameters p that
# minimize the least squares error of the _eNegX_ function with respect to the original
# x and y coordinate vectors that are sent to it.
p, cov, infodict, mesg, ier = scipy.optimize.leastsq(
_eNegX_residuals,p_guess,args=(x,y),full_output=1,warning=True)
# Define the optimal values for each element of p that were returned by the leastsq() function.
x0,y0,c,k=p
print('''Reference data:\
x0 = {x0}
y0 = {y0}
c = {c}
k = {k}
'''.format(x0=x0,y0=y0,c=c,k=k))
print 'x.min() is: ',x.min()
print 'x.max() is: ',x.max()
# Create a numpy array of x-values
numPoints = np.floor((x.max()-x.min())*100)
xp = np.linspace(x.min(), x.max(), numPoints)
print 'numPoints is: ',numPoints
print 'xp is: ',xp
print 'p is: ',p
pxp=_eNegX_(p,xp)
print 'pxp is: ',pxp
# Plot the results
plt.plot(x, y, '>', xp, pxp, 'g-')
plt.xlabel('BPM%Rest')
plt.ylabel('LVET/BPM',rotation='vertical')
plt.xlim(0,3)
plt.ylim(0,4)
plt.grid(True)
plt.show()
return p
# Declare raw data for use in creating regression equation
x = np.array([1,1.425,1.736,2.178,2.518],dtype='float')
y = np.array([3.489,2.256,1.640,1.043,0.853],dtype='float')
p=Get_eNegX_Coefficients(x,y)
It looks like it's a problem with your initial guesses; something like (1, 1, 1, 1) works fine:
You have
p_guess=(np.median(x),np.min(y),np.max(y),.01)
对于函数
def _eNegX_(p,x):
x0,y0,c,k=p
y = (c * np.exp(-k*(x-x0))) + y0
return y
这就是 test_data_maxe^( -.01(x - test_data_median)) + test_data_min
I don't know much about the art of choosing good starting parameters, but I can say a few things. leastsq is finding a local minimum here - the key in choosing these values is to find the right mountain to climb, not to try to cut down on the work that the minimization algorithm has to do. Your initial guess looks like this (green):
(1.736, 0.85299999999999998, 3.4889999999999999, 0.01)
这会导致你的平线（蓝色）：(-59.20295956, 1.8562 , 1.03477144, 0.69483784)
调整线的高度比增加 k 值获得的收益更大。如果您知道自己适合此类数据，请使用更大的 k。如果您不知道，我想您可以尝试通过对数据进行采样来找到合适的 k 值，或者从前半部分和后半部分的平均值之间的斜率返回，但我不知道该怎么做关于那个。
编辑：您也可以从几次猜测开始，运行几次最小化，然后选择残差最低的线。