溢出错误与熊猫

Question

我下面这个教程：https://www.youtube.com/watch?v=wfTABU8VeoY&list=PLQVvvaa0QuDfHt4XU7vTm22xDegR0v0fQ&index=7与大熊猫的数据分析，但是当我想下面的代码运行

import datetime 
import pandas as pd 
from pandas import DataFrame 
import pandas.io.data 
import matplotlib.pyplot as plt 
from mpl_toolkits.mplot3d import Axes3D 

sp500 = pd.io.data.get_data_yahoo('%5EGSPC', start = datetime.datetime(2015, 10, 15), 
            end = datetime.datetime(2016, 10, 15)) 
sp500.to_csv('sp500.csv') 

df = pd.read_csv('sp500.csv', index_col = 'Date', parse_dates=True) 

df['H-L'] = df['High'] - df.Low 
df['100MA'] = pd.rolling_mean(df['Close'], 100) 
df['Difference'] = df['Close'].diff() 

threedee = plt.figure().gca(projection='3d') 
threedee.scatter(df.index, df['H-L'], df['Close']) 
threedee.set_xlabel('Index') 
threedee.set_zlabel('Close') 
threedee.set_ylabel('H-L') 

plt.show() 

它无论是在Jupyter笔记本电脑和PyCharm错误产生如下：

OverflowError        Traceback (most recent call last)  
C:\Program Files\Anaconda2\lib\site-packages\IPython\core\formatters.py in __call__(self, obj) 
     305     pass 
     306    else: 
    --> 307     return printer(obj) 
     308    # Finally look for special method names 
     309    method = get_real_method(obj, self.print_method) 

C:\Program Files\Anaconda2\lib\site-packages\IPython\core\pylabtools.py in <lambda>(fig) 
    225 
    226  if 'png' in formats: 
--> 227   png_formatter.for_type(Figure, lambda fig: print_figure(fig, 'png', **kwargs)) 
    228  if 'retina' in formats or 'png2x' in formats: 
    229   png_formatter.for_type(Figure, lambda fig: retina_figure(fig, **kwargs)) 

C:\Program Files\Anaconda2\lib\site-packages\IPython\core\pylabtools.py in print_figure(fig, fmt, bbox_inches, **kwargs) 
    117 
    118  bytes_io = BytesIO() 
--> 119  fig.canvas.print_figure(bytes_io, **kw) 
    120  data = bytes_io.getvalue() 
    121  if fmt == 'svg': 

与许多其他各种路径，包括matplotlib.py等。 有什么问题？加载的数据不是太多，是吗？

Answer 1

您是否尝试过更换此行

threedee.scatter(df.index, df['H-L'], df['Close']) 

用下面的？

threedee.scatter(range(len(df.index)), df['H-L'], df['Close']) 

您正在密谋时间戳的值。 matplotlib可能不知道时间戳记携带的数值。

编辑：不幸的是，这种解决方法此变通办法将xaxis滴答变成了一个数字范围。但是我们总是可以手动设置刻度：

threedee.scatter(df.index, df['H-L'], df['Close']) 

renderer = fig.canvas.get_renderer() 
threedee.draw(renderer) 
old_xticks = [t.get_text() for t in threedee.xaxis.get_ticklabels()] 
new_xticks = [df.index[int(t)].strftime("%Y-%m-%d") 
       if t is not '' else '' for t in old_xticks] 
threedee.xaxis.set_ticklabels(new_xticks) 

threedee.set_xlabel('Index') 
threedee.set_zlabel('Close') 
threedee.set_ylabel('H-L') 

plt.show()