如何快速对多个数据集执行最小二乘拟合？
Mic*_*ael 11 python gaussian curve-fitting scipy least-squares
我试图在许多数据点上进行高斯拟合.例如,我有一个256 x 262144数据阵列.256点需要拟合高斯分布,我需要262144.
有时高斯分布的峰值在数据范围之外,因此要获得准确的平均结果,曲线拟合是最好的方法.即使峰值在范围内,曲线拟合也会提供更好的西格玛,因为其他数据不在范围内.
我使用http://www.scipy.org/Cookbook/FittingData中的代码为一个数据点工作.
我试图重复这个算法,但看起来需要大约43分钟才能解决这个问题.是否有一种已经写好的快速方法可以并行或更有效地执行此操作？
from scipy import optimize
from numpy import *
import numpy
# Fitting code taken from: http://www.scipy.org/Cookbook/FittingData
class Parameter:
def __init__(self, value):
self.value = value
def set(self, value):
self.value = value
def __call__(self):
return self.value
def fit(function, parameters, y, x = None):
def f(params):
i = 0
for p in parameters:
p.set(params[i])
i += 1
return y - function(x)
if x is None: x = arange(y.shape[0])
p = [param() for param in parameters]
optimize.leastsq(f, p)
def nd_fit(function, parameters, y, x = None, axis=0):
"""
Tries to an n-dimensional array to the data as though each point is a new dataset valid across the appropriate axis.
"""
y = y.swapaxes(0, axis)
shape = y.shape
axis_of_interest_len = shape[0]
prod = numpy.array(shape[1:]).prod()
y = y.reshape(axis_of_interest_len, prod)
params = numpy.zeros([len(parameters), prod])
for i in range(prod):
print "at %d of %d"%(i, prod)
fit(function, parameters, y[:,i], x)
for p in range(len(parameters)):
params[p, i] = parameters[p]()
shape[0] = len(parameters)
params = params.reshape(shape)
return params
请注意,数据不一定是256x262144,我已经在nd_fit中做了一些捏造以使其工作.
我用来使这个工作的代码是
from curve_fitting import *
import numpy
frames = numpy.load("data.npy")
y = frames[:,0,0,20,40]
x = range(0, 512, 2)
mu = Parameter(x[argmax(y)])
height = Parameter(max(y))
sigma = Parameter(50)
def f(x): return height() * exp (-((x - mu()) / sigma()) ** 2)
ls_data = nd_fit(f, [mu, sigma, height], frames, x, 0)
注意:@JoeKington在下面发布的解决方案很棒,解决得非常快.然而,除非高斯的重要区域在适当的区域内,否则它似乎不起作用.我将不得不测试平均值是否仍然准确,因为这是我使用它的主要内容.
Joe*_*ton 18
最简单的方法是将问题线性化.您使用的是非线性迭代方法,它比线性最小二乘解法慢.
基本上,你有:
y = height * exp(-(x - mu)^2 / (2 * sigma^ 2))
要使这成为线性方程,请采用两边的(自然)对数:
ln(y) = ln(height) - (x - mu)^2 / (2 * sigma^2)
然后,这简化为多项式:
ln(y) = -x^2 / (2 * sigma^2) + x * mu / sigma^2 - mu^2 / sigma^2 + ln(height)
我们可以用更简单的形式重新设计:
ln(y) = A * x^2 + B * x + C
哪里:
A = 1 / (2 * sigma^2)
B = mu / (2 * sigma^2)
C = mu^2 / sigma^2 + ln(height)
然而,有一个问题.在分布的"尾部"存在噪声时,这将变得不稳定.
因此,我们只需要使用分布"峰值"附近的数据.只需在拟合中包含超过某个阈值的数据就足够了.在这个例子中,我只包括大于我们拟合的给定高斯曲线的最大观测值的20%的数据.
但是,一旦我们完成了这项工作,它就会非常快.求解262144个不同的高斯曲线只需要约1分钟(如果你在大的东西上运行代码,请务必删除代码的绘图部分.).如果你想要,它也很容易并行化......
import numpy as np
import matplotlib.pyplot as plt
import matplotlib as mpl
import itertools
def main():
x, data = generate_data(256, 6)
model = [invert(x, y) for y in data.T]
sigma, mu, height = [np.array(item) for item in zip(*model)]
prediction = gaussian(x, sigma, mu, height)
plot(x, data, linestyle='none', marker='o')
plot(x, prediction, linestyle='-')
plt.show()
def invert(x, y):
# Use only data within the "peak" (20% of the max value...)
key_points = y > (0.2 * y.max())
x = x[key_points]
y = y[key_points]
# Fit a 2nd order polynomial to the log of the observed values
A, B, C = np.polyfit(x, np.log(y), 2)
# Solve for the desired parameters...
sigma = np.sqrt(-1 / (2.0 * A))
mu = B * sigma**2
height = np.exp(C + 0.5 * mu**2 / sigma**2)
return sigma, mu, height
def generate_data(numpoints, numcurves):
np.random.seed(3)
x = np.linspace(0, 500, numpoints)
height = 100 * np.random.random(numcurves)
mu = 200 * np.random.random(numcurves) + 200
sigma = 100 * np.random.random(numcurves) + 0.1
data = gaussian(x, sigma, mu, height)
noise = 5 * (np.random.random(data.shape) - 0.5)
return x, data + noise
def gaussian(x, sigma, mu, height):
data = -np.subtract.outer(x, mu)**2 / (2 * sigma**2)
return height * np.exp(data)
def plot(x, ydata, ax=None, **kwargs):
if ax is None:
ax = plt.gca()
colorcycle = itertools.cycle(mpl.rcParams['axes.color_cycle'])
for y, color in zip(ydata.T, colorcycle):
ax.plot(x, y, color=color, **kwargs)
main()
对于并行版本,我们唯一需要更改的是主要功能.(我们还需要一个虚函数,因为multiprocessing.Pool.imap它不能为其函数提供额外的参数......)它看起来像这样:
def parallel_main():
import multiprocessing
p = multiprocessing.Pool()
x, data = generate_data(256, 262144)
args = itertools.izip(itertools.repeat(x), data.T)
model = p.imap(parallel_func, args, chunksize=500)
sigma, mu, height = [np.array(item) for item in zip(*model)]
prediction = gaussian(x, sigma, mu, height)
def parallel_func(args):
return invert(*args)
编辑:如果简单的多项式拟合不能很好地工作,请尝试用y值加权问题,如 @tslisten共享的链接/论文中所述(和Stefan van der Walt实现,尽管我的实现有点不同).
import numpy as np
import matplotlib.pyplot as plt
import matplotlib as mpl
import itertools
def main():
def run(x, data, func, threshold=0):
model = [func(x, y, threshold=threshold) for y in data.T]
sigma, mu, height = [np.array(item) for item in zip(*model)]
prediction = gaussian(x, sigma, mu, height)
plt.figure()
plot(x, data, linestyle='none', marker='o', markersize=4)
plot(x, prediction, linestyle='-', lw=2)
x, data = generate_data(256, 6, noise=100)
threshold = 50
run(x, data, weighted_invert, threshold=threshold)
plt.title('Weighted by Y-Value')
run(x, data, invert, threshold=threshold)
plt.title('Un-weighted Linear Inverse'
plt.show()
def invert(x, y, threshold=0):
mask = y > threshold
x, y = x[mask], y[mask]
# Fit a 2nd order polynomial to the log of the observed values
A, B, C = np.polyfit(x, np.log(y), 2)
# Solve for the desired parameters...
sigma, mu, height = poly_to_gauss(A,B,C)
return sigma, mu, height
def poly_to_gauss(A,B,C):
sigma = np.sqrt(-1 / (2.0 * A))
mu = B * sigma**2
height = np.exp(C + 0.5 * mu**2 / sigma**2)
return sigma, mu, height
def weighted_invert(x, y, weights=None, threshold=0):
mask = y > threshold
x,y = x[mask], y[mask]
if weights is None:
weights = y
else:
weights = weights[mask]
d = np.log(y)
G = np.ones((x.size, 3), dtype=np.float)
G[:,0] = x**2
G[:,1] = x
model,_,_,_ = np.linalg.lstsq((G.T*weights**2).T, d*weights**2)
return poly_to_gauss(*model)
def generate_data(numpoints, numcurves, noise=None):
np.random.seed(3)
x = np.linspace(0, 500, numpoints)
height = 7000 * np.random.random(numcurves)
mu = 1100 * np.random.random(numcurves)
sigma = 100 * np.random.random(numcurves) + 0.1
data = gaussian(x, sigma, mu, height)
if noise is None:
noise = 0.1 * height.max()
noise = noise * (np.random.random(data.shape) - 0.5)
return x, data + noise
def gaussian(x, sigma, mu, height):
data = -np.subtract.outer(x, mu)**2 / (2 * sigma**2)
return height * np.exp(data)
def plot(x, ydata, ax=None, **kwargs):
if ax is None:
ax = plt.gca()
colorcycle = itertools.cycle(mpl.rcParams['axes.color_cycle'])
for y, color in zip(ydata.T, colorcycle):
#kwargs['color'] = kwargs.get('color', color)
ax.plot(x, y, color=color, **kwargs)
main()
如果这仍然给你带来麻烦,那么尝试迭代重新加权最小二乘问题(在@tslisten链接中提到的最终"最佳"推荐方法).但请记住,这会慢得多.
def iterative_weighted_invert(x, y, threshold=None, numiter=5):
last_y = y
for _ in range(numiter):
model = weighted_invert(x, y, weights=last_y, threshold=threshold)
last_y = gaussian(x, *model)
return model
- http://scipy-central.org/item/28/2/fitting-a-gaussian-to-noisy-data-points了解更多信息. (2认同)