Question

这个问题不同于post，后者正在讨论其他设置网格的方法，而不是使用pyplot.setp（）。

matplotlib.pyplot.setp（）可用于为所有子图设置一些属性。

这段代码

xlim = (-2,2)
ylim = (-2,2)
f, axs = plt.subplots(2, 2)
a = plt.setp(axs, xlim=xlim, ylim=ylim)
plt.show()

将所有子图的限制设置为（-2,2）。

AxesSubplot.grid可用于设置网格线

f, axs = plt.subplots(2, 2)
axs[0,0].grid(True)
plt.show()

我正在尝试使用pyplot.setp（）设置网格线。

f, axs = plt.subplots(2, 2)
plt.setp(axs, grid=True)
plt.show()

遇到此错误

--------------------------------------------------------------------------
AttributeError                           Traceback (most recent call last)
<ipython-input-10-5014b2a61ed3> in <module>()
      1 f, axs = plt.subplots(2, 2)
----> 2 plt.setp(axs, grid=True)
      3 plt.show()

~/anaconda3/envs/tf11/lib/python3.6/site-packages/matplotlib/pyplot.py in setp(obj, *args, **kwargs)
    340 @docstring.copy(_setp)
    341 def setp(obj, *args, **kwargs):
--> 342     return _setp(obj, *args, **kwargs)
    343 
    344 

~/anaconda3/envs/tf11/lib/python3.6/site-packages/matplotlib/artist.py in setp(obj, *args, **kwargs)
   1505     # put args into ordereddict to maintain order
   1506     funcvals = OrderedDict((k, v) for k, v in zip(args[::2], args[1::2]))
-> 1507     ret = [o.update(funcvals) for o in objs] + [o.set(**kwargs) for o in objs]
   1508     return list(cbook.flatten(ret))
   1509 

~/anaconda3/envs/tf11/lib/python3.6/site-packages/matplotlib/artist.py in <listcomp>(.0)
   1505     # put args into ordereddict to maintain order
   1506     funcvals = OrderedDict((k, v) for k, v in zip(args[::2], args[1::2]))
-> 1507     ret = [o.update(funcvals) for o in objs] + [o.set(**kwargs) for o in objs]
   1508     return list(cbook.flatten(ret))
   1509 

~/anaconda3/envs/tf11/lib/python3.6/site-packages/matplotlib/artist.py in set(self, **kwargs)
   1013                    key=lambda x: (self._prop_order.get(x[0], 0), x[0])))
   1014 
-> 1015         return self.update(props)
   1016 
   1017     def findobj(self, match=None, include_self=True):

~/anaconda3/envs/tf11/lib/python3.6/site-packages/matplotlib/artist.py in update(self, props)
    914 
    915         with cbook._setattr_cm(self, eventson=False):
--> 916             ret = [_update_property(self, k, v) for k, v in props.items()]
    917 
    918         if len(ret):

~/anaconda3/envs/tf11/lib/python3.6/site-packages/matplotlib/artist.py in <listcomp>(.0)
    914 
    915         with cbook._setattr_cm(self, eventson=False):
--> 916             ret = [_update_property(self, k, v) for k, v in props.items()]
    917 
    918         if len(ret):

~/anaconda3/envs/tf11/lib/python3.6/site-packages/matplotlib/artist.py in _update_property(self, k, v)
    910                 func = getattr(self, 'set_' + k, None)
    911                 if not callable(func):
--> 912                     raise AttributeError('Unknown property %s' % k)
    913                 return func(v)
    914 

AttributeError: Unknown property grid

我要求解释为什么此代码会导致错误，而不是为所有子图设置网格的解决方案。

Answer 1

可以使用rcParams。您可以使用以下

import matplotlib.pyplot as plt

rc = {"axes.grid" : True}
plt.rcParams.update(rc)

xlim = (-2,2)
ylim = (-2,2)
f, axs = plt.subplots(2, 2)
a = plt.setp(axs, xlim=xlim, ylim=ylim)
plt.show()

Answer 2

否，您不能使用plt.setp来打开网格。

原因是轴没有属性grid并且没有set_grid方法，而只有方法.grid(..)。

您需要调用该方法，

for ax in axs.flat:
    ax.grid(True)

其他打开网格的选项显示在How do I draw a grid onto a plot in Python?

中

matplotlib：为什么用pyplot.setp（）设置网格会导致错误？

2 个答案: