python - 高斯拟合无法使用 Python

Question

我正在尝试将此矩阵 data 放入列中使用此 Python 代码:

#!/usr/local/bin/env python
import numpy as np
import Tkinter #Used for file import
import tkFileDialog #Used for file import
import os
import scipy
import scipy.optimize as optimize


root = Tkinter.Tk()
root.withdraw() #use to hide tkinter window

filename = os.getcwd()
background = os.getcwd()
filename = tkFileDialog.askopenfile(parent=root,mode='rb',title='Choose a file') 
background = tkFileDialog.askopenfile(parent=root,mode='rb',title='Choose a background') 
filename = filename.name
#filename = r'bb1e03'
#background = r'bb1e03_background'


T0 = np.loadtxt(filename, unpack=False)
bg = np.loadtxt(background, unpack=False)

T = T0-bg # background subtraction?
#T = T.clip(min=0)
T[T<0]=0
T = np.flipud(T)
N, M = T.shape
datax = np.arange(N)


def gaussian(x, height, center, width, offset):
    return height*np.exp(-(x - center)**2/(2*width**2)) + offset

def three_gaussians(x, h1, c1, w1, h2, c2, w2, h3, c3, w3, offset):
    return (gaussian(x, h1, c1, w1, offset=0) +
        gaussian(x, h2, c2, w2, offset=0) +
        gaussian(x, h3, c3, w3, offset=0) + offset)

def two_gaussians(x, h1, c1, w1, h2, c2, w2, offset):
    return three_gaussians(x, h1, c1, w1, h2, c2, w2, 0,0,1, offset)

def one_gaussian(x,h1,c1,w1, offset):
    return (gaussian(x, h1, c1, w1, offset=0)+offset)

#errfunc3 = lambda p, x, y: (three_gaussians(x, *p) - y)**2
#errfunc2 = lambda p, x, y: (two_gaussians(x, *p) - y)**2
#errfunc1 = lambda p, x, y: (one_gaussian(x, *p) - y)**2
 #output files for fit parameters
outfile1 = open('results_1gau.txt', 'w')
outfile2 = open('results_2gau.txt', 'w')
outfile3 = open('results_3gau.txt', 'w')
outfile1.write('column\th1\tc1\tw1\toffset\n')
outfile2.write('column\th1\tc1\tw1\th2\tc2\tw2\toffset\n')
outfile3.write('column\th1\tc1\tw1\th2\tc2\tw2\th3\tc3\tw3\toffset\n')

# new matrices for fitted data
datafit1 = np.empty_like(T)
datafit2 = np.empty_like(T)
datafit3 = np.empty_like(T)

for n in xrange(M):

    Mmax = T[:,n].max()
    guess1 = [0.5*Mmax, N/10., 10., 0.]
    guess2 = [0.5*Mmax, N/10., 10., 0.5*Mmax, N/10., 10., 0.]
    guess3 = [0.5*Mmax, N/10., 10., 0.5*Mmax, N/10., 10.,
              0.5*Mmax, N/10., 10., 0]
    #optim3, success = optimize.leastsq(errfunc3, guess3[:],
    #                                   args=(datax, data[:,n]))
    #optim2, success = optimize.leastsq(errfunc2, guess2[:],
    #                                   args=(datax, data[:,n]))

    try:
        optim1, pcov = optimize.curve_fit(one_gaussian, datax, T[:,n], guess1)
    except:
        optim1 = [0, 0, 1, 0]

    try:
        optim2, pcov = optimize.curve_fit(two_gaussians, datax, T[:,n], guess2)
    except:
        optim2 = [0, 0, 1, 0, 0, 1, 0]

    try:
        optim3, pcov = optimize.curve_fit(three_gaussians, datax, T[:,n], guess3)
    except:
        optim3 = [0, 0, 1, 0, 0, 1, 0, 0, 1, 0]

   # write parameters to file (1 gau)
    s = '{}'.format(n)
    for x in guess1:
        s += '\t{:g}'.format(x)
    outfile1.write(s + '\n')

   # write parameters to file (2 gau)
    s = '{}'.format(n)
    for x in guess2:
        s += '\t{:g}'.format(x)
    outfile2.write(s + '\n')

    # write parameters to file (3 gau)
    s = '{}'.format(n)
    for x in guess3:
        s += '\t{:g}'.format(x)
    outfile3.write(s + '\n')
    # fill new matrices with fitted data
    datafit1[:,n] = one_gaussian(datax, *optim1)
    datafit2[:,n] = two_gaussians(datax, *optim2)
    datafit3[:,n] = three_gaussians(datax, *optim3)

T = datafit1 

我已经阅读了大部分与拟合相关的帖子，但我找不到我的代码有什么问题。它应该可以工作，但最终的矩阵“T”仅显示具有常数的列，而不是漂亮的平滑高斯形状曲线。请看看并告诉我我做错了什么。我在其他程序中尝试过，例如 OriginLab，并且拟合效果很好。

谢谢。

Answer 1

您遇到了向曲线拟合算法提供错误猜测 的经典问题。这完全是由于您不必要地颠倒矩阵 T 然后没有考虑高斯的新位置(称为 center 的参数，传递给 gaussian() - 我记得 this code)。

你看，当我对你的原始数据进行拟合时会发生以下情况:

T = T0-bg # background subtraction?
fitparams_me, fitparams_you = [], []

for colind in xrange(16,19):
    column = T[:,colind]
    guess = column.max(), column.argmax(), 3, 0  # Good guess for a SINGLE gaussian

    popt, pcov = optimize.curve_fit(one_gaussian, datax, column, p0=guess)
    fitparams_me.append(popt)
print(fitparams_me)

显示:

[array([ 365.40098996,   91.24095009,    1.11390434,   -0.99632476]),
 array([ 348.4327168 ,   92.0262556 ,    1.26650618,   -1.08018819]),
 array([ 413.21526868,   90.8569241 ,    1.0445618 ,   -1.0565371 ])]

这些导致非常适合。

现在这就是您正在做的事情:您首先将矩阵上下颠倒，但您一直假设峰值位于第一行。然而，情况已不再如此，这段代码强调了这一点:

T = np.flipud(T)

for colind in xrange(16,19):
    column = T[:,colind]
    guess = column.max(), column.argmax(), 3, 0  # Good guess for a SINGLE gaussian
    your_guess = [0.5*Mmax, N/10., 10., 0.]
    print guess[1], your_guess[1]
    popt, pcov = optimize.curve_fit(one_gaussian, datax, column, p0=your_guess)
    fitparams_you.append(popt)

# printed results:
932 102.4
931 102.4
932 102.4

因此，每次我仍然正确猜测最大值出现的位置，但您假设它始终在数据的第 102 行附近(形状为 1024, 1024)。

你的曲线拟合结果与我的大不相同也就不足为奇了:

>>> print(fitparams_you)  

[array([ -1.640e-07,   1.024e+02,   1.000e+01,   2.046e-10]),
 array([ -1.640e-07,   1.024e+02,   1.000e+01,   2.046e-10]),
 array([ -1.640e-07,   1.024e+02,   1.000e+01,   2.046e-10])]

只需翻转您的列即可轻松解决:

popt, pcov = optimize.curve_fit(one_gaussian, datax, column[::-1], p0=your_guess) 

或者您可以尝试使用 argmax 等技巧使您的算法更健壮。