如何获得熊猫系列的按元素逻辑非？

I have a pandas Series object containing boolean values. How can I get a series containing the logical NOT of each value?

For example, consider a series containing:

True
True
True
False

The series I’d like to get would contain:

False
False
False
True

This seems like it should be reasonably simple, but apparently I’ve misplaced my mojo =(

To invert a boolean Series, use ~s:

In [7]: s = pd.Series([True, True, False, True])

In [8]: ~s
Out[8]: 
0    False
1    False
2     True
3    False
dtype: bool

Using Python2.7, NumPy 1.8.0, Pandas 0.13.1:

In [119]: s = pd.Series([True, True, False, True]*10000)

In [10]:  %timeit np.invert(s)
10000 loops, best of 3: 91.8 µs per loop

In [11]: %timeit ~s
10000 loops, best of 3: 73.5 µs per loop

In [12]: %timeit (-s)
10000 loops, best of 3: 73.5 µs per loop

As of Pandas 0.13.0, Series are no longer subclasses of numpy.ndarray; they are now subclasses of pd.NDFrame. This might have something to do with why np.invert(s) is no longer as fast as ~s or -s.

Caveat: timeit results may vary depending on many factors including hardware, compiler, OS, Python, NumPy and Pandas versions.

@unutbu’s answer is spot on, just wanted to add a warning that your mask needs to be dtype bool, not ‘object’. Ie your mask can’t have ever had any nan’s. See here – even if your mask is nan-free now, it will remain ‘object’ type.

The inverse of an ‘object’ series won’t throw an error, instead you’ll get a garbage mask of ints that won’t work as you expect.

In[1]: df = pd.DataFrame({'A':[True, False, np.nan], 'B':[True, False, True]})
In[2]: df.dropna(inplace=True)
In[3]: df['A']
Out[3]:
0    True
1   False
Name: A, dtype object
In[4]: ~df['A']
Out[4]:
0   -2
0   -1
Name: A, dtype object

After speaking with colleagues about this one I have an explanation: It looks like pandas is reverting to the bitwise operator:

In [1]: ~True
Out[1]: -2

As @geher says, you can convert it to bool with astype before you inverse with ~

~df['A'].astype(bool)
0    False
1     True
Name: A, dtype: bool
(~df['A']).astype(bool)
0    True
1    True
Name: A, dtype: bool

I just give it a shot:

In [9]: s = Series([True, True, True, False])

In [10]: s
Out[10]: 
0     True
1     True
2     True
3    False

In [11]: -s
Out[11]: 
0    False
1    False
2    False
3     True

You can also use numpy.invert:

In [1]: import numpy as np

In [2]: import pandas as pd

In [3]: s = pd.Series([True, True, False, True])

In [4]: np.invert(s)
Out[4]: 
0    False
1    False
2     True
3    False

EDIT: The difference in performance appears on Ubuntu 12.04, Python 2.7, NumPy 1.7.0 – doesn’t seem to exist using NumPy 1.6.2 though:

In [5]: %timeit (-s)
10000 loops, best of 3: 26.8 us per loop

In [6]: %timeit np.invert(s)
100000 loops, best of 3: 7.85 us per loop

In [7]: %timeit ~s
10000 loops, best of 3: 27.3 us per loop

NumPy is slower because it casts the input to boolean values (so None and 0 becomes False and everything else becomes True).

import pandas as pd
import numpy as np
s = pd.Series([True, None, False, True])
np.logical_not(s)

gives you

0    False
1     True
2     True
3    False
dtype: object

whereas ~s would crash. In most cases tilde would be a safer choice than NumPy.

Pandas 0.25, NumPy 1.17