如何使用Scipy和1D数组正确地重新塑造值以进行N维插值?

3
给定一些数据。
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d.axes3d import Axes3D

x = [0.02,0.02,0.02,0.02,0.02,0.02,0.02,0.02,0.02,0.46,0.46,0.46,0.46,0.46,0.46,0.46,0.46,0.46,0.9,0.9,0.9,0.9,0.9,0.9,0.9,0.9,0.9,]
y = [0.01,0.01,0.01,0.255,0.255,0.255,0.5,0.5,0.5,0.01,0.01,0.01,0.255,0.255,0.255,0.5,0.5,0.5,0.01,0.01,0.01,0.255,0.255,0.255,0.5,0.5,0.5,]
z = [0.,0.1,0.2,0.,0.1,0.2,0.,0.1,0.2,0.,0.1,0.2,0.,0.1,0.2,0.,0.1,0.2,0.,0.1,0.2,0.,0.1,0.2,0.,0.1,0.2]
v = [0.,0.,1.,0.,1.,1.,0.,1.,1.,0.,0.,0.,0.,0.5,1.,0.,1.,1.,0.,0.,0.,0.,0.,0.,0.,0.5,1.,]

fig, ax = plt.subplots(subplot_kw=dict(projection='3d'))
c = ax.scatter(x, y, z, c = v, s = 100, cmap = plt.cm.bwr, edgecolor = "black", alpha = 0.2)

ax.scatter(0.51, 0.32, 0.12, s = 100, c = "black", edgecolor = "black")

plt.show()

我想获得一个函数f(x,y,z),以找出任意位置v的值。很简单吧?那我为什么死活搞不定这个scipy问题呢?

我找到的示例定义了x、y、z,定义了某种漂亮的网格,并进行评估以获得正确形状和顺序的v。这就是我的尝试失败的地方。如果所有数据最初都格式化为1D,怎么办?

我以为我可以这样做

V = zeros((len(x),len(y),len(z)))
for i in range(len(x)):
    V[i, None, None] = v[i]
    for j in range(len(y)):
        V[None, j, None] = v[j]
        for k in range(len(z)):
            V[None, None, k] = v[k]

fn = RegularGridInterpolator((x,y,z), V)

但是这会返回 ValueError: 维度0中的点必须严格递增

enter image description here

1个回答

2
您想要使用griddata,这里有一个使用您的数据的示例:
import matplotlib.pyplot as plt
from mpl_toolkits.mplot3d.axes3d import Axes3D
from scipy.interpolate import griddata
import numpy as np

x = [0.02,0.02,0.02,0.02,0.02,0.02,0.02,0.02,0.02,0.46,0.46,0.46,0.46,0.46,0.46,0.46,0.46,0.46,0.9,0.9,0.9,0.9,0.9,0.9,0.9,0.9,0.9,]
y = [0.01,0.01,0.01,0.255,0.255,0.255,0.5,0.5,0.5,0.01,0.01,0.01,0.255,0.255,0.255,0.5,0.5,0.5,0.01,0.01,0.01,0.255,0.255,0.255,0.5,0.5,0.5,]
z = [0.,0.1,0.2,0.,0.1,0.2,0.,0.1,0.2,0.,0.1,0.2,0.,0.1,0.2,0.,0.1,0.2,0.,0.1,0.2,0.,0.1,0.2,0.,0.1,0.2]
v = [0.,0.,1.,0.,1.,1.,0.,1.,1.,0.,0.,0.,0.,0.5,1.,0.,1.,1.,0.,0.,0.,0.,0.,0.,0.,0.5,1.,]
points = np.array([x, y, z])

fig, ax = plt.subplots(subplot_kw=dict(projection='3d'))
c = ax.scatter(x, y, z, c=v, s = 100, cmap = plt.cm.bwr, edgecolor = "black", alpha = 0.2)

# This is the interpolation, use existing points and values (1D)
# to get the value at 0.51, 0.32, 0.12
p = griddata(points.T, v, (0.51, 0.32, 0.12))
ax.scatter(0.51, 0.32, 0.12, s=100, c=[p], cmap = plt.cm.bwr)

#Plot a meshgrid of interpolated values (optional)
pad = 0.02
xg = np.linspace(min(x)-pad, max(x)+pad,10)
yg = np.linspace(min(y)-pad, max(y)+pad,10)
zg = np.linspace(min(z)-pad, max(z)+pad,10)
X, Y, Z = np.meshgrid(xg, yg, zg)
vinterp = griddata(points.T, v, (X, Y, Z))
ci = ax.scatter(X.ravel(), Y.ravel(), Z.ravel(), c=vinterp.ravel(), s=10, cmap = plt.cm.bwr, edgecolor = "black", alpha = 0.2)

plt.show()

看起来像这样,

enter image description here

使用数值网格绘制图表,以展示插值和您的示例点。

啊,就是这个。我也用__call__进行了封装,使它更加方便携带,就像其他Scipy函数一样:class Interpolate3D(): def __init__(self, x, y, z, v): self.x = x self.y = y self.z = z self.v = v self.pts = np.array([self.x, self.y, self.z]).T def __call__(self, *args): return scipy.interpolate.griddata(self.pts, self.v, args, method="linear") - komodovaran_

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