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为什么插值单变量样条线返回nan值

  •  2
  • jakebrinkmann  · 技术社区  · 9 年前

    我有一些数据, y x ,我希望以更精细的分辨率进行插值 xx 使用三次样条曲线。

    这是我的数据集:

    import numpy as np
    print np.version.version
    import scipy
    print scipy.version.version   
    

    1.9.2
    0.15.1

    x = np.array([0.5372973, 0.5382103, 0.5392305, 0.5402197, 0.5412042, 0.54221, 0.543209,
                  0.5442277, 0.5442277, 0.5452125, 0.546217, 0.5472153, 0.5482086,
                  0.5492241, 0.5502117, 0.5512249, 0.5522136, 0.5532056, 0.5532056,
                  0.5542281, 0.5552039, 0.5562125, 0.5567836])
    y = np.array([0.01, 0.03108, 0.08981, 0.18362, 0.32167, 0.50941, 0.72415, 0.90698,
                  0.9071, 0.97955, 0.99802, 1., 0.97863, 0.9323, 0.85344, 0.72936,
                  0.56413, 0.36997, 0.36957, 0.17623, 0.05922, 0.0163, 0.01, ])
    
    xx = np.array([0.5372981, 0.5374106, 0.5375231, 0.5376356, 0.5377481, 0.5378606,
                   0.5379731, 0.5380856, 0.5381981, 0.5383106, 0.5384231, 0.5385356,
                   0.5386481, 0.5387606, 0.5388731, 0.5389856, 0.5390981, 0.5392106,
                   0.5393231, 0.5394356, 0.5395481, 0.5396606, 0.5397731, 0.5398856,
                   0.5399981, 0.5401106, 0.5402231, 0.5403356, 0.5404481, 0.5405606,
                   0.5406731, 0.5407856, 0.5408981, 0.5410106, 0.5411231, 0.5412356,
                   0.5413481, 0.5414606, 0.5415731, 0.5416856, 0.5417981, 0.5419106,
                   0.5420231, 0.5421356, 0.5422481, 0.5423606, 0.5424731, 0.5425856,
                   0.5426981, 0.5428106, 0.5429231, 0.5430356, 0.5431481, 0.5432606,
                   0.5433731, 0.5434856, 0.5435981, 0.5437106, 0.5438231, 0.5439356,
                   0.5440481, 0.5441606, 0.5442731, 0.5443856, 0.5444981, 0.5446106,
                   0.5447231, 0.5448356, 0.5449481, 0.5450606, 0.5451731, 0.5452856,
                   0.5453981, 0.5455106, 0.5456231, 0.5457356, 0.5458481, 0.5459606,
                   0.5460731, 0.5461856, 0.5462981, 0.5464106, 0.5465231, 0.5466356,
                   0.5467481, 0.5468606, 0.5469731, 0.5470856, 0.5471981, 0.5473106,
                   0.5474231, 0.5475356, 0.5476481, 0.5477606, 0.5478731, 0.5479856,
                   0.5480981, 0.5482106, 0.5483231, 0.5484356, 0.5485481, 0.5486606,
                   0.5487731, 0.5488856, 0.5489981, 0.5491106, 0.5492231, 0.5493356,
                   0.5494481, 0.5495606, 0.5496731, 0.5497856, 0.5498981, 0.5500106,
                   0.5501231, 0.5502356, 0.5503481, 0.5504606, 0.5505731, 0.5506856,
                   0.5507981, 0.5509106, 0.5510231, 0.5511356, 0.5512481, 0.5513606,
                   0.5514731, 0.5515856, 0.5516981, 0.5518106, 0.5519231, 0.5520356,
                   0.5521481, 0.5522606, 0.5523731, 0.5524856, 0.5525981, 0.5527106,
                   0.5528231, 0.5529356, 0.5530481, 0.5531606, 0.5532731, 0.5533856,
                   0.5534981, 0.5536106, 0.5537231, 0.5538356, 0.5539481, 0.5540606,
                   0.5541731, 0.5542856, 0.5543981, 0.5545106, 0.5546231, 0.5547356,
                   0.5548481, 0.5549606, 0.5550731, 0.5551856, 0.5552981, 0.5554106,
                   0.5555231, 0.5556356, 0.5557481, 0.5558606, 0.5559731, 0.5560856,
                   0.5561981, 0.5563106, 0.5564231, 0.5565356, 0.5566481, 0.5567606])
    

    我在试着使用scipy InterpolatedUnivariateSpline 方法,用三阶样条插值 k=3 ,并外推为零 ext='zeros' :

    import scipy.interpolate as interp
    yspline = interp.InterpolatedUnivariateSpline(x,y, k=3, ext='zeros')
    yvals = yspline(xx)
    
    import matplotlib.pyplot as plt
    fig = plt.figure()
    ax = fig.add_subplot(111)
    ax.plot(x, y, 'ko', label='Values')
    ax.plot(xx, yvals, 'b-.', lw=2, label='Spline')
    plt.xlim([min(x), max(x)])
    

    Y vs X

    然而,正如您在这张图片中看到的,我的样条线返回 NaN 值:( 有什么原因吗?我很确定我的x值都在增加,所以我很困惑为什么会发生这种情况。我有许多其他数据集正在使用此方法拟合,但它仅在这组特定数据上失败。

    非常感谢您的帮助。 感谢您的阅读。

    编辑!

    解决办法是我有两个 x(x) 值,不同 y 价值观

    2 回复  |  直到 9 年前
        1
  •  2
  •   rth    9 年前

    对于此插值,您应该使用 scipy.interpolate.interp1d 带着争论 kind='cubic' (请参见 related SO question )

    我还没有找到一个用例 InterpolatedUnivariateSpline 可以在实践中使用(或者我只是不明白它的目的)。我得到了你的代码, spline interpolation

    因此,插值有效,但显示出极强的振荡,使其无法使用,这通常是我过去使用这种插值方法得到的结果。使用低阶样条(例如。 k=1 )这样做效果更好,但这样就失去了三次插值的优势。

        2
  •  0
  •   Georgy rassa45    6 年前

    我也遇到了 InterpolatedUnivariateSpline 返回NaN值。但在我的案例中,原因并不在于 x 数组,但因为中的值 x(x) 减少 什么时候 docs 声明值“必须 增加的 ".
    因此,在这种情况下 x(x) y 必须反向提供: x[::-1] y[::-1] .