如何在NumPy中将N-D切片转换为索引? [英] How to convert N-D slice to indexes in NumPy?

问题描述

给出 NumPy 中的任何N个元组切片(也称为ND切片)，如何将其转换为ND数组表示为一维数组的元组(沿每个轴的索引)?例如.如果我们有np.nd_slice_to_indexes下一个代码:

Given any N-tuple of slices (aka N-D slice) in NumPy how to convert it to corresponding indexes of N-D array represented as tuple of 1D arrays (indexes along each axes)? E.g. if we have np.nd_slice_to_indexes next code:

import numpy as np
print(np.nd_slice_to_indexes(np.s_[1 : 3]))
print(np.nd_slice_to_indexes(np.s_[1 : 3, 5 : 11 : 2]))

应打印

(array([1, 2]),)
(array([1, 1, 1, 2, 2, 2]), array([5, 7, 9, 5, 7, 9]))

NumPy通常将N-D数组的索引表示为相同长度的一维数组的N-元组(元组中k-th数组的每个元素表示沿第k维的下一个索引).例如. np.nonzero在代码中返回这样的N元组

It is common for NumPy to represent indexes of N-D array as N-tuple of 1-D arrays of same length (each element of k-th array in tuple represents next index along k-th dimension). E.g. np.nonzero returns such N-tuple in code

print(np.nonzero([[0, 1, 1], [1, 1, 0]])) # Non-zero elements in 2D array.
# (array([0, 0, 1, 1], dtype=int64), array([1, 2, 0, 1], dtype=int64))

应该像下面的Pythonic函数一样实现相同的行为，但是要采用更有效的方式:

Same behavior should be achieved like in Pythonic function below, but in a more efficient (performant) way:

在线试用！

import numpy as np

def nd_slice_to_indexes(nd_slice):
    assert type(nd_slice) in [tuple, slice], type(nd_slice)
    if type(nd_slice) is not tuple:
        nd_slice = (nd_slice,)
    def iter_slices(slices):
        if len(slices) == 0:
            yield ()
        else:
            for i in range(slices[0].start, slices[0].stop, slices[0].step or 1):
                for r in iter_slices(slices[1:]):
                    yield (i,) + r
    *res, = np.vstack(list(iter_slices(nd_slice))).T
    return tuple(res)

print(nd_slice_to_indexes(np.s_[1 : 3]))
print(nd_slice_to_indexes(np.s_[1 : 3, 5 : 11 : 2]))
print(nd_slice_to_indexes(np.s_[1 : 3, 5 : 11 : 2, 8 : 14 : 3]))
# (array([1, 2]),)
# (array([1, 1, 1, 2, 2, 2]), array([5, 7, 9, 5, 7, 9]))
# (array([1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2]), array([5, 5, 7, 7, 9, 9, 5, 5, 7, 7, 9, 9]), array([ 8, 11,  8, 11,  8, 11,  8, 11,  8, 11,  8, 11]))

推荐答案

感谢 np.mgrid .

在线试用！

import numpy as np

def nd_slice_to_indexes(nd_slice):
    grid = np.mgrid[{tuple: nd_slice, slice: (nd_slice,)}[type(nd_slice)]]
    return tuple(grid[i].ravel() for i in range(grid.shape[0]))
    
print(nd_slice_to_indexes(np.s_[1 : 3]))
print(nd_slice_to_indexes(np.s_[1 : 3, 5 : 11 : 2]))
print(nd_slice_to_indexes(np.s_[1 : 3, 5 : 11 : 2, 8 : 14 : 3]))
# (array([1, 2]),)
# (array([1, 1, 1, 2, 2, 2]), array([5, 7, 9, 5, 7, 9]))
# (array([1, 1, 1, 1, 1, 1, 2, 2, 2, 2, 2, 2]), array([5, 5, 7, 7, 9, 9, 5, 5, 7, 7, 9, 9]), array([ 8, 11,  8, 11,  8, 11,  8, 11,  8, 11,  8, 11]))

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