Develop Reference

Python Spinmob curve_fit works but fitter does not

I'm trying to fit data with a Gaussian.
The raw data itself displays a very obvious peak.
When I attempt fitting using curve_fit, the fit identifies the peak but it does not have a curved top.
I am trying to fit the data now with spinmob's fitter as well. However, this fitting just gives a straight line.
I've tried changing several parameters of the fitter, the Gaussian function definition, and the initial parameters for the fit but nothing seems to work.
Here is the code:
from scipy.optimize import curve_fit
from scipy import asarray as ar,exp
import spinmob as s
x = x30
y = ydata
def gaussian(x, A, mu, sig): # See http://mathworld.wolfram.com/GaussianFunction.html
return A/(sig * np.sqrt(2*np.pi)) * np.exp(-np.power(x-mu, 2) / (2 * np.power(sig, 2)))
popt,pcov = curve_fit(gaussian,x,y,p0=[1,7.688,0.005])
FWHM = 2*np.sqrt(2*np.log(2))*popt[2]
print("FWHM: {}".format(FWHM))
plt.plot(x,y,'bo',label='data')
plt.plot(x,gaussian(x,*popt),'r+-',label='fit')
plt.legend()
fitter = s.data.fitter()
fitter.set(subtract_bg=True, plot_guess_zoom=True)
fitter.set_functions(f=gaussian, p='A=1,mu=8.688,sig=0.001')
fitter.set_data(x, y, eydata = 0.03)
fitter.fit()
The curve_fit returns this plot:
Curve_fit plot
The spinmob fitter plot gives this:
Spinmob Fitter Plot

Assuming that spinmob actually uses scipy.curve_fit under the hood, I would guess (sorry) that the problem is that the initial values you give to it are so far off that it cannot possibly find a solution.
For sure, A=1 is not a very good guess for either scipy.curve_fit() or spinmob.fitter(). The peak is definitely negative, and you should be guessing a value more like -0.1 than +1. In fact you could probably assert that A must be < 0.
The initial value of 7.688 for mu that you give to curve_fit() is pretty good, and will allow a solution. I do not know whether it is a typo or not, but the initial value of 8.688 for mu that you give to spinmob.fitter() is very far off (that is, way outside the data range), and the fit will never be able to refine its way to the correct solution from there.
Initial values matter for curve-fitting and poor initial values can lead to bad results.
It might be viewed by some as a shameless plug, but allow me to encourage you to try lmfit (https://lmfit.github.io/lmfit-py/) (I am a lead author) for this kind of problem. Lmfit replaces the array of parameter values with named Parameter objects for better organization of fits. It also has a built-in Gaussian model (which also calculates FWHM, including an uncertainty). That is, with Lmfit, your script might look like:
import numpy as np
import matplotlib.pyplot as plt
from lmfit.models import GaussianModel
from lmfit.lineshapes import gaussian
# create fake data that looks like yours
xdata = 7.670 + np.arange(41)*0.0010
ydata = gaussian(xdata, amplitude=-0.196, center=7.6881, sigma=0.001)
ydata += np.random.normal(size=41, scale=10.0)
# create gaussian model
gmodel = GaussianModel()
# fit data, giving initial values for amplitude, center, and sigma
result = gmodel.fit(ydata, x=xdata, amplitude=-0.1, center=7.688, sigma=0.005)
# show results
print(result.fit_report())
plt.plot(xdata, ydata, 'bo', label='data')
plt.plot(xdata, result.best_fit, 'r+-', label='fit')
plt.legend()
plt.show()
This will print out a report like
[Model]]
Model(gaussian)
[[Fit Statistics]]
# fitting method = leastsq
# function evals = 21
# data points = 41
# variables = 3
chi-square = 5114.87632
reduced chi-square = 134.602009
Akaike info crit = 203.879794
Bayesian info crit = 209.020510
[[Variables]]
sigma: 9.7713e-04 +/- 1.5456e-04 (15.82%) (init = 0.005)
center: 7.68822727 +/- 1.5484e-04 (0.00%) (init = 7.688)
amplitude: -0.19273945 +/- 0.02643400 (13.71%) (init = -0.1)
fwhm: 0.00230096 +/- 3.6396e-04 (15.82%) == '2.3548200*sigma'
height: -78.6917624 +/- 10.7894236 (13.71%) == '0.3989423*amplitude/max(1.e-15, sigma)'
[[Correlations]] (unreported correlations are < 0.100)
C(sigma, amplitude) = -0.577
and produce a plot of data and best fit like
which should be close to what you are trying to do.

how to set up the initial value for curve_fit to find the best optimizing, not just local optimizing?

I am trying to fit a power-law function, and in order to find the best fit parameter. However, I find that if the initial guess of parameter is different, the "best fit" output is different. Unless I find the right initial guess, I can get the best optimizing, instead of local optimizing. Is there any way to find the **appropriate initial guess ** ????. My code is listed below. Please feel free make any input. Thanks!
import numpy as np
import pandas as pd
from scipy.optimize import curve_fit
import matplotlib.pyplot as plt
%matplotlib inline
# power law function
def func_powerlaw(x,a,b,c):
return a*(x**b)+c
test_X = [1.0,2,3,4,5,6,7,8,9,10]
test_Y =[3.0,1.5,1.2222222222222223,1.125,1.08,1.0555555555555556,1.0408163265306123,1.03125, 1.0246913580246915,1.02]
predict_Y = []
for x in test_X:
predict_Y.append(2*x**-2+1)
If I align with default initial guess, which p0 = [1,1,1]
popt, pcov = curve_fit(func_powerlaw, test_X[1:], test_Y[1:], maxfev=2000)
plt.figure(figsize=(10, 5))
plt.plot(test_X, func_powerlaw(test_X, *popt),'r',linewidth=4, label='fit: a=%.4f, b=%.4f, c=%.4f' % tuple(popt))
plt.plot(test_X[1:], test_Y[1:], '--bo')
plt.plot(test_X[1:], predict_Y[1:], '-b')
plt.legend()
plt.show()
The fit is like below, which is not the best fit.
If I change the initial guess to p0 = [0.5,0.5,0.5]
popt, pcov = curve_fit(func_powerlaw, test_X[1:], test_Y[1:], p0=np.asarray([0.5,0.5,0.5]), maxfev=2000)
I can get the best fit
---------------------Updated in 10.7.2018-------------------------------------------------------------------------------------------------------------------------
As I need to run thousands to even millions of Power Law function, using #James Phillips's method is too expensive. So what method is appropriate besides curve_fit? such as sklearn, np.linalg.lstsq etc.

Here is example code using the scipy.optimize.differential_evolution genetic algorithm, with your data and equation. This scipy module uses the Latin Hypercube algorithm to ensure a thorough search of parameter space and so requires bounds within which to search - in this example, those bounds are based on the data maximum and minimum values. For other problems you might need to supply different search bounds if you know what range of parameter values to expect.
import numpy, scipy, matplotlib
import matplotlib.pyplot as plt
from scipy.optimize import curve_fit
from scipy.optimize import differential_evolution
import warnings
# power law function
def func_power_law(x,a,b,c):
return a*(x**b)+c
test_X = [1.0,2,3,4,5,6,7,8,9,10]
test_Y =[3.0,1.5,1.2222222222222223,1.125,1.08,1.0555555555555556,1.0408163265306123,1.03125, 1.0246913580246915,1.02]
# function for genetic algorithm to minimize (sum of squared error)
def sumOfSquaredError(parameterTuple):
warnings.filterwarnings("ignore") # do not print warnings by genetic algorithm
val = func_power_law(test_X, *parameterTuple)
return numpy.sum((test_Y - val) ** 2.0)
def generate_Initial_Parameters():
# min and max used for bounds
maxX = max(test_X)
minX = min(test_X)
maxY = max(test_Y)
minY = min(test_Y)
maxXY = max(maxX, maxY)
parameterBounds = []
parameterBounds.append([-maxXY, maxXY]) # seach bounds for a
parameterBounds.append([-maxXY, maxXY]) # seach bounds for b
parameterBounds.append([-maxXY, maxXY]) # seach bounds for c
# "seed" the numpy random number generator for repeatable results
result = differential_evolution(sumOfSquaredError, parameterBounds, seed=3)
return result.x
# generate initial parameter values
geneticParameters = generate_Initial_Parameters()
# curve fit the test data
fittedParameters, pcov = curve_fit(func_power_law, test_X, test_Y, geneticParameters)
print('Parameters', fittedParameters)
modelPredictions = func_power_law(test_X, *fittedParameters)
absError = modelPredictions - test_Y
SE = numpy.square(absError) # squared errors
MSE = numpy.mean(SE) # mean squared errors
RMSE = numpy.sqrt(MSE) # Root Mean Squared Error, RMSE
Rsquared = 1.0 - (numpy.var(absError) / numpy.var(test_Y))
print('RMSE:', RMSE)
print('R-squared:', Rsquared)
print()
##########################################################
# graphics output section
def ModelAndScatterPlot(graphWidth, graphHeight):
f = plt.figure(figsize=(graphWidth/100.0, graphHeight/100.0), dpi=100)
axes = f.add_subplot(111)
# first the raw data as a scatter plot
axes.plot(test_X, test_Y, 'D')
# create data for the fitted equation plot
xModel = numpy.linspace(min(test_X), max(test_X))
yModel = func_power_law(xModel, *fittedParameters)
# now the model as a line plot
axes.plot(xModel, yModel)
axes.set_xlabel('X Data') # X axis data label
axes.set_ylabel('Y Data') # Y axis data label
plt.show()
plt.close('all') # clean up after using pyplot
graphWidth = 800
graphHeight = 600
ModelAndScatterPlot(graphWidth, graphHeight)

There is no simple answer: if there was, it would be implemented in curve_fit and then it would not have to ask you for the starting point. One reasonable approach is to fit the homogeneous model y = a*x**b first. Assuming positive y (which is usually the case when you work with power law), this can be done in a rough and quick way: on the log-log scale, log(y) = log(a) + b*log(x) which is linear regression which can be solved with np.linalg.lstsq. This gives candidates for log(a) and for b; the candidate for c with this approach is 0.
test_X = np.array([1.0,2,3,4,5,6,7,8,9,10])
test_Y = np.array([3.0,1.5,1.2222222222222223,1.125,1.08,1.0555555555555556,1.0408163265306123,1.03125, 1.0246913580246915,1.02])
rough_fit = np.linalg.lstsq(np.stack((np.ones_like(test_X), np.log(test_X)), axis=1), np.log(test_Y))[0]
p0 = [np.exp(rough_fit[0]), rough_fit[1], 0]
The result is the good fit you see in the second picture.
By the way, it's better to make test_X a NumPy array at once. Otherwise, you are slicing X[1:] first, this gets NumPy-fied as an array of integers, and then an error is thrown with negative exponents. (And I suppose the purpose of 1.0 was to make it a float array? This is what dtype=np.float parameter should be used for.)

