我认为我有一个相对简单的问题,但我现在已经尝试了几个小时而没有运气。我试图拟合线性函数(linearf)或幂律函数(plaw),其中我已经知道这些函数的斜率(b,我必须在本研究中保持它不变)。结果应该给出1.8左右的截距,这是我无法得到的。我必须做错事,但我不能指责它。有人知道如何解决这个问题吗?
提前谢谢!
import numpy as np
from scipy import optimize
p2 = np.array([ 8.08543600e-06, 1.61708700e-06, 1.61708700e-05,
4.04271800e-07, 4.04271800e-06, 8.08543600e-07])
pD = np.array([ 12.86156, 16.79658, 11.52103, 21.092 , 14.47469, 18.87318])
# Power-law function
def plaw(a,x):
b=-0.1677 # known slope
y = a*(x**b)
return y
# linear function
def linearf(a,x):
b=-0.1677 # known slope
y = b*x + a
return y
## First way, via power-law function ##
popt, pcov = optimize.curve_fit(plaw,p2,pD,p0=1.8)
# array([ 7.12248200e-37]) wrong
popt, pcov = optimize.curve_fit(plaw,p2,pD)
# >>> return 0.9, it is wrong too (the results should be around 1.8)
## Second way, via log10 and linear function ##
x = np.log10(p2)
y = np.log10(pD)
popt, pcov = optimize.curve_fit(linearf,x,y,p0=0.3)
K = 10**popt[0]
## >>>> return 3.4712954470408948e-41, it is wrong
答案 0 :(得分:0)
我发现函数中存在错误:
应该是:
def plaw(x,a):
b=-0.1677 # known slope
y = a*(x**b)
return y
而不是
def plaw(a,x):
b=-0.1677 # known slope
y = a*(x**b)
return y
愚蠢的错误。