numpy数组的第一个最小值或鞍点(微积分导数)是什么?

这就是我如何检测局部最大值/最小值、拐点和鞍座的方法。

首先定义以下函数

import numpy as np


def n_derivative(arr, degree=1):
    """Compute the n-th derivative."""
    result = arr.copy()
    for i in range(degree):
        result = np.gradient(result)
    return result


def sign_change(arr):
    """Detect sign changes."""
    sign = np.sign(arr)
    result = ((np.roll(sign, 1) - sign) != 0).astype(bool)
    result[0] = False
    return result


def zeroes(arr, threshold=1e-8):
    """Find zeroes of an array."""
    return sign_change(arr) | (abs(arr) < threshold)

我们现在可以利用 derivative test

临界点的一阶导数等于零。

def critical_points(arr):
    return zeroes(n_derivative(arr, 1))

如果临界点具有非零的二阶导数,则该点为最大值或最小值:

def maxima_minima(arr):
    return zeroes(n_derivative(arr, 1)) & ~zeroes(n_derivative(arr, 2))


def maxima(arr):
    return zeroes(n_derivative(arr, 1)) & (n_derivative(arr, 2) < 0)


def minima(arr):
    return zeroes(n_derivative(arr, 1)) & (n_derivative(arr, 2) > 0)

如果二阶导数等于零,但三阶导数非零,则该点为拐点:

def inflections(arr):
    return zeroes(n_derivative(arr, 2)) & ~zeroes(n_derivative(arr, 3))

如果临界点的二阶导数等于零,但三阶导数非零,则这是一个鞍:

def inflections(arr):
    return zeroes(n_derivative(arr, 1)) & zeroes(n_derivative(arr, 2)) & ~zeroes(n_derivative(arr, 3))

注意,该方法在数值上不稳定,一方面在某些任意阈值定义上检测到零,另一方面不同的采样可能导致函数/数组不可微。 因此,根据这个定义,您期望的实际上不是鞍点。

为了更好地逼近连续函数,可以在很大程度上过采样(根据 K 在代码)函数中,例如:

import scipy as sp
import scipy.interpolate

data = [
    1.04814804, 0.90445908, 0.62026396, 0.60566623, 0.32295758, 0.26658469, 0.19059289,
    0.10281547, 0.08582772, 0.05091265, 0.03391474, 0.03844931, 0.03315003, 0.02838656,
    0.03420759, 0.03567401, 0.038203, 0.03530763, 0.04394316, 0.03876966, 0.04156067,
    0.03937291, 0.03966426, 0.04438747, 0.03690863, 0.0363976, 0.03171374, 0.03644719,
    0.02989291, 0.03166156, 0.0323875, 0.03406287, 0.03691943, 0.02829374, 0.0368121,
    0.02971704, 0.03427005, 0.02873735, 0.02843848, 0.02101889, 0.02114978, 0.02128403,
    0.0185619, 0.01749904, 0.01441699, 0.02118773, 0.02091855, 0.02431763, 0.02472427,
    0.03186318, 0.03205664, 0.03135686, 0.02838413, 0.03206674, 0.02638371, 0.02048122,
    0.01502128, 0.0162665, 0.01331485, 0.01569286, 0.00901017, 0.01343558, 0.00908635,
    0.00990869, 0.01041151, 0.01063606, 0.00822482, 0.01312368, 0.0115005, 0.00620334,
    0.0084177, 0.01058152, 0.01198732, 0.01451455, 0.01605602, 0.01823713, 0.01685975,
    0.03161889, 0.0216687, 0.03052391, 0.02220871, 0.02420951, 0.01651778, 0.02066987,
    0.01999613, 0.02532265, 0.02589186, 0.02748692, 0.02191687, 0.02612152, 0.02309497,
    0.02744753, 0.02619196, 0.02281516, 0.0254296, 0.02732746, 0.02567608, 0.0199178,
    0.01831929, 0.01776025]
samples = np.arange(len(data))
f = sp.interpolate.interp1d(samples, data, 'cubic')

