python-2.7 - .ix (Pandas) 上的按位运算

Question

为什么开发人员不允许对 .ix 进行按位运算？好奇这是技术限制问题还是我忽略的逻辑问题。

df.ix[df["ptdelta"]<=0 & df["ptdelta"]>5]

回溯是：

TypeError: ufunc 'bitwise_and' not supported for the input types, and the inputs could not be safely coerced to any supported types according to the casting rule 'safe'

'''

注意：从 Pandas v0.20 开始，.ix 不推荐使用索引器以支持.iloc/ .loc。

Answer 1

我认为您误解了方括号内.ix发生的事情 - 与它无关。如果你适当地括号，这有效：

>>> df = pd.DataFrame({"a": [-1,2,-3,4,6,-8.2]})
>>> df.ix[(df['a'] <= 0) | (df['a'] > 5)]
     a
0 -1.0
2 -3.0
4  6.0
5 -8.2

否则，您将尝试对（大概）浮点数执行按位运算。例如，如果它是一个 int，那么它会“工作”：

>>> df['a'] <= 0 & df['a']
Traceback (most recent call last):
  File "<ipython-input-40-9173361ec31b>", line 1, in <module>
    df['a'] <= 0 & df['a']
TypeError: ufunc 'bitwise_and' not supported for the input types, and the inputs could not be safely coerced to any supported types according to the casting rule 'safe'

>>> df['a'] <= 0 & df['a'].astype(int)
0     True
1    False
2     True
3    False
4    False
5     True
Name: a, Dtype: bool