In addition to the very fine answers from Welcome to Stack Overflow that "there is no easy, universal approach and James Phillips that "differential evolution often
helps find good starting points (or even good solutions!) if somewhat slower than curve_fit()", allow me to give a separate answer that you may find helpful.
First, the fact that curve_fit() defaults to any parameter values is soul-crushingly bad idea. There is no possible justification for this behavior, and you and everyone else should treat the fact that there are default values for parameters as a serious error in the implementation of curve_fit() and pretend this bug does not exist. NEVER believe these defaults are reasonable.
From a simple plot of data, it should be obvious that a=1, b=1, c=1 are very, very bad starting values. The function decays, so b < 0. In fact, if you had started with a=1, b=-1, c=1 you would have found the correct solution.
It may have also helped to place sensible bounds on the parameters. Even setting bounds of c of (-100, 100) may have helped. As with the sign of b, I think you could have seen that boundary from a simple plot of the data. When I try this for your problem, bounds on c do not help if the initial value is b=1, but it does for b=0 or b=-5.
More importantly, although you print the best-fit params popt in the plot, you do not print the uncertainties or correlations between variables held in pcov, and thus your interpretation of the results is incomplete. If you had looked at these values, you would have seen that starting with b=1 leads not only to bad values but also to huge uncertainties in the parameters and very, very high correlation. This is the fit telling you that it found a poor solution. Unfortunately, the return pcov from curve_fit is not very easy to unpack.
Allow me to recommend lmfit (https://lmfit.github.io/lmfit-py/) (disclaimer: I'm a lead developer). Among other features, this module forces you to give non-default starting values, and to more easily a more complete report. For your problem, even starting with a=1, b=1, c=1 would have given a more meaningful indication that something was wrong:
from lmfit import Model
mod = Model(func_powerlaw)
params = mod.make_params(a=1, b=1, c=1)
ret = mod.fit(test_Y[1:], params, x=test_X[1:])
print(ret.fit_report())
which would print out:
[[Model]]
Model(func_powerlaw)
[[Fit Statistics]]
# fitting method = leastsq
# function evals = 1318
# data points = 9
# variables = 3
chi-square = 0.03300395
reduced chi-square = 0.00550066
Akaike info crit = -44.4751740
Bayesian info crit = -43.8835003
[[Variables]]
a: -1319.16780 +/- 6892109.87 (522458.92%) (init = 1)
b: 2.0034e-04 +/- 1.04592341 (522076.12%) (init = 1)
c: 1320.73359 +/- 6892110.20 (521839.55%) (init = 1)
[[Correlations]] (unreported correlations are < 0.100)
C(a, c) = -1.000
C(b, c) = -1.000
C(a, b) = 1.000
That is a = -1.3e3 +/- 6.8e6 -- not very well defined! In addition all parameters are completely correlated.
Changing the initial value for b to -0.5:
params = mod.make_params(a=1, b=-0.5, c=1) ## Note !
ret = mod.fit(test_Y[1:], params, x=test_X[1:])
print(ret.fit_report())
gives
[[Model]]
Model(func_powerlaw)
[[Fit Statistics]]
# fitting method = leastsq
# function evals = 31
# data points = 9
# variables = 3
chi-square = 4.9304e-32
reduced chi-square = 8.2173e-33
Akaike info crit = -662.560782
Bayesian info crit = -661.969108
[[Variables]]
a: 2.00000000 +/- 1.5579e-15 (0.00%) (init = 1)
b: -2.00000000 +/- 1.1989e-15 (0.00%) (init = -0.5)
c: 1.00000000 +/- 8.2926e-17 (0.00%) (init = 1)
[[Correlations]] (unreported correlations are < 0.100)
C(a, b) = -0.964
C(b, c) = -0.880
C(a, c) = 0.769
which is somewhat better.
In short, initial values always matter, and the result is not only the best-fit values, but includes the uncertainties and correlations.