K = 10
N = len(data) * K

x = np.linspace(min(samples), max(samples), N)
y = f(x)

然后,所有这些定义都可以通过以下方式进行视觉测试:

import matplotlib.pyplot as plt

plt.figure()
plt.plot(samples, data, label='data')
plt.plot(x, y, label='f')
plt.plot(x, n_derivative(y, 1), label='d1f')
plt.plot(x, n_derivative(y, 2), label='d2f')
plt.plot(x, n_derivative(y, 3), label='d3f')
plt.legend()
for w in np.where(inflections(y))[0]:
    plt.axvline(x=x[w])
plt.show()

但即使在这种情况下,这一点也不是马鞍。

你可以使用 np.gradient 或 np.diff 要评估差异(第一个计算中心差异,第二个仅为x[1:]-x[:-1]),然后使用 np.sign 得到梯度符号和另一个 NP.DIFF 看看标志在哪里改变。然后过滤正符号变化(对应于最小值):

 np.where(np.diff(np.sign(np.gradient(x))) > 0)[0][0]+2   #add 2 as each time you call np.gradient or np.diff you are substracting 1 in size, the first [0] is to get the positions, the second [0] is to get the "first" element
>> 8
x[np.where(np.diff(np.sign(np.gradient(x))) > 0)[0][0]+2]
>> 0.03420759

我环顾了一下周围,并从给出的两个建议(到目前为止)中得出以下结论:

import scipy
from scipy import interpolate

x = np.arange(0, 100)
spl = scipy.interpolate.splrep(x,arr,k=3) # no smoothing, 3rd order spline
ddy = scipy.interpolate.splev(x,spl,der=2) # use those knots to get second derivative 
print(ddy)

asign = np.sign(ddy)
signchange = ((np.roll(asign, 1) - asign) != 0).astype(int)
print(signchange)

这给了我 second derivative ,然后我可以分析,例如,查看符号更改发生的位置:

[-0.894053   -0.14050616  0.61304067 -0.69407217  0.55458251 -0.16624336
 -0.0073225   0.12481963 -0.067218    0.03648846  0.02876712 -0.02236204
  0.00167794  0.01886512 -0.0136314   0.00953279 -0.01812436  0.03041855
 -0.03436446  0.02418512 -0.01458896  0.00429809  0.01227133 -0.02679232
  0.02168571 -0.0181437   0.02585209 -0.02876075  0.0214645  -0.00715966
  0.0009179   0.00918466 -0.03056938  0.04419937 -0.0433638   0.03557532
 -0.02904901  0.02010647 -0.0199739   0.0170648  -0.00298236 -0.00511529
  0.00630525 -0.01015011  0.02218007 -0.01945341  0.01339405 -0.01211326
  0.01710444 -0.01591092  0.00486652 -0.00891456  0.01715403 -0.01976949
  0.00573004 -0.00446743  0.01479495 -0.01448144  0.01794968 -0.02533936
  0.02904355 -0.02418628  0.01505374 -0.00499926  0.00302616 -0.00877499
  0.01625907 -0.01240068 -0.00578862  0.01351128 -0.00318733 -0.0010652
  0.0029     -0.0038062   0.0064102  -0.01799678  0.04422601 -0.0620881
  0.05587037 -0.04856099  0.03535114 -0.03094757  0.03028399 -0.01912546
  0.01726283 -0.01392421  0.00989012 -0.01948119  0.02504401 -0.02204667
  0.0197554  -0.01270022 -0.00260326  0.01038581 -0.00299247 -0.00271539
 -0.00744152  0.00784016  0.00103947 -0.00576122]

[0 0 1 1 1 1 0 1 1 1 0 1 1 0 1 1 1 1 1 1 1 1 0 1 1 1 1 1 1 1 1 0 1 1 1 1 1
 1 1 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 0 1 1
 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 0 1 1 0 0 1 0 1]