Python 如何将缺少的点绘制成一个完整的圆？

我有9个温度点。1在中心，8在圆上。我需要在一个圆圈中创建一个热图。我设置了要执行计算的点，并使用scipy.interpolate.griddata，但没有绘制完整的圆，程序绘制了一个八角形。我该如何填写缺失的部分

径向基函数（Rbf）可用于插值/外推数据。 这是一个修改过的代码，可以生成所需的绘图

import numpy as np
import matplotlib
import matplotlib.pyplot as plt
import math
from scipy.interpolate import Rbf

# some parameters
xy_center = [2,2]   # center of the plot
radius = 2          # radius

# Data part
# ---------
# mostly original code
meanR = [33.9, 34.2, 33.1, 33.5, 33., 32.7, 32.3, 31.8, 35.]  #9 points data
x = np.array([2, 2, 2+math.sqrt(2), 4, 2+math.sqrt(2), 2, 2+(-math.sqrt(2)), 0, 2+(-math.sqrt(2))])
y = np.array([2, 4, 2+math.sqrt(2), 2, 2+(-math.sqrt(2)), 0, 2+(-math.sqrt(2)), 2, 2+math.sqrt(2)])
z = meanR

# use RBF (Radial basis functions) that allows extrapolation
rbf = Rbf(x, y, z, epsilon=radius+1)  #epsilon is based on some parameters of the data

# Interpolation/extrapolation
# ---------------------------
xi, yi = np.mgrid[x.min():x.max():500j, y.min():y.max():500j]
# applies and get inter/extra-polated values
zi = rbf(xi, yi)

# make zi outside circle --> np.none
midr,midc = zi.shape[0]/2, zi.shape[1]/2
for er in range(zi.shape[0]):
    for ec in range(zi.shape[1]):
        if np.abs(math.sqrt((er-midr)**2 + (ec-midc)**2))>zi.shape[0]/2:
            # outside the circle, dont plot this pixel
            zi[er][ec] = np.nan
        pass
    pass

# make figure
fig = plt.figure(figsize=(8, 8))

# set aspect = 1 to make it a circle
ax = fig.add_subplot(111, aspect = 1)

# add the data points
ax.scatter(x, y, marker = 'o', c = 'b', s = 15, zorder = 3)

for i in range(9):
    ax.annotate(str(z[i]), (x[i],y[i]))

# draw a circle
circle = matplotlib.patches.Circle(xy = xy_center, radius = radius, edgecolor = "k", facecolor = "none")
ax.add_patch(circle)

CS = ax.contourf(xi, yi, zi, 300, cmap=plt.cm.viridis, zorder=1)
cbar = fig.colorbar(CS, ax=ax, shrink=0.7) # make a color bar

# remove the ticks
ax.set_xticks([])
ax.set_yticks([])

# set axes limits
ax.set_xlim(-0.5, 4.5)
ax.set_ylim(-0.5, 4.5)
plt.show() 

结果是：

import numpy as np
import matplotlib
import matplotlib.pyplot as plt
import math
from scipy.interpolate import Rbf

# some parameters
xy_center = [2,2]   # center of the plot
radius = 2          # radius

# Data part
# ---------
# mostly original code
meanR = [33.9, 34.2, 33.1, 33.5, 33., 32.7, 32.3, 31.8, 35.]  #9 points data
x = np.array([2, 2, 2+math.sqrt(2), 4, 2+math.sqrt(2), 2, 2+(-math.sqrt(2)), 0, 2+(-math.sqrt(2))])
y = np.array([2, 4, 2+math.sqrt(2), 2, 2+(-math.sqrt(2)), 0, 2+(-math.sqrt(2)), 2, 2+math.sqrt(2)])
z = meanR

# use RBF (Radial basis functions) that allows extrapolation
rbf = Rbf(x, y, z, epsilon=radius+1)  #epsilon is based on some parameters of the data

# Interpolation/extrapolation
# ---------------------------
xi, yi = np.mgrid[x.min():x.max():500j, y.min():y.max():500j]
# applies and get inter/extra-polated values
zi = rbf(xi, yi)

# make zi outside circle --> np.none
midr,midc = zi.shape[0]/2, zi.shape[1]/2
for er in range(zi.shape[0]):
    for ec in range(zi.shape[1]):
        if np.abs(math.sqrt((er-midr)**2 + (ec-midc)**2))>zi.shape[0]/2:
            # outside the circle, dont plot this pixel
            zi[er][ec] = np.nan
        pass
    pass

# make figure
fig = plt.figure(figsize=(8, 8))

# set aspect = 1 to make it a circle
ax = fig.add_subplot(111, aspect = 1)

# add the data points
ax.scatter(x, y, marker = 'o', c = 'b', s = 15, zorder = 3)

for i in range(9):
    ax.annotate(str(z[i]), (x[i],y[i]))

# draw a circle
circle = matplotlib.patches.Circle(xy = xy_center, radius = radius, edgecolor = "k", facecolor = "none")
ax.add_patch(circle)

CS = ax.contourf(xi, yi, zi, 300, cmap=plt.cm.viridis, zorder=1)
cbar = fig.colorbar(CS, ax=ax, shrink=0.7) # make a color bar

# remove the ticks
ax.set_xticks([])
ax.set_yticks([])

# set axes limits
ax.set_xlim(-0.5, 4.5)
ax.set_ylim(-0.5, 4.5)
plt.show()