ast*_*ath 9 python gaussian curve-fitting python-2.7
我一直在寻找一种方法来对我的数据进行多次高斯拟合.到目前为止,我发现的大多数例子都使用正态分布来制作随机数.但我有兴趣查看我的数据图并检查是否有1-3个峰.
我可以为一个高峰做到这一点,但我不知道如何做更多.
例如,我有这些数据:http://www.filedropper.com/data_11
我尝试过使用lmfit,当然还有scipy,但没有很好的结果.
谢谢你的帮助!
Oli*_* W. 14
简单地制作单个高斯之和的参数化模型函数.为您的初始猜测选择一个好的值(这是一个非常关键的步骤),然后scipy.optimize
稍微调整这些数字.
这是你如何做到这一点:
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')
Run Code Online (Sandbox Code Playgroud)
如您所见,这两种拟合(视觉上)几乎没有区别.因此,您无法确定源中是否存在3个高斯或只有2个.但是,如果您必须进行猜测,则检查最小残差:
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
Run Code Online (Sandbox Code Playgroud)
在这种情况下,3个高斯提供了更好的结果,但我也使我的初始猜测相当准确.
归档时间: |
|
查看次数: |
11222 次 |
最近记录: |