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Python find curve fit of slightly inclided vertical

I have a set of data points on the two-axis scale. I want to fit the points. The number of datapoints runs into a few thousand. Here, I have given reproducible shape data points.
My code:
from scipy.optimize import curve_fit
x1 = [327.98999023, 327.98999023, 328.02999878, 367.6499939 ,
374.72000122, 372.73001099, 372.95001221, 373.02999878,
379.39001465, 375.58999634, 378.97000122, 378.1499939 ,
380.70001221, 379.42999268, 379.5 , 379.5 ,
392.22000122, 379.57998657, 376.67999268, 376.67999268]
ym1 = [2.54999995, 2.73999989, 2.91000009, 2.96999991, 3.17000002,
3.40000004, 3.60000014, 3.77999991, 3.98999989, 4.21000004,
4.44000006, 4.62000012, 4.83999997, 5.19999981, 5.32999992,
5.59000015, 5.88999987, 6.20000005, 6.46000028, 6.66999996]
def testfit(x, *p):
''' function to fit the indentation curve
p = [x0,c, poly1d_coeffs ]'''
x = x.astype(float)
y = p[1]*(1-sigmoid(x-p[0],k=1)) + np.poly1d(p[2:])(x) * sigmoid(x-p[0],k=1)
return y
def sigmoid(x, k=1):
return 1/(1+np.exp(-k*x))
p0_guess = (30, 5, 0.3, -10 )
popt, pcov = curve_fit(testfit, x1, ym1, p0=p0_guess) # find optimal parameters
# calculate prediction
yp1 = testfit(x1,popt[0],popt[1],popt[2])
# calculate r^2
r1 = r2_score(ym1,yp1)
print(r1)
plt.plot(x1,ym1,'.')
plt.plot(x1,yp1,'-')
plt.show()
Present output:
r1
Out[128]: -2.1490993028157854 # Negative fit score, Really bad fit.

scipy.optimize.curve_fit return inf pcov

I use curve_fit to fit a very simple line as below code:
from scipy.optimize import curve_fit
def func(x, a, b):
return a * x + b
x = [6.6000000000000005, 7.599]
y = [123.9835274456227, 144.9319749893788]
popt, pcov = curve_fit(func,x,y,method='dogbox',p0=[20,-15])
print(popt) # get [ 20.96941696 -14.4146245 ]
print(pcov) # get [[inf inf], [inf inf]]
But the pcov result is inf. How can I get correct pcov values ?

The result should be no value, since two points fit curve should no parameter error. So the covariance of parameters is zero.

Using scipy.optimize.curve_fit with weights

According to the documentation, the argument sigma can be used to set the weights of the data points in the fit. These "describe" 1-sigma errors when the argument absolute_sigma=True.
I have some data with artificial normally-distributed noise which varies:
n = 200
x = np.linspace(1, 20, n)
x0, A, alpha = 12, 3, 3
def f(x, x0, A, alpha):
return A * np.exp(-((x-x0)/alpha)**2)
noise_sigma = x/20
noise = np.random.randn(n) * noise_sigma
yexact = f(x, x0, A, alpha)
y = yexact + noise
If I want to fit the noisy y to f using curve_fit to what should I set sigma? The documentation isn't very specific here, but I would usually use 1/noise_sigma**2 as the weight:
p0 = 10, 4, 2
popt, pcov = curve_fit(f, x, y, p0)
popt2, pcov2 = curve_fit(f, x, y, p0, sigma=1/noise_sigma**2, absolute_sigma=True)
It doesn't seem to improve the fit much, though.
Is this option only used to better interpret the fit uncertainties through the covariance matrix? What is the difference between these two telling me?
In [249]: pcov
Out[249]:
array([[ 1.10205238e-02, -3.91494024e-08, 8.81822412e-08],
[ -3.91494024e-08, 1.52660426e-02, -1.05907265e-02],
[ 8.81822412e-08, -1.05907265e-02, 2.20414887e-02]])
In [250]: pcov2
Out[250]:
array([[ 0.26584674, -0.01836064, -0.17867193],
[-0.01836064, 0.27833 , -0.1459469 ],
[-0.17867193, -0.1459469 , 0.38659059]])

