Element-wise XOR in pandas

Question

I know that logical AND is &, and logical OR is | in a Pandas Series, but I was looking for an element-wise logical XOR. I could express it in terms of AND and OR, I suppose, but I'd prefer to use an XOR if one is available.

Thank you!

Solution

Python XOR: a ^ b

Numpy logical XOR: np.logical_xor(a,b)

Testing performance - result are equal:

1. Sequence of random booleans with size 10000

In [7]: a = np.random.choice([True, False], size=10000)
In [8]: b = np.random.choice([True, False], size=10000)

In [9]: %timeit a ^ b
The slowest run took 7.61 times longer than the fastest. This could mean that an intermediate result is being cached
100000 loops, best of 3: 11 us per loop

In [10]: %timeit np.logical_xor(a,b)
The slowest run took 6.25 times longer than the fastest. This could mean that an intermediate result is being cached
100000 loops, best of 3: 11 us per loop

2. Sequence of random booleans with size 1000

In [11]: a = np.random.choice([True, False], size=1000)
In [12]: b = np.random.choice([True, False], size=1000)

In [13]: %timeit a ^ b
The slowest run took 21.52 times longer than the fastest. This could mean that an intermediate result is being cached
1000000 loops, best of 3: 1.58 us per loop

In [14]: %timeit np.logical_xor(a,b)
The slowest run took 19.45 times longer than the fastest. This could mean that an intermediate result is being cached
1000000 loops, best of 3: 1.58 us per loop

3. Sequence of random booleans with size 100

In [15]: a = np.random.choice([True, False], size=100)
In [16]: b = np.random.choice([True, False], size=100)

In [17]: %timeit a ^ b
The slowest run took 33.43 times longer than the fastest. This could mean that an intermediate result is being cached
1000000 loops, best of 3: 614 ns per loop

In [18]: %timeit np.logical_xor(a,b)
The slowest run took 45.49 times longer than the fastest. This could mean that an intermediate result is being cached
1000000 loops, best of 3: 616 ns per loop

4. Sequence of random booleans with size 10

In [19]: a = np.random.choice([True, False], size=10)
In [20]: b = np.random.choice([True, False], size=10)

In [21]: %timeit a ^ b
The slowest run took 86.10 times longer than the fastest. This could mean that an intermediate result is being cached
1000000 loops, best of 3: 509 ns per loop

In [22]: %timeit np.logical_xor(a,b)
The slowest run took 40.94 times longer than the fastest. This could mean that an intermediate result is being cached
1000000 loops, best of 3: 511 ns per loop