a = numpy.array([0,1,2,3,4,5,6,7,8,9])
我想指定位置5，并从两侧获取三个相邻元素。这可以实现。
index = 5
num_neighbor=3
left = a[index-num_neighbor:index]
right= a[num_neighbor+1:num_neighbor+index+1]
上述代码未考虑边界问题...我希望能够获取数组边界内的邻居。以以下示例为例，如果索引是1，则左侧邻居仅有一个元素，即0。
非常感谢
left = a[max(0,index-num_neighbor):index]
import numpy as np
a = np.array([0,1,2,3,4,5,6,7,8,9])
num_neighbor=3
for index in range(len(a)):
left = a[:index][-num_neighbor:]
right= a[index+1:num_neighbor+index+1]
print(index,left,right)
(0, array([], dtype=int32), array([1, 2, 3]))
(1, array([0]), array([2, 3, 4]))
(2, array([0, 1]), array([3, 4, 5]))
(3, array([0, 1, 2]), array([4, 5, 6]))
(4, array([1, 2, 3]), array([5, 6, 7]))
(5, array([2, 3, 4]), array([6, 7, 8]))
(6, array([3, 4, 5]), array([7, 8, 9]))
(7, array([4, 5, 6]), array([8, 9]))
(8, array([5, 6, 7]), array([9]))
(9, array([6, 7, 8]), array([], dtype=int32))
当
index<num_neighbor时，a[index-num_neighbor:index]不起作用的原因是由于切片规则＃3和＃4：给定
s [i：j]：如果i或j为负，则索引相对于字符串末尾：将替换为len（s）+ i或len（s）+ j。
从i到j的s的片段被定义为具有索引k的项目序列，使得i <= k < j。如果i或j大于len（s），则使用len（s）。如果省略i或None，则使用0。如果省略j或None，则使用len（s）。如果i大于或等于j，则切片为空。
因此，当
index = 1时，a [index-num_neighbor：index] = a [-2：1] = a [10-2：1] = a [8：1] = []。
def surrounding(x, idx, radius=1, fill=0):
"""
Gets surrounding elements from a numpy array
Parameters:
x (ndarray of rank N): Input array
idx (N-Dimensional Index): The index at which to get surrounding elements. If None is specified for a particular axis,
the entire axis is returned.
radius (array-like of rank N or scalar): The radius across each axis. If None is specified for a particular axis,
the entire axis is returned.
fill (scalar or None): The value to fill the array for indices that are out-of-bounds.
If value is None, only the surrounding indices that are within the original array are returned.
Returns:
ndarray: The surrounding elements at the specified index
"""
assert len(idx) == len(x.shape)
if np.isscalar(radius): radius = tuple([radius for i in range(len(x.shape))])
slices = []
paddings = []
for axis in range(len(x.shape)):
if idx[axis] is None or radius[axis] is None:
slices.append(slice(0, x.shape[axis]))
paddings.append((0, 0))
continue
r = radius[axis]
l = idx[axis] - r
r = idx[axis] + r
pl = 0 if l > 0 else abs(l)
pr = 0 if r < x.shape[axis] else r - x.shape[axis] + 1
slices.append(slice(max(0, l), min(x.shape[axis], r+1)))
paddings.append((pl, pr))
if fill is None: return x[slices]
return np.pad(x[slices], paddings, 'constant', constant_values=fill)
Python会为您处理边界问题：
>>> a = [0,1,2,3,4,5,6,7,8,9]
>>> a[-100 : 1000]
[0, 1, 2, 3, 4, 5, 6, 7, 8, 9]
>>> a[-100:3]
[0, 1, 2]
a [0]将返回0，a [-1]将返回9。 - talonmies
left和right的类比是什么？ - unutbu