如何根据条件选择数组元素？

Question

假设我有一个numpy数组 x = [5, 2, 3, 1, 4, 5]，还有一个数组 y = ['f', 'o', 'o', 'b', 'a', 'r']。我想从 y 中选出那些和 x 中大于1且小于5的元素对应的元素。

我试过这样做：

x = array([5, 2, 3, 1, 4, 5])
y = array(['f','o','o','b','a','r'])
output = y[x > 1 & x < 5] # desired output is ['o','o','a']

但是这样不行。我该怎么做呢？

Answer 1

补充一下@J.F. Sebastian和@Mark Mikofski的回答：
如果你想获取对应的索引（而不是数组的实际值），可以使用以下代码：

满足多个（全部）条件的情况：

select_indices = np.where( np.logical_and( x > 1, x < 5) )[0] #   1 < x <5

满足多个（任意）条件的情况：

select_indices = np.where( np.logical_or( x < 1, x > 5 ) )[0] # x <1 or x >5

Answer 2

我觉得提问者其实并不想要使用np.bitwise_and()（也就是&），而是想要用np.logical_and()，因为他们在比较的是逻辑值，比如True和False。想了解这两者的区别，可以看看这个关于逻辑与位运算的帖子。

>>> x = array([5, 2, 3, 1, 4, 5])
>>> y = array(['f','o','o','b','a','r'])
>>> output = y[np.logical_and(x > 1, x < 5)] # desired output is ['o','o','a']
>>> output
array(['o', 'o', 'a'],
      dtype='|S1')

还有一种等效的方法是使用np.all()，只需要适当地设置axis参数。

>>> output = y[np.all([x > 1, x < 5], axis=0)] # desired output is ['o','o','a']
>>> output
array(['o', 'o', 'a'],
      dtype='|S1')

具体来说：

>>> %timeit (a < b) & (b < c)
The slowest run took 32.97 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 1.15 µs per loop

>>> %timeit np.logical_and(a < b, b < c)
The slowest run took 32.59 times longer than the fastest. This could mean that an intermediate result is being cached.
1000000 loops, best of 3: 1.17 µs per loop

>>> %timeit np.all([a < b, b < c], 0)
The slowest run took 67.47 times longer than the fastest. This could mean that an intermediate result is being cached.
100000 loops, best of 3: 5.06 µs per loop

使用np.all()会稍微慢一些，但&和logical_and的速度差不多。

Answer 3

你的表达式如果加上括号就能正常工作：

>>> y[(1 < x) & (x < 5)]
array(['o', 'o', 'a'], 
      dtype='|S1')