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Today, I will talk to you about the example analysis of Python data fitting to achieve the least square method. Many people may not know much about it. In order to make you understand better, the editor has summarized the following content for you. I hope you can get something according to this article.

Linear fitting

This expression is still very simple.

For some cases, we often choose the natural sequence as the independent variable. At this time, we can use some elementary mathematical inferences when finding the value of the independent variable. For the natural sequence of x ∈ [m, n], there are

# filename core.pyimport numpy as npdef leastSquare (x Y): if len (x) = = 2: # then x is the natural sequence sx = 0.5* (x [1]-x [0] + 1) * (x [1] + x [0]) ex = sx/ (x [1]-x [0] + 1) sx2 = ((x [1] * (x [1] + 1) * (2x1 [1] + 1)-(x [0]) * (x [0]-1) * (2x [0]-1)) / 6 x = np.array (range (x [0]) X [1] + 1)) else: sx = sum (x) ex = sx/len (x) sx2 = sum (Xeroy) sxy = sum (Xeroy) ey = np.mean (y) a = (sxy-ey*sx) / (sx2-ex*sx) b = (ey*sx2-sxy*ex) / (sx2-ex*sx) return

Test it

> x = np.arange (25) > y = x*15+20+np.random.randn (len (x)) * 5 # randn generates normal distribution noise > a core.leastSquare b = core.leastSquare (xpeny) > plt.scatter (xmagin y) # Raw data scatter plot > plt.plot (x A*x+b) # fit Line [] > plt.show ()

Get

Higher order polynomial

As before, the agreement

The code is as follows

# the format of the input parameters is np.array,n as order def leastSquareMulti (xjinyjinn): X = [np.sum (xmagedi) for i in range (2*n+1)] Y = np.array ([[np.sum (y*x**i)] for i in range (nasty 1)]) S = np.array ([X [I: i+n+1] for i in range (nasty 1)]) return np.linalg.solve (SMagy) #

The test results are as follows:

> x = np.arange (25) > y = x**3+3*x**2+2*x+12 > import core > core.leastSquareMulti ([[12.], # this is a constant term [2.], [3.], [1.]) Multiple independent variables

For samples

Then the corresponding error equations can be expressed as

Exponential function

Its code is

Def expFit (XMague y): Y0 = y [0:-3] y1 = y [1:-2] y2 = y [2:-1] BMague C = leastSquare (y2/y0 Y1/y0) b1 = np.log ((B-np.sqrt (Background2Benz4C)) / 2) b2 = np.log ((B+np.sqrt (Bathy4C) / 2) X = np.exp (b1-b2) * x = y/np.exp (b2yogx) A1 Magi a2 = leastSquare (XMY) return a1Magi a2menb1Personb2

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