Gaussian fit returning negative sigma

One of my algorithms performs automatic peak detection based on a Gaussian function, and then later determines the the edges based either on a multiplier (user setting) of the sigma or the 'full width at half maximum'. In the scenario where a user specified that he/she wants the peak limited at 2 Sigma, the algorithm takes -/+ 2*sigma from the peak center (mu). However, I noticed that the sigma returned by curve_fit can be negative, which is something that has been noticed before as can be seen here. However, as I determine the border by doing -/+ this can lead to the algorithm 'failing' (due to a - - scenario) as can be seen in the following code.
MVCE
#! /usr/bin/env python
from scipy.optimize import curve_fit
import bisect
import numpy as np
X = [16.4697402328,16.4701402404,16.4705402481,16.4709402557,16.4713402633,16.4717402709,16.4721402785,16.4725402862,16.4729402938,16.4733403014,16.473740309,16.4741403166,16.4745403243,16.4749403319,16.4753403395,16.4757403471,16.4761403547,16.4765403623,16.47694037,16.4773403776,16.4777403852,16.4781403928,16.4785404004,16.4789404081,16.4793404157,16.4797404233,16.4801404309,16.4805404385,16.4809404462,16.4813404538,16.4817404614,16.482140469,16.4825404766,16.4829404843,16.4833404919,16.4837404995,16.4841405071,16.4845405147,16.4849405224,16.48534053,16.4857405376,16.4861405452,16.4865405528,16.4869405604,16.4873405681,16.4877405757,16.4881405833,16.4885405909,16.4889405985,16.4893406062,16.4897406138,16.4901406214,16.490540629,16.4909406366,16.4913406443,16.4917406519,16.4921406595,16.4925406671,16.4929406747,16.4933406824,16.49374069,16.4941406976,16.4945407052,16.4949407128,16.4953407205,16.4957407281,16.4961407357,16.4965407433,16.4969407509,16.4973407585,16.4977407662,16.4981407738,16.4985407814,16.498940789,16.4993407966,16.4997408043,16.5001408119,16.5005408195,16.5009408271,16.5013408347,16.5017408424,16.50214085,16.5025408576,16.5029408652,16.5033408728,16.5037408805,16.5041408881,16.5045408957,16.5049409033,16.5053409109,16.5057409186,16.5061409262,16.5065409338,16.5069409414,16.507340949,16.5077409566,16.5081409643,16.5085409719,16.5089409795,16.5093409871,16.5097409947,16.5101410024,16.51054101,16.5109410176,16.5113410252,16.5117410328,16.5121410405,16.5125410481,16.5129410557,16.5133410633,16.5137410709,16.5141410786,16.5145410862,16.5149410938,16.5153411014,16.515741109,16.5161411166,16.5165411243,16.5169411319,16.5173411395,16.5177411471,16.5181411547,16.5185411624,16.51894117,16.5193411776,16.5197411852,16.5201411928,16.5205412005,16.5209412081,16.5213412157,16.5217412233,16.5221412309,16.5225412386,16.5229412462,16.5233412538,16.5237412614,16.524141269,16.5245412767,16.5249412843,16.5253412919,16.5257412995,16.5261413071,16.5265413147,16.5269413224,16.52734133,16.5277413376,16.5281413452,16.5285413528,16.5289413605,16.5293413681,16.5297413757,16.5301413833,16.5305413909,16.5309413986,16.5313414062,16.5317414138,16.5321414214,16.532541429,16.5329414367,16.5333414443,16.5337414519,16.5341414595,16.5345414671,16.5349414748,16.5353414824,16.53574149,16.5361414976,16.5365415052,16.5369415128,16.5373415205,16.5377415281,16.5381415357,16.5385415433,16.5389415509,16.5393415586,16.5397415662,16.5401415738,16.5405415814,16.540941589,16.5413415967,16.5417416043,16.5421416119,16.5425416195,16.5429416271,16.5433416348,16.5437416424,16.54414165,16.5445416576,16.5449416652,16.5453416729,16.5457416805,16.5461416881,16.5465416957,16.5469417033,16.5473417109,16.5477417186,16.5481417262,16.5485417338,16.5489417414,16.549341749,16.5497417567,16.5501417643,16.5505417719,16.5509417795,16.5513417871,16.5517417948,16.5521418024,16.55254181,16.5529418176,16.5533418252,16.5537418329,16.5541418405,16.5545418481,16.5549418557,16.5553418633,16.5557418709,16.5561418786,16.5565418862,16.5569418938,16.5573419014,16.557741909,16.5581419167,16.5585419243,16.5589419319,16.5593419395,16.5597419471,16.5601419548,16.5605419624,16.56094197,16.5613419776,16.5617419852,16.5621419929,16.5625420005,16.5629420081,16.5633420157,16.5637420233,16.564142031]
Y = [11579127.8554,11671781.7263,11764419.0191,11857026.0444,11949589.1124,12042094.5338,12134528.6188,12226877.6781,12319128.0219,12411265.9609,12503277.8053,12595149.8657,12686868.4525,12778419.8762,12869790.334,12960965.209,13051929.5278,13142668.3154,13233166.5969,13323409.3973,13413381.7417,13503068.6552,13592455.1627,13681526.2894,13770267.0602,13858662.5004,13946697.6348,14034357.4886,14121627.0868,14208491.4544,14294935.6166,14380944.5984,14466503.4248,14551597.1208,14636210.7116,14720329.3102,14803938.4081,14887023.5981,14969570.4732,15051564.6263,15132991.6503,15213837.1383,15294086.683,15373725.8775,15452740.3147,15531115.5875,15608837.2888,15685891.0116,15762262.3488,15837936.8934,15912900.2382,15987137.9762,16060635.7004,16133379.0036,16205353.4789,16276544.72,16346938.7731,16416522.8674,16485284.4226,16553210.8587,16620289.5956,16686508.0531,16751853.6511,16816313.8096,16879875.9485,16942527.4876,17004255.8468,17065048.446,17124892.7052,17183776.0442,17241685.8829,17298609.6412,17354534.739,17409448.5962,17463338.6327,17516192.2683,17567996.9463,17618741.7702,17668418.588,17717019.5043,17764536.6238,17810962.0514,17856287.8916,17900506.2493,17943609.2292,17985588.936,18026437.4744,18066146.9493,18104709.4653,18142117.1271,18178362.0396,18213436.3074,18247332.0352,18280041.3279,18311556.2901,18341869.0265,18370971.642,18398856.332,18425517.6188,18450952.493,18475158.064,18498131.4412,18519869.7341,18540370.0523,18559629.505,18577645.202,18594414.2525,18609933.7661,18624200.8523,18637212.6205,18648966.1802,18659458.6408,18668687.1119,18676648.7029,18683340.5233,18688759.6825,18692903.29,18695768.4553,18697352.5327,18697655.9558,18696681.2608,18694431.0245,18690907.8241,18686114.2363,18680052.838,18672726.2063,18664136.918,18654287.5501,18643180.6795,18630818.883,18617204.7377,18602340.8204,18586229.7081,18568873.9777,18550276.2061,18530438.9703,18509364.8471,18487056.4135,18463516.2464,18438747.4526,18412756.9228,18385553.1936,18357144.808,18327540.3094,18296748.2409,18264777.1456,18231635.5669,18197332.0479,18161875.1318,18125273.3619,18087535.2812,18048669.4331,18008684.3606,17967588.6071,17925390.7158,17882099.2297,17837722.6922,17792269.6464,17745748.6355,17698168.2027,17649537.512,17599868.3744,17549173.3069,17497464.8262,17444755.4492,17391057.6927,17336384.0736,17280747.1087,17224159.3148,17166633.2088,17108181.3075,17048816.1277,16988550.1864,16927396.0002,16865366.0862,16802472.961,16738729.1416,16674147.1447,16608739.4873,16542518.6861,16475497.2591,16407688.2541,16339106.0951,16269765.4262,16199680.8916,16128867.1358,16057338.8029,15985110.5372,15912196.9829,15838612.7844,15764372.5859,15689491.0316,15613982.7659,15537862.4329,15461144.6771,15383844.1425,15305975.4735,15227553.3143,15148592.3093,15069107.1026,14989112.3386,14908622.6595,14827652.5673,14746216.3337,14664328.209,14582002.4435,14499253.2874,14416094.9911,14332541.8049,14248607.9791,14164307.764,14079655.4098,13994665.1668,13909351.2855,13823728.016,13737809.6086,13651610.3137,13565144.3816,13478426.0625,13391469.6068,13304289.2646,13216899.2865,13129313.8865,13041546.3657,12953609.0623,12865514.2686,12777274.277,12688901.3798,12600407.8693,12511806.0378,12423108.1777,12334326.5812,12245473.5407,12156561.3486,12067602.297,11978608.6785,11889592.7852]
def gaussFunction(x, *p):
"""Define and return a Gaussian function.
This function returns the value of a Gaussian function, using the
A, mu and sigma value that is provided as *p.
Keyword arguments:
x -- number
p -- A, mu and sigma numbers
"""
A, mu, sigma = p
return A*np.exp(-(x-mu)**2/(2.*sigma**2))
newGaussX = np.linspace(10, 25, 2500*(X[-1]-X[0]))
p0 = [np.max(Y), X[np.argmax(Y)],0.1]
coeff, var_matrix = curve_fit(gaussFunction, X, Y, p0)
newGaussY = gaussFunction(newGaussX, *coeff)
print "Sigma is "+str(coeff[2])
# Original
low = bisect.bisect_left(newGaussX,coeff[1]-2*coeff[2])
high = bisect.bisect_right(newGaussX,coeff[1]+2*coeff[2])
print newGaussX[low], newGaussX[high]
# Absolute
low = bisect.bisect_left(newGaussX,coeff[1]-2*abs(coeff[2]))
high = bisect.bisect_right(newGaussX,coeff[1]+2*abs(coeff[2]))
print newGaussX[low], newGaussX[high]
Bottom-line, is taking the abs() of the sigma 'correct' or should this problem be solved in a different way?

