scipy UnivariateSpline因多值X失败

Question

scipy UnivariateSpline不允许多值X。我读到它已经更改，但似乎对我无效。我正在使用最新版本，刚刚使用pip尝试下载，这表明我拥有最新版本。

我尝试将s（平滑度）从0更改为None（定义为X必须严格增加），但是这不能解决问题。

import matplotlib.pyplot as plt
from scipy.interpolate import UnivariateSpline

x=[152,152,152,152,152,159,159,159,159,159,166,166,166,166,166,174,174,174,174,174,181,181,181,181,181,188,188,188,188,188,194,194,194,194,194,202,202,202,202,202,208,208,208,208,208,215,215,215,215,215,222,222,222,222,222,229,229,229,229,229,236,236,236,236,236,243,243,243,243,243,250,250,250,250,250,258,258,258,258]

y=[-1.31639523,-1.90045889,-1.81769285,-1.25702203,-1.31975784,-0.76206863,-0.74170737,-0.66029284,-0.58124809,-0.49593701,-0.19309943,0.02254396,-0.04614866,0.06709774,0.10436002,0.577175,0.56809403,0.89547559,0.60922195,0.76220672,1.0461253,1.1304339,1.56360338,1.34189828,1.41658105,1.98677786,2.40487089,2.20431052,1.91072699,2.49328809,2.670556,2.85024397,3.24333426,2.44841554,3.14604703,3.39128172,3.78063788,3.21446612,3.07158159,3.79503965,3.40717945,4.02417242,3.70708767,4.00729682,4.25504517,4.28874564,3.9356614,4.30337567,4.02388633,4.65376986,4.33884509,4.68839858,4.10508666,4.26236997,4.53098529,5.03443645,4.07940011,4.3033351,4.43476139,4.80221614,4.49558967,4.5052504,4.40289487,5.15433152,5.1330299,4.30299696,4.47974301,5.34886789,4.60896298,5.35997675,4.40204983,5.50162549,4.3056854,4.87120463,5.36265274,4.33578634,5.06347439,4.46811258,5.30920785]

s = 0.1 # set smoothing to non-zero
spl = UnivariateSpline(x, y, s=s)

我收到此错误消息：- spl = UnivariateSpline(x, y, s=s) File "C:\Python37\lib\site-packages\scipy\interpolate\fitpack2.py", line 177, in __init__ raise ValueError('x must be strictly increasing') ValueError: x must be strictly increasing.

任何帮助或建议都将受到欢迎！

Answer 1

这不适用于您的数据集。为了进行插值，需要连续增加x变量。这是实施的要求。

• 您可以编辑数据集以删除重复的x值，它将起作用。或者，您可以尝试以下涉及更多方法的方法。

@ xdze2给出的答案通过另一种方法使用单个坐标平滑曲线。

fit a spline separately on each coordinates of the given curve

Answer 2

知道了！ 经过数小时的研究，我发现了一个链接https://github.com/kawache/Python-B-spline-examples，该链接为我提供了一个线索，现在我可以产生与原始FORTRAN代码（使用MG Cox原始代码1编写的代码）相似的结果，然后修改为Cox＆de Boor 2）。

也许我应该将其写成“对连续不规则时间点的多个观测数据进行样条拟合”。

感谢所有帮助。

这是我的代码：-

import matplotlib.pyplot as plt
from scipy import interpolate
import numpy as np

x=[152.0,152,152,152,152,159,159,159,159,159,166,166,166,166,166,174,174,174,174,174,181,181,181,181,181,188,188,188,188,188,194,194,194,194,194,202,202,202,202,202,208,208,208,208,208,215,215,215,215,215,222,222,222,222,222,229,229,229,229,229,236,236,236,236,236,243,243,243,243,243,250,250,250,250,250,258,258,258,258]

y=[-1.31639523,-1.90045889,-1.81769285,-1.25702203,-1.31975784,-0.76206863,-0.74170737,-0.66029284,-0.58124809,-0.49593701,-0.19309943,0.02254396,-0.04614866,0.06709774,0.10436002,0.577175,0.56809403,0.89547559,0.60922195,0.76220672,1.0461253,1.1304339,1.56360338,1.34189828,1.41658105,1.98677786,2.40487089,2.20431052,1.91072699,2.49328809,2.670556,2.85024397,3.24333426,2.44841554,3.14604703,3.39128172,3.78063788,3.21446612,3.07158159,3.79503965,3.40717945,4.02417242,3.70708767,4.00729682,4.25504517,4.28874564,3.9356614,4.30337567,4.02388633,4.65376986,4.33884509,4.68839858,4.10508666,4.26236997,4.53098529,5.03443645,4.07940011,4.3033351,4.43476139,4.80221614,4.49558967,4.5052504,4.40289487,5.15433152,5.1330299,4.30299696,4.47974301,5.34886789,4.60896298,5.35997675,4.40204983,5.50162549,4.3056854,4.87120463,5.36265274,4.33578634,5.06347439,4.46811258,5.30920785]

plt.plot(x, y, 'ro', ms=5)

tck,u = interpolate.splprep([x,y],k=3,s=32)

u=np.linspace(0,1,num=50,endpoint=True)
out = interpolate.splev(u,tck)

plt.plot(x, y, 'ro', out[0], out[1], 'b' )

plt.show()

和结果（接着是Cox 1和de Boor 2的原始样条图）。

1：M。G. Cox，“ b样条的数值评估”，J。Inst。 Maths Applics，第10页，第134-149页，1972年。

2：C。de Boor，“关于用b样条进行计算”，J。逼近理论，第6页，第50-62页，1972年。