scipy UnivariateSpline因多值X而失败

Question

scipy单变量pline不允许多值X。我读到这已经被改变了，但似乎不起作用。我使用的是最新版本，刚刚用pip下载，上面说我有最新版本。在

我尝试过将s（平滑）从0改为None（定义为X必须严格递增），但这并不能解决问题。在

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)

我得到这个错误消息：- 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.

欢迎任何帮助或建议！在

Answer 1

明白了！ 在看了好几个小时的代码之后，我发现了一个链接https://github.com/kawache/Python-B-spline-examples，它给了我一个线索，现在我可以生成与我最初的FORTRAN代码类似的结果（它是用M.G.Cox原始代码1）编写的，后来又改为Cox&deboor2）。在

也许我应该把它写成“对连续不规则时间点的多个观测数据的样条拟合”。在

谢谢你的帮助。在

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()

{3{decox}的原始结果。在

考克斯，“b样条函数的数值计算”，数学应用学报，第10期，第134-1491972页。在

[3}:C.de Boor，“关于用b样条函数进行计算”，J.近似理论，第6期，第50-62页，1972年。

Answer 2

这对你的数据集无效。它需要一个不断增加的x变量来进行插值。这是实现的要求。在

• 您可以编辑数据集以删除重复的x值，它将起作用。或者，您可以尝试下面的方法，其中涉及更多。在

@xdze2给出的答案通过另一种方法，使用单独的坐标平滑曲线。在

fit a spline separately on each coordinates of the given curve