在numpy数组中一次访问块

问题描述：

提供了一个numpy数组:

Provided a numpy array:

arr = np.array([0,1,2,3,4,5,6,7,8,9,10,11,12])

我想知道如何访问选定大小的块以及选定的分隔，包括串联和切片:

I wonder how access chosen size chunks with chosen separation, both concatenated and in slices:

例如:获取大小为3的块，并用两个值分隔:

E.g.: obtain chunks of size 3 separated by two values:

arr_chunk_3_sep_2 = np.array([0,1,2,5,6,7,10,11,12])
arr_chunk_3_sep_2_in_slices = np.array([[0,1,2],[5,6,7],[10,11,12])

哪种方法最有效?如果可能的话，我想尽可能避免复制或创建新对象.也许 Memoryviews 在这里可能有帮助?

Wha is the most efficient way to do it? If possible, I would like to avoid copying or creating new objects as much as possible. Maybe Memoryviews could be of help here?

答

方法1

这里是masking-

def slice_grps(a, chunk, sep):
    N = chunk + sep
    return a[np.arange(len(a))%N < chunk]

样品运行-

In [223]: arr
Out[223]: array([ 0,  1,  2,  3,  4,  5,  6,  7,  8,  9, 10, 11, 12])

In [224]: slice_grps(arr, chunk=3, sep=2)
Out[224]: array([ 0,  1,  2,  5,  6,  7, 10, 11, 12])

方法2

如果输入数组使得最后一块具有足够的跑道，我们可以利用 np.lib.stride_tricks.as_strided ，受 this post 的启发，选择m每个n元素块中的所有元素-

If the input array is such that the last chunk would have enough runway, we could , we could leverage np.lib.stride_tricks.as_strided, inspired by this post to select m elements off each block of n elements -

# https://stackoverflow.com/a/51640641/ @Divakar
def skipped_view(a, m, n):
    s = a.strides[0]
    strided = np.lib.stride_tricks.as_strided
    shp = ((a.size+n-1)//n,n)
    return strided(a,shape=shp,strides=(n*s,s), writeable=False)[:,:m]

out = skipped_view(arr,chunk,chunk+sep)

请注意，输出将是输入数组的视图，因此不会产生额外的内存开销，并且实际上是免费的！

Note that the output would be a view into the input array and as such no extra memory overhead and virtually free!

运行样本以使事情变得清晰-

Sample run to make things clear -

In [255]: arr
Out[255]: array([ 0,  1,  2,  3,  4,  5,  6,  7,  8,  9, 10, 11, 12])

In [256]: chunk = 3

In [257]: sep = 2

In [258]: skipped_view(arr,chunk,chunk+sep)
Out[258]: 
array([[ 0,  1,  2],
       [ 5,  6,  7],
       [10, 11, 12]])

# Let's prove that the output is a view indeed
In [259]: np.shares_memory(arr, skipped_view(arr,chunk,chunk+sep))
Out[259]: True

在numpy数组中一次访问块

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