numpy的数组分片 [英] Numpy Array Slicing

问题描述

我有一个一维数组numpy的，有些偏移/长度值。我想从这个数组中提取落入内的偏移的所有条目，偏移+长度，然后将其用于建立从原来一个新的降低数组，即只包括由偏移/长度成对拾取这些值的

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.

对于单个偏移/长度对这个是微不足道的标准阵列切片 [偏移：偏移+长度] 。但是，如何才能有效地我这样做（即没有任何循环）为许多偏移/长度值？

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?

谢谢，
马克

推荐答案

有是天真的方法;只是在做切片：

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])

---

下面的可能的比较快，但是你必须测试/基准。

---

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

它的工作原理构造的所需索引的列表，这是编入索引numpy的阵列的有效方法。

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])

目前尚不清楚这是否会实际上比天真的方法快，但如果你有很多非常短的间隔可能是。但我不知道。

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.

（这最后的方法是基本相同@飞梭的溶液，只是使用使得索引列表的不同的方法。）

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

我测试了几个不同的情况：短短的时间间隔，几个长的时间间隔，大量的短间隔。我用下面的脚本：

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)

此输出：

** 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

这表明，我的第一个方法是最快的，当你有几个区间（它是显著更快），和我的第二个是最快的，当你有很多时间间隔。

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.

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