Curve fitting in Python with constraint of zero slope at edges - python

I am looking to curve fit the following data, such that I get it to fit a trend with the condition of zero slope at the edges. The output of polyfit fits that data, but not with zero slopes at the edges.
Here is what I'm looking to output - pardon my Paint job. I need to it to fit like this so I can properly remove this sine/cosine bias of the data that isn't real towards the center.
Here is the data:
[0.23353535 0.25586247 0.26661164 0.26410896 0.24963951 0.22670266
0.19955422 0.17190263 0.1598439 0.17351905 0.18212444 0.18438673
0.17952432 0.18314894 0.19265689 0.19432385 0.19605163 0.20326011
0.20890851 0.20590997 0.21856518 0.23771665 0.24530019 0.23940831
0.22078396 0.23075128 0.2346082 0.22466281 0.24384843 0.26339594
0.26414153 0.24664183 0.24278978 0.31023648 0.3614195 0.37773436
0.3505998 0.28893167 0.23965877 0.24063917 0.27922502 0.32716477
0.36553767 0.42293146 0.50968856 0.5458872 0.52192533 0.45243764
0.36313155 0.3683921 0.40942553 0.4420537 0.46145585 0.4648034
0.4523771 0.4272876 0.39404616 0.3570107 0.35060245 0.3860975
0.3996996 0.44551122 0.46611032 0.45998383 0.4309985 0.38563925
0.37105605 0.4074444 0.48815584 0.5704579 0.6448988 0.7018853
0.73397845 0.73739105 0.7122451 0.6618154 0.591451 0.5076601
0.48578677 0.47347385 0.4791471 0.48306277 0.47025493 0.43479836
0.44380915 0.45868078 0.5341566 0.57549906 0.55790776 0.56244135
0.57668275 0.561856 0.67564166 0.7512851 0.76957643 0.7266262
0.734133 0.7231936 0.6776926 0.60511285 0.51599765 0.5579323
0.56723005 0.5440337 0.5775593 0.5950776 0.5722321 0.57858473
0.5652703 0.54723704 0.59561515 0.7071321 0.8169259 0.91443264
0.9883759 1.0275097 1.0235045 0.9737119 1.029139 1.1354861
1.1910824 1.1826864 1.1092159 0.9832138 0.9643041 0.92324203
0.9093703 0.88915515 1.0007693 1.0542978 1.0857164 1.0211861
0.88474303 0.8458009 0.76522666 0.7478076 0.90081936 1.0690157
1.1569089 1.1493248 1.0622779 1.0327609 0.9805119 0.9583969
0.8973544 0.9543319 0.9777171 0.94951093 0.97323567 1.0244237
1.0569099 1.0951824 1.0771195 1.3078191 1.7212077 2.09409
2.320331 2.3279085 2.125451 1.7908521 1.4180487 1.0744424
1.0218129 1.0916439 1.1255138 1.125803 1.1139745 1.2187989
1.300092 1.3025533 1.2312403 1.221301 1.2535597 1.2298189
1.1458241 1.1012102 1.0889369 1.1558667 1.3051153 1.4143198
1.6345526 1.8093723 1.9037704 1.8961821 1.7866236 1.5958548
1.3865516 1.5308585 1.6140417 1.627337 1.5733193 1.4981418
1.5048542 1.4935548 1.4798748 1.4131776 1.3792214 1.3728334
1.3683671 1.3593615 1.2995907 1.2965002 1.366058 1.4795257
1.5462885 1.61591 1.5968509 1.5222199 1.6210756 1.7074443
1.8351102 2.3187535 2.6568012 2.7676315 2.6480794 2.3636303
2.0673316 1.9607923 1.8074365 1.713272 1.5893831 1.4734347
1.507817 1.5213271 1.6091452 1.7162323 1.7608733 1.7497622
1.9187828 2.0197518 2.0487514 2.01107 1.9193696 1.7904462
1.8558109 2.1955926 2.4700975 2.6562278 2.675197 2.6645825
2.6295316 2.4182043 2.2114453 2.2506614 2.2086055 2.0497518
1.9557768 1.901191 2.067513 2.1077373 2.0159333 1.8138607
1.5413624 1.600069 1.7631899 1.9541935 1.9340311 1.805134
2.0671906 2.2247658 2.2641945 2.3594956 2.2504601 1.9749025
1.8905054 2.0679731 2.1193469 2.0307171 2.0717037 2.0340347
1.925536 1.7820351 1.9467943 2.315468 2.4834185 2.3751369
2.0240622 1.9363666 2.1732547 2.3113241 2.3264208 2.22015
2.0187428 1.7619076 1.796859 1.8757095 2.0501778 2.44711
2.6179967 2.508112 2.1694388 1.7242104 1.7671669 1.862043
1.8392721 1.7120028 1.6650634 1.6319505 1.482931 1.5240219
1.5815579 1.5691646 1.4766116 1.3731087 1.4666644 1.4061015
1.3652745 1.425564 1.4006845 1.5000012 1.581379 1.6329607
1.6444355 1.6098644 1.5300899 1.6876912 1.8968476 2.048039
2.1006014 2.0271482 1.8300935 1.6986666 1.9628603 2.0521066
1.9337255 1.6407858 1.2583638 1.2110122 1.2476432 1.2360718
1.2886397 1.2862154 1.2343681 1.1458222 1.209224 1.2475786
1.2353342 1.1797879 1.0963987 1.0928186 1.1553882 1.1569618
1.1932304 1.3002363 1.3386917 1.2973225 1.1816871 1.0557054
0.9350373 0.896656 0.8565816 0.90168726 0.9897751 1.02342
1.0232298 1.1199353 1.1466643 1.1081418 1.0377598 1.0348651
1.0223045 1.0607077 1.0089502 0.885213 1.023178 1.1131796
1.1331098 1.0779471 0.9626393 0.81472665 0.85455835 0.87542623
0.87286425 0.89130884 0.9545931 1.0355722 1.0201533 0.93568784
0.9180018 0.8202782 0.7450139 0.72550577 0.68578506 0.6431666
0.66193295 0.6386373 0.7060119 0.7650972 0.80093855 0.803342
0.76590335 0.7151591 0.6946282 0.7136788 0.7714012 0.8022328
0.79840165 0.8543819 0.8586749 0.8028453 0.7383879 0.73423904
0.65107304 0.61139977 0.5940311 0.6151931 0.59349155 0.54995483
0.5837645 0.5891752 0.56406695 0.5638191 0.5762535 0.58305734
0.5830114 0.57470953 0.5568098 0.52852243 0.49031836 0.45275375
0.47168964 0.46634504 0.4600581 0.45332378 0.41508177 0.3834329
0.4137769 0.41392407 0.3824464 0.36310086 0.434278 0.48041886
0.49433306 0.475708 0.43060693 0.36886734 0.34740242 0.34108457
0.36160505 0.40907663 0.43613982 0.4394311 0.42070773 0.38575593
0.3827834 0.4338096 0.46581286 0.45669746 0.40830874 0.3505502
0.32584783 0.3381971 0.33949164 0.36409503 0.3759155 0.3610108
0.37174097 0.39990777 0.38925973 0.34376588 0.32478797 0.32705626
0.3228174 0.30941254 0.28542265 0.2687348 0.25517422 0.26127565
0.27331188 0.3028561 0.31277937 0.29953563 0.2660389 0.27051866
0.2913383 0.30363902 0.30684754 0.3011791 0.28737035 0.26648855
0.26413882 0.25501928 0.23947525 0.21937743 0.19659272 0.18965112
0.21511254 0.23329383 0.24157354 0.2391297 0.22697571 0.20739041
0.1855308 0.18856761 0.19565174 0.20542233 0.21473111 0.22244582
0.22726117 0.22789808 0.22336568 0.21322969 0.20314343 0.2031754
0.19738965 0.1959791 0.20284075 0.20859875 0.21363212 0.21804498
0.22160804 0.22381367]
This came close, but not exactly it as the edges aren't zero slope: How do I fit a sine curve to my data with pylab and numpy?
Is there anything available that will let me do this without having to write up a custom algorithm to handle this? Thanks.

