下面的代码为我提供了最佳拟合线的平线,而不是沿着适合数据的 e^(-x) 模型的漂亮曲线。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)
最佳答案
看来是你一开始的猜测有问题;像 (1, 1, 1, 1) 这样的东西很好用:
你有
p_guess=(np.median(x),np.min(y),np.max(y),.01)
函数
def _eNegX_(p,x):
x0,y0,c,k=p
y = (c * np.exp(-k*(x-x0))) + y0
return y
所以这是 test_data_maxe^( -.01(x - test_data_median)) + test_data_min
我不太了解选择好的起始参数的技巧,但我可以说几句。 leastsq
在这里找到一个局部最小值 - 选择这些值的关键是找到合适的山峰来攀登,而不是试图减少最小化算法必须做的工作。您的初始猜测如下所示(绿色
):
(1.736,0.85299999999999998,3.4889999999999999,0.01)
这导致你的扁平线(蓝色):
(-59.20295956, 1.8562, 1.03477144, 0.69483784)
调整线的高度比增加k值获得更大的 yield 。如果您知道自己适合这种数据,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,或者从上半场和下半场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:您也可以从几个猜测开始,多次运行最小化,然后选择残差最小的那条线。
关于python - 使用 scipy、python、numpy 进行非线性 e^(-x) 回归,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/4495127/