At least with scipy version 1.1.0 the parameter sigma should be equal to the error on each parameter. Specifically the documentation says:
A 1-d sigma should contain values of standard deviations of errors in
ydata. In this case, the optimized function is chisq = sum((r / sigma)
** 2).
In your case that would be:
curve_fit(f, x, y, p0, sigma=noise_sigma, absolute_sigma=True)
I looked through the source code and verified that when you specify sigma this way it minimizes ((f-data)/sigma)**2.
As a side note, this is in general what you want to be minimizing when you know the errors. The likelihood of observing points data given a model f is given by:
L(data|x0,A,alpha) = product over i Gaus(data_i, mean=f(x_i,x0,A,alpha), sigma=sigma_i)
which if you take the negative log becomes (up to constant factors that don't depend on the parameters):
-log(L) = sum over i (f(x_i,x0,A,alpha)-data_i)**2/(sigma_i**2)
which is just the chisquare.
I wrote a test program to verify that curve_fit was indeed returning the correct values with the sigma specified correctly:
from __future__ import print_function
import numpy as np
from scipy.optimize import curve_fit, fmin
np.random.seed(0)
def make_chi2(x, data, sigma):
def chi2(args):
x0, A, alpha = args
return np.sum(((f(x,x0,A,alpha)-data)/sigma)**2)
return chi2
n = 200
x = np.linspace(1, 20, n)
x0, A, alpha = 12, 3, 3
def f(x, x0, A, alpha):
return A * np.exp(-((x-x0)/alpha)**2)
noise_sigma = x/20
noise = np.random.randn(n) * noise_sigma
yexact = f(x, x0, A, alpha)
y = yexact + noise
p0 = 10, 4, 2
# curve_fit without parameters (sigma is implicitly equal to one)
popt, pcov = curve_fit(f, x, y, p0)
# curve_fit with wrong sigma specified
popt2, pcov2 = curve_fit(f, x, y, p0, sigma=1/noise_sigma**2, absolute_sigma=True)
# curve_fit with correct sigma
popt3, pcov3 = curve_fit(f, x, y, p0, sigma=noise_sigma, absolute_sigma=True)
chi2 = make_chi2(x,y,noise_sigma)
# double checking that we get the correct answer
xopt = fmin(chi2,p0,xtol=1e-10,ftol=1e-10)
print("popt = %s, chi2 = %.2f" % (popt,chi2(popt)))
print("popt2 = %s, chi2 = %.2f" % (popt2, chi2(popt2)))
print("popt3 = %s, chi2 = %.2f" % (popt3, chi2(popt3)))
print("xopt = %s, chi2 = %.2f" % (xopt, chi2(xopt)))
which outputs:
popt = [ 11.93617403 3.30528488 2.86314641], chi2 = 200.66
popt2 = [ 11.94169083 3.30372955 2.86207253], chi2 = 200.64
popt3 = [ 11.93128545 3.333727 2.81403324], chi2 = 200.44
xopt = [ 11.93128603 3.33373094 2.81402741], chi2 = 200.44
As you can see the chi2 is indeed minimized correctly when you specify sigma=sigma as an argument to curve_fit.
As to why the improvement isn't "better", I'm not really sure. My only guess is that without specifying a sigma value you implicitly assume they are equal and over the part of the data where the fit matters (the peak), the errors are "approximately" equal.
To answer your second question, no the sigma option is not only used to change the output of the covariance matrix, it actually changes what is being minimized.

2D Gaussian Fit for intensities at certain coordinates in Python

I have a set of coordinates (x, y, z(x, y)) which describe intensities (z) at coordinates x, y. For a set number of these intensities at different coordinates, I need to fit a 2D Gaussian that minimizes the mean squared error.
The data is in numpy matrices and for each fitting session I will have either 4, 9, 16 or 25 coordinates. Ultimately I just need to get the central position of the gaussian (x_0, y_0) that has smallest MSE.
All of the examples that I have found use scipy.optimize.curve_fit but the input data they have is over an entire mesh rather than a few coordinates.
Any help would be appreciated.

