这个问题可能更适合天文堆栈交换网站，但我想我会在这里问。假设我有一些物体的测量值作为半径的函数。
这是一个使事情变得具体的示例：
# Object 1
In [92]: r1
Out[92]:
array([ 0.05 , 0.07040816, 0.09081633, 0.11122449, 0.13163265,
0.15204082, 0.17244898, 0.19285714, 0.21326531, 0.23367347,
0.25408163, 0.2744898 , 0.29489796, 0.31530612, 0.33571429,
0.35612245, 0.37653061, 0.39693878, 0.41734694, 0.4377551 ,
0.45816327, 0.47857143, 0.49897959, 0.51938776, 0.53979592,
0.56020408, 0.58061224, 0.60102041, 0.62142857, 0.64183673,
0.6622449 , 0.68265306, 0.70306122, 0.72346939, 0.74387755,
0.76428571, 0.78469388, 0.80510204, 0.8255102 , 0.84591837,
0.86632653, 0.88673469, 0.90714286, 0.92755102, 0.94795918,
0.96836735, 0.98877551, 1.00918367, 1.02959184, 1.05 ])
# Object 2
In [93]: r2
Out[93]:
array([ 0.04 , 0.06081633, 0.08163265, 0.10244898, 0.12326531,
0.14408163, 0.16489796, 0.18571429, 0.20653061, 0.22734694,
0.24816327, 0.26897959, 0.28979592, 0.31061224, 0.33142857,
0.3522449 , 0.37306122, 0.39387755, 0.41469388, 0.4355102 ,
0.45632653, 0.47714286, 0.49795918, 0.51877551, 0.53959184,
0.56040816, 0.58122449, 0.60204082, 0.62285714, 0.64367347,
0.6644898 , 0.68530612, 0.70612245, 0.72693878, 0.7477551 ,
0.76857143, 0.78938776, 0.81020408, 0.83102041, 0.85183673,
0.87265306, 0.89346939, 0.91428571, 0.93510204, 0.95591837,
0.97673469, 0.99755102, 1.01836735, 1.03918367, 1.06 ])
# Object 3
In [94]: r3
Out[94]:
array([ 0.045 , 0.06540816, 0.08581633, 0.10622449, 0.12663265,
0.14704082, 0.16744898, 0.18785714, 0.20826531, 0.22867347,
0.24908163, 0.2694898 , 0.28989796, 0.31030612, 0.33071429,
0.35112245, 0.37153061, 0.39193878, 0.41234694, 0.4327551 ,
0.45316327, 0.47357143, 0.49397959, 0.51438776, 0.53479592,
0.55520408, 0.57561224, 0.59602041, 0.61642857, 0.63683673,
0.6572449 , 0.67765306, 0.69806122, 0.71846939, 0.73887755,
0.75928571, 0.77969388, 0.80010204, 0.8205102 , 0.84091837,
0.86132653, 0.88173469, 0.90214286, 0.92255102, 0.94295918,
0.96336735, 0.98377551, 1.00418367, 1.02459184, 1.045 ])
其中 r1、r2 和 r3 是我进行测量的径向列表（分别为对象 1、2 和 3）。测量结果如下所示：
In [95]: out1
Out[95]:
array([ 0.39919261, 0.40078876, 0.40335025, 0.40702841, 0.41194375,
0.41818517, 0.42580846, 0.43483499, 0.44525116, 0.45700891,
0.4700276 , 0.48419736, 0.49938363, 0.51543283, 0.53217868,
0.54944865, 0.56707028, 0.58487678, 0.60271173, 0.62043267,
0.63791355, 0.65504607, 0.67174007, 0.6879231 , 0.70353938,
0.71854842, 0.73292329, 0.74664884, 0.75971994, 0.77213972,
0.783918 , 0.79506992, 0.80561461, 0.81557416, 0.82497273,
0.83383575, 0.84218935, 0.85005984, 0.85747334, 0.86445547,
0.87103113, 0.87722433, 0.88305811, 0.88855443, 0.89373415,
0.898617 , 0.90322162, 0.90756555, 0.91166525, 0.91553618])
In [96]: out2
Out[96]:
array([ 0.31626381, 0.31638513, 0.31667084, 0.31721226, 0.31811294,
0.31948646, 0.32145416, 0.32414259, 0.32768059, 0.33219587,
0.33781106, 0.34463926, 0.35277915, 0.36230989, 0.37328607,
0.38573305, 0.39964318, 0.41497335, 0.43164433, 0.449542 ,
0.46852083, 0.48840914, 0.50901598, 0.53013886, 0.55157189,
0.57311338, 0.59457264, 0.61577537, 0.63656761, 0.65681803,
0.6764189 , 0.69528578, 0.71335626, 0.73058811, 0.74695691,
0.76245356, 0.77708177, 0.79085565, 0.80379747, 0.81593569,
0.82730323, 0.83793606, 0.84787193, 0.85714946, 0.86580729,
0.87388357, 0.88141541, 0.88843862, 0.89498747, 0.90109452])
In [97]: out3
Out[97]:
array([ 0.6325231 , 0.63424383, 0.63647521, 0.63916579, 0.64227168,
0.64575317, 0.64957324, 0.65369686, 0.65809065, 0.6627228 ,
0.66756296, 0.67258231, 0.67775356, 0.68305097, 0.68845041,
0.69392934, 0.69946681, 0.70504349, 0.71064159, 0.71624488,
0.72183859, 0.72740939, 0.7329453 , 0.73843566, 0.74387099,
0.749243 , 0.75454444, 0.75976907, 0.76491154, 0.76996738,
0.77493287, 0.77980499, 0.78458137, 0.78926022, 0.79384025,
0.79832065, 0.80270103, 0.80698134, 0.81116187, 0.81524321,
0.81922618, 0.82311181, 0.82690134, 0.83059613, 0.83419771,
0.83770771, 0.84112783, 0.84445988, 0.84770571, 0.8508672 ])
对不起所有的输出。无论如何，问题是，“添加”这些曲线的正确技术是什么？我正在寻找平均测量与半径曲线（带有误差线），我不完全确定这对于不同系统的不同径向测量意味着什么。我的第一个想法是在对象 1 的值之间进行插值，并在对象 2 和 3 的位置评估它们，以便可以在相同的基础上评估它们。但是，我可能有数千个这样的对象，这不喜欢为那么多对象做正确的事情（另外，这也带来了我的插值方法——三次、线性等）。任何想法，将不胜感激！让我知道是否需要进一步澄清。谢谢！