python - numpy.polyfit vs numpy.polynomial.polynomial.polyfit

Question

为什么要 numpy.polyfit 和 numpy.polynomial.polynomial.polyfit
在下面的测试中产生不同的图？

import numpy as np
from numpy.polynomial.polynomial import polyfit
import matplotlib.pyplot as plt

x = np.linspace(0, 10, 50)
y = 5 * x + 10 + (np.random.random(len(x)) - 0.5) * 5

plt.scatter(x, y,marker='.', label='Data for regression')
plt.plot(np.unique(x), np.poly1d(np.polyfit(x, y, 1))(np.unique(x)), 
         label='numpy.polyfit')
plt.plot(np.unique(x), np.poly1d(polyfit(x, y, 1))(np.unique(x)), 
         label='polynomial.polyfit')
plt.legend()
plt.show()

Answer 1

乍一看，文档似乎表明它们应该给出相同的结果 -

numpy.polyfit(x, y, deg, rcond=None, full=False, w=None, cov=False)

Least squares polynomial fit.

Fit a polynomial p(x) = p[0] * x**deg + ... + p[deg] of degree deg to points (x, y). Returns a vector of coefficients p that minimises the squared error in the order deg, deg-1, … 0.

和

numpy.polynomial.polynomial.polyfit(x, y, deg, rcond=None, full=False, w=None)

Least-squares fit of a polynomial to data.

Return the coefficients of a polynomial of degree deg that is the least squares fit to the data values y given at points x. If y is 1-D the returned coefficients will also be 1-D. If y is 2-D multiple fits are done, one for each column of y, and the resulting coefficients are stored in the corresponding columns of a 2-D return. The fitted polynomial(s) are in the form

p(x) = c₀ + c₁ * x + ... + c_n * xⁿ

但区别在于从两种方法返回的系数的顺序，至少对于所讨论的用例是这样。

numpy.polyfit根据生成方程，按度数降序返回系数

p(x) = cn * xn + c(n-1) * x(n-1) + ... + c1 * x + c0

numpy.polynomial.polynomial.polyfit根据生成方程，按度数升序返回系数

p(x) = c0 + c1 * x + ... + c(n-1) * x(n-1) + cn * xn

尽管在数学上相同，但这两个方程在 ndarray 中并不相同。表示。在文档中使用不同的符号可能会混淆这一点。为了演示，请考虑以下内容

import numpy as np

x = np.linspace(0, 10, 50)
y = x**2 + 5 * x + 10

print(np.polyfit(x, y, 2))
print(np.polynomial.polynomial.polyfit(x, y, 2))

[ 1.  5. 10.]
[10.  5.  1.]

两种方法得到的结果相同，但顺序相反，前者是 np.poly1d() 预计，

print(np.poly1d(np.polyfit(x, y, 2)))
print(np.poly1d(np.polynomial.polynomial.polyfit(x, y, 2)))

   2
1 x + 5 x + 10
    2
10 x + 5 x + 1

后者是 np.polynomial.polynomial.Polynomial() 构造函数期望。，

print(np.polynomial.polynomial.Polynomial(np.polynomial.polynomial.polyfit(x, y, 2)))
print(np.polynomial.polynomial.Polynomial(np.polyfit(x, y, 2)))

poly([10.  5.  1.])  # 10 + 5 * x + 1 * x**2
poly([ 1.  5. 10.])  # 1 + 5 * x + 10 * x**2

Flipping来自 np.polynomial.polynomial.polyfit 的结果在将其传递给 poly1d() 之前或使用 np.polynomial.polynomial.Polynomial将产生预期的结果: