我有一个点，logR，logT和logX的数据集，其中X是R和T的函数。它只是一个数据集，我没有为X定义的函数。数据列在表中，其中logR对应于列，logT对应于logT。我试图使用插值函数在logR和logT的两个输入处评估此网格。我发现我的情况与这篇文章最相关：
How to pass arrays into Scipy Interpolate RectBivariateSpline?
但我无法安排事情，所以我的功能可以在我的输入上进行评估。以下是我的尝试：
import numpy as np
from scipy.interpolate import RectBivariateSpline, interp2d
op_r = np.array([1e-8, 3.1622e-8, 1e-7, 3.1622e-7, 1e-6, 3.1622e-6, 1e-5,
3.1622e-5, 1e-4, 3.1622e-4, 1e-3, 3.1622e-3, 1e-2, 3.1622e-2, 0.1, .31622, 1,
3.1622, 10])
op_T = np.array([17782.794, 19952.623, 22387.211, 25118.864, 28183.829,
31622.777, 35481.339, 39810.717, 44668.359, 50118.723, 56234.133, 63095.734,
79432.823, 89125.094])
log_op_val = np.array([[-0.598, -0.593, -0.583, -0.568, -0.539, -0.477,
-0.353, -0.142, 0.168, 0.558, 0.990, 1.443, 1.915, 2.407, 2.866, 3.239, 3.517,
3.725, 3.896], [-0.597, -0.592, -0.580, -0.561, -0.532, -0.474, -0.362,
-0.165, 0.138, 0.539, 1.001, 1.476, 1.942, 2.426, 2.912, 3.352, 3.702, 3.968,
4.175], [-0.588, -0.588, -0.578, -0.555, -0.520, -0.462, -0.357, -0.171,
0.124, 0.529, 1.009, 1.507, 2.001, 2.487, 2.979, 3.453, 3.856, 4.176, 4.422],
[-0.545, -0.559, -0.563, -0.546, -0.506, -0.442, -0.338, -0.159, 0.132, 0.538,
1.015, 1.525, 2.051, 2.565, 3.072, 3.563, 3.996, 4.356, 4.634], [-0.520,
-0.521, -0.519, -0.509, -0.475, -0.409, -0.301, -0.122, 0.167, 0.571, 1.052,
1.570, 2.106, 2.642, 3.176, 3.684, 4.136, 4.517, 4.822], [-0.518, -0.514,
-0.504, -0.478, -0.425, -0.344, -0.232, -0.056, 0.226, 0.629, 1.111, 1.631,
2.169, 2.719, 3.276, 3.804, 4.275, 4.672, 4.990], [-0.517, -0.513, -0.504,
-0.479, -0.417, -0.297, -0.129, 0.074, 0.353, 0.734, 1.202, 1.715, 2.250,
2.800, 3.364, 3.907, 4.394, 4.802, 5.127], [-0.518, -0.514, -0.505, -0.484,
-0.429, -0.311, -0.104, 0.185, 0.521, 0.894, 1.329, 1.818, 2.341, 2.883,
3.441, 3.986, 4.481, 4.894, 5.218], [-0.517, -0.514, -0.507, -0.490, -0.443,
-0.337, -0.142, 0.169, 0.588, 1.039, 1.480, 1.936, 2.431, 2.955, 3.496, 4.031,
4.521, 4.934, 5.253], [-0.516, -0.513, -0.507, -0.492, -0.453, -0.361, -0.184,
0.103, 0.510, 1.009, 1.531, 2.022, 2.502, 3.002, 3.519, 4.035, 4.513, 4.920,
5.235], [-0.515, -0.511, -0.506, -0.493, -0.460, -0.381, -0.225, 0.036, 0.409,
0.877, 1.415, 1.973, 2.502, 3.005, 3.505, 4.002, 4.468, 4.868, 5.183],
[-0.515, -0.511, -0.503, -0.490, -0.462, -0.394, -0.257, -0.022, 0.321, 0.759,
1.269, 1.827, 2.403, 2.949, 3.458, 3.948, 4.405, 4.802, 5.113], [-0.516,
-0.512, -0.502, -0.487, -0.460, -0.400, -0.279, -0.066, 0.254, 0.672, 1.164,
1.701, 2.278, 2.851, 3.388, 3.889, 4.347, 4.741, 5.047], [-0.517, -0.512,
-0.503, -0.485, -0.454, -0.397, -0.287, -0.092, 0.211, 0.620, 1.101, 1.628,
2.190, 2.762, 3.322, 3.841, 4.305, 4.695, 4.989], [-0.516, -0.512, -0.503,
-0.484, -0.449, -0.388, -0.283, -0.099, 0.192, 0.596, 1.071, 1.596, 2.148,
2.714, 3.281, 3.811, 4.280, 4.661, 4.937]])
T_1a = 22100.
R_a = rho_ta /(((T_1a)/(1e6))**3)
gri_chi_a = RectBivariateSpline(op_r, op_T, op_val)
chi_a = RectBivariateSpline(R_a, T_1a)
print chi_a
这是我得到的错误：
Traceback (most recent call last):
File "model.py", line 279, in <module>
gri_chi_a = RectBivariateSpline(op_r, op_T, op_val)
"/System/Library/Frameworks/Python.framework/Versions/2.7/Extras/lib/python/scipy/interpolate/fitpack2.py", line 882, in __init__
raise TypeError('x dimension of z must have same number of '
TypeError: x dimension of z must have same number of elements as x
就像我使用interp2d函数一样。任何帮助将不胜感激。
答案 0 :(得分：0)
复制粘贴数组我得到：
In [391]: op_r.shape
Out[391]: (19,)
In [393]: op_T.shape
Out[393]: (14,)
In [395]: log_op_val.shape
Out[395]: (15, 19)
15不等于14或19！