blocks|key|1665770|text|是的，这通常是通过使用最小二乘来完成的。还有其他方法可以指定多项式的拟合程度，但这个理论对于最小二乘来说是最简单的。一般的理论称为线性回归。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1665771|您最好的选择可能是从Numerical+Recipes开始。|offset|length|1665772|R是免费的，可以做你想做的任何事情，甚至更多，但它有一个很大的学习曲线。|1665773|如果您可以访问Mathematica，则可以使用fit函数进行最小二乘拟合。我想Matlab和它的开源对应物Octave也有类似的功能。|1665774|entityMap|0|LINK|mutability|MUTABLE|url|http://www.numerical.recipes/oldverswitcher.html|1|http://www.r-project.org/^0|0|A|H|0|0|0|1|1|0|0^^$0|@$1|2|3|4|5|6|7|T|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|U|8|@]|9|@$D|V|E|W|1|X]]|A|$]]|$1|F|3|G|5|6|7|Y|8|@]|9|@$D|Z|E|10|1|11]]|A|$]]|$1|H|3|I|5|6|7|12|8|@]|9|@]|A|$]]|$1|J|3|-4|5|6|7|13|8|@]|9|@]|A|$]]]|K|$L|$5|M|N|O|A|$P|Q]]|R|$5|M|N|O|A|$P|S]]]]

Yes, the way this is typically done is by using least squares. There are other ways of specifying how well a polynomial fits, but the theory is simplest for least squares. The general theory is called linear regression. 

Your best bet is probably to start with <a href="http://www.numerical.recipes/oldverswitcher.html" rel="nofollow noreferrer">Numerical Recipes</a>.

<a href="http://www.r-project.org/" rel="nofollow noreferrer">R</a> is free and will do everything you want and more, but it has a big learning curve. 

If you have access to Mathematica, you can use the Fit function to do a least squares fit. I imagine Matlab and its open source counterpart Octave have a similar function.

blocks|key|1665849|text|对于(x，f(x))情形：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1665850|import+numpy

x+=+numpy.arange(10)
y+=+x**2

coeffs+=+numpy.polyfit(x,+y,+deg=2)
poly+=+numpy.poly1d(coeffs)
print+poly
yp+=+numpy.polyval(poly,+x)
print+(yp-y)|code-block|syntax|javascript|1665851|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

For (x, f(x)) case:

<pre><code>import numpy

x = numpy.arange(10)
y = x**2

coeffs = numpy.polyfit(x, y, deg=2)
poly = numpy.poly1d(coeffs)
print poly
yp = numpy.polyval(poly, x)
print (yp-y)
</code></pre>

blocks|key|1450941|text|请记住，高次多项式总是能更好地拟合数据。然而，高次多项式通常会导致非常不可能的函数(参见Occam's+Razor)+(过拟合)。你想在简单性(多项式的次数)和拟合(例如最小二乘误差)之间找到一个平衡点。在数量上，有对此的测试，Akaike+Information+Criterion或Bayesian+Information+Criterion。这些测试给出了优先选择哪种模型的分数。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1450942|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/Occams_razor|1|http://en.wikipedia.org/wiki/Akaike_information_criterion|2|http://en.wikipedia.org/wiki/Bayesian_information_criterion^0|18|D|0|36|S|1|3Z|U|2|0^^$0|@$1|2|3|4|5|6|7|P|8|@]|9|@$A|Q|B|R|1|S]|$A|T|B|U|1|V]|$A|W|B|X|1|Y]]|C|$]]|$1|D|3|-4|5|6|7|Z|8|@]|9|@]|C|$]]]|E|$F|$5|G|H|I|C|$J|K]]|L|$5|G|H|I|C|$J|M]]|N|$5|G|H|I|C|$J|O]]]]

Bare in mind that a polynomial of higher degree ALWAYS fits the data better. Polynomials of higher degree typically leads to highly improbable functions (see <a href="http://en.wikipedia.org/wiki/Occams_razor" rel="nofollow noreferrer">Occam's Razor</a>), though (overfitting). You want to find a balance between simplicity (degree of polynomial) and fit (e.g. least square error). Quantitatively, there are tests for this, the <a href="http://en.wikipedia.org/wiki/Akaike_information_criterion" rel="nofollow noreferrer">Akaike Information Criterion</a> or the <a href="http://en.wikipedia.org/wiki/Bayesian_information_criterion" rel="nofollow noreferrer">Bayesian Information Criterion</a>. These tests give a score which model is to be prefered.

