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我对 Python 很陌生。我有一个数据集，我正在尝试使用 numPy/sciPy 来预测/推断未来的数据点。有没有一种简单的方法可以提出适合我当前数据的数学函数（例如，正弦函数），然后我可以将新值传递给该函数以获得我的预测？

这是我所拥有的，但我不认为它在做我想要的：

import numpy as np
from scipy.optimize import curve_fit
import matplotlib.pyplot as plt

def main():

    y = [8.3, 8.3, 8.3, 8.3, 7.2, 7.8, 7.8, 8.3, 9.4, 10.6, 10.0, 10.6, 11.1, 12.8,
         12.8, 12.8, 11.7, 10.6, 10.6, 10.0, 10.0, 8.9, 8.9, 8.3, 7.2, 6.7, 6.7, 6.7,
         7.2, 8.3, 7.2, 10.6, 11.1, 11.7, 12.8, 13.3, 15.0, 15.6, 13.3, 15.0, 13.3,
         11.7, 11.1, 10.0, 10.6, 9.4, 8.9, 8.3, 8.9, 6.7, 6.7, 6.0, 6.1, 8.3, 8.3,
         10.6, 11.1, 11.1, 11.7, 12.2, 13.3, 14.4, 16.7, 14.4, 13.3, 12.2, 11.7,
         11.1, 10.0, 8.3, 7.8, 7.2, 8.0, 6.7, 7.2, 7.2, 7.8, 10.0, 12.2, 12.8,
         12.8, 13.9, 15.0, 16.7, 16.7, 16.7, 15.6, 13.9, 12.8, 12.2, 10.6, 9.0,
         8.9, 8.9, 8.9, 8.9, 8.9, 8.9, 8.9, 8.9, 10.0, 10.6, 11.1, 12.0, 11.7,
         11.1, 13.0, 13.3, 13.0, 11.1, 10.6, 10.6, 10.0, 10.0, 10.0, 9.4, 9.4,
         8.9, 8.3, 9.0, 8.9, 9.4, 9.0, 9.4, 10.6, 11.7, 11.1, 11.7, 12.8, 12.8,
         12.8, 13.0, 11.7, 10.6, 10.0, 10.0, 8.9, 9.4, 7.8, 7.8, 8.3, 7.8, 8.9,
         8.9, 8.9, 9.4, 10.0, 10.0, 10.6, 11.0, 11.1, 11.1, 12.2, 10.6, 10.0, 8.9,
         8.9, 9.0, 8.9, 8.3, 8.9, 8.9, 9.4, 9.4, 9.4, 8.9, 8.9, 8.9, 9.4, 10.0,
         11.1, 11.7, 11.7, 11.7, 11.7, 12.0, 11.7, 11.7, 12.0, 11.7, 11.0, 10.6,
         9.4, 10.0, 8.3, 8.0, 7.2, 5.6, 6.1, 5.6, 6.1, 6.7, 8.0, 10.0, 10.6, 11.1,
         13.3, 12.8, 12.8, 12.2, 11.1, 10.0, 10.0, 10.0, 10.0, 9.4, 8.3]    
    x = np.array(np.arange(len(y)))        

    fitting_parameters, covariance = curve_fit(fit, x, y)
    a = fitting_parameters[0]
    b = fitting_parameters[1]
    c = fitting_parameters[2]
    d = fitting_parameters[3]

    for x_predict in range(len(y) + 1, len(y) + 24):
        next_x = x_predict
        next_y = fit(next_x, a, b, c, d)

        print("next_x: " + str(next_x))
        print("next_y: " + str(next_y))
        y.append(next_y)

    plt.plot(y)
    plt.show()

def fit(x, a, b, c, d):
    return a*np.sin(b*x + c) + d

我尝试对我的数据进行曲线拟合和单变量样条化，但这仅适合我当前的数据并分别平滑我的点。我的观点是，这些工具只是“拟合”我的数据，但实际上并没有给我一个可以用来获得未来分数的函数。

我认为我可以使用离散傅立叶变换，因为我的数据是周期性的，并且看起来可以描述为正弦和余弦的总和。但是，一旦我从时域获得频域，我就被困在如何“外推”以预测时域中的未来周期和点：

import numpy as np
import matplotlib.pyplot as plt

mydata = [8.3, 8.3, 8.3, 8.3, 7.2, 7.8, 7.8, 8.3, 9.4, 10.6, 10.0, 10.6, 11.1, 12.8,
         12.8, 12.8, 11.7, 10.6, 10.6, 10.0, 10.0, 8.9, 8.9, 8.3, 7.2, 6.7, 6.7, 6.7,
         7.2, 8.3, 7.2, 10.6, 11.1, 11.7, 12.8, 13.3, 15.0, 15.6, 13.3, 15.0, 13.3,
         11.7, 11.1, 10.0, 10.6, 9.4, 8.9, 8.3, 8.9, 6.7, 6.7, 6.0, 6.1, 8.3, 8.3,
         10.6, 11.1, 11.1, 11.7, 12.2, 13.3, 14.4, 16.7, 14.4, 13.3, 12.2, 11.7,
         11.1, 10.0, 8.3, 7.8, 7.2, 8.0, 6.7, 7.2, 7.2, 7.8, 10.0, 12.2, 12.8,
         12.8, 13.9, 15.0, 16.7, 16.7, 16.7, 15.6, 13.9, 12.8, 12.2, 10.6, 9.0,
         8.9, 8.9, 8.9, 8.9, 8.9, 8.9, 8.9, 8.9, 10.0, 10.6, 11.1, 12.0, 11.7,
         11.1, 13.0, 13.3, 13.0, 11.1, 10.6, 10.6, 10.0, 10.0, 10.0, 9.4, 9.4,
         8.9, 8.3, 9.0, 8.9, 9.4, 9.0, 9.4, 10.6, 11.7, 11.1, 11.7, 12.8, 12.8,
         12.8, 13.0, 11.7, 10.6, 10.0, 10.0, 8.9, 9.4, 7.8, 7.8, 8.3, 7.8, 8.9,
         8.9, 8.9, 9.4, 10.0, 10.0, 10.6, 11.0, 11.1, 11.1, 12.2, 10.6, 10.0, 8.9,
         8.9, 9.0, 8.9, 8.3, 8.9, 8.9, 9.4, 9.4, 9.4, 8.9, 8.9, 8.9, 9.4, 10.0,
         11.1, 11.7, 11.7, 11.7, 11.7, 12.0, 11.7, 11.7, 12.0, 11.7, 11.0, 10.6,
         9.4, 10.0, 8.3, 8.0, 7.2, 5.6, 6.1, 5.6, 6.1, 6.7, 8.0, 10.0, 10.6, 11.1,
         13.3, 12.8, 12.8, 12.2, 11.1, 10.0, 10.0, 10.0, 10.0, 9.4, 8.3] 

sp = np.fft.rfft(mydata)
freq = np.fft.rfftfreq(len(mydata), d= 1.0)

plt.subplot(211)
plt.plot(mydata)
plt.subplot(212)
plt.plot(freq, sp, 'r')
plt.show()

我知道外推可能是危险且不可靠的，但出于本项目的目的，我只是想获得一个可以绘制的有效预测函数。

非常感谢您的帮助。

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1 回答 1

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这是一种通过将周期性数据表示为傅立叶级数来进行插值的方法。傅里叶级数中使用的系数是通过离散 FFT 获得的。

我不推荐这个——你可以在下面看到插值并不是直觉上认为非常好的——但由于我在评论中提到了它，我将继续并显示一些代码:)

import numpy as np
import matplotlib.pyplot as plt
import scipy.fftpack as fftpack

def fft_inverse(Yhat, x):
    """Based on http://stackoverflow.com/a/4452499/190597 (mtrw)"""
    Yhat = np.asarray(Yhat)
    x = np.asarray(x).reshape(-1, 1)
    N = len(Yhat)
    k = np.arange(N)
    total = Yhat * np.exp(1j * x * k * 2 * np.pi / N)
    return np.real(total.sum(axis=1))/N

mydata = [8.3, 8.3, 8.3, 8.3, 7.2, 7.8, 7.8, 8.3, 9.4, 10.6, 10.0, 10.6, 11.1, 12.8,
         12.8, 12.8, 11.7, 10.6, 10.6, 10.0, 10.0, 8.9, 8.9, 8.3, 7.2, 6.7, 6.7, 6.7,
         7.2, 8.3, 7.2, 10.6, 11.1, 11.7, 12.8, 13.3, 15.0, 15.6, 13.3, 15.0, 13.3,
         11.7, 11.1, 10.0, 10.6, 9.4, 8.9, 8.3, 8.9, 6.7, 6.7, 6.0, 6.1, 8.3, 8.3,
         10.6, 11.1, 11.1, 11.7, 12.2, 13.3, 14.4, 16.7, 14.4, 13.3, 12.2, 11.7,
         11.1, 10.0, 8.3, 7.8, 7.2, 8.0, 6.7, 7.2, 7.2, 7.8, 10.0, 12.2, 12.8,
         12.8, 13.9, 15.0, 16.7, 16.7, 16.7, 15.6, 13.9, 12.8, 12.2, 10.6, 9.0,
         8.9, 8.9, 8.9, 8.9, 8.9, 8.9, 8.9, 8.9, 10.0, 10.6, 11.1, 12.0, 11.7,
         11.1, 13.0, 13.3, 13.0, 11.1, 10.6, 10.6, 10.0, 10.0, 10.0, 9.4, 9.4,
         8.9, 8.3, 9.0, 8.9, 9.4, 9.0, 9.4, 10.6, 11.7, 11.1, 11.7, 12.8, 12.8,
         12.8, 13.0, 11.7, 10.6, 10.0, 10.0, 8.9, 9.4, 7.8, 7.8, 8.3, 7.8, 8.9,
         8.9, 8.9, 9.4, 10.0, 10.0, 10.6, 11.0, 11.1, 11.1, 12.2, 10.6, 10.0, 8.9,
         8.9, 9.0, 8.9, 8.3, 8.9, 8.9, 9.4, 9.4, 9.4, 8.9, 8.9, 8.9, 9.4, 10.0,
         11.1, 11.7, 11.7, 11.7, 11.7, 12.0, 11.7, 11.7, 12.0, 11.7, 11.0, 10.6,
         9.4, 10.0, 8.3, 8.0, 7.2, 5.6, 6.1, 5.6, 6.1, 6.7, 8.0, 10.0, 10.6, 11.1,
         13.3, 12.8, 12.8, 12.2, 11.1, 10.0, 10.0, 10.0, 10.0, 9.4, 8.3] 

