I am trying to fit a top hat function to some data, ie. f(x) is constant for the entire real line, except for one segment of finite length which is equal to another constant. My parameters are the two constants of the tophat function, the midpoint, and the width and I'm trying to use scipy.optimize.curve_fit to get all of these. Unfortunately, curve_fit is having trouble obtaining the width of the hat. No matter what I do, it refuses to test any value of the width other than the one I start with, and fits the rest of the data very badly. The following code snippet illustrates the problem:
我试图将一个大礼帽函数适用于某些数据,即。 f(x)对于整个实线是恒定的,除了一个有限长度的段等于另一个常数。我的参数是tophat函数的两个常量,中点和宽度,我正在尝试使用scipy.optimize.curve_fit来获取所有这些。不幸的是,curve_fit无法获得帽子的宽度。无论我做什么,它都拒绝测试除我开始之外的任何宽度值,并且非常严重地适应其余数据。以下代码段说明了此问题:
import numpy as np
import matplotlib.pyplot as plt
from scipy.optimize import curve_fit
def tophat(x, base_level, hat_level, hat_mid, hat_width):
ret=[]
for xx in x:
if hat_mid-hat_width/2. < xx < hat_mid+hat_width/2.:
ret.append(hat_level)
else:
ret.append(base_level)
return np.array(ret)
x = np.arange(-10., 10., 0.01)
y = tophat(x, 1.0, 5.0, 0.0, 1.0)+np.random.rand(len(x))*0.2-0.1
guesses = [ [1.0, 5.0, 0.0, 1.0],
[1.0, 5.0, 0.0, 0.1],
[1.0, 5.0, 0.0, 2.0] ]
plt.plot(x,y)
for guess in guesses:
popt, pcov = curve_fit( tophat, x, y, p0=guess )
print popt
plt.plot( x, tophat(x, popt[0], popt[1], popt[2], popt[3]) )
plt.show()
Why is curve_fit so extremely terrible at getting this right, and how can I fix it?
为什么在实现这一目标时,curve_fit非常糟糕,我该如何解决?
2 个解决方案
#1
2
First, the definition of tophat could use numpy.where instead of a loop:
首先,tophat的定义可以使用numpy.where而不是循环:
def tophat(x, base_level, hat_level, hat_mid, hat_width):
return np.where((hat_mid-hat_width/2. < x) & (x < hat_mid+hat_width/2.), hat_level, base_level)
Second, the tricky discontinuous objective function resists the optimization algorithms that curve_fit calls. The Nelder-Mead method is usually preferable for rough functions, but it looks like curve_fit cannot use it. So I set up an objective function (just the sum of absolute values of deviations) and minimize that:
其次,棘手的不连续目标函数抵抗curve_fit调用的优化算法。 Nelder-Mead方法通常适用于粗糙函数,但看起来像curve_fit不能使用它。所以我设置了一个目标函数(只是偏差绝对值的总和)并最小化:
def objective(params, x, y):
return np.sum(np.abs(tophat(x, *params) - y))
plt.plot(x,y)
for guess in guesses:
res = minimize(objective, guess, args=(x, y), method='Nelder-Mead')
print(res.x)
plt.plot(x, tophat(x, *(res.x)))
The results are better, in that starting with a too-wide hat of width 2 makes it shrink down to the correct size (see the last of three guesses).
结果更好,因为从宽度为2的太宽的帽子开始使其缩小到正确的大小(参见三个猜测中的最后一个)。
[9.96041297e-01 5.00035502e+00 2.39462103e-04 9.99759984e-01]
[ 1.00115808e+00 4.94088711e+00 -2.21340843e-05 1.04924153e-01]
[9.95947108e-01 4.99871040e+00 1.26575116e-03 9.97908018e-01]
Unfortunately, when the starting guess is a too-narrow hat, the optimizer is still stuck.
不幸的是,当开始猜测是一个太窄的帽子时,优化器仍然卡住了。
You can try other optimization method / objective function combinations but I haven't found one that makes the hat reliably expand.
您可以尝试其他优化方法/目标函数组合,但我没有找到一个使帽子可靠地扩展。
One thing to try is not to use the parameters that are too close to the true levels; this sometimes might hurt. With
要尝试的一件事是不要使用太接近真实水平的参数;这有时可能会受到伤害同
guesses = [ [1.0, 1.0, 0.0, 1.0],
[1.0, 1.0, 0.0, 0.1],
[1.0, 1.0, 0.0, 2.0] ]
I once managed to get
我曾经设法得到
[ 1.00131181 4.99156649 -0.01109271 0.96822019]
[ 1.00137925 4.97879423 -0.05091561 1.096166 ]
[ 1.00130568 4.98679988 -0.01133717 0.99339777]
which is correct for all three widths. However, this was only on some of several tries (there is some randomness in the initialization of the optimizing procedure). Some other attempts with the same initial points failed; the process is not robust enough.
这对于所有三个宽度都是正确的。但是,这只是在几次尝试中的一些尝试(在优化过程的初始化中存在一些随机性)。其他一些具有相同初始点的尝试失败了;这个过程不够健壮。
#2
1
By its nature, non-linear least-squares fitting as with curve_fit() works with real, floating-point numbers and is not good at dealing with discrete variables. In the fit process, small changes (like, at the 1e-7 level) are made to each variable, and the effect of that small change on the fit result is used to decide how to change that variable to improve the fit. With discretely sampled data, small changes to your hat_mid and/or hat_width could easily be smaller than the spacing of data points and so have no effect at all on the fit. That is why curve_fit is "extremely terrible" at this problem.
