What's the error of numpy.polyfit?

Question

I want to use numpy.polyfit for physical calculations, therefore I need the magnitude of the error.

Answer 1

If you specify full=True in your call to polyfit, it will include extra information:

>>> x = np.arange(100) >>> y = x**2 + 3*x + 5 + np.random.rand(100) >>> np.polyfit(x, y, 2) array([ 0.99995888,  3.00221219,  5.56776641]) >>> np.polyfit(x, y, 2, full=True) (array([ 0.99995888,  3.00221219,  5.56776641]), # coefficients  array([ 7.19260721]), # residuals  3, # rank  array([ 11.87708199,   3.5299267 ,   0.52876389]), # singular values  2.2204460492503131e-14) # conditioning threshold 

The residual value returned is the sum of the squares of the fit errors, not sure if this is what you are after:

>>> np.sum((np.polyval(np.polyfit(x, y, 2), x) - y)**2) 7.1926072073491056 

In version 1.7 there is also a cov keyword that will return the covariance matrix for your coefficients, which you could use to calculate the uncertainty of the fit coefficients themselves.

Answer 2

As you can see in the documentation:

Returns ------- p : ndarray, shape (M,) or (M, K)     Polynomial coefficients, highest power first.     If `y` was 2-D, the coefficients for `k`-th data set are in ``p[:,k]``.  residuals, rank, singular_values, rcond : present only if `full` = True     Residuals of the least-squares fit, the effective rank of the scaled     Vandermonde coefficient matrix, its singular values, and the specified     value of `rcond`. For more details, see `linalg.lstsq`. 

Which means that if you can do a fit and get the residuals as:

 import numpy as np  x = np.arange(10)  y = x**2 -3*x + np.random.random(10)   p, res, _, _, _ = numpy.polyfit(x, y, deg, full=True) 

Then, the p are your fit parameters, and the res will be the residuals, as described above. The _'s are because you don't need to save the last three parameters, so you can just save them in the variable _ which you won't use. This is a convention and is not required.

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@Jaime's answer explains what the residual means. Another thing you can do is look at those squared deviations as a function (the sum of which is res). This is particularly helpful to see a trend that didn't fit sufficiently. res can be large because of statistical noise, or possibly systematic poor fitting, for example:

x = np.arange(100) y = 1000*np.sqrt(x) + x**2 - 10*x + 500*np.random.random(100) - 250  p = np.polyfit(x,y,2) # insufficient degree to include sqrt  yfit = np.polyval(p,x)  figure() plot(x,y, label='data') plot(x,yfit, label='fit') plot(x,yfit-y, label='var') 

So in the figure, note the bad fit near x = 0: