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python - TypeError: &: 'float' 和 'numpy.float64' 不支持的操作数类型

转载 作者:太空狗 更新时间:2023-10-29 20:47:12 24 4
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<分区>

我正在尝试使用以下代码将连续变量转换为分类变量:

def score_to_categorical(x):
if x<0.25:
return 'very bad'
if x>=0.25 & x<0.5:
return 'bad'
if x>=0.5 & x<0.75:
return 'good'
else:
return 'very good'

ConceptTemp['Score'] = ConceptTemp['Score'].apply(score_to_categorical)
ConceptTemp1['Score'] = ConceptTemp1['Score'].apply(score_to_categorical)

但我收到以下错误:

---------------------------------------------------------------------------
TypeError Traceback (most recent call last)
<ipython-input-72-7ec42b055d4f> in <module>()
----> 1 ConceptTemp['Score'] = ConceptTemp['Score'].apply(score_to_categorical)
2 ConceptTemp1['Score'] = ConceptTemp1['Score'].apply(score_to_categorical)

E:\Anaconda2\lib\site-packages\pandas\core\series.pyc in apply(self, func, convert_dtype, args, **kwds)
2167 values = lib.map_infer(values, lib.Timestamp)
2168
-> 2169 mapped = lib.map_infer(values, f, convert=convert_dtype)
2170 if len(mapped) and isinstance(mapped[0], Series):
2171 from pandas.core.frame import DataFrame

pandas\src\inference.pyx in pandas.lib.map_infer (pandas\lib.c:62578)()

<ipython-input-11-1c4f9c7bfafe> in score_to_categorical(x)
10 if x<0.25:
11 return 'very bad'
---> 12 if x>=0.25 & x<0.5:
13 return 'bad'
14 if x>=0.5 & x<0.75:

TypeError: unsupported operand type(s) for &: 'float' and 'numpy.float64'

我本来认为 floatnumpy.float64 是兼容的,但事实似乎并非如此。

在这方面的任何帮助将不胜感激。

TIA。

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