matplotlibでコンタープロットの凡例を作成するにはどうすればよいですか？

Question

私はどこでも答えを見つけることができないようです！私は議論hereを見つけましたが、私はTypeError: 'NoneType' object is not iterableを取得し、これをしよう：

>>> import numpy as np 
>>> import matplotlib.pyplot as plt 
>>> x, y = np.meshgrid(np.arange(10),np.arange(10)) 
>>> z = x + y 
>>> cs = plt.contourf(x,y,z,levels=[2,3]) 
>>> cs.collections[0].set_label('test') 
>>> plt.legend() 
Traceback (most recent call last): 
    File "<stdin>", line 1, in <module> 
    File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/matplotlib/pyplot.py", line 2791, in legend 
    ret = gca().legend(*args, **kwargs) 
    File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/matplotlib/axes.py", line 4475, in legend 
    self.legend_ = mlegend.Legend(self, handles, labels, **kwargs) 
    File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/matplotlib/legend.py", line 365, in __init__ 
    self._init_legend_box(handles, labels) 
    File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/matplotlib/legend.py", line 627, in _init_legend_box 
    handlebox) 
    File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/matplotlib/legend_handler.py", line 110, in __call__ 
    handlebox.get_transform()) 
    File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/matplotlib/legend_handler.py", line 352, in create_artists 
    width, height, fontsize) 
    File "/opt/local/Library/Frameworks/Python.framework/Versions/2.7/lib/python2.7/site-packages/matplotlib/legend_handler.py", line 307, in get_sizes 
    size_max = max(orig_handle.get_sizes())*legend.markerscale**2 
TypeError: 'NoneType' object is not iterable 

EDIT：あなたが作るために、プロキシのアーティストを作成することができます

Answer 1

：私はこのような何かを探しています伝説：

import numpy as np 
import matplotlib.pyplot as plt 
x, y = np.meshgrid(np.arange(10),np.arange(10)) 
z = np.sqrt(x**2 + y**2) 
cs = plt.contourf(x,y,z,levels=[2,3,4,6]) 

proxy = [plt.Rectangle((0,0),1,1,fc = pc.get_facecolor()[0]) 
    for pc in cs.collections] 

plt.legend(proxy, ["range(2-3)", "range(3-4)", "range(4-6)"]) 
plt.show() 

Answer 2

プロキシアーティストを使用せずに輪郭線で直接行うこともできます。

import matplotlib 
import numpy as np 
import matplotlib.cm as cm 
import matplotlib.mlab as mlab 
import matplotlib.pyplot as plt 

matplotlib.rcParams['xtick.direction'] = 'out' 
matplotlib.rcParams['ytick.direction'] = 'out' 

delta = 0.025 
x = np.arange(-3.0, 3.0, delta) 
y = np.arange(-2.0, 2.0, delta) 
X, Y = np.meshgrid(x, y) 
Z1 = mlab.bivariate_normal(X, Y, 1.0, 1.0, 0.0, 0.0) 
Z2 = mlab.bivariate_normal(X, Y, 1.5, 0.5, 1, 1) 
# difference of Gaussians 
Z = 10.0 * (Z2 - Z1) 



# Create a simple contour plot with labels using default colors. The 
# inline argument to clabel will control whether the labels are draw 
# over the line segments of the contour, removing the lines beneath 
# the label 
plt.figure() 
CS = plt.contour(X, Y, Z) 
plt.clabel(CS, inline=1, fontsize=10) 
plt.title('Simplest default with labels') 

labels = ['line1', 'line2','line3','line4', 
      'line5', 'line6'] 
for i in range(len(labels)): 
    CS.collections[i].set_label(labels[i]) 

plt.legend(loc='upper left') 

が生成されます。

しかし、あなたはまた、独自の必要性のための注釈に見たいと思うかもしれません。私の意見では、それはあなたと何をイメージに書き込むの詳細きめの細かい制御を与えるだろう、ここでいくつかの注釈と同じ例です。

### better with annotation, more flexible 
plt.figure(2) 
CS = plt.contour(X, Y, Z) 
plt.clabel(CS, inline=1, fontsize=10) 
plt.title('Simplest default with labels') 

plt.annotate('some text here',(1.4,1.6)) 
plt.annotate('some text there',(-2,-1.5))