Introduction
There are multiple ways to approach this. You can use non-linear methods (e.g. scipy.optimize.curve_fit), but they'll be slow and aren't guaranteed to converge. You can linearize the problem (fast, unique solution), but any noise in the "tails" of the distribution will cause issues. There are actually a few tricks you can apply to this particular case to avoid the latter issue. I'll show some examples, but I don't have time to demonstrate all of the "tricks" right now.
Just as a side note, a general 2D guassian has 6 parameters, so you won't be able to fully fit things with 4 points. However, it sounds like you might be assuming that there's no covariance between x and y and that the variances are the same in each direction (i.e. a perfectly "round" bell curve). If that's the case, then you only need four parameters. If you know the amplitude of the guassian, you'll only need three. However, I'm going to start with the general solution, and you can simplify it later on, if you want to.
For the moment, let's focus on solving this problem using non-linear methods (e.g. scipy.optimize.curve_fit).
The general equation for a 2D guassian is (directly from wikipedia):
where:
is essentially 0.5 over the covariance matrix, A is the amplitude,
and (X₀, Y₀) is the center
Generate simplified sample data
Let's write the equation above out:
import numpy as np
import matplotlib.pyplot as plt
def gauss2d(x, y, amp, x0, y0, a, b, c):
inner = a * (x - x0)**2
inner += 2 * b * (x - x0)**2 * (y - y0)**2
inner += c * (y - y0)**2
return amp * np.exp(-inner)
And then let's generate some example data. To start with, we'll generate some data that will be easy to fit:
np.random.seed(1977) # For consistency
x, y = np.random.random((2, 10))
x0, y0 = 0.3, 0.7
amp, a, b, c = 1, 2, 3, 4
zobs = gauss2d(x, y, amp, x0, y0, a, b, c)
fig, ax = plt.subplots()
scat = ax.scatter(x, y, c=zobs, s=200)
fig.colorbar(scat)
plt.show()
Note that we haven't added any noise, and the center of the distribution is within the range that we have data (i.e. center at 0.3, 0.7 and a scatter of x,y observations between 0 and 1). For the moment, let's stick with this, and then we'll see what happens when we add noise and shift the center.
Non-linear fitting
To start with, let's use scpy.optimize.curve_fit to preform a non-linear least-squares fit to the gaussian function. (On a side note, you can play around with the exact minimization algorithm by using some of the other functions in scipy.optimize.)
The scipy.optimize functions expect a slightly different function signature than the one we originally wrote above. We could write a wrapper to "translate", but let's just re-write the gauss2d function instead:
def gauss2d(xy, amp, x0, y0, a, b, c):
x, y = xy
inner = a * (x - x0)**2
inner += 2 * b * (x - x0)**2 * (y - y0)**2
inner += c * (y - y0)**2
return amp * np.exp(-inner)
All we did was have the function expect the independent variables (x & y) as a single 2xN array.
Now we need to make an initial guess at what the guassian curve's parameters actually are. This is optional (the default is all ones, if I recall correctly), but you're likely to have problems converging if 1, 1 is not particularly close to the "true" center of the gaussian curve. For that reason, we'll use the x and y values of our largest observed z-value as a starting point for the center. I'll leave the rest of the parameters as 1, but if you know that they're likely to consistently be significantly different, change them to something more reasonable.
Here's the full, stand-alone example:
import numpy as np
import scipy.optimize as opt
import matplotlib.pyplot as plt
def main():
x0, y0 = 0.3, 0.7
amp, a, b, c = 1, 2, 3, 4
true_params = [amp, x0, y0, a, b, c]
xy, zobs = generate_example_data(10, true_params)
x, y = xy
i = zobs.argmax()
guess = [1, x[i], y[i], 1, 1, 1]
pred_params, uncert_cov = opt.curve_fit(gauss2d, xy, zobs, p0=guess)
zpred = gauss2d(xy, *pred_params)
print 'True parameters: ', true_params
print 'Predicted params:', pred_params
print 'Residual, RMS(obs - pred):', np.sqrt(np.mean((zobs - zpred)**2))
plot(xy, zobs, pred_params)
plt.show()
def gauss2d(xy, amp, x0, y0, a, b, c):
x, y = xy
inner = a * (x - x0)**2
inner += 2 * b * (x - x0)**2 * (y - y0)**2
inner += c * (y - y0)**2
return amp * np.exp(-inner)
def generate_example_data(num, params):
np.random.seed(1977) # For consistency
xy = np.random.random((2, num))
zobs = gauss2d(xy, *params)
return xy, zobs
def plot(xy, zobs, pred_params):
x, y = xy
yi, xi = np.mgrid[:1:30j, -.2:1.2:30j]
xyi = np.vstack([xi.ravel(), yi.ravel()])
zpred = gauss2d(xyi, *pred_params)
zpred.shape = xi.shape
fig, ax = plt.subplots()
ax.scatter(x, y, c=zobs, s=200, vmin=zpred.min(), vmax=zpred.max())
im = ax.imshow(zpred, extent=[xi.min(), xi.max(), yi.max(), yi.min()],
aspect='auto')
fig.colorbar(im)
ax.invert_yaxis()
return fig
main()
In this case, we exactly(ish) recover our original "true" parameters.
True parameters: [1, 0.3, 0.7, 2, 3, 4]
Predicted params: [ 1. 0.3 0.7 2. 3. 4. ]
Residual, RMS(obs - pred): 1.01560615193e-16
As we'll see in a second, this won't always be the case...
Adding Noise
Let's add some noise to our observations. All I've done here is change the generate_example_data function:
def generate_example_data(num, params):
np.random.seed(1977) # For consistency
xy = np.random.random((2, num))
noise = np.random.normal(0, 0.3, num)
zobs = gauss2d(xy, *params) + noise
return xy, zobs
However, the result looks quite different:
And as far as the parameters go:
True parameters: [1, 0.3, 0.7, 2, 3, 4]
Predicted params: [ 1.129 0.263 0.750 1.280 32.333 10.103 ]
Residual, RMS(obs - pred): 0.152444640098
The predicted center hasn't changed much, but the b and c parameters have changed quite a bit.
If we change the center of the function to somewhere slightly outside of our scatter of points:
x0, y0 = -0.3, 1.1
We'll wind up with complete nonsense as a result in the presence of noise! (It still works correctly without noise.)
True parameters: [1, -0.3, 1.1, 2, 3, 4]
Predicted params: [ 0.546 -0.939 0.857 -0.488 44.069 -4.136]
Residual, RMS(obs - pred): 0.235664449826
This is a common problem when fitting a function that decays to zero. Any noise in the "tails" can result in a very poor result. There are a number of strategies to deal with this. One of the easiest is to weight the inversion by the observed z-values. Here's an example for the 1D case: (focusing on linearized the problem) How can I perform a least-squares fitting over multiple data sets fast? If I have time later, I'll add an example of this for the 2D case.

