使用numpy.polynomial.legendre时，如何获取将输入转换为Legendre多项式的参数的函数？ [英] How do I get the function which transforms an input to be the argument of a Legendre polynomial when using numpy.polynomial.legendre?
问题描述
＃个导入包，我们以后需要
import matplotlib.pyplot as plt
import numpy as np
我在做什么
受此
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有什么问题
print（legendrefit_curve1）返回：
leg（[36823.85778316 96929.13731379 123557.55165344 112110.13559758
75345.0434688 32377.19460001 -182.38440131 -15562.47475287
-16142.22533582 -8379.06875482 -744.73929814]）
b $ c> 但是，我使用的是Jupyter笔记本，所以如果我只写 legendrefit_curve1 ，而没有
（什么区别 print（） 使Jupyter的输出与此问题有关。）
很显然， print（legendrefit_curve1）仅给出了每个勒让德多项式的系数（相同与 legendrefit_curve1.coef ）。
如何获取将x转换为每个Legendre多项式的参数的值？
即如何从ex获取值压缩： -1.0512820512820513 + 0.05128205128205128x ： -1.0512820512820513 和 0.05128205128205128 （不只是手动复制它们）？
什么不起作用
依靠此线程我运行：
用于目录中的attr（legendrefit_curve1）：
print（'###'+ attr +'###'）
print（getattr（legendrefit_curve1，attr））
此文本输出很长，但是我没有在其中找到 -1.05 （ ctrl-f ），因此表明未返回 -1.0512820512512820513 值，因此该方法不会
通过查看这些数字，我意识到我可以用数学来构造它们。
1 /（len（curve1）-1）* 2 ，即 1/39 * 2 返回： 0.05128205128205128
1 + 1 /（len（curve1）-1 ）* 2 即 1 + 1/39 * 2 返回：`1.05
其中是我们要寻找的数字。
我仍然不知道执行 legendrefit_curve1时如何显示它在Jupyter Notebook单元中，但这没什么意义。
我不知道为什么上面的公式有效，可能是在 math.stackexchange.com 上的问题。
# import packages we need later
import matplotlib.pyplot as plt
import numpy as np
What I am doing
Inspired by this question & answer, I am fitting a series of Legendre polynomials to a time series:
curve1 = \
np.asarray([942.153,353.081,53.088,125.110,140.851,188.170,70.536,-122.473,-369.061,-407.945,88.734,484.334,267.762,65.831,74.010,-55.781,-260.024,-466.830,-524.511,-76.833,-36.779,-117.366,218.578,175.662,185.653,299.285,215.276,546.048,1210.132,3087.326,7052.849,13867.824,27156.939,51379.664,91908.266,148874.563,215825.031,290073.219,369567.781,437031.688])
The time values:
tvals = \
np.asarray([1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19,20,21,22,23,24,25,26,27,28,29,30,31,32,33,34,35,36,37,38,39,40])
Using a numpy's function:
degree=10
legendrefit_curve1 = np.polynomial.legendre.Legendre.fit(tvals, curve1, deg=degree)
The fit seems pretty good:
# generate points of fitted curve
n=100
fitted_vals_curve1 = legendrefit_curve1.linspace(n=n)
# plot data and fitted curve
plt.scatter(tvals, curve1)
plt.plot(fitted_vals_curve1[0],fitted_vals_curve1[1],c='r')
What's the question
print(legendrefit_curve1) returns:
leg([ 36823.85778316 96929.13731379 123557.55165344 112110.13559758
75345.0434688 32377.19460001 -182.38440131 -15562.47475287
-16142.22533582 -8379.06875482 -744.73929814])
However, I am using a Jupyter notebook, so if I just write legendrefit_curve1, without print(), I get an output:
(What difference print() makes to Jupyter's output is related to this question.)
Clearly, print(legendrefit_curve1) only gave the coefficients of each Legendre polynomial (same with legendrefit_curve1.coef).
How do I get the values which transform x to be the argument of each Legendre polynomial?
ie how to obtain the values from the expression: -1.0512820512820513+0.05128205128205128x: -1.0512820512820513 and 0.05128205128205128 (without just copying them manually)?
What didn't work
Relying on this thread I run:
for attr in dir(legendrefit_curve1):
print('###'+attr+'###')
print(getattr(legendrefit_curve1, attr))
This had a long text output, but I did not find -1.05 in it (ctrl-f), so that suggest that the -1.0512820512820513 value did not get returned, so this method doesn't work.
By looking at those numbers I realized I can construct them from math.
1/(len(curve1)-1)*2, ie 1/39*2 returns: 0.05128205128205128
1+1/(len(curve1)-1)*2 ie 1+1/39*2 returns: `1.05
Which are the numbers we were looking for.
I still don't know how it is displayed when executing legendrefit_curve1 in a Jupyter Notebook cell, but that is less of the point.
I don't know why the formula above works, it'll probably be a question on math.stackexchange.com.
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