如何在matplotlib中为轮廓图创建图例？
如何在matplotlib中为轮廓图创建图例？
提问于 2012-05-08 07:03:54
我似乎到处都找不到答案！我找到了一个讨论here，但是尝试一下我得到了一个TypeError: 'NoneType' object is not iterable
>>> 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编辑：我正在寻找类似这样的东西：
回答 3
Stack Overflow用户
回答已采纳
发布于 2012-05-08 09:00:13
您可以创建代理艺术家来制作图例：
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()
Stack Overflow用户
发布于 2012-05-08 15:05:51
您也可以直接使用等高线，而无需使用代理美工人员。
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')将产生：
但是，您可能还希望根据自己的需要查看注释。在我看来，它将给你一个更细粒度的控制，控制你在图像上写的位置和内容，下面是带有一些注释的相同示例：
### 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))
Stack Overflow用户
发布于 2019-07-15 04:16:33
我有一个类似的question，但需要超越HYRY's answer。为了使包对用户友好，我希望ax.legend()在不需要用户传递任何句柄的情况下工作，这可以通过将标签传递给代理来实现
proxy = plt.Rectangle((0, 0), 1, 1, fc='red', label='some label')然后将代理添加到轴的面片：
ax.patches += [proxy](执行ax = plt.gca()以获取当前轴)
这在this answer中有更详细的描述。
页面原文内容由Stack Overflow提供。腾讯云小微IT领域专用引擎提供翻译支持
原文链接：
https://stackoverflow.com/questions/10490302
复制相关文章
相似问题
腾讯云开发者
