试图在python中线性插值

时间:2016-09-12 19:30:46

标签: python scipy interpolation

我有3个数组:a,b,c都是长度为15.

a=[950, 850, 750, 675, 600, 525, 460, 400, 350, 300, 250, 225, 200, 175, 150] 

b = [16, 12, 9, -35, -40, -40, -40, -45, -50, -55, -60, -65, -70, -75, -80]

c=[32.0, 22.2, 12.399999999999999, 2.599999999999998, -7.200000000000003, -17.0, -26.800000000000004, -36.60000000000001, -46.400000000000006, -56.2, -66.0, -75.80000000000001, -85.60000000000001, -95.4, -105.20000000000002] 

我试图在索引处找到a的值,其中b = c。 Ť

问题在于没有b = c的地方,所以我需要在数组中的值之间进行线性插值,以找到a b = c的值。那有意义吗?

我正在考虑使用 scipy.interpolate 进行插值。

我很难解决如何解决这个问题。对此的任何想法都会很棒!

3 个答案:

答案 0 :(得分:3)

以下是来自another answer of mine的函数的简单变体:

from __future__ import division

import numpy as np


def find_roots(t, y):
    """
    Given the input signal `y` with samples at times `t`,
    find the times where `y` is 0.

    `t` and `y` must be 1-D numpy arrays.

    Linear interpolation is used to estimate the time `t` between
    samples at which sign changes in `y` occur.
    """
    # Find where y crosses 0.
    transition_indices = np.where(np.sign(y[1:]) != np.sign(y[:-1]))[0]

    # Linearly interpolate the time values where the transition occurs.
    t0 = t[transition_indices]
    t1 = t[transition_indices + 1]
    y0 = y[transition_indices]
    y1 = y[transition_indices + 1]
    slope = (y1 - y0) / (t1 - t0)
    transition_times = t0 - y0/slope

    return transition_times

该功能可与t = ay = b - c一起使用。例如,这是您的数据,输入为numpy数组:

In [354]: a = np.array([950, 850, 750, 675, 600, 525, 460, 400, 350, 300, 250, 225, 200, 175, 150])

In [355]: b = np.array([16, 12, 9, -35, -40, -40, -40, -45, -50, -55, -60, -65, -70, -75, -80])

In [356]: c = np.array([32.0, 22.2, 12.399999999999999, 2.599999999999998, -7.200000000000003, -17.0, -26.800000000000004, -3
     ...: 6.60000000000001, -46.400000000000006, -56.2, -66.0, -75.80000000000001, -85.60000000000001, -95.4, -105.2000000000
     ...: 0002])

“b = c”的地方是“b - c = 0”的地方,所以我们将b - c传递给y

In [357]: find_roots(a, b - c)
Out[357]: array([ 312.5])

因此a的线性插值为312.5。

使用以下matplotlib命令:

In [391]: plot(a, b, label="b")
Out[391]: [<matplotlib.lines.Line2D at 0x11eac8780>]

In [392]: plot(a, c, label="c")
Out[392]: [<matplotlib.lines.Line2D at 0x11f23aef0>]

In [393]: roots = find_roots(a, b - c)

In [394]: [axvline(root, color='k', alpha=0.2) for root in roots]
Out[394]: [<matplotlib.lines.Line2D at 0x11f258208>]

In [395]: grid()

In [396]: legend(loc="best")
Out[396]: <matplotlib.legend.Legend at 0x11f260ba8>

In [397]: xlabel("a")
Out[397]: <matplotlib.text.Text at 0x11e71c470>

我得到了情节

plot

答案 1 :(得分:1)

这不一定是您问题的解决方案,因为您的数据似乎不是线性的,但它可能会给您一些想法。如果你假设你的线a,b和c是线性的,那么下面的想法是有效的:

对线a,b和c进行线性回归以得到它们各自的斜率(m_a,m_b,m_c)和y-截距(b_a,b_b,b_c)。然后求解x的等式'y_b = y_c',找到y = m_a * x + b_a得到你的结果。

由于线性回归近似求解y = m * x + b,因此可以通过手动求解方程y_b = y_c:x =(b_b-b_c)/(m_c-m_b)。

使用python,你得到:

>> m_a, b_a, r_a, p_a, err_a = stats.linregress(range(15), a)
>> m_b, b_b, r_b, p_b, err_b = stats.linregress(range(15), b)
>> m_c, b_c, r_c, p_c, err_c = stats.linregress(range(15), c)
>> x = (b_b-b_c) / (m_c-m_b)
>> m_a * x + b_a
379.55151515151516

由于您的数据不是线性的,您可能需要逐个浏览矢量并搜索重叠的y间隔。然后,您可以应用上述方法,但仅使用两个区间的端点来构建线性回归的b和c输入。在这种情况下,你应该得到一个精确的结果,因为最小二乘法将只用两个点完美插值(尽管有更有效的方法来做到这一点,因为交叉点可以在这个简单的情况下完全解决,其中有两个直线线)。

干杯。

答案 2 :(得分:0)

使用的另一个简单解决方案:

  • 每个载体的一个线性回归量(使用scikit-learn做scipy-docs对我来说很容易;很容易切换到基于numpy / scipy的线性回归)
  • 使用scipy.optimize.minimize
  • 的通用最小化

代码

a=[950, 850, 750, 675, 600, 525, 460, 400, 350, 300, 250, 225, 200, 175, 150]
b = [16, 12, 9, -35, -40, -40, -40, -45, -50, -55, -60, -65, -70, -75, -80]
c=[32.0, 22.2, 12.399999999999999, 2.599999999999998, -7.200000000000003, -17.0, -26.800000000000004, -36.60000000000001, -46.400000000000006, -56.2, -66.0, -75.80000000000001, -85.60000000000001, -95.4, -105.20000000000002]

from sklearn.linear_model import LinearRegression
from scipy.optimize import minimize
import numpy as np

reg_a = LinearRegression().fit(np.arange(len(a)).reshape(-1,1), a)
reg_b = LinearRegression().fit(np.arange(len(b)).reshape(-1,1), b)
reg_c = LinearRegression().fit(np.arange(len(c)).reshape(-1,1), c)

funA = lambda x: reg_a.predict(x.reshape(-1,1))
funB = lambda x: reg_b.predict(x.reshape(-1,1))
funC = lambda x: reg_c.predict(x.reshape(-1,1))

opt_crossing = lambda x: (funB(x) - funC(x))**2
x0 = 1
res = minimize(opt_crossing, x0, method='SLSQP', tol=1e-6)
print(res)
print('Solution: ', funA(res.x))

import matplotlib.pyplot as plt

x = np.linspace(0, 15, 100)
a_ = reg_a.predict(x.reshape(-1,1))
b_ = reg_b.predict(x.reshape(-1,1))
c_ = reg_c.predict(x.reshape(-1,1))

plt.plot(x, a_, color='blue')
plt.plot(x, b_, color='green')
plt.plot(x, c_, color='cyan')
plt.scatter(np.arange(15), a, color='blue')
plt.scatter(np.arange(15), b, color='green')
plt.scatter(np.arange(15), c, color='cyan')

plt.axvline(res.x, color='red', linestyle='solid')
plt.axhline(funA(res.x), color='red', linestyle='solid')

plt.show()

输出

fun: array([  7.17320622e-15])
jac: array([ -3.99479864e-07,   0.00000000e+00])
message: 'Optimization terminated successfully.'
nfev: 8
nit: 2
njev: 2
status: 0
success: True
  x: array([ 8.37754008])
Solution:  [ 379.55151658]

剧情

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