就其本质而言,与curve_fit()一样的非线性最小二乘拟合与实数浮点数一起使用,并且不善于处理离散变量。在拟合过程中,对每个变量进行小的变化(例如,在1e-7级别),并且使用该小变化对拟合结果的影响来决定如何改变该变量以改善拟合。对于离散采样数据,对hat_mid和/或hat_width的小变化可能很容易小于数据点的间距,因此对拟合没有任何影响。这就是为什么curve_fit在这个问题上“极其糟糕”的原因。
You may find that giving a finite width (that is, comparable to the step size of your discrete data) to the steps helps to better find where the edges of you hat are.
您可能会发现给步骤提供有限的宽度(即与离散数据的步长相当)有助于更好地找到帽子边缘的位置。
#1
2
First, the definition of tophat could use numpy.where instead of a loop:
首先,tophat的定义可以使用numpy.where而不是循环:
def tophat(x, base_level, hat_level, hat_mid, hat_width):
return np.where((hat_mid-hat_width/2. < x) & (x < hat_mid+hat_width/2.), hat_level, base_level)
Second, the tricky discontinuous objective function resists the optimization algorithms that curve_fit calls. The Nelder-Mead method is usually preferable for rough functions, but it looks like curve_fit cannot use it. So I set up an objective function (just the sum of absolute values of deviations) and minimize that:
其次,棘手的不连续目标函数抵抗curve_fit调用的优化算法。 Nelder-Mead方法通常适用于粗糙函数,但看起来像curve_fit不能使用它。所以我设置了一个目标函数(只是偏差绝对值的总和)并最小化:
def objective(params, x, y):
return np.sum(np.abs(tophat(x, *params) - y))
plt.plot(x,y)
for guess in guesses:
res = minimize(objective, guess, args=(x, y), method='Nelder-Mead')
print(res.x)
plt.plot(x, tophat(x, *(res.x)))
The results are better, in that starting with a too-wide hat of width 2 makes it shrink down to the correct size (see the last of three guesses).
结果更好,因为从宽度为2的太宽的帽子开始使其缩小到正确的大小(参见三个猜测中的最后一个)。
[9.96041297e-01 5.00035502e+00 2.39462103e-04 9.99759984e-01]
[ 1.00115808e+00 4.94088711e+00 -2.21340843e-05 1.04924153e-01]
[9.95947108e-01 4.99871040e+00 1.26575116e-03 9.97908018e-01]
Unfortunately, when the starting guess is a too-narrow hat, the optimizer is still stuck.
不幸的是,当开始猜测是一个太窄的帽子时,优化器仍然卡住了。
You can try other optimization method / objective function combinations but I haven't found one that makes the hat reliably expand.
您可以尝试其他优化方法/目标函数组合,但我没有找到一个使帽子可靠地扩展。
One thing to try is not to use the parameters that are too close to the true levels; this sometimes might hurt. With
要尝试的一件事是不要使用太接近真实水平的参数;这有时可能会受到伤害同
guesses = [ [1.0, 1.0, 0.0, 1.0],
[1.0, 1.0, 0.0, 0.1],
[1.0, 1.0, 0.0, 2.0] ]
I once managed to get
我曾经设法得到
[ 1.00131181 4.99156649 -0.01109271 0.96822019]
[ 1.00137925 4.97879423 -0.05091561 1.096166 ]
[ 1.00130568 4.98679988 -0.01133717 0.99339777]
which is correct for all three widths. However, this was only on some of several tries (there is some randomness in the initialization of the optimizing procedure). Some other attempts with the same initial points failed; the process is not robust enough.
这对于所有三个宽度都是正确的。但是,这只是在几次尝试中的一些尝试(在优化过程的初始化中存在一些随机性)。其他一些具有相同初始点的尝试失败了;这个过程不够健壮。
#2
1
By its nature, non-linear least-squares fitting as with curve_fit() works with real, floating-point numbers and is not good at dealing with discrete variables. In the fit process, small changes (like, at the 1e-7 level) are made to each variable, and the effect of that small change on the fit result is used to decide how to change that variable to improve the fit. With discretely sampled data, small changes to your hat_mid and/or hat_width could easily be smaller than the spacing of data points and so have no effect at all on the fit. That is why curve_fit is "extremely terrible" at this problem.
就其本质而言,与curve_fit()一样的非线性最小二乘拟合与实数浮点数一起使用,并且不善于处理离散变量。在拟合过程中,对每个变量进行小的变化(例如,在1e-7级别),并且使用该小变化对拟合结果的影响来决定如何改变该变量以改善拟合。对于离散采样数据,对hat_mid和/或hat_width的小变化可能很容易小于数据点的间距,因此对拟合没有任何影响。这就是为什么curve_fit在这个问题上“极其糟糕”的原因。
You may find that giving a finite width (that is, comparable to the step size of your discrete data) to the steps helps to better find where the edges of you hat are.
您可能会发现给步骤提供有限的宽度(即与离散数据的步长相当)有助于更好地找到帽子边缘的位置。