You are fitting a function gaussFunction that does not care whether sigma is positive or negative. So whether you get a positive or negative result is mostly a matter of luck, and taking the absolute value of the returned sigma is fine. Also consider other possibilities:
(Suggested by Thomas Kühn): modify the model function so that it cares about the sign of sigma. Bringing it closer to the normalized Gaussian form would be enough: the formula A/np.sqrt(sigma)*np.exp(-(x-mu)**2/(2.*sigma**2)) would ensure that you get positive sigma only. A possible, mild downside is that the function takes a bit longer to compute.
Use the variance, sigma_squared, as a parameter:
A, mu, sigma_squared = p
return A*np.exp(-(x-mu)**2/(2.*sigma_squared))
This is probably easiest in terms of keeping the model equation simple. You will need to square your initial guess for that parameter, and take square root when you need sigma itself.
Aside: you hardcoded 0.1 as a guess for standard deviation. This probably should be based on data, like this:
peak = X[Y > np.exp(-0.5)*Y.max()]
guess_sigma = 0.5*(peak.max() - peak.min())
The idea is that within one standard deviation of the mean, the values of the Gaussian are greater than np.exp(-0.5) times the maximum value. So the first line locates this "peak" and the second takes half of its width as the guess for sigma.
For the above to work, X and Y should be already converted to NumPy arrays, e.g., X = np.array([16.4697402328,16.4701402404,..... This is a good idea in general: otherwise, you are making each NumPy method that receives X or Y make this conversion again.

