python - 检查 numpy 数组是否为二进制的快速方法(仅包含 0 和 1)

Question

给定一个 numpy 数组，如果它只包含 0 和 1，我如何快速计算出它？有实现方法吗？

Answer 1

一些方法-

((a==0) | (a==1)).all()
~((a!=0) & (a!=1)).any()
np.count_nonzero((a!=0) & (a!=1))==0
a.size == np.count_nonzero((a==0) | (a==1))

运行时测试-

In [313]: a = np.random.randint(0,2,(3000,3000)) # Only 0s and 1s

In [314]: %timeit ((a==0) | (a==1)).all()
     ...: %timeit ~((a!=0) & (a!=1)).any()
     ...: %timeit np.count_nonzero((a!=0) & (a!=1))==0
     ...: %timeit a.size == np.count_nonzero((a==0) | (a==1))
     ...: 
10 loops, best of 3: 28.8 ms per loop
10 loops, best of 3: 29.3 ms per loop
10 loops, best of 3: 28.9 ms per loop
10 loops, best of 3: 28.8 ms per loop

In [315]: a = np.random.randint(0,3,(3000,3000)) # Contains 2 as well

In [316]: %timeit ((a==0) | (a==1)).all()
     ...: %timeit ~((a!=0) & (a!=1)).any()
     ...: %timeit np.count_nonzero((a!=0) & (a!=1))==0
     ...: %timeit a.size == np.count_nonzero((a==0) | (a==1))
     ...: 
10 loops, best of 3: 28 ms per loop
10 loops, best of 3: 27.5 ms per loop
10 loops, best of 3: 29.1 ms per loop
10 loops, best of 3: 28.9 ms per loop

它们的运行时间似乎相当。