Here is a Lorentzian type of peak equation fitted to your data, and for the "x" values I used an index similar to what I see on the Example Output plot in your post. I have also zoomed in on the peak center to better display the sinusoidal shapes you mention. You may be able to subtract the predicted value from this peak equation to condition or pre-process the raw data as you discuss.
a = 1.7056067124682076E+02
b = 7.2900803359572393E+01
c = 2.5047064423525464E+02
d = 1.4184767800540945E+01
Offset = -2.4940994412221318E-01
y = a/ (b + pow((x-c)/d, 2.0)) + Offset

Starting from you own example based on a sine fit, I added the constraints such that the derivatives of the model have to be zero at the end points. I did this using symfit, a package I wrote to make this kind of thing easier. If you prefer to do this using scipy you can adapt the example to that syntax if you want, symfit is just a wrapper around their minimizers that adds symbolical manipulations using sympy.
# Make variables and parameters
x, y = variables('x, y')
a, b, c, d = parameters('a, b, c, d')
# Initial guesses
b.value = 1e-2
c.value = 100
# Create a model object
model = Model({y: a * sin(b * x + c) + d})
# Take the derivative and constrain the end-points to be equal to zero.
dydx = D(model[y], x).doit()
constraints = [Eq(dydx.subs(x, xdata[0]), 0),
Eq(dydx.subs(x, xdata[-1]), 0)]
# Do the fit!
fit = Fit(model, x=xdata, y=ydata, constraints=constraints)
fit_result = fit.execute()
print(fit_result)
plt.plot(xdata, ydata)
plt.plot(xdata, model(x=xdata, **fit_result.params).y)
plt.show()
This prints: (from current symfit PR#221, which has better reporting of the results.)
Parameter Value Standard Deviation
a 8.790393e-01 1.879788e-02
b 1.229586e-02 3.824249e-04
c 9.896017e+01 1.011472e-01
d 1.001717e+00 2.928506e-02
Status message Optimization terminated successfully.
Number of iterations 10
Objective <symfit.core.objectives.LeastSquares object at 0x0000016F670DF080>
Minimizer <symfit.core.minimizers.SLSQP object at 0x0000016F78057A58>
Goodness of fit qualifiers:
chi_squared 29.72125657199736
objective_value 14.86062828599868
r_squared 0.8695978050586373
Constraints:
--------------------
Question: a*b*cos(c) == 0?
Answer: 1.5904051811454707e-17
Question: a*b*cos(511*b + c) == 0?
Answer: -6.354261416082215e-17

Related

lmfit - SineModel+ConstantModel appears inaccurate fit

I'm trying to fit a simple sine function to some experimental data using lmfit and I find that the SineModel with a constant model offset returns, what looks like an inaccurate fit to the data (to me). I suppose it may be helpful to highlight that I am most interested in the frequency of the peaks (and I appreciate that I can simply use a scipy.find_peaks() but would prefer to show a fit to the data).
I use the function below for lmfit model:
def Sine(self, x_axis, y_axis):
sine = SineModel()
const = ConstantModel()
x_fit = np.linspace(min(x_axis), max(x_axis), x_axis.size)
guess_sine = sine.guess(y_axis, x=x_fit)
pars = sine.guess(y_axis, x=x_fit)
sine_offset = SineModel() + ConstantModel()
pars.add('c', value=1, vary=True)
result = sine_offset.fit(y_axis, pars, x=x_fit)
return result
Sine function output (graph and report results) are provided here:
SineModel+ConstModel
I then tried to define my own function, defining my own parameters and evaluating in the same lmfit method, providing sensible "guess" initial values etc.
def Sine_User2(self, x_axis, y_axis):
def sine_func(x, amplitude, freq, shift, c):
return amplitude * np.sin(freq * x + shift) + c
sinemodel = Model(sine_func)
# Take a FFT of the data to provide a guess starting location for the curve fitting
x = np.array(x_axis)
y = np.array(y_axis)
ff = np.fft.fftfreq(len(x), (x[1] - x[0])) # assume uniform spacing
Fyy = abs(np.fft.fft(y))
guess_freq = abs(ff[np.argmax(Fyy[1:]) + 1]) * 2. * np.pi
guess_amp = np.std(y) * 2.**0.5
guess_offset = np.mean(y)
x_fit = np.linspace(min(x_axis), max(x_axis), x_axis.size)
params = sinemodel.make_params(amplitude = guess_amp, freq = guess_freq, shift = 0, c = guess_offset )
result = sinemodel.fit(y_axis, params, x = x_fit)
return result
The output of the user defined model appears to provide a much closer fit to the data, however, the report does not provide uncertainties citing a warning that the "Uncertainties could not be estimated":
SineUser2 function outputs (graph and report results) are provided here: User Defined Model
I then tried to include min/max values to the parameters by replacing the "sinmodel.make_params" line with:
params = Parameters()
params.add('amplitude', value=guess_amp, min = 0)
params.add('freq', value=guess_freq, min=0)
params.add('shift', value=0, min=-2*np.pi, max=2*np.pi)
params.add('c', value=guess_offset)
But the results resort back to the SineModel+ConstModel results seen in the first linked graph/report results. Therefore it must be something to do with the way I'm setting initial values.
The fit using the "SineUser2" function appears to be better. Is there a way to improve the fit for "Sine" function in the first block of code.
Why are the uncertainties not calculated in the second function "Sine_User2"?
Data (.csv):
Wavelength (nm),Power (dBm),,,,,
1549.9,-13.76008731,,,,,
1549.905,-13.69423162,,,,,
1549.91,-12.59004339,,,,,
1549.915,-11.31061848,,,,,
1549.92,-10.58731809,,,,,
1549.925,-10.19024329,,,,,
1549.93,-10.07301418,,,,,
1549.935,-10.19513172,,,,,
1549.94,-10.45582159,,,,,
1549.945,-11.15984161,,,,,
1549.95,-12.15876596,,,,,
1549.955,-13.44674933,,,,,
1549.96,-13.56388277,,,,,
1549.965,-12.2513065,,,,,
1549.97,-11.08699015,,,,,
1549.975,-10.43829185,,,,,
1549.98,-10.12861158,,,,,
1549.985,-10.0962929,,,,,
1549.99,-10.1852173,,,,,
1549.995,-10.55438183,,,,,
1550,-11.19555345,,,,,
1550.005,-12.28715299,,,,,
1550.01,-13.5153863,,,,,
1550.015,-13.47019261,,,,,
1550.02,-12.12394732,,,,,
1550.025,-11.01946751,,,,,
1550.03,-10.42138778,,,,,
1550.035,-10.14438079,,,,,
1550.04,-10.05681218,,,,,
1550.045,-10.17148605,,,,,
1550.05,-10.56046759,,,,,
1550.055,-11.11621478,,,,,
1550.06,-12.19930263,,,,,
1550.065,-13.48428349,,,,,
1550.07,-13.43424913,,,,,
1550.075,-12.08019952,,,,,
1550.08,-11.08731704,,,,,
1550.085,-10.45730899,,,,,
1550.09,-10.11278169,,,,,
1550.095,-10.00651194,,,,,
,,,,,,

Exponential fit in pandas

I have this data:
puf = pd.DataFrame({'id':[1,2,3,4,5,6,7,8],
'val':[850,1889,3289,6083,10349,17860,28180,41236]})
The data seems to follow an exponential curve. Let's see the plot:
puf.plot('id','val')
I want to fit an exponential curve ($$ y = Ae^{Bx} $$, A times e to the B*X)and add it as a column in Pandas. Firstly I tried to log the values:
puf['log_val'] = np.log(puf['val'])
And then to use Numpy to fit the equation:
puf['fit'] = np.polyfit(puf['id'],puf['log_val'],1)
But I get an error:
ValueError: Length of values (2) does not match length of index (8)
My expected result is the fitted values as a new column in Pandas. I attach an image with the column fitted values I want (in orange):
I'm stuck in this code. I'm not sure what I am doing wrong. How can I create a new column with my fitted values?
Note that you asked for an exponential model yet you have the results for log-linear model.
Check out the work below:
For log-linear, we are fitting E(log(Y))ie log(y) - (log(b[0]) +b[1]*x):
from scipy.optimize import least_squares
least_squares(lambda b: np.log(puf['val']) -(np.log(b[0]) + b[1] * puf['id']),
[1,1])['x']
array([5.99531305e+02, 5.51106793e-01])
These are the values that excel gives.
On the other hand to fit an exponential curve, the randomness is on Y and not on its logarithm, E(Y)=b[0]*exp(b[1] *x) Hence we have:
least_squares(lambda b: puf['val'] - b[0]*exp(b[1] * puf['id']), [0,1])['x']
array([1.08047304e+03, 4.58116127e-01]) # correct results for exponential fit
Depending on your model choice, the values are alittle different.
Better Model? Since you have same number of parameters, consider the one that gives you lower deviance or better out of sample prediction
Note that the ideal exponential model is E(Y) = A'B'^X which for comparison can be written as log(E(Y)) = A + XB while log-linear model will be E(log(Y) = A + XB. Note the difference in Expectation.
From the two models we have:
Notice how when we go to higher numbers the log-linear overestimates. While in the lower numbers the exponential overestimates.
Code for image:
from scipy.optimize import least_squares
log_lin = least_squares(lambda b: np.log(puf['val']) -(np.log(b[0]) + b[1] * puf['id']),
[1,1])['x']
expo = least_squares(lambda b: puf['val'] - b[0]*exp(b[1] * puf['id']), [0,1])['x']
exp_fun = lambda x: expo[0] * exp(expo[1]*x)
log_lin_fun = lambda x:log_lin[0] * exp(log_lin[1]*x)
plt.plot(puf.id, puf.val, label = 'original')
plt.plot(puf.id, exp_fun(puf.id), label='exponential')
plt.plot(puf.id, log_lin_fun(puf.id), label='log-linear')
plt.legend()
Your getting that error because np.polyfit(puf['id'],puf['log_val'],1) returns two values array([0.55110679, 6.39614819]) which isn't the shape of your dataframe.
This is what you want
y = a* exp (b*x) -> ln(y)=ln(a)+bx
f = np.polyfit(df['id'], np.log(df['val']), 1)
where
a = np.exp(f[1]) -> 599.5313046712091
b = f[0] -> 0.5511067934637022
Giving
puf['fit'] = a * np.exp(b * puf['id'])
id val fit
0 1 850 1040.290193
1 2 1889 1805.082864
2 3 3289 3132.130026
3 4 6083 5434.785677
4 5 10349 9430.290286
5 6 17860 16363.179739
6 7 28180 28392.938399
7 8 41236 49266.644002

