How can I obtain the element-wise logical NOT of a pandas Series?

Question

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 =(

Solution 1

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.

Solution 2

@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

Solution 3

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

Solution 4

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

Solution 5

In support to the excellent answers here, and for future convenience, there may be a case where you want to flip the truth values in the columns and have other values remain the same (nan values for instance)

In[1]: series = pd.Series([True, np.nan, False, np.nan])
In[2]: series = series[series.notna()] #remove nan values
 
In[3]: series # without nan                                            
Out[3]: 
0     True
2    False
dtype: object

# Out[4] expected to be inverse of Out[3], pandas applies bitwise complement 
# operator instead as in `lambda x : (-1*x)-1`

In[4]: ~series
Out[4]: 
0    -2
2    -1
dtype: object

as a simple non-vectorized solution you can just, 1. check types2. inverse bools

In[1]: series = pd.Series([True, np.nan, False, np.nan])

In[2]: series = series.apply(lambda x : not x if x is bool else x)
Out[2]: 
Out[2]: 
0     True
1      NaN
2    False
3      NaN
dtype: object

View more solutions

Comments

[LearnOPhile]

It is important that the data does not contain object types for the answers below to work, so use: ~ df.astype('bool')

[cs95]

I've written about all of the logical operators in this post. The post also includes alternatives.

[Louis Thibault]

I literally tried every operator other than -! I'll keep this in mind for next time.

[Louis Thibault]

Duly noted. Other than being much slower, what's the difference between the tilde and - ?

[root]

Wierd, I actually tested the tilde as it was mentioned in the documentation, but it didn't perform the same as np.invert :S

[unutbu]

@blz: At least on my Ubuntu machine, running NumPy 1.6.2, the performance of np.invert(s), ~s and -s are all the same.

[unutbu]

@root: I'm not sure why there is such a great discrepancy in our timeit results, but it certainly can happen. What OS and version of NumPy are you using?

[root]

Also on Ubuntu, but using NumPy 1.7.0...(np.bitwise_not(s) performs the same as np.inverse).

[Louis Thibault]

@root, @unutbu, I can confirm that np.invert and the ~ operator have identical performance on my machine as well: numpy 1.6.2 on Ubuntu latest.

[Robert Pollak]

Where does the Pandas documentation tell about the ~ operator?

[unutbu]

@RobertPollak: It is mentioned here.

[gaozhidf]

it may not be correct on a different platform. Win 7, python 3.6.3 numpy 1.13.3, pandas 0.20.3, (-s) will be the fastest, (~s) is the second, and np.invert(s) is the slowest one

[geher]

in your example, the output ints mask can be converted to the bool series you want with .astype(bool) e.g. ~df['A'].astype(bool)

[JSharm]

This is working because astype(bool) is happening before the ~ ~df['A'].astype(bool) vs (~df['A']).astype(bool)

[Olsgaard]

For completeness sake, consider adding %timeit s == False - it's about twice as slow as the slowest contender, but it paints a more complete picture, IMHO :)