我有一些一维数据,并用样条曲线拟合。然后我想在其中找到拐点(忽略鞍点)。现在我通过在splev生成的很多值上使用scipy.signal.argrelmin(和argrelmax)来搜索它的第一个推导的极值。
import scipy.interpolate
import scipy.optimize
import scipy.signal
import numpy as np
import matplotlib.pyplot as plt
import operator
y = [-1, 5, 6, 4, 2, 5, 8, 5, 1]
x = np.arange(0, len(y))
tck = scipy.interpolate.splrep(x, y, s=0)
print 'roots', scipy.interpolate.sproot(tck)
# output:
# [0.11381478]
xnew = np.arange(0, len(y), 0.01)
ynew = scipy.interpolate.splev(xnew, tck, der=0)
ynew_deriv = scipy.interpolate.splev(xnew, tck, der=1)
min_idxs = scipy.signal.argrelmin(ynew_deriv)
max_idxs = scipy.signal.argrelmax(ynew_deriv)
mins = zip(xnew[min_idxs].tolist(), ynew_deriv[min_idxs].tolist())
maxs = zip(xnew[max_idxs].tolist(), ynew_deriv[max_idxs].tolist())
inflection_points = sorted(mins + maxs, key=operator.itemgetter(0))
print 'inflection_points', inflection_points
# output:
# [(3.13, -2.9822449358974357),
# (5.03, 4.3817785256410255)
# (7.13, -4.867132628205128)]
plt.legend(['data','Cubic Spline', '1st deriv'])
plt.plot(x, y, 'o',
xnew, ynew, '-',
xnew, ynew_deriv, '-')
plt.show()
但这感觉非常错误。我想有可能在没有产生这么多价值的情况下找到我要找的东西。像sproot这样的东西,但也许适用于二次推导?
答案 0 :(得分:4)
derivative of a B-spline is also a B-spline。因此,您可以首先为样本拟合样条曲线,然后使用导数公式构造导数样条曲线的系数,最后使用样条曲线根查找来获得导数样条曲线的根。这些是原始曲线的最大值/最小值。
以下是代码:https://gist.github.com/pv/5504366
系数的相关计算是:
t, c, k = scipys_spline_representation
# Compute the denominator in the differentiation formula.
dt = t[k+1:-1] - t[1:-k-1]
# Compute the new coefficients
d = (c[1:-1-k] - c[:-2-k]) * k / dt
# Adjust knots
t2 = t[1:-1]
# Pad coefficient array to same size as knots (FITPACK convention)
d = np.r_[d, [0]*k]
# Done, a new spline
new_spline_repr = t2, d, k-1