pandas 熊猫中的逐元素异或
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Element-wise XOR in pandas
提问by SulfoCyaNate
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.
我知道逻辑 AND 是 &，逻辑 OR 是 | 在 Pandas 系列中，但我正在寻找一个元素明智的逻辑异或。我想我可以用 AND 和 OR 来表达它，但我更愿意使用 XOR，如果有的话。
Thank you!
谢谢！
回答by jezrael
Python XOR: a ^ b
Python异或： a ^ b
Numpy logical XOR: np.logical_xor(a,b)
Numpy 逻辑异或：np.logical_xor(a,b)
Testing performance - result are equal:
测试性能 - 结果相等：
1. Sequence of random booleans with size 10000
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
2. Sequence of random booleans with size 1000
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
3. Sequence of random booleans with size 100
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
4. Sequence of random booleans with size 10
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