Strange result from Fast Fourier transform signal reconstruction

I have some data which is shown in the below figure and am interested in finding some of its Fourier series coefficients.
r = np.array([119.80601628, 119.84629291, 119.85290735, 119.45778804,
115.64497439, 105.58519852, 100.72765819, 100.04327702,
100.08590518, 100.35824977, 101.58424993, 105.47976376,
112.27556007, 117.07679226, 118.99998888, 119.60458086,
119.78624424, 119.83022022, 119.36116943, 115.72323767,
106.58946834, 101.19479124, 100.11537349, 100.13313755,
100.41846106, 101.42255377, 104.33650237, 109.73625492,
115.14763728, 118.24665037, 119.35359999, 119.68061835])
z = np.array([-411.42980545, -384.98596279, -358.13032372, -330.89578468,
-303.39129113, -275.76248957, -248.24478443, -221.07069838,
-194.33260984, -168.05271807, -142.19357982, -116.62090103,
-91.15354178, -65.56745626, -39.65284757, -13.29632162,
13.54374939, 40.84929432, 68.50496394, 96.33720787,
124.08525182, 151.36802193, 177.98791952, 204.0805317 ,
229.85399128, 255.44727674, 281.02166554, 306.75399703,
332.74638285, 359.05528646, 385.74336711, 412.8189858 ])
plt.plot(z, r, label='data')
plt.legend()
Then I calculate the average sampling period, since it is not constant as seen in the Z variable:
l = []
for i in range(32-1):
l.append(z[i]-z[i+1])
Ts = np.mean(l)
Then I calculate the fft:
from scipy.fftpack import fft
rf = scipy.fftpack.fft(r)
For reconstruction of the signal then:
fs = 1/Ts
amp = np.abs(rf)/r.shape[0]
n = r.shape[0]
s = 0
for i in range(n//2):
phi = np.angle(rf[i], deg=False)
a = amp[i]
k = i*fs/n
s += a*np.cos(2*np.pi*k *(z) +phi)
plt.plot(z, s, label='fft result')
plt.plot(z, r, label='data')
plt.legend()
The result is strange however both in terms of amplitude and frequency.
The complex spectrum is a symmetric spectrum with the range of (-fMax/2, ..., +fMax/2).
You only used the right hand positive part of the spectrum. This means, your reconstructed signal contains only half of the spectrums frequencies.
Because the spectrum is symmetric, all you have to do is to double the calculated absolute values. However, there is an important exception. The DC value amplitude[0] must not be doubled.

How to make scipy.optimize.curve_fit result in a better sine regression fit?