You might find lmfit (http://lmfit.github.io/lmfit-py/) useful for this. It includes a Gaussian Model for curve-fitting that does normalize the Gaussian and also restricts sigma to be positive using a parameter transformation that is more gentle than abs(sigma). Your example would look like this
from lmfit.models import GaussianModel
xdat = np.array(X)
ydat = np.array(Y)
model = GaussianModel()
params = model.guess(ydat, x=xdat)
result = model.fit(ydat, params, x=xdat)
print(result.fit_report())
which will print a report with best-fit values and estimated uncertainties for all the parameters, and include FWHM.
[[Model]]
Model(gaussian)
[[Fit Statistics]]
# function evals = 31
# data points = 237
# variables = 3
chi-square = 95927408861.607
reduced chi-square = 409946191.716
Akaike info crit = 4703.055
Bayesian info crit = 4713.459
[[Variables]]
sigma: 0.04880178 +/- 1.57e-05 (0.03%) (init= 0.0314006)
center: 16.5174203 +/- 8.01e-06 (0.00%) (init= 16.51754)
amplitude: 2.2859e+06 +/- 586.4103 (0.03%) (init= 670578.1)
fwhm: 0.11491942 +/- 3.51e-05 (0.03%) == '2.3548200*sigma'
height: 1.8687e+07 +/- 910.0152 (0.00%) == '0.3989423*amplitude/max(1.e-15, sigma)'
[[Correlations]] (unreported correlations are < 0.100)
C(sigma, amplitude) = 0.949
The values for center +/- 2*sigma would be found with
xlo = result.params['center'].value - 2 * result.params['sigma'].value
xhi = result.params['center'].value + 2 * result.params['sigma'].value
You can use the result to evaluate the model with fitted parameters and different X values:
newGaussX = np.linspace(10, 25, 2500*(X[-1]-X[0]))
newGaussY = result.eval(x=newGaussX)
I would also recommend using numpy.where to find the location of center+/-2*sigma instead of bisect:
low = np.where(newGaussX > xlo)[0][0] # replace bisect_left
high = np.where(newGaussX <= xhi)[0][-1] + 1 # replace bisect_right

I got the same problem and I came up with a trivial but effective solution, which is basically to use the variance in the gaussian function definition instead of the standard deviation, since the variance is always positive. Then, you get the std_dev by square rooting the variance, obtaining a positive value i.e., the std_dev will always be positive. So, problem solved easily ;)
I mean, create the function this way:
def gaussian(x, Heigh, Mean, Variance):
return Heigh * np.exp(- (x-Mean)**2 / (2 * Variance))
Instead of:
def gaussian(x, Heigh, Mean, Std_dev):
return Heigh * np.exp(- (x-Mean)**2 / (2 * Std_dev**2))
And then do the fit as usual.

How to compute standard deviation errors with scipy.optimize.least_squares

I compare fitting with optimize.curve_fit and optimize.least_squares. With curve_fit I get the covariance matrix pcov as an output and I can calculate the standard deviation errors for my fitted variables by that:
perr = np.sqrt(np.diag(pcov))
If I do the fitting with least_squares, I do not get any covariance matrix output and I am not able to calculate the standard deviation errors for my variables.
Here's my example:
#import modules
import matplotlib
import numpy as np
import matplotlib.pyplot as plt
from scipy.optimize import curve_fit
from scipy.optimize import least_squares
noise = 0.5
N = 100
t = np.linspace(0, 4*np.pi, N)
# generate data
def generate_data(t, freq, amplitude, phase, offset, noise=0, n_outliers=0, random_state=0):
#formula for data generation with noise and outliers
y = np.sin(t * freq + phase) * amplitude + offset
rnd = np.random.RandomState(random_state)
error = noise * rnd.randn(t.size)
outliers = rnd.randint(0, t.size, n_outliers)
error[outliers] *= 10
return y + error
#generate data
data = generate_data(t, 1, 3, 0.001, 0.5, noise, n_outliers=10)
#initial guesses
p0=np.ones(4)
x0=np.ones(4)
# create the function we want to fit
def my_sin(x, freq, amplitude, phase, offset):
return np.sin(x * freq + phase) * amplitude + offset
# create the function we want to fit for least-square
def my_sin_lsq(x, t, y):
# freq=x[0]
# phase=x[1]
# amplitude=x[2]
# offset=x[3]
return (np.sin(t*x[0]+x[2])*x[1]+ x[3]) - y
# now do the fit for curve_fit
fit = curve_fit(my_sin, t, data, p0=p0)
print 'Curve fit output:'+str(fit[0])
#now do the fit for least_square
res_lsq = least_squares(my_sin_lsq, x0, args=(t, data))
print 'Least_squares output:'+str(res_lsq.x)
# we'll use this to plot our first estimate. This might already be good enough for you
data_first_guess = my_sin(t, *p0)
#data_first_guess_lsq = x0[2]*np.sin(t*x0[0]+x0[1])+x0[3]
data_first_guess_lsq = my_sin(t, *x0)
# recreate the fitted curve using the optimized parameters
data_fit = my_sin(t, *fit[0])
data_fit_lsq = my_sin(t, *res_lsq.x)
#calculation of residuals
residuals = data - data_fit
residuals_lsq = data - data_fit_lsq
ss_res = np.sum(residuals**2)
ss_tot = np.sum((data-np.mean(data))**2)
ss_res_lsq = np.sum(residuals_lsq**2)
ss_tot_lsq = np.sum((data-np.mean(data))**2)
#R squared
r_squared = 1 - (ss_res/ss_tot)
r_squared_lsq = 1 - (ss_res_lsq/ss_tot_lsq)
print 'R squared curve_fit is:'+str(r_squared)
print 'R squared least_squares is:'+str(r_squared_lsq)
plt.figure()
plt.plot(t, data)
plt.title('curve_fit')
plt.plot(t, data_first_guess)
plt.plot(t, data_fit)
plt.plot(t, residuals)
plt.figure()
plt.plot(t, data)
plt.title('lsq')
plt.plot(t, data_first_guess_lsq)
plt.plot(t, data_fit_lsq)
plt.plot(t, residuals_lsq)
#error
perr = np.sqrt(np.diag(fit[1]))
print 'The standard deviation errors for curve_fit are:' +str(perr)
I would be very thankful for any help, best wishes
ps: I got a lot of input from this source and used part of the code Robust regression

