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

时间:2016-06-14 17:12:20

标签: python numpy typeerror

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)

但是我收到以下错误:

---------------------------------------------------------------------------
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'

我会认为floatnumpy.float64是兼容的,但似乎并非如此。

非常感谢这方面的任何帮助。

TIA。

1 个答案:

答案 0 :(得分:12)

此处x>=0.25 & x<0.5 &执行bitwise AND operation(例如,1 & 52为零,将被视为False),而您当然打算检查x>=0.25 and x<0.5是否都为真。

所以,这样做:

x>=0.25 and x<0.5

同样的错误在下一行。