我有一个绘制线条的函数,类似的东西:
def tmp_plot(*args, **kwargs):
plt.plot([1,2,3,4,5],[1,2,3,4,5], *args, **kwargs)
当我通过将行作为关键字参数传递来调用它时:
tmp_plot(line = '-')
我收到此错误:
TypeError: set_lineprops() got multiple values for keyword argument 'line'
但它可以使用颜色参数。
我使用matplotlib 1.4.3和python 2.7.7
任何线索?
答案 0 :(得分:3)
您可以在下面的Traceback中看到Matplotlib在哪里添加自己的line
参数。这意味着您自己的关键字参数与Matplotlib在set_lineprops
调用中拥有的关键字参数重复:
In [1]: import matplotlib.pyplot as plt
In [2]: plt.plot([1,2,3], [1,4,9], line='-')
---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
<ipython-input-82-f298702afcfe> in <module>()
----> 1 plt.plot([1,2,3], [1,4,9], line='-')
/Users/xnx/anaconda/envs/py33/lib/python3.3/site-packages/matplotlib/pyplot.py in plot(*args, **kwargs)
2985 ax.hold(hold)
2986 try:
-> 2987 ret = ax.plot(*args, **kwargs)
2988 draw_if_interactive()
2989 finally:
/Users/xnx/anaconda/envs/py33/lib/python3.3/site-packages/matplotlib/axes.py in plot(self, *args, **kwargs)
4137 lines = []
4138
-> 4139 for line in self._get_lines(*args, **kwargs):
4140 self.add_line(line)
4141 lines.append(line)
/Users/xnx/anaconda/envs/py33/lib/python3.3/site-packages/matplotlib/axes.py in _grab_next_args(self, *args, **kwargs)
317 return
318 if len(remaining) <= 3:
--> 319 for seg in self._plot_args(remaining, kwargs):
320 yield seg
321 return
/Users/xnx/anaconda/envs/py33/lib/python3.3/site-packages/matplotlib/axes.py in _plot_args(self, tup, kwargs)
305 ncx, ncy = x.shape[1], y.shape[1]
306 for j in range(max(ncx, ncy)):
--> 307 seg = func(x[:, j % ncx], y[:, j % ncy], kw, kwargs)
308 ret.append(seg)
309 return ret
/Users/xnx/anaconda/envs/py33/lib/python3.3/site-packages/matplotlib/axes.py in _makeline(self, x, y, kw, kwargs)
257 **kw
258 )
--> 259 self.set_lineprops(seg, **kwargs)
260 return seg
261
TypeError: set_lineprops() got multiple values for argument 'line'
在任何情况下,您的意思是ls
或linestyle
而不是line
?
In [83]: plt.plot([1,2,3], [1,4,9], ls='-')
Out[83]: [<matplotlib.lines.Line2D at 0x10ed65610>]
答案 1 :(得分:0)
我猜想matplotlib的内部结构除了调用者提供的参数外,还解压缩参数的内部字典,而不删除重复项,因此你和matplot lib内部都通过两条并行路由提供相同名称的单独关键字参数。