在一维中外推 2d numpy 数组

Question

我有 numpy.array 个模拟数据集，但我缺少边缘点 (x=0.1)，我如何 interpolate/extrapolate z 中的数据到边缘？我有：

x = [ 0.  0.00667  0.02692  0.05385  0.08077]
y = [ 0.     10.     20.     30.     40.     50.]

#       0.      0.00667 0.02692 0.05385 0.08077 
z = [[ 25.     25.     25.     25.     25.   ]   # 0.
     [ 25.301  25.368  25.617  26.089  26.787]   # 10.
     [ 25.955  26.094  26.601  27.531  28.861]   # 20.
     [ 26.915  27.126  27.887  29.241  31.113]   # 30.
     [ 28.106  28.386  29.378  31.097  33.402]   # 40.
     [ 29.443  29.784  30.973  32.982  35.603]]  # 50.

我想在 z 中添加一个对应于 x = 0.1 的新列，这样我的新 x 将是

x_new = [ 0.  0.00667  0.02692  0.05385  0.08077  0.1]

#       0.      0.00667 0.02692 0.05385 0.08077 0.01
z = [[ 25.     25.     25.     25.     25.       ?   ]   # 0.
     [ 25.301  25.368  25.617  26.089  26.787    ?   ]   # 10.
     [ 25.955  26.094  26.601  27.531  28.861    ?   ]   # 20.
     [ 26.915  27.126  27.887  29.241  31.113    ?   ]   # 30.
     [ 28.106  28.386  29.378  31.097  33.402    ?   ]   # 40.
     [ 29.443  29.784  30.973  32.982  35.603    ?   ]]  # 50.

哪里都是'?'替换为 interpolated/extrapolated 数据。感谢您的帮助！

Answer 1

您看过 scipy.interpolate2d.interp2d（使用样条曲线）吗？

from scipy.interpolate import interp2d
fspline = interp2d(x,y,z) # maybe need to switch x and y around
znew = fspline([0.1], y)
z = np.c_[[z, znew] # to join arrays

编辑:

我和@dnalow设想的方法是这样的：

import numpy as np
import matplotlib.pyplot as plt

# make some test data
def func(x, y):
    return np.sin(np.pi*x) + np.sin(np.pi*y)

xx, yy = np.mgrid[0:2:20j, 0:2:20j]
zz = func(xx[:], yy[:]).reshape(xx.shape)

fig, (ax1, ax2, ax3, ax4) = plt.subplots(1,4, figsize=(13, 3))
ax1.imshow(zz, interpolation='nearest')
ax1.set_title('Original')

# remove last column
zz[:,-1] = np.nan
ax2.imshow(zz, interpolation='nearest')
ax2.set_title('Missing data')

# compute missing column using simplest imaginable model: first order Taylor
gxx, gyy = np.gradient(zz[:, :-1])
zz[:, -1] =  zz[:, -2] + gxx[:, -1] + gyy[:,-1]
ax3.imshow(zz, interpolation='nearest')
ax3.set_title('1st order Taylor approx')

# add curvature to estimate
ggxx, _ = np.gradient(gxx)
_, ggyy = np.gradient(gyy)
zz[:, -1] = zz[:, -2] + gxx[:, -1] + gyy[:,-1] + ggxx[:,-1] + ggyy[:, -1]
ax4.imshow(zz, interpolation='nearest')
ax4.set_title('2nd order Taylor approx')

fig.tight_layout()
fig.savefig('extrapolate_2d.png')

plt.show()

您可以通过
改进估算 (a) 添加高阶导数（又名泰勒展开），或
(b) 计算更多方向的梯度，而不仅仅是 x 和 y（然后相应地对梯度进行加权）。

此外，如果您预先平滑图像（现在我们有一个完整的 Sobel 过滤器...），您将获得更好的渐变。

在一维中外推 2d numpy 数组

Extrapolate 2d numpy array in one dimension

python

arrays

interpolation

numpy

extrapolation