I have a problem where I am using scipy.optimize.curve_fit to do a regression fit to a sine/cosine function but the fit does not seem as optimized as I want it to be. How can I change my code to make the fitting better?
I have already tried changing how parameters are tried for the dataset and there is always seemingly a difference in phase-offset of my generated fit or the fitting function is not fitting to the proper minima/maxima.
Here is the code I am using to generate the regression fit. The output (fitfunc) can be plotted to show the result.
def sin_regress(data_x, data_y):
"""Function regression fits data to SIN function; does not need guess of freq.
Parameters
----------
data_x :
Data for X values, most likely a set of voltages.
data_y :
Data for Y values, most likely the resulting powers from voltages.
Returns
-------
__ :
Dictionary containing values for amplitude, angular frequency, phase, offset, frequency, period, fit function, max covariance, initial guess.
"""
data_x = np.array(data_x)
data_y = np.array(data_y)
freqz = np.fft.rfftfreq(len(data_x), (data_x[1] - data_x[0])) # uniform spacing
freq_y = abs(np.fft.rfft(data_y))
guess_freq = abs(freqz[np.argmax(freq_y[1:])+1]) # exclude offset peak
guess_amp = np.std(data_y) * 2.**0.5
guess_offset = np.mean(data_y)
guess = np.array([guess_amp, 2.*np.pi*guess_freq, 0., guess_offset])
def sinfunc(t, A, w, p, c):
"""Raw function to be used to fit data.
Parameters
----------
t :
Voltage array
A :
Amplitude
w :
Angular frequency
p :
Phase
c :
Constant value
Returns
-------
__ :
Formed fit function with provided values.
"""
return A * np.sin(w*t + p) + c
popt, pcov = scipy.optimize.curve_fit(sinfunc, data_x, data_y, p0=guess)
A, w, p, c = popt
f = w/(2.*np.pi)
fitfunc = lambda t: A * np.sin(w*t + p) + c
return {"amp": A, "omega": w, "phase": p, "offset": c, "freq": f, "period": 1./f, "fitfunc": fitfunc, "maxcov": np.max(pcov), "rawres": (guess,popt,pcov)}
With my trial dataset being:
x = np.linspace(3.5,9.5,(9.5-3.5)/0.00625 + 1)
pow1 = [1.8262110863, 1.80944546009, 1.7970185646900003, 1.77120336754, 1.7458101235699999, 1.73597098224, 1.7122529922799998, 1.70015674142, 1.68968617429, 1.6989396515, 1.69760676076, 1.6946375613599998, 1.6895321899, 1.68145658386, 1.68581793183, 1.6920468775900002, 1.6865452951599997, 1.68570953338, 1.6922784791700003, 1.70958957412, 1.71683408637, 1.70360183933, 1.6919669752199997, 1.6669487117300001, 1.6351298032300001, 1.6061729066600001, 1.57344333403, 1.54723708217, 1.5277773737599998, 1.5122628414300001, 1.4962354965200002, 1.4873367459, 1.47567715522, 1.4696584634, 1.46159565032, 1.45320592315, 1.4487225244200002, 1.44572887186, 1.44089260198, 1.4367157657399998, 1.4349226211, 1.43614316806, 1.4381950627400002, 1.43947658627, 1.4483572314200002, 1.4504305909200002, 1.44436990692, 1.43367609757, 1.42637295252, 1.41197427963, 1.4067529511399999, 1.39714414185, 1.38309980493, 1.3730701362500004, 1.3693239836499997, 1.3729558979599998, 1.38291189477, 1.3988274622900003, 1.42112832324, 1.44217266068, 1.4578792438300001, 1.46478639274, 1.46676801398, 1.4646383458800003, 1.45918801344, 1.44561402809, 1.4212145146499997, 1.4012453921299999, 1.38070199226, 1.36215759642, 1.3540496661500003, 1.35470913884, 1.3481165993199997, 1.34059081754, 1.332964567, 1.33426054366, 1.34052562222, 1.3343255632100002, 1.3310385903, 1.33044179339, 1.32827462527, 1.3356201140500001, 1.3400144893900001, 1.3157198001600001, 1.27716313727, 1.2517667292400003, 1.2406836620500001, 1.2354036030700002, 1.23110776291, 1.22492582889, 1.22074838719, 1.21816502762, 1.21015135518, 1.20038737012, 1.1920263929700001, 1.18723010357, 1.19656731125, 1.2237068834899998, 1.2373841696199999, 1.2251076648299999, 1.1963014909299998, 1.16152861736, 1.13940556893, 1.12839812676, 1.12368066547, 1.1190219542100002, 1.11384679759, 1.10555781262, 1.0977575386300003, 1.0901734365399998, 1.0824275375699999, 1.07552931443, 1.0696565210100002, 1.06481394254, 1.0578173014299999, 1.05204230102, 1.0482530038799998, 1.04237087457, 1.0361766944300002, 1.0297906393, 1.0240842912299999, 1.01250548183, 0.9964340353700001, 0.9859450307400002, 0.98614987451, 0.9826424718800002, 0.9739505767299999, 0.9578738177999998, 0.9416973908799999, 0.92975112051, 0.9204409049900001, 0.91821299468, 0.9100360995600001, 0.89589154778, 0.8799530701000002, 0.8640439088, 0.8500274234399999, 0.8428500205999999, 0.8358678326, 0.8333072464999999, 0.83420148485, 0.8362578717, 0.83608947323, 0.83035464861, 0.82315039029, 0.81220152235, 0.80169300598, 0.7918658959, 0.7808782388700001, 0.77684747687, 0.7743299962, 0.76797978094, 0.7591097217, 0.7520710688500001, 0.7452609707, 0.73562753255, 0.7256206568399999, 0.71663518742, 0.70951165178, 0.7035884873, 0.6973768853, 0.6900439160299999, 0.68062538021, 0.67096725454, 0.66585371901, 0.6663177033900001, 0.67214877804, 0.6787934074299999, 0.68365489213, 0.68581510712, 0.6820892084400001, 0.67805153237, 0.67540688376, 0.6724865515, 0.6674502035, 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0.93015244864, 0.9343631130099999, 0.93763016402, 0.9384009648400001, 0.93879867973, 0.93652442175, 0.93662918739, 0.9331820972899999, 0.93503584744, 0.9360406912399999, 0.93994795716, 0.9444487777899999, 0.95150762595, 0.9574753021500001, 0.9659650293199998, 0.9757605964, 0.9878513785299999, 0.99883880117, 1.01323052095, 1.0311493112499999, 1.04763474212, 1.0677277318200002, 1.086237323, 1.0988490621599998, 1.10287175775, 1.11006095748, 1.1203823058799998, 1.1266948453599999, 1.1295011150999998, 1.13468379124, 1.13839008058, 1.1417559206699999, 1.1386140845, 1.1368738695300002, 1.13791410398, 1.1443759989699998, 1.1533826011700001, 1.16127430094, 1.1771807669, 1.19318348288, 1.2014892452, 1.20715822998, 1.21764737132, 1.23158125907, 1.2387470993899998, 1.2441262208700001, 1.2562376475, 1.2682344256899998, 1.28293907518, 1.2903573374300001, 1.3040509126199997, 1.3260814219800001, 1.3595052134299999, 1.3870089263099998, 1.4040962907899999, 1.4190098465199998, 1.43005375357, 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0.8947884227199999, 0.8843405703999998, 0.8769049253500001, 0.8719632452999999, 0.86833484662, 0.8680955887799999, 0.86604049098, 0.86558996362, 0.86372701427, 0.85893691627, 0.85435131048, 0.84886228665, 0.8409088095199999, 0.82732292967, 0.8182398235399999, 0.81298593645, 0.8065804672500001, 0.7963832009099999, 0.7813524576499999, 0.7642633939500001, 0.74891606863, 0.73387495429, 0.72021307831, 0.70711249145, 0.6972523931, 0.68836254874, 0.6789805168, 0.66917573095, 0.65520369872, 0.6405349086200001, 0.6262600443299999, 0.6128265668199999, 0.6004827768800001, 0.58821246352, 0.5763513298499999, 0.56580466895, 0.55820613325, 0.5498382224900001, 0.5432313079700001, 0.5383656045, 0.53169802591];
Here are some additional values for the pow dataset:
(Link to pastebin to not exceed post length limit)
https://pastebin.com/5GP8sj4N
The resulting fit that from the trial dataset (x, pow1) I get is shown here (orange) with the original (pow1) data (blue)
As mentioned, there is an issue with how the phase fits the minima and maxima. Unfortunately the application of getting this fit function correct has very little room for error.
Please help out if you have an idea of how to make this fit the data better!
Edit:
I tried what #Joe mentioned in the comments, with first filtering the data. I utilized a Savitzky-Golay filter and recieved the following result, Original data (blue), the filtered data (green), and the fit to the filtered data (orange). Again the same shift in minima and maxima is still present in the fit function to the filtered data.
Here are my results with more aggressive clipping bounds of 0.5 to 1.75 for each data set.
for pow1:
A = 9.6711505138648990E-01
c = 9.7613787086912507E-01
p = 4.0262076448344617E+00
w = 1.2654001570670070E+00
for pow2:
A = 9.4894637490866129E-01
c = 9.6733405789489280E-01
p = 4.0892433833755097E+00
w = 1.2578627414445132E+00
for pow3:
A = 9.8595630272060597E-01
c = 9.6749868212694512E-01
p = 4.0859456191316230E+00
w = 1.2598547148182329E+00
for pow4:
A = -9.4636707498392481E-01
c = 9.5047597808408602E-01
p = -4.2643913461857056E+02
w = 1.2761107231684055E+00
I think I have this figured out - your data is not a mathematically perfect sine wave + noise, so the fitting software can only come close to modeling a sine function to this data. If you must have more accuracy, try splitting the model into different segments and use a piecewise fit. Here is a close-up of the problem area:

Calculating medoid of a cluster (Python)

So I'm running a KNN in order to create clusters. From each cluster, I would like to obtain the medoid of the cluster.
I'm employing a fractional distance metric in order to calculate distances:
where d is the number of dimensions, the first data point's coordinates are x^i, the second data point's coordinates are y^i, and f is an arbitrary number between 0 and 1
I would then calculate the medoid as:
where S is the set of datapoints, and δ is the absolute value of the distance metric used above.
I've looked online to no avail trying to find implementations of medoid (even with other distance metrics, but most thing were specifically k-means or k-medoid which [I think] is relatively different from what I want.
Essentially this boils down to me being unable to translate the math into effective programming. Any help would or pointers in the right direction would be much appreciated! Here's a short list of what I have so far:
I have figured out how to calculate the fractional distance metric (the first equation) so I think I'm good there.
I know numpy has an argmin() function (documented here).
Extra points for increased efficiency without lack of accuracy (I'm trying not to brute force by calculating every single fractional distance metric (because the number of point pairs might lead to a factorial complexity...).
compute pairwise distance matrix
compute column or row sum
argmin to find medoid index
i.e. numpy.argmin(distMatrix.sum(axis=0)) or similar.
So I've accepted the answer here, but I thought I'd provide my implementation if anyone else was trying to do something similar:
(1) This is the distance function:
def fractional(p_coord_array, q_coord_array):
# f is an arbitrary value, but must be greater than zero and
# less than one. In this case, I used 3/10. I took advantage
# of the difference of cubes in this case, so that I wouldn't
# encounter an overflow error.
a = np.sum(np.array(p_coord_array, dtype=np.float64))
b = np.sum(np.array(q_coord_array, dtype=np.float64))
a2 = np.sum(np.power(p_coord_array, 2))
ab = np.sum(p_coord_array) * np.sum(q_coord_array)
b2 = np.sum(np.power(p_coord_array, 2))
diffab = a - b
suma2abb2 = a2 + ab + b2
temp_dist = abs(diffab * suma2abb2)
temp_dist = np.power(temp_dist, 1./10)
dist = np.power(temp_dist, 10./3)
return dist
(2) The medoid function (if the length of the dataset was less than 6000 [if greater than that, I ran into overflow errors... I'm still working on that bit to be perfectly honest...]):
def medoid(dataset):
point = []
w = len(dataset)
if(len(dataset) < 6000):
h = len(dataset)
dist_matrix = [[0 for x in range(w)] for y in range(h)]
list_combinations = [(counter_1, counter_2, data_1, data_2) for counter_1, data_1 in enumerate(dataset) for counter_2, data_2 in enumerate(dataset) if counter_1 < counter_2]
for counter_3, tuple in enumerate(list_combinations):
temp_dist = fractional(tuple[2], tuple[3])
dist_matrix[tuple[0]][tuple[1]] = abs(temp_dist)
dist_matrix[tuple[1]][tuple[0]] = abs(temp_dist)
Any questions, feel free to comment!
If you don't mind using brute force this might help:
def calc_medoid(X, Y, f=2):
n = len(X)
m = len(Y)
dist_mat = np.zeros((m, n))
# compute distance matrix
for j in range(n):
center = X[j, :]
for i in range(m):
if i != j:
dist_mat[i, j] = np.linalg.norm(Y[i, :] - center, ord=f)
medoid_id = np.argmin(dist_mat.sum(axis=0)) # sum over y
return medoid_id, X[medoid_id, :]
Here is an example of computing a medoid for a single cluster with Euclidean distance.
import numpy as np, pandas as pd, matplotlib.pyplot as plt
a, b, c, d = np.array([0,1]), np.array([1, 3]), np.array([4,2]), np.array([3, 1.5])
vCenroid = np.mean([a, b, c, d], axis=0)
def GetMedoid(vX):
vMean = np.mean(vX, axis=0) # compute centroid
return vX[np.argmin([sum((x - vMean)**2) for x in vX])] # pick a point closest to centroid
vMedoid = GetMedoid([a, b, c, d])
print(f'centroid = {vCenroid}')
print(f'medoid = {vMedoid}')
df = pd.DataFrame([a, b, c, d], columns=['x', 'y'])
ax = df.plot.scatter('x', 'y', grid=True, title='Centroid in 2D plane', s=100);
plt.plot(vCenroid[0], vCenroid[1], 'ro', ms=10); # plot centroid as red circle
plt.plot(vMedoid[0], vMedoid[1], 'rx', ms=20); # plot medoid as red star
You can also use the following package to compute medoid for one or more clusters
!pip -q install scikit-learn-extra > log
from sklearn_extra.cluster import KMedoids
GetMedoid = lambda vX: KMedoids(n_clusters=1).fit(vX).cluster_centers_
GetMedoid([a, b, c, d])[0]
I would say that you just need to compute the median.
np.median(np.asarray(points), axis=0)
Your median is the point with the biggest centrality.
Note: if you are using distances different than Euclidean this doesn't hold.

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