Numpy Array Slicing | 易学教程

问题

I have a 1D numpy array, and some offset/length values. I would like to extract from this array all entries which fall within offset, offset+length, which are then used to build up a new 'reduced' array from the original one, that only consists of those values picked by the offset/length pairs.

For a single offset/length pair this is trivial with standard array slicing [offset:offset+length]. But how can I do this efficiently (i.e. without any loops) for many offset/length values?

Thanks, Mark

回答1:

There is the naive method; just doing the slices:

>>> import numpy as np
>>> a = np.arange(100)
>>> 
>>> offset_length = [(3,10),(50,3),(60,20),(95,1)]
>>>
>>> np.concatenate([a[offset:offset+length] for offset,length in offset_length])
array([ 3,  4,  5,  6,  7,  8,  9, 10, 11, 12, 50, 51, 52, 60, 61, 62, 63,
       64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 95])

The following might be faster, but you would have to test/benchmark.

It works by constructing a list of the desired indices, which is valid method of indexing a numpy array.

>>> indices = [offset + i for offset,length in offset_length for i in xrange(length)]
>>> a[indices]
array([ 3,  4,  5,  6,  7,  8,  9, 10, 11, 12, 50, 51, 52, 60, 61, 62, 63,
       64, 65, 66, 67, 68, 69, 70, 71, 72, 73, 74, 75, 76, 77, 78, 79, 95])

It's not clear if this would actually be faster than the naive method but it might be if you have a lot of very short intervals. But I don't know.

(This last method is basically the same as @fraxel's solution, just using a different method of making the index list.)

Performance testing

I've tested a few different cases: a few short intervals, a few long intervals, lots of short intervals. I used the following script:

import timeit

setup = 'import numpy as np; a = np.arange(1000); offset_length = %s'

for title, ol in [('few short', '[(3,10),(50,3),(60,10),(95,1)]'),
                  ('few long', '[(3,100),(200,200),(600,300)]'),
                  ('many short', '[(2*x,1) for x in range(400)]')]:
  print '**',title,'**'
  print 'dbaupp 1st:', timeit.timeit('np.concatenate([a[offset:offset+length] for offset,length in offset_length])', setup % ol, number=10000)
  print 'dbaupp 2nd:', timeit.timeit('a[[offset + i for offset,length in offset_length for i in xrange(length)]]', setup % ol, number=10000)
  print '    fraxel:', timeit.timeit('a[np.concatenate([np.arange(offset,offset+length) for offset,length in offset_length])]', setup % ol, number=10000)

This outputs:

** few short **
dbaupp 1st: 0.0474979877472
dbaupp 2nd: 0.190793991089
    fraxel: 0.128381967545
** few long **
dbaupp 1st: 0.0416231155396
dbaupp 2nd: 1.58000087738
    fraxel: 0.228138923645
** many short **
dbaupp 1st: 3.97210478783
dbaupp 2nd: 2.73584890366
    fraxel: 7.34302687645

This suggests that my first method is the fastest when you have a few intervals (and it is significantly faster), and my second is the fastest when you have lots of intervals.

回答2:

>>> import numpy as np
>>> a = np.arange(100)
>>> ind = np.concatenate((np.arange(5),np.arange(10,15),np.arange(20,30,2),np.array([8])))
>>> a[[ind]]
array([ 0,  1,  2,  3,  4, 10, 11, 12, 13, 14, 20, 22, 24, 26, 28,  8])

来源：https://stackoverflow.com/questions/11061928/numpy-array-slicing

标签

python

arrays

numpy

scipy

numeric