matplotlib.pyplot.contourfの結果を円に合わせるにはどうすればいいですか？ここ

Question

は、いくつかのデータをプロットするために私のコードです：

from scipy.interpolate import griddata 
from numpy import linspace 
import matplotlib.pyplot as plt 
meanR = [9.95184937, 9.87947708, 9.87628496, 9.78414422, 
     9.79365258, 9.96168969, 9.87537519, 9.74536093, 
     10.16686878, 10.04425475, 10.10444126, 10.2917172 , 
     10.16745917, 10.0235203 , 9.89914 , 10.11263505, 
     9.99756449, 10.17861254, 10.04704248] 
koord = [[1,4],[3,4],[1,3],[3,3],[2,3],[1,2],[3,2],[2,2],[1,1],[3,1],[2,1],[1,0],[3,0],[0,3],[4,3],[0,2],[4,2],[0,1],[4,1]] 
x,y=[],[] 
for i in koord: 
    x.append(i[0]) 
    y.append(i[1]) 
z = meanR 
xi = linspace(-2,6,300); 
yi = linspace(-2,6,300); 
zi = griddata((x, y), z, (xi[None,:], yi[:,None]), method='cubic') 
CS = plt.contourf(xi,yi,zi,15,cmap=plt.cm.jet) 
plt.scatter(x,y,marker='o',c='b',s=15) 
plt.xlim(min(x),max(x)) 
plt.ylim(min(y),max(y)) 
plt.show() 

結果では、我々が持っている：

を私は円形にそれを刻むことができますどのように？この

Answer 1

通常の投影法を使うこともできるので、軸を必要としないので、サークル。私は楽しい時間を過ごし、ボーナス耳、鼻、カラーバーを追加しました。私はコードに注釈をつけました、私はそれがはっきりしていることを望みます。

from __future__ import print_function 
from __future__ import division 
from __future__ import absolute_import 

import scipy.interpolate 
import numpy 
import matplotlib 
import matplotlib.pyplot as plt 

# close old plots 
plt.close("all") 

# some parameters 
N = 300    # number of points for interpolation 
xy_center = [2,2] # center of the plot 
radius = 2   # radius 

# mostly original code 
meanR = [9.95184937, 9.87947708, 9.87628496, 9.78414422, 
     9.79365258, 9.96168969, 9.87537519, 9.74536093, 
     10.16686878, 10.04425475, 10.10444126, 10.2917172 , 
     10.16745917, 10.0235203 , 9.89914 , 10.11263505, 
     9.99756449, 10.17861254, 10.04704248] 

koord = [[1,4],[3,4],[1,3],[3,3],[2,3],[1,2],[3,2],[2,2],[1,1],[3,1],[2,1],[1,0],[3,0],[0,3],[4,3],[0,2],[4,2],[0,1],[4,1]] 

x,y = [],[] 
for i in koord: 
    x.append(i[0]) 
    y.append(i[1]) 

z = meanR 

xi = numpy.linspace(-2, 6, N) 
yi = numpy.linspace(-2, 6, N) 
zi = scipy.interpolate.griddata((x, y), z, (xi[None,:], yi[:,None]), method='cubic') 

# set points > radius to not-a-number. They will not be plotted. 
# the dr/2 makes the edges a bit smoother 
dr = xi[1] - xi[0] 
for i in range(N): 
    for j in range(N): 
     r = numpy.sqrt((xi[i] - xy_center[0])**2 + (yi[j] - xy_center[1])**2) 
     if (r - dr/2) > radius: 
      zi[j,i] = "nan" 

# make figure 
fig = plt.figure() 

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

# use different number of levels for the fill and the lines 
CS = ax.contourf(xi, yi, zi, 60, cmap = plt.cm.jet, zorder = 1) 
ax.contour(xi, yi, zi, 15, colors = "grey", zorder = 2) 

# make a color bar 
cbar = fig.colorbar(CS, ax=ax) 

# add the data points 
# I guess there are no data points outside the head... 
ax.scatter(x, y, marker = 'o', c = 'b', s = 15, zorder = 3) 

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

# make the axis invisible 
for loc, spine in ax.spines.iteritems(): 
    # use ax.spines.items() in Python 3 
    spine.set_linewidth(0) 

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

# Add some body parts. Hide unwanted parts by setting the zorder low 
# add two ears 
circle = matplotlib.patches.Ellipse(xy = [0,2], width = 0.5, height = 1.0, angle = 0, edgecolor = "k", facecolor = "w", zorder = 0) 
ax.add_patch(circle) 
circle = matplotlib.patches.Ellipse(xy = [4,2], width = 0.5, height = 1.0, angle = 0, edgecolor = "k", facecolor = "w", zorder = 0) 
ax.add_patch(circle) 
# add a nose 
xy = [[1.5,3], [2,4.5],[2.5,3]] 
polygon = matplotlib.patches.Polygon(xy = xy, facecolor = "w", zorder = 0) 
ax.add_patch(polygon) 

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

plt.show() 

Answer 2

のようなもの、あなたがしてプロットしない部品を交換する場合：

fig = plt.figure() 
ax = fig.add_subplot(111, polar=True) 
CS = ax.contourf(xi,yi,zi,15,cmap=plt.cm.jet) 
ax.scatter(x,y,marker='o',c='b',s=15) 
ax.set_xlim(min(x),max(x)) 
ax.set_ylim(min(y),max(y)) 

あなたはこの

はあなたが望む結果を得るために取得するには、あなたがxを再スケールする必要があり、y、 xi,yiであり、画像はゼロの中心にある。極座標に変換する必要があるかもしれません。今、私はもっと情報を提供する時間がありませんが、これはあなたが始めるのに役立つことを願っています