Python 多维切片的紧凑表示法

假设我有3（或更多）维数组

A

和两个数组，它们具有最小和最大边

左

，

右

，用于切片

是否有比此更紧凑的符号来选择由“我的边”分隔的视图

V = A[left[0]:right[0], left[1]:right[1], left[2]:right[2])

也许这样已经更好了

view = [ slice(a, b) for a,b in zip(left, right) ]
V = A[view]

但我觉得有一种更为自然的方式我看不到。。。不知道，用

np.s.

的东西

编辑：预期结果的示例

A = np.arange(1000).reshape(10,10,10)
left = np.array([0, 0, 0])
right = np.array([2, 4, 3])
view = [ slice(a, b) for a,b in zip(left, right) ]

In [32]: A[view]
Out[32]: 
array([[[  0,   1,   2],
        [ 10,  11,  12],
        [ 20,  21,  22],
        [ 30,  31,  32]],

       [[100, 101, 102],
        [110, 111, 112],
        [120, 121, 122],
        [130, 131, 132]]])

---

您可以子类化

np.s\uu

以支持“矢量化”切片表示法：

import numpy as np
import types

class Block_Slice(np.lib.index_tricks.IndexExpression):
    @staticmethod
    def expand_slice(slice_, result_type=types.GeneratorType, force_type=False):
        bc = np.broadcast(slice_.start, slice_.stop, slice_.step)
        if bc.size == 1 and not force_type:
            return slice_
        if result_type in ('generator', types.GeneratorType):
            return (slice(*idx) for idx in bc)
        result_type = tuple if result_type=='tuple' else result_type
        return result_type(slice(*idx) for idx in bc)
    def __getitem__(self, item):
        if isinstance(item, slice):
            item = self.expand_slice(item, tuple)
        elif isinstance(item, tuple):
            item = tuple(j for i in item for j in
                         (self.expand_slice(i, force_type=True)
                          if isinstance(i, slice) else (i,)))
        return super().__getitem__(item)

bs_ = Block_Slice(maketuple=False)


a = sum(np.ogrid[:2000:1000, :300:100, :40:10, :6])
left = [1, 0, 0]
right = [2, 2, 4]
a[bs_[left:right]]
# array([[[[1000, 1001, 1002, 1003, 1004, 1005],
#          [1010, 1011, 1012, 1013, 1014, 1015],
#          [1020, 1021, 1022, 1023, 1024, 1025],
#          [1030, 1031, 1032, 1033, 1034, 1035]],

#         [[1100, 1101, 1102, 1103, 1104, 1105],
#          [1110, 1111, 1112, 1113, 1114, 1115],
#          [1120, 1121, 1122, 1123, 1124, 1125],
#          [1130, 1131, 1132, 1133, 1134, 1135]]]])
a[bs_[..., range(3):5]]
# array([[[[  12,   13,   14],
#          [  22,   23,   24],
#          [  32,   33,   34]],

#         [[ 112,  113,  114],
#          [ 122,  123,  124],
#          [ 132,  133,  134]],

#         [[ 212,  213,  214],
#          [ 222,  223,  224],
#          [ 232,  233,  234]]],


#        [[[1012, 1013, 1014],
#          [1022, 1023, 1024],
#          [1032, 1033, 1034]],

#         [[1112, 1113, 1114],
#          [1122, 1123, 1124],
#          [1132, 1133, 1134]],

#         [[1212, 1213, 1214],
#          [1222, 1223, 1224],
#          [1232, 1233, 1234]]]])
a[bs_[:right, [4, 2, 1]]]
# array([[[[   4,    2,    1],
#          [  14,   12,   11],
#          [  24,   22,   21],
#          [  34,   32,   31]],

#         [[ 104,  102,  101],
#          [ 114,  112,  111],
#          [ 124,  122,  121],
#          [ 134,  132,  131]]],


#        [[[1004, 1002, 1001],
#          [1014, 1012, 1011],
#          [1024, 1022, 1021],
#          [1034, 1032, 1031]],

#         [[1104, 1102, 1101],
#          [1114, 1112, 1111],
#          [1124, 1122, 1121],
#          [1134, 1132, 1131]]]])

---

您有示例数组和预期的输出吗？当然，请参阅编辑

np.s

，它只允许您从第一个表达式

np.s_[left[0]：right[0]，…]生成切片的元组。不管怎样，您都需要生成切片的元组。numpythonic方法只是一个Python编码任务。如果你想让它变得漂亮，把它包装在一个函数（或类）中。也许你是对的，它已经是这样了。。。在编写python/numpy代码时，我总是觉得自己比我应该/能够做的更冗长，而且我缺少一些使事情变得优雅的符号技巧
A[…]
是
A的缩写。uu getitem\uuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuuu
atuple
有三个
slice
对象。每个
slice
都有3个属性。我不知道下一票是什么，我喜欢它，可能对我的需要有点过分，但它在我的生活中可能会很有用toolbox@filippo仅供参考，我已经发布了一个正确的版本。