在Python中使用matplotlib设置刻度位置时出错
我正在尝试在matplotlib中设置刻度标记的位置。运行下面这个简单的例子时,我遇到了错误:
import matplotlib.pyplot as plt
import numpy as np
from matplotlib.ticker import MultipleLocator
x = 10.*np.random.randn(1000)
y = 10.*np.random.randn(1000)
fig = plt.figure()
ax = fig.add_subplot(1,1,1)
ax.scatter(x, y)
ax.xaxis.set_major_formatter(MultipleLocator(1.))
ax.yaxis.set_major_formatter(MultipleLocator(1.))
plt.show()
错误出现在设置x轴和y轴刻度标记的那两行。如果我用NullFormatter()代替,或者干脆省略这两行,代码就能正常运行,并且生成预期的图表。不过,上面的代码却返回了以下错误:
Exception in Tkinter callback
Traceback (most recent call last):
File "/usr/lib/python2.7/lib-tk/Tkinter.py", line 1413, in __call__
return self.func(*args)
File "/usr/lib/pymodules/python2.7/matplotlib/backends/backend_tkagg.py", line 245, in resize
self.show()
File "/usr/lib/pymodules/python2.7/matplotlib/backends/backend_tkagg.py", line 248, in draw
FigureCanvasAgg.draw(self)
File "/usr/lib/pymodules/python2.7/matplotlib/backends/backend_agg.py", line 394, in draw
self.figure.draw(self.renderer)
File "/usr/lib/pymodules/python2.7/matplotlib/artist.py", line 55, in draw_wrapper
draw(artist, renderer, *args, **kwargs)
File "/usr/lib/pymodules/python2.7/matplotlib/figure.py", line 798, in draw
func(*args)
File "/usr/lib/pymodules/python2.7/matplotlib/artist.py", line 55, in draw_wrapper
draw(artist, renderer, *args, **kwargs)
File "/usr/lib/pymodules/python2.7/matplotlib/axes.py", line 1946, in draw
a.draw(renderer)
File "/usr/lib/pymodules/python2.7/matplotlib/artist.py", line 55, in draw_wrapper
draw(artist, renderer, *args, **kwargs)
File "/usr/lib/pymodules/python2.7/matplotlib/axis.py", line 971, in draw
tick_tups = [ t for t in self.iter_ticks()]
File "/usr/lib/pymodules/python2.7/matplotlib/axis.py", line 906, in iter_ticks
self.major.formatter.set_locs(majorLocs)
AttributeError: MultipleLocator instance has no attribute 'set_locs'
我试着在网上搜索这个错误,但找不到其他人有类似的问题。有没有人知道为什么使用定位器会导致错误呢?
1 个回答
8
MultipleLocator 是一个定位器,而不是格式化工具。你应该使用
ax.xaxis.set_major_locator(MultipleLocator(1.))
ax.yaxis.set_major_locator(MultipleLocator(1.))
这个对我来说有效(虽然用1看起来不太好,但确实有效)。