Question

我正在阅读一些关于使用 poly 函数的R代码，但我无法弄清楚它是如何工作的。

我知道：

ttt = 1:10
tttMod = poly(ttt, degree = 3, raw = FALSE)
tttMod
            1           2          3
[1,] -0.49543369  0.52223297 -0.4534252
[2,] -0.38533732  0.17407766  0.1511417
[3,] -0.27524094 -0.08703883  0.3778543
[4,] -0.16514456 -0.26111648  0.3346710
[5,] -0.05504819 -0.34815531  0.1295501
[6,]  0.05504819 -0.34815531 -0.1295501
[7,]  0.16514456 -0.26111648 -0.3346710
[8,]  0.27524094 -0.08703883 -0.3778543
[9,]  0.38533732  0.17407766 -0.1511417
[10,]  0.49543369  0.52223297  0.4534252
attr(,"coefs")
attr(,"coefs")$alpha
[1] 5.5 5.5 5.5

attr(,"coefs")$norm2
[1]    1.0   10.0   82.5  528.0 3088.8

attr(,"degree")
[1] 1 2 3
attr(,"class")
[1] "poly"   "matrix"

我无法理解以下代码

ttt2 = 1:5
ttt2Predicted = predict(tttMod, ttt2)

ttt2Predicted
           1           2          3
[1,] -0.49543369  0.52223297 -0.4534252
[2,] -0.38533732  0.17407766  0.1511417
[3,] -0.27524094 -0.08703883  0.3778543
[4,] -0.16514456 -0.26111648  0.3346710
[5,] -0.05504819 -0.34815531  0.1295501

我不知道来自tttMod的系数是什么（以及如何）预测函数用于生成ttt2Predicted。如何通过不使用 predict（tttMod，ttt2）获得相同的结果（即ttt2Predicted）？例如，进行等效的线性代数计算。

我尝试使用 poly ，但似乎没有以这种方式工作

ttt2Predicted_test = poly(ttt2, degree = 3, raw = FALSE)
          1          2             3
[1,] -0.6324555  0.5345225 -3.162278e-01
[2,] -0.3162278 -0.2672612  6.324555e-01
[3,]  0.0000000 -0.5345225 -4.095972e-16
[4,]  0.3162278 -0.2672612 -6.324555e-01
[5,]  0.6324555  0.5345225  3.162278e-01
attr(,"coefs")
attr(,"coefs")$alpha
[1] 3 3 3

attr(,"coefs")$norm2
[1]  1.0  5.0 10.0 14.0 14.4

attr(,"degree")
[1] 1 2 3
attr(,"class")
[1] "poly"   "matrix"

更新

虽然问题已被标记为重复，但我提供了R＆＃39;的Python实现。多项函数及其预测。希望你也会觉得这很有帮助。

在Stack Overflow社区的帮助下，我已经为以下R代码实现了等效的Python代码

ttt = 1:10
tttMod = poly(ttt, degree = 3, raw = FALSE)
ttt2 = 1:5
ttt2Predicted = predict(tttMod, ttt2)

等效的Python代码段是从here翻译而来，在下面给出，

import numpy as np
import sys

def poly(x, degree):
    '''x: ndarray
    degree: scalar'''
    ## check feasibility
    if len(set(x)) < degree:
        print('Insufficient unique data points for specified degree!')
        sys.exit()
    centre = np.mean(x)
    x = x - centre
    Q, R = np.linalg.qr(np.polynomial.polynomial.polyvander(x, degree))
    h, tau = np.linalg.qr(np.polynomial.polynomial.polyvander(x, degree), 'raw')
    ## rescaling of Q by h.T
    X = (Q * np.diag(h.T))[:, 1:]
    X2 = X ** 2
    norm2 = X2.sum(0)
    alpha = np.dot(X2.T, x) / norm2
    beta = norm2 / (np.append([len(x)], norm2[:-1]))
    scale = np.sqrt(norm2)
    X = X * (1 / scale)
    return X, centre, scale, alpha[:-1], beta[:-1]

def predict_poly(X, centre, scale, alpha, beta, x):
    '''
        X, centre, scale, alpha, beta: returns from poly function
        x: new data to be predicted
    '''
    degree = X.shape[1]
    x = x - centre
    X = np.array(x.tolist() * degree).reshape((len(x), degree), order='F')
    if degree > 1:
        ## second polynomial
        X[:, 1] = (x - alpha[0]) * X[:, 0] - beta[0]
        ## further polynomials obtained from recursion
        i = 2
        while i < degree:
            X[:, i] = (x - alpha[i-1]) * X[:, i-1] - beta[i-1] * X[:, i-2]
            i += 1
        return X * (1 / scale)
    else:
        return None

#### test 1
x = np.array([1.160, 1.143, 1.126, 1.109, 1.079, 1.053, 1.040, 1.027, 1.015, 1.004, 0.994, 0.985, 0.977])
X, centre, scale, alpha, beta = poly(x, 5)

x = np.array([1.141096, 1.025588, 1.045099, 1.081832, 1.143202])
predict_poly(X, centre, scale, alpha, beta, x)

#### test 2
x = np.arange(1, 11, 1)
X, centre, scale, alpha, beta = poly(x, 3)

x = np.arange(1, 6, 1)
predict_poly(X, centre, scale, alpha, beta, x)

#### result of test 2
array([[-0.49543369,  0.52223297, -0.45342519],
   [-0.38533732,  0.17407766,  0.15114173],
   [-0.27524094, -0.08703883,  0.37785433],
   [-0.16514456, -0.26111648,  0.33467098],
   [-0.05504819, -0.34815531,  0.12955006]])

如何解释R中多边形模型的预测并编写等效的Python代码？

更新

0 个答案: