这是一个矢量化的方法 –

def slicer_vectorized(a,start,end):

b = a.view('S1').reshape(len(a),-1)[:,start:end]

return np.fromstring(b.tostring(),dtype='S'+str(end-start))

样品运行 –

In [68]: a = np.array(['hello', 'how', 'are', 'you'])

In [69]: slicer_vectorized(a,1,3)

Out[69]:

array(['el', 'ow', 're', 'ou'],

dtype='|S2')

In [70]: slicer_vectorized(a,0,3)

Out[70]:

array(['hel', 'how', 'are', 'you'],

dtype='|S3')

运行时测试 –

测试其他作者发布的所有方法,我可以在最后运行,还包括本文前面的向量化方法.

这是时间 –

In [53]: # Setup input array

...: a = np.array(['hello', 'how', 'are', 'you'])

...: a = np.repeat(a,10000)

...:

# @Alberto Garcia-Raboso's answer

In [54]: %timeit slicer(1, 3)(a)

10 loops, best of 3: 23.5 ms per loop

# @hapaulj's answer

In [55]: %timeit np.frompyfunc(lambda x:x[1:3],1,1)(a)

100 loops, best of 3: 11.6 ms per loop

# Using loop-comprehension

In [56]: %timeit np.array([i[1:3] for i in a])

100 loops, best of 3: 12.1 ms per loop

# From this post

In [57]: %timeit slicer_vectorized(a,1,3)

1000 loops, best of 3: 787 µs per loop