19

My knowledge of maths is limited which is why I am probably stuck. I have a spectra to which I am trying to fit two Gaussian peaks. I can fit to the largest peak, but I cannot fit to the smallest peak. I understand that I need to sum the Gaussian function for the two peaks but I do not know where I have gone wrong. An image of my current output is shown:

Current Output

The blue line is my data and the green line is my current fit. There is a shoulder to the left of the main peak in my data which I am currently trying to fit, using the following code:

import matplotlib.pyplot as pt
import numpy as np
from scipy.optimize import leastsq
from pylab import *

time = []
counts = []


for i in open('/some/folder/to/file.txt', 'r'):
    segs = i.split()
    time.append(float(segs[0]))
    counts.append(segs[1])

time_array = arange(len(time), dtype=float)
counts_array = arange(len(counts))
time_array[0:] = time
counts_array[0:] = counts


def model(time_array0, coeffs0):
    a = coeffs0[0] + coeffs0[1] * np.exp( - ((time_array0-coeffs0[2])/coeffs0[3])**2 )
    b = coeffs0[4] + coeffs0[5] * np.exp( - ((time_array0-coeffs0[6])/coeffs0[7])**2 ) 
    c = a+b
    return c


def residuals(coeffs, counts_array, time_array):
    return counts_array - model(time_array, coeffs)

# 0 = baseline, 1 = amplitude, 2 = centre, 3 = width
peak1 = np.array([0,6337,16.2,4.47,0,2300,13.5,2], dtype=float)
#peak2 = np.array([0,2300,13.5,2], dtype=float)

x, flag = leastsq(residuals, peak1, args=(counts_array, time_array))
#z, flag = leastsq(residuals, peak2, args=(counts_array, time_array))

plt.plot(time_array, counts_array)
plt.plot(time_array, model(time_array, x), color = 'g') 
#plt.plot(time_array, model(time_array, z), color = 'r')
plt.show()
4

3 回答 3

18

如果您只拟合一个由两个高斯分布组合而成的函数，则此代码对我有用。

我刚刚做了一个残差函数，它添加了两个高斯函数，然后从真实数据中减去它们。

我传递给 Numpy 最小二乘函数的参数（p）包括：第一个高斯函数的均值（m），第一和第二个高斯函数的均值之差（dm，即水平位移），标准差第一个 (sd1) 的标准差和第二个 (sd2) 的标准差。

import numpy as np
from scipy.optimize import leastsq
import matplotlib.pyplot as plt

######################################
# Setting up test data
def norm(x, mean, sd):
  norm = []
  for i in range(x.size):
    norm += [1.0/(sd*np.sqrt(2*np.pi))*np.exp(-(x[i] - mean)**2/(2*sd**2))]
  return np.array(norm)

mean1, mean2 = 0, -2
std1, std2 = 0.5, 1 

x = np.linspace(-20, 20, 500)
y_real = norm(x, mean1, std1) + norm(x, mean2, std2)

######################################
# Solving
m, dm, sd1, sd2 = [5, 10, 1, 1]
p = [m, dm, sd1, sd2] # Initial guesses for leastsq
y_init = norm(x, m, sd1) + norm(x, m + dm, sd2) # For final comparison plot

def res(p, y, x):
  m, dm, sd1, sd2 = p
  m1 = m
  m2 = m1 + dm
  y_fit = norm(x, m1, sd1) + norm(x, m2, sd2)
  err = y - y_fit
  return err

plsq = leastsq(res, p, args = (y_real, x))

y_est = norm(x, plsq[0][0], plsq[0][2]) + norm(x, plsq[0][0] + plsq[0][1], plsq[0][3])

plt.plot(x, y_real, label='Real Data')
plt.plot(x, y_init, 'r.', label='Starting Guess')
plt.plot(x, y_est, 'g.', label='Fitted')
plt.legend()
plt.show()

代码的结果。

于 2012-04-13T23:36:22.480 回答
17

您可以使用scikit-learn中的高斯混合模型：

from sklearn import mixture
import matplotlib.pyplot
import matplotlib.mlab
import numpy as np
clf = mixture.GMM(n_components=2, covariance_type='full')
clf.fit(yourdata)
m1, m2 = clf.means_
w1, w2 = clf.weights_
c1, c2 = clf.covars_
histdist = matplotlib.pyplot.hist(yourdata, 100, normed=True)
plotgauss1 = lambda x: plot(x,w1*matplotlib.mlab.normpdf(x,m1,np.sqrt(c1))[0], linewidth=3)
plotgauss2 = lambda x: plot(x,w2*matplotlib.mlab.normpdf(x,m2,np.sqrt(c2))[0], linewidth=3)
plotgauss1(histdist[1])
plotgauss2(histdist[1])

在此处输入图像描述

您还可以使用下面的函数通过 ncomp 参数拟合所需的高斯数：

from sklearn import mixture
%pylab

def fit_mixture(data, ncomp=2, doplot=False):
    clf = mixture.GMM(n_components=ncomp, covariance_type='full')
    clf.fit(data)
    ml = clf.means_
    wl = clf.weights_
    cl = clf.covars_
    ms = [m[0] for m in ml]
    cs = [numpy.sqrt(c[0][0]) for c in cl]
    ws = [w for w in wl]
    if doplot == True:
        histo = hist(data, 200, normed=True)
        for w, m, c in zip(ws, ms, cs):
            plot(histo[1],w*matplotlib.mlab.normpdf(histo[1],m,np.sqrt(c)), linewidth=3)
    return ms, cs, ws
于 2013-10-04T13:48:09.153 回答
4

系数 0 和 4 是退化的 - 数据中绝对没有任何东西可以在它们之间做出决定。您应该使用单个零级参数而不是两个（即从您的代码中删除其中一个）。这可能是阻止你适应的原因（忽略这里的评论说这是不可能的 - 该数据中显然至少有两个峰值，你当然应该能够适应那个）。

（可能不清楚我为什么要建议这个，但发生的事情是系数 0 和 4 可以相互抵消。它们都可以是零，或者一个可以是 100，另一个可以是 -100 - 无论哪种方式，适合一样好。这会“混淆”拟合程序，当没有一个正确答案时，它会花时间试图找出它们应该是什么，因为无论一个是什么值，另一个可能只是那个的否定，并且适合将是相同的）。

事实上，从剧情上看，可能根本不需要零级。我会尝试放弃这两个，看看合身的样子。

此外，无需在最小二乘中拟合系数 1 和 5（或零点）。相反，因为模型是线性的，你可以在每个循环中计算它们的值。这将使事情变得更快，但并不重要。我刚刚注意到你说你的数学不太好，所以可能忽略这个。

于 2012-04-14T01:05:34.507 回答