scipy UnivariateSpline 因多值 X 而失败

Question

scipy UnivariateSpline 不允许多值 X。我读到这已更改，但似乎对我不起作用。我用的是最新版本，刚刚用pip尝试下载，说我有最新版本。

我曾尝试将 s（平滑）从 0 和 None（定义为 X 必须严格增加）更改，但这并不能解决问题。

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.

任何帮助或建议将是最受欢迎的！

Answer 1

明白了！\n经过几个小时的查看代码后，我发现了一个链接https://github.com/kawache/Python-B-spline-examples，它给了我一个线索，现在我可以生成与原始 FORTRAN 类似的结果代码（使用 MG Cox 原始代码1编写，随后修改为 Cox & de Boor 2）。

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也许我应该将其写为“对连续不规则时间点的多个观测数据进行样条拟合”。

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感谢所有对此的帮助。

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这是我的代码：-

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import matplotlib.pyplot as plt\nfrom scipy import interpolate\nimport numpy as np\n\nx=[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]\n\ny=[-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]\n\nplt.plot(x, y, \'ro\', ms=5)\n\ntck,u = interpolate.splprep([x,y],k=3,s=32)\n\nu=np.linspace(0,1,num=50,endpoint=True)\nout = interpolate.splev(u,tck)\n\nplt.plot(x, y, \'ro\', out[0], out[1], \'b\' )\n\nplt.show()\n

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和结果（后面是 Cox 1和 de Boor 2的原始样条曲线图）。

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1：MG Cox，\xe2\x80\x9cb 样条的数值评估\xe2\x80\x9d，J. Inst。数学应用，10，第 134-149 页，1972 年。

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2 : C. de Boor, \xe2\x80\x9c关于使用 b 样条计算\xe2\x80\x9d, J. 逼近理论, 6, p.50-62, 1972。\n\n

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