在执行范围分块时处理python中的值错误

Handling Value error in Python while performing Range binning

我正在尝试将pandas列值分类为范围值。但是当我使用二分法时会得到值错误。

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from pandas_datareader import data
import pandas
import bisect
import fix_yahoo_finance as yf

yf.pdr_override()
df = data.get_data_yahoo('SPY', '2015-01-01', '2018-04-05')
df.tail(2)

def Daily_Returns(A, B):
return (B - A)*100/A

df['OC_Return_%'] = Daily_Returns(df['Open'], df['Close'])

def b(value):
intervals = ['Less Than -10 %','-10% to -5%','-5% to -2.5%','-2.5% to -2%','-2% to -1.5%','-1.5% to -1%','-1% to -0.5%','-0.5% to 0%','0% to 0.5%','0.5% to 1%','1% to 1.5%','1.5% to 2%','2% to 2.5%','2.5% to 5%','5% to 10%','Greater Than 10 %']
return intervals[bisect.bisect_left([-float('inf'),-10,-5,-2.5,-2,-1.5,-1,-0.5,0,0.5,1,1.5,2,2.5,5,10,float('inf')], value)-1]

df['OC_Return_Bin'] = b(df["OC_Return_%"])
df

如果使用a.any()或a.all()，则错误将消失。但它用错误的值填充结果列。

这是注释中要求的完整跟踪。

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ValueError Traceback (most recent call last)
<ipython-input-80-a571e502f6a6> in <module>()
17 return intervals[bisect.bisect_left([-float('inf'),-10,-5,-2.5,-2,-1.5,-1,-0.5,0,0.5,1,1.5,2,2.5,5,10,float('inf')], value)-1]
18
19 df['OC_Return_Bin'] = b(df["OC_Return_%"])
20 df

<ipython-input-80-a571e502f6a6> in b(value)
15 def b(value):
16 intervals = ['Less Than -10 %','-10% to -5%','-5% to -2.5%','-2.5% to -2%','-2% to -1.5%','-1.5% to -1%','-1% to -0.5%','-0.5% to 0%','0% to 0.5%','0.5% to 1%','1% to 1.5%','1.5% to 2%','2% to 2.5%','2.5% to 5%','5% to 10%','Greater Than 10 %']
17 return intervals[bisect.bisect_left([-float('inf'),-10,-5,-2.5,-2,-1.5,-1,-0.5,0,0.5,1,1.5,2,2.5,5,10,float('inf')], value)-1]
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19 df['OC_Return_Bin'] = b(df["OC_Return_%"])

C:\Users\USER\Anaconda2\lib\site-packages\pandas\core\generic.pyc in __nonzero__(self)
953 raise ValueError("The truth value of a {0} is ambiguous."
954 "Use a.empty, a.bool(), a.item(), a.any() or a.all()."
955 .format(self.__class__.__name__))
956
957 __bool__ = __nonzero__

ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().