大熊猫的元素XOR

Sul*_*ate 7 python logic xor pandas

我知道逻辑AND是&,逻辑OR是| 在Pandas系列中,但我一直在寻找符合逻辑的逻辑异或.我想,我可以用AND和OR来表达它,但如果有可用的话,我更愿意使用XOR.

谢谢!

jez*_*ael 10

Python异或: a ^ b

Numpy逻辑异或:np.logical_xor(a,b)

测试性能 - 结果相同:

1.大小为10000的随机布尔序列

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
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2.大小为1000的随机布尔序列

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
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3.大小为100的随机布尔序列

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
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4.大小为10的随机布尔序列

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
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