问题描述
我一直在寻找对数据进行多次高斯拟合的方法。到目前为止，我发现的大多数示例都使用正态分布来生成随机数。但是我有兴趣查看数据图并检查是否有1-3个峰值。
I've been looking for a way to do multiple Gaussian fitting to my data. Most of the examples I've found so far use a normal distribution to make random numbers. But I am interested in looking at the plot of my data and checking if there are 1-3 peaks.
我可以在一个峰值上执行此操作，但是我没有知道如何做得更多。
I can do this for one peak, but I don't know how to do it for more.
例如，我有以下数据：
For example, I have this data: http://www.filedropper.com/data_11
我尝试使用lmfit，当然也可以使用scipy，但效果不佳。
I have tried using lmfit, and of course scipy, but with no nice results.
感谢您的帮助！
推荐答案
简单地使单个高斯和的参数化模型函数生效。为您的初始猜测选择一个好的值（这是非常关键的一步），然后让 scipy.optimize 稍微调整一下这些数字。
Simply make parameterized model functions of the sum of single Gaussians. Choose a good value for your initial guess (this is a really critical step) and then have scipy.optimize tweak those numbers a bit.
这是您可能的操作方式：
Here's how you might do it:
import numpy as np
import matplotlib.pyplot as plt
from scipy import optimize
data = np.genfromtxt('data.txt')
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)
errfunc3 = lambda p, x, y: (three_gaussians(x, *p) - y)**2
errfunc2 = lambda p, x, y: (two_gaussians(x, *p) - y)**2
guess3 = [0.49, 0.55, 0.01, 0.6, 0.61, 0.01, 1, 0.64, 0.01, 0] # I guess there are 3 peaks, 2 are clear, but between them there seems to be another one, based on the change in slope smoothness there
guess2 = [0.49, 0.55, 0.01, 1, 0.64, 0.01, 0] # I removed the peak I'm not too sure about
optim3, success = optimize.leastsq(errfunc3, guess3[:], args=(data[:,0], data[:,1]))
optim2, success = optimize.leastsq(errfunc2, guess2[:], args=(data[:,0], data[:,1]))
optim3
plt.plot(data[:,0], data[:,1], lw=5, c='g', label='measurement')
plt.plot(data[:,0], three_gaussians(data[:,0], *optim3),
lw=3, c='b', label='fit of 3 Gaussians')
plt.plot(data[:,0], two_gaussians(data[:,0], *optim2),
lw=1, c='r', ls='--', label='fit of 2 Gaussians')
plt.legend(loc='best')
plt.savefig('result.png')
您可以看到，几乎这两个拟合之间（在视觉上）没有区别。因此，您不能确定源中是否存在3个高斯或仅存在2个。但是，如果您必须进行猜测，请检查最小的残差：
As you can see, there is almost no difference between these two fits (visually). So you can't know for sure if there were 3 Gaussians present in the source or only 2. However, if you had to make a guess, then check for the smallest residual:
err3 = np.sqrt(errfunc3(optim3, data[:,0], data[:,1])).sum()
err2 = np.sqrt(errfunc2(optim2, data[:,0], data[:,1])).sum()
print('Residual error when fitting 3 Gaussians: {}\n'
'Residual error when fitting 2 Gaussians: {}'.format(err3, err2))
# Residual error when fitting 3 Gaussians: 3.52000910965
# Residual error when fitting 2 Gaussians: 3.82054499044
在这种情况下，3个高斯人给出了更好的结果，但我也使最初的猜测相当准确。
In this case, 3 Gaussians gives a better result, but I also made my initial guess fairly accurate.
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