The result of optimize.least_squares has a parameter inside of it called jac. From the documentation:
jac : ndarray, sparse matrix or LinearOperator, shape (m, n)
Modified Jacobian matrix at the solution, in the sense that J^T J is a Gauss-Newton approximation of the Hessian of the cost function. The type is the same as the one used by the algorithm.
This can be used to estimate the Covariance Matrix of the parameters using the following formula: Sigma = (J'J)^-1.
J = res_lsq.jac
cov = np.linalg.inv(J.T.dot(J))
To find the variance of the parameters one can then use:
var = np.sqrt(np.diagonal(cov))

The SciPy program optimize.least_squares requires the user to provide in input a function fun(...) which returns a vector of residuals. This is typically defined as
residuals = (data - model)/sigma
where data and model are vectors with the data to fit and the corresponding model predictions for each data point, while sigma is the 1σ uncertainty in each data value.
In this situation, and assuming one can trust the input sigma uncertainties, one can use the output Jacobian matrix jac returned by least_squares to estimate the covariance matrix. Moreover, assuming the covariance matrix is diagonal, or simply ignoring non-diagonal terms, one can also obtain the 1σ uncertainty perr in the model parameters (often called "formal errors") as follows (see Section 15.4.2 of Numerical Recipes 3rd ed.)
import numpy as np
from scipy import linalg, optimize
res = optimize.least_squares(...)
U, s, Vh = linalg.svd(res.jac, full_matrices=False)
tol = np.finfo(float).eps*s[0]*max(res.jac.shape)
w = s > tol
cov = (Vh[w].T/s[w]**2) # Vh[w] # robust covariance matrix
perr = np.sqrt(np.diag(cov)) # 1sigma uncertainty on fitted parameters
The above code to obtain the covariance matrix is formally the same as the following simpler one (as suggested by Alex), but the above has the major advantage that it works even when the Jacobian is close to degenerate, which is a common occurrence in real-world least-squares fits
cov = linalg.inv(res.jac.T # res.jac) # covariance matrix when jac not degenerate
If one does not trust the input uncertainties sigma, one can still assume that the fit is good, to estimate the data uncertainties from the fit itself. This corresponds to assuming chi**2/DOF=1, where DOF is the number of degrees of freedom. In this case, one can use the following lines to rescale the covariance matrix before computing the uncertainties
chi2dof = np.sum(res.fun**2)/(res.fun.size - res.x.size)
cov *= chi2dof
perr = np.sqrt(np.diag(cov)) # 1sigma uncertainty on fitted parameters

How to do linear regression, taking errorbars into account?

I am doing a computer simulation for some physical system of finite size, and after this I am doing extrapolation to the infinity (Thermodynamic limit). Some theory says that data should scale linearly with system size, so I am doing linear regression.
The data I have is noisy, but for each data point I can estimate errorbars. So, for example data points looks like:
x_list = [0.3333333333333333, 0.2886751345948129, 0.25, 0.23570226039551587, 0.22360679774997896, 0.20412414523193154, 0.2, 0.16666666666666666]
y_list = [0.13250359351851854, 0.12098339583333334, 0.12398501145833334, 0.09152715, 0.11167239583333334, 0.10876248333333333, 0.09814170444444444, 0.08560799305555555]
y_err = [0.003306749165349316, 0.003818446389148108, 0.0056036878203831785, 0.0036635292592592595, 0.0037034897788415424, 0.007576672222222223, 0.002981084130692832, 0.0034913019065973983]
Let's say I am trying to do this in Python.
First way that I know is:
m, c, r_value, p_value, std_err = scipy.stats.linregress(x_list, y_list)
I understand this gives me errorbars of the result, but this does not take into account errorbars of the initial data.
Second way that I know is:
m, c = numpy.polynomial.polynomial.polyfit(x_list, y_list, 1, w = [1.0 / ty for ty in y_err], full=False)
Here we use the inverse of the errorbar for the each point as a weight that is used in the least square approximation. So if a point is not really that reliable it will not influence result a lot, which is reasonable.
But I can not figure out how to get something that combines both these methods.
What I really want is what second method does, meaning use regression when every point influences the result with different weight. But at the same time I want to know how accurate my result is, meaning, I want to know what are errorbars of the resulting coefficients.
How can I do this?

