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高斯拟合不使用Python
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Stack Overflow用户
提问于 2015-02-08 13:23:50
回答 1查看 616关注 0票数 4

我正在尝试使用下面的数据代码在列上安装这个矩阵:

代码语言:javascript
复制
#!/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。它应该工作,但最后的矩阵,"T“只显示列与常量,而不是一个光滑的高斯形状曲线。请看一看,告诉我我做错了什么。我曾尝试过其他程序,如OriginLab和配件工作良好。

谢谢。

EN

回答 1

Stack Overflow用户

回答已采纳

发布于 2015-02-08 15:53:28

你正在经历一个经典的问题,就是给曲线拟合算法提供一个不正确的猜测。这完全是由于你不必要的颠倒翻转矩阵T,然后没有考虑到高斯人的新位置(称为center的参数,传递给gaussian() -我记得这段代码)。

你看,当我对你的原始数据进行拟合时,会发生这样的情况:

代码语言:javascript
复制
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)

这表明:

代码语言:javascript
复制
[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 ])]

这些会让你很合身。

下面是您要做的:首先将矩阵倒转,但始终假设峰值位于第一行。然而,情况已不再如此,这段代码着重指出:

代码语言:javascript
复制
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,每次我仍然正确地猜测最大值出现在哪里,但是您假设它总是在数据的第102行(形状为1024, 1024)附近。

你的曲线拟合结果与我的结果大不相同,这一点也不足为奇:

代码语言:javascript
复制
>>> 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])]

您可以轻松地通过翻转您的专栏来解决这个问题:

代码语言:javascript
复制
popt, pcov = optimize.curve_fit(one_gaussian, datax, column[::-1], p0=your_guess) 

或者,您可以尝试通过使用像argmax这样的技巧来使算法更加健壮。

票数 3
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页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接:

https://stackoverflow.com/questions/28394406

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