blocks|key|1666073|text|在大学时，我们有一本书，我仍然觉得这本书非常有用:+Conte，de+Boor；初等数值分析；Mc。相关段落为6.2:数据拟合。|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1666074|示例代码是用FORTRAN编写的，清单的可读性也不是很好，但是解释很深刻，同时也很清楚。你最终会明白你在做什么，而不仅仅是做它(就像我在数字配方方面的经验一样)。|1666075|我通常从数字Recipes开始，但对于这样的事情，我很快就必须抓住Conte-de+Boor。|1666076|也许更好的方法是发布一些代码...它有点精简了，但最相关的部分都在那里。显然，它依赖于numpy！|1666077|def+Tn(n,+x):
++if+n==0:
++++return+1.0
++elif+n==1:
++++return+float(x)
++else:
++++return+(2.0+*+x+*+Tn(n+-+1,+x))+-+Tn(n+-+2,+x)

class+ChebyshevFit:

++def+__init__(self):
++++self.Tn+=+Memoize(Tn)

++def+fit(self,+data,+degree=None):
++++"""fit+the+data+by+a+'minimal+squares'+linear+combination+of+chebyshev+polinomials.

++++cfr:+Conte,+de+Boor;+elementary+numerical+analysis;+Mc+Grow+Hill+(6.2:+Data+Fitting)
++++"""

++++if+degree+is+None:
++++++degree+=+5

++++data+=+sorted(data)
++++self.range+=+start,+end+=+(min(data)[0],+max(data)[0])
++++self.halfwidth+=+(end+-+start)+/+2.0
++++vec_x+=+[(x+-+start+-+self.halfwidth)/self.halfwidth+for+(x,+y)+in+data]
++++vec_f+=+[y+for+(x,+y)+in+data]

++++mat_phi+=+[numpy.array([self.Tn(i,+x)+for+x+in+vec_x])+for+i+in+range(degree%2B1)]
++++mat_A+=+numpy.inner(mat_phi,+mat_phi)
++++vec_b+=+numpy.inner(vec_f,+mat_phi)

++++self.coefficients+=+numpy.linalg.solve(mat_A,+vec_b)
++++self.degree+=+degree

++def+evaluate(self,+x):
++++"""use+Clenshaw+algorithm

++++http://en.wikipedia.org/wiki/Clenshaw_algorithm
++++"""

++++x+=+(x-self.range[0]-self.halfwidth)+/+self.halfwidth

++++b_2+=+float(self.coefficients[self.degree])
++++b_1+=+2+*+x+*+b_2+%2B+float(self.coefficients[self.degree+-+1])

++++for+i+in+range(2,+self.degree):
++++++b_1,+b_2+=+2.0+*+x+*+b_1+%2B+self.coefficients[self.degree+-+i]+-+b_2,+b_1
++++else:
++++++b_0+=+x*b_1+%2B+self.coefficients[0]+-+b_2

++++return+b_0|code-block|syntax|javascript|1666078|entityMap^0|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|O|8|@]|9|@]|A|$]]|$1|B|3|C|5|6|7|P|8|@]|9|@]|A|$]]|$1|D|3|E|5|6|7|Q|8|@]|9|@]|A|$]]|$1|F|3|G|5|6|7|R|8|@]|9|@]|A|$]]|$1|H|3|I|5|J|7|S|8|@]|9|@]|A|$K|L]]|$1|M|3|-4|5|6|7|T|8|@]|9|@]|A|$]]]|N|$]]

at college we had this book which I still find extremely useful: Conte, de Boor; elementary numerical analysis; Mc Grow Hill. The relevant paragraph is 6.2: Data Fitting. 
example code comes in FORTRAN, and the listings are not very readable either, but the explanations are deep and clear at the same time. you end up understanding what you are doing, not just doing it (as is my experience of Numerical Recipes). 
I usually start with Numerical Recipes but for things like this I quickly have to grab Conte-de Boor.

maybe better posting some code... it's a bit stripped down, but the most relevant parts are there. it relies on numpy, obviously!

<pre><code>def Tn(n, x):
 if n==0:
 return 1.0
 elif n==1:
 return float(x)
 else:
 return (2.0 * x * Tn(n - 1, x)) - Tn(n - 2, x)

class ChebyshevFit:

 def __init__(self):
 self.Tn = Memoize(Tn)

 def fit(self, data, degree=None):
 """fit the data by a 'minimal squares' linear combination of chebyshev polinomials.