Yhat = fftpack.fft(mydata)

fig, ax = plt.subplots(nrows=2, sharex=True)
xs = np.arange(len(mydata))
ax[0].plot(xs, mydata)

new_xs = np.linspace(xs.min(), xs.max(), len(mydata)*1.5)
new_ys = fft_inverse(Yhat, new_xs)
ax[1].plot(new_xs, new_ys)

plt.xlim(xs.min(), xs.max())
plt.show()

在此处输入图像描述

以下是如何使用scipy.optimize来查找参数以拟合模型函数，然后可以使用该函数在任意 x 坐标处进行插值。使用单个的拟合sin仍然非常糟糕，但我将发布代码只是为了展示如何使用scipy.optimize：

import numpy as np
import matplotlib.pyplot as plt
import scipy.optimize as optimize

mydata = np.array(
    [8.3, 8.3, 8.3, 8.3, 7.2, 7.8, 7.8, 8.3, 9.4, 10.6, 10.0, 10.6, 11.1, 12.8,
     12.8, 12.8, 11.7, 10.6, 10.6, 10.0, 10.0, 8.9, 8.9, 8.3, 7.2, 6.7, 6.7, 6.7,
     7.2, 8.3, 7.2, 10.6, 11.1, 11.7, 12.8, 13.3, 15.0, 15.6, 13.3, 15.0, 13.3,
     11.7, 11.1, 10.0, 10.6, 9.4, 8.9, 8.3, 8.9, 6.7, 6.7, 6.0, 6.1, 8.3, 8.3,
     10.6, 11.1, 11.1, 11.7, 12.2, 13.3, 14.4, 16.7, 14.4, 13.3, 12.2, 11.7,
     11.1, 10.0, 8.3, 7.8, 7.2, 8.0, 6.7, 7.2, 7.2, 7.8, 10.0, 12.2, 12.8,
     12.8, 13.9, 15.0, 16.7, 16.7, 16.7, 15.6, 13.9, 12.8, 12.2, 10.6, 9.0,
     8.9, 8.9, 8.9, 8.9, 8.9, 8.9, 8.9, 8.9, 10.0, 10.6, 11.1, 12.0, 11.7,
     11.1, 13.0, 13.3, 13.0, 11.1, 10.6, 10.6, 10.0, 10.0, 10.0, 9.4, 9.4,
     8.9, 8.3, 9.0, 8.9, 9.4, 9.0, 9.4, 10.6, 11.7, 11.1, 11.7, 12.8, 12.8,
     12.8, 13.0, 11.7, 10.6, 10.0, 10.0, 8.9, 9.4, 7.8, 7.8, 8.3, 7.8, 8.9,
     8.9, 8.9, 9.4, 10.0, 10.0, 10.6, 11.0, 11.1, 11.1, 12.2, 10.6, 10.0, 8.9,
     8.9, 9.0, 8.9, 8.3, 8.9, 8.9, 9.4, 9.4, 9.4, 8.9, 8.9, 8.9, 9.4, 10.0,
     11.1, 11.7, 11.7, 11.7, 11.7, 12.0, 11.7, 11.7, 12.0, 11.7, 11.0, 10.6,
     9.4, 10.0, 8.3, 8.0, 7.2, 5.6, 6.1, 5.6, 6.1, 6.7, 8.0, 10.0, 10.6, 11.1,
     13.3, 12.8, 12.8, 12.2, 11.1, 10.0, 10.0, 10.0, 10.0, 9.4, 8.3]) 


def fit(x, a, b, c, d):
    return a*np.sin(b*x + c) + d

xs = np.linspace(0, 2*np.pi, len(mydata))

guess = (mydata.ptp()/2, 10, 0, mydata.mean())
fitting_parameters, covariance = optimize.curve_fit(fit, xs, mydata, p0=guess)
a, b, c, d = fitting_parameters
print(a, b, c, d)

fig, ax = plt.subplots(nrows=2, sharex=True)
ax[0].plot(xs, mydata)

new_xs = np.linspace(xs.min(), xs.max(), len(mydata)*1.5)
new_ys = fit(new_xs, a, b, c, d)
ax[1].plot(new_xs, new_ys)

plt.xlim(xs.min(), xs.max())
plt.show()

在此处输入图像描述

您可以通过选择更好的模型函数（代替fit）来改进拟合。选择什么取决于您的问题领域的先验知识所引导的创造力和直觉。更好的方法不仅取决于拟合的优劣，还取决于您希望模型的简单或复杂程度，和/或应用于新数据集时的预测能力。

于 2013-12-29T10:22:07.797 回答