Python - calculating trendlines with errors

So I've got some data stored as two lists, and plotted them using
plot(datasetx, datasety)
Then I set a trendline
trend = polyfit(datasetx, datasety)
trendx = []
trendy = []
for a in range(datasetx[0], (datasetx[-1]+1)):
trendx.append(a)
trendy.append(trend[0]*a**2 + trend[1]*a + trend[2])
plot(trendx, trendy)
But I have a third list of data, which is the error in the original datasety. I'm fine with plotting the errorbars, but what I don't know is using this, how to find the error in the coefficients of the polynomial trendline.
So say my trendline came out to be 5x^2 + 3x + 4 = y, there needs to be some sort of error on the 5, 3 and 4 values.
Is there a tool using NumPy that will calculate this for me?

I think you can use the function curve_fit of scipy.optimize (documentation). A basic example of the usage:
import numpy as np
from scipy.optimize import curve_fit
def func(x, a, b, c):
return a*x**2 + b*x + c
x = np.linspace(0,4,50)
y = func(x, 5, 3, 4)
yn = y + 0.2*np.random.normal(size=len(x))
popt, pcov = curve_fit(func, x, yn)
Following the documentation, pcov gives:
The estimated covariance of popt. The diagonals provide the variance
of the parameter estimate.
So in this way you can calculate an error estimate on the coefficients. To have the standard deviation you can take the square root of the variance.
Now you have an error on the coefficients, but it is only based on the deviation between the ydata and the fit. In case you also want to account for an error on the ydata itself, the curve_fit function provides the sigma argument:
sigma : None or N-length sequence
If not None, it represents the standard-deviation of ydata. This
vector, if given, will be used as weights in the least-squares
problem.
A complete example:
import numpy as np
from scipy.optimize import curve_fit
def func(x, a, b, c):
return a*x**2 + b*x + c
x = np.linspace(0,4,20)
y = func(x, 5, 3, 4)
# generate noisy ydata
yn = y + 0.2 * y * np.random.normal(size=len(x))
# generate error on ydata
y_sigma = 0.2 * y * np.random.normal(size=len(x))
popt, pcov = curve_fit(func, x, yn, sigma = y_sigma)
# plot
import matplotlib.pyplot as plt
fig = plt.figure()
ax = fig.add_subplot(111)
ax.errorbar(x, yn, yerr = y_sigma, fmt = 'o')
ax.plot(x, np.polyval(popt, x), '-')
ax.text(0.5, 100, r"a = {0:.3f} +/- {1:.3f}".format(popt[0], pcov[0,0]**0.5))
ax.text(0.5, 90, r"b = {0:.3f} +/- {1:.3f}".format(popt[1], pcov[1,1]**0.5))
ax.text(0.5, 80, r"c = {0:.3f} +/- {1:.3f}".format(popt[2], pcov[2,2]**0.5))
ax.grid()
plt.show()
Then something else, about using numpy arrays. One of the main advantages of using numpy is that you can avoid for loops because operations on arrays apply elementwise. So the for-loop in your example can also be done as following:
trendx = arange(datasetx[0], (datasetx[-1]+1))
trendy = trend[0]*trendx**2 + trend[1]*trendx + trend[2]
Where I use arange instead of range as it returns a numpy array instead of a list.
In this case you can also use the numpy function polyval:
trendy = polyval(trend, trendx)

I have not been able to find any way of getting the errors in the coefficients that is built in to numpy or python. I have a simple tool that I wrote based on Section 8.5 and 8.6 of John Taylor's An Introduction to Error Analysis. Maybe this will be sufficient for your task (note the default return is the variance, not the standard deviation). You can get large errors (as in the provided example) because of significant covariance.
def leastSquares(xMat, yMat):
'''
Purpose
-------
Perform least squares using the procedure outlined in 8.5 and 8.6 of Taylor, solving
matrix equation X a = Y
Examples
--------
>>> from scipy import matrix
>>> xMat = matrix([[ 1, 5, 25],
[ 1, 7, 49],
[ 1, 9, 81],
[ 1, 11, 121]])
>>> # matrix has rows of format [constant, x, x^2]
>>> yMat = matrix([[142],
[168],
[211],
[251]])
>>> a, varCoef, yRes = leastSquares(xMat, yMat)
>>> # a is a column matrix, holding the three coefficients a, b, c, corresponding to
>>> # the equation a + b*x + c*x^2
Returns
-------
a: matrix
best fit coefficients
varCoef: matrix
variance of derived coefficents
yRes: matrix
y-residuals of fit
'''
xMatSize = xMat.shape
numMeas = xMatSize[0]
numVars = xMatSize[1]
xxMat = xMat.T * xMat
xyMat = xMat.T * yMat
xxMatI = xxMat.I
aMat = xxMatI * xyMat
yAvgMat = xMat * aMat
yRes = yMat - yAvgMat
var = (yRes.T * yRes) / (numMeas - numVars)
varCoef = xxMatI.diagonal() * var[0, 0]
return aMat, varCoef, yRes