numpy数组，如何选择满足多个条件的索引？

Question

假设我有一个numpy数组x = [5, 2, 3, 1, 4, 5]，y = ['f', 'o', 'o', 'b', 'a', 'r']。我要选择与大于1小于5 的元素y相对应的元素x。我试过了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']但这不起作用。我该怎么做？python numpy的

Answer 1

如果添加括号，则表达式有效：

>>> y[(1 < x) & (x < 5)]

array(['o', 'o', 'a'], 

      dtype='|S1')

Answer 2

IMO OP实际上并不需要np.bitwise_and()（aka &），但实际上是需要的，np.logical_and()因为它们正在比较逻辑值，例如True和False-请参阅此SO 逻辑与按位比较，以了解区别。

>>> 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')

同样的方法是np.all()通过axis适当设置参数。

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

>>> output

array(['o', 'o', 'a'],

      dtype='|S1')

通过数字：

>>> %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

所以使用np.all()比较慢，但&和logical_and大致相同。