Not entirely sure if this is what you mean, but…using pandas, statsmodels, and patsy, we can compare an ordinary least-squares fit and a weighted least-squares fit which uses the inverse of the noise you provided as a weight matrix (statsmodels will complain about sample sizes < 20, by the way).
import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
import matplotlib as mpl
mpl.rcParams['figure.dpi'] = 300
import statsmodels.formula.api as sm
x_list = [0.3333333333333333, 0.2886751345948129, 0.25, 0.23570226039551587, 0.22360679774997896, 0.20412414523193154, 0.2, 0.16666666666666666]
y_list = [0.13250359351851854, 0.12098339583333334, 0.12398501145833334, 0.09152715, 0.11167239583333334, 0.10876248333333333, 0.09814170444444444, 0.08560799305555555]
y_err = [0.003306749165349316, 0.003818446389148108, 0.0056036878203831785, 0.0036635292592592595, 0.0037034897788415424, 0.007576672222222223, 0.002981084130692832, 0.0034913019065973983]
# put x and y into a pandas DataFrame, and the weights into a Series
ws = pd.DataFrame({
'x': x_list,
'y': y_list
})
weights = pd.Series(y_err)
wls_fit = sm.wls('x ~ y', data=ws, weights=1 / weights).fit()
ols_fit = sm.ols('x ~ y', data=ws).fit()
# show the fit summary by calling wls_fit.summary()
# wls fit r-squared is 0.754
# ols fit r-squared is 0.701
# let's plot our data
plt.clf()
fig = plt.figure()
ax = fig.add_subplot(111, facecolor='w')
ws.plot(
kind='scatter',
x='x',
y='y',
style='o',
alpha=1.,
ax=ax,
title='x vs y scatter',
edgecolor='#ff8300',
s=40
)
# weighted prediction
wp, = ax.plot(
wls_fit.predict(),
ws['y'],
color='#e55ea2',
lw=1.,
alpha=1.0,
)
# unweighted prediction
op, = ax.plot(
ols_fit.predict(),
ws['y'],
color='k',
ls='solid',
lw=1,
alpha=1.0,
)
leg = plt.legend(
(op, wp),
('Ordinary Least Squares', 'Weighted Least Squares'),
loc='upper left',
fontsize=8)
plt.tight_layout()
fig.set_size_inches(6.40, 5.12)
plt.show()
WLS residuals:
[0.025624005084707302,
0.013611438189866154,
-0.033569595462217161,
0.044110895217014695,
-0.025071632845910546,
-0.036308252199571928,
-0.010335514810672464,
-0.0081511479431851663]
The mean squared error of the residuals for the weighted fit (wls_fit.mse_resid or wls_fit.scale) is 0.22964802498892287, and the r-squared value of the fit is 0.754.
You can obtain a wealth of data about the fits by calling their summary() method, and/or doing dir(wls_fit), if you need a list of every available property and method.

I wrote a concise function to perform the weighted linear regression of a data set, which is a direct translation of GSL's "gsl_fit_wlinear" function. This is useful if you want to know exactly what your function is doing when it performs the fit
def wlinear_fit (x,y,w) :
"""
Fit (x,y,w) to a linear function, using exact formulae for weighted linear
regression. This code was translated from the GNU Scientific Library (GSL),
it is an exact copy of the function gsl_fit_wlinear.
"""
# compute the weighted means and weighted deviations from the means
# wm denotes a "weighted mean", wm(f) = (sum_i w_i f_i) / (sum_i w_i)
W = np.sum(w)
wm_x = np.average(x,weights=w)
wm_y = np.average(y,weights=w)
dx = x-wm_x
dy = y-wm_y
wm_dx2 = np.average(dx**2,weights=w)
wm_dxdy = np.average(dx*dy,weights=w)
# In terms of y = a + b x
b = wm_dxdy / wm_dx2
a = wm_y - wm_x*b
cov_00 = (1.0/W) * (1.0 + wm_x**2/wm_dx2)
cov_11 = 1.0 / (W*wm_dx2)
cov_01 = -wm_x / (W*wm_dx2)
# Compute chi^2 = \sum w_i (y_i - (a + b * x_i))^2
chi2 = np.sum (w * (y-(a+b*x))**2)
return a,b,cov_00,cov_11,cov_01,chi2
To perform your fit, you would do
a,b,cov_00,cov_11,cov_01,chi2 = wlinear_fit(x_list,y_list,1.0/y_err**2)
Which will return the best estimate for the coefficients a (the intercept) and b (the slope) of the linear regression, along with the elements of the covariance matrix cov_00, cov_01 and cov_11. The best estimate on the error on a is then the square root of cov_00 and the one on b is the square root of cov_11. The weighted sum of the residuals is returned in the chi2 variable.
IMPORTANT: this function accepts inverse variances, not the inverse standard deviations as the weights for the data points.

sklearn.linear_model.LinearRegression supports specification of weights during fit:
x_data = np.array(x_list).reshape(-1, 1) # The model expects shape (n_samples, n_features).
y_data = np.array(y_list)
y_err = np.array(y_err)
model = LinearRegression()
model.fit(x_data, y_data, sample_weight=1/y_err)
Here the sample weight is specified as 1 / y_err. Different versions are possible and often it's a good idea to clip these sample weights to a maximum value in case the y_err varies strongly or has small outliers:
sample_weight = 1 / y_err
sample_weight = np.minimum(sample_weight, MAX_WEIGHT)
where MAX_WEIGHT should be determined from your data (by looking at the y_err or 1 / y_err distributions, e.g. if they have outliers they can be clipped).

I found this document helpful in understanding and setting up my own weighted least squares routine (applicable for any programming language).
Typically learning and using optimized routines is the best way to go but there are times where understanding the guts of a routine is important.