 cfr: Conte, de Boor; elementary numerical analysis; Mc Grow Hill (6.2: Data Fitting)
 """

 if degree is None:
 degree = 5

 data = sorted(data)
 self.range = start, end = (min(data)[0], max(data)[0])
 self.halfwidth = (end - start) / 2.0
 vec_x = [(x - start - self.halfwidth)/self.halfwidth for (x, y) in data]
 vec_f = [y for (x, y) in data]

 mat_phi = [numpy.array([self.Tn(i, x) for x in vec_x]) for i in range(degree+1)]
 mat_A = numpy.inner(mat_phi, mat_phi)
 vec_b = numpy.inner(vec_f, mat_phi)

 self.coefficients = numpy.linalg.solve(mat_A, vec_b)
 self.degree = degree

 def evaluate(self, x):
 """use Clenshaw algorithm

 http://en.wikipedia.org/wiki/Clenshaw_algorithm
 """

 x = (x-self.range[0]-self.halfwidth) / self.halfwidth

 b_2 = float(self.coefficients[self.degree])
 b_1 = 2 * x * b_2 + float(self.coefficients[self.degree - 1])

 for i in range(2, self.degree):
 b_1, b_2 = 2.0 * x * b_1 + self.coefficients[self.degree - i] - b_2, b_1
 else:
 b_0 = x*b_1 + self.coefficients[0] - b_2

 return b_0
</code></pre>

blocks|key|1666013|text|如果您知道如何将最小二乘问题表示为线性代数问题，那么使用Excel的矩阵函数就很容易得到一个快速拟合。(这取决于您认为Excel作为线性代数求解器的可靠性。)|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1666014|entityMap^0|0^^$0|@$1|2|3|4|5|6|7|D|8|@]|9|@]|A|$]]|$1|B|3|-4|5|6|7|E|8|@]|9|@]|A|$]]]|C|$]]

It's rather easy to scare up a quick fit using Excel's matrix functions if you know how to represent the least squares problem as a linear algebra problem. (That depends on how reliable you think Excel is as a linear algebra solver.)

blocks|key|1450874|text|在某种意义上，lagrange+polynomial是拟合一组给定数据点的“最简单”插值多项式。|type|unstyled|depth|inlineStyleRanges|entityRanges|offset|length|data|1450875|它有时是有问题的，因为它可能在数据点之间变化很大。|1450876|entityMap|0|LINK|mutability|MUTABLE|url|http://en.wikipedia.org/wiki/Lagrange_polynomial^0|7|J|0|0|0^^$0|@$1|2|3|4|5|6|7|N|8|@]|9|@$A|O|B|P|1|Q]]|C|$]]|$1|D|3|E|5|6|7|R|8|@]|9|@]|C|$]]|$1|F|3|-4|5|6|7|S|8|@]|9|@]|C|$]]]|G|$H|$5|I|J|K|C|$L|M]]]]

The <a href="http://en.wikipedia.org/wiki/Lagrange_polynomial" rel="nofollow noreferrer">lagrange polynomial</a> is in some sense the "simplest" interpolating polynomial that fits a given set of data points.

It is sometimes problematic because it can vary wildly between data points.

Is there a way, given a set of values <code>(x,f(x))</code>, to find the polynomial of a given degree that best fits the data? 

I know <a href="http://en.wikipedia.org/wiki/Polynomial_interpolation" rel="noreferrer">polynomial interpolation</a>, which is for finding a polynomial of degree <code>n</code> given <code>n+1</code> data points, but here there are a large number of values and we want to find a low-degree polynomial (find best linear fit, best quadratic, best cubic, etc.). It might be related to <a href="http://en.wikipedia.org/wiki/Least_squares" rel="noreferrer">least squares</a>...

More generally, I would like to know the answer when we have a multivariate function -- points like <code>(x,y,f(x,y))</code>, say -- and want to find the best polynomial (<code>p(x,y)</code>) of a given degree in the variables. (Specifically a polynomial, not splines or Fourier series.) 

Both theory and code/libraries (preferably in Python, but any language is okay) would be useful.

Fitting polynomials to data

Python

在给定一组值(x,f(x))的情况下，有没有办法找到最适合数据的给定次数的多项式？我知道，它用于找到给定n+1数据点的n次多项式，但这里有大量的值，我们想要找到一个低次多项式(找到最佳线性拟合、最佳二次、最佳三次等)。这可能与有关。更广泛地说，我想知道当我们有一个多元函数时的答案--比如像(x,y,f(x,y))这样的点--并且想在变量中找到给定次数的最佳多项式(p(x,y))。(具体地说是多项式

问对数据进行多项式拟合
EN

回答 6

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问对数据进行多项式拟合EN

回答 6

Stack Overflow用户

Stack Overflow用户

Stack Overflow用户

社区

活动

资源

关于

腾讯云开发者

热门产品

热门推荐

更多推荐

问对数据进行多项式拟合
EN