假设我有一个名为array
的2d数组和一个2d索引:(x,y)
,我想获取最接近(x,y)
的非零元素,并获取np.nonzero(array)
中元素的对应索引
def nearest_nonzero_idx(a,x,y):
idx = np.argwhere(a)
# If (x,y) itself is also non-zero, we want to avoid those, so delete that
# But, if we are sure that (x,y) won't be non-zero, skip the next step
idx = idx[~(idx == [x,y]).all(1)]
return idx[((idx - [x,y])**2).sum(1).argmin()]
样品运行-
In [64]: a
Out[64]:
array([[0, 0, 1, 1, 0, 1, 1],
[0, 0, 1, 0, 0, 0, 1],
[1, 1, 0, 0, 0, 0, 0],
[0, 1, 1, 0, 0, 0, 0],
[0, 1, 0, 0, 0, 0, 0],
[1, 1, 0, 0, 1, 1, 0],
[0, 1, 0, 0, 1, 0, 1],
[1, 0, 0, 1, 1, 1, 0]])
In [65]: x,y =(3,5)
In [66]: nearest_nonzero(a,x,y)
Out[66]: array([5, 5])
方法#2:这是另一种注重性能的方法,该方法避免了通过将点临时设置为(x,y)
来从非零索引数组中跳过0
点的先前方法的第二步,获取那些非零索引,然后将原始值放回原值。同样,我们可以使用np.nonzero
存储row, col
索引,然后使用它们进行距离计算。因此,实现将是-
def nearest_nonzero_idx_v2(a,x,y): tmp = a[x,y] a[x,y] = 0 r,c = np.nonzero(a) a[x,y] = tmp min_idx = ((r - x)**2 + (c - y)**2).argmin() return r[min_idx], c[min_idx]
运行时测试
In [110]: a Out[110]: array([[3, 2, 3, 3, 0, 2, 4, 2, 1], [0, 3, 4, 3, 4, 3, 3, 2, 0], [1, 3, 0, 0, 0, 0, 0, 0, 0], [0, 1, 2, 0, 0, 2, 0, 0, 2], [3, 0, 0, 0, 0, 0, 0, 0, 1], [0, 0, 2, 2, 4, 4, 3, 4, 3], [2, 2, 2, 1, 0, 0, 1, 1, 1], [3, 4, 3, 1, 0, 4, 0, 4, 2]]) In [111]: x,y =(3,5) In [112]: nearest_nonzero_idx(a,x,y) Out[112]: array([1, 5]) In [113]: nearest_nonzero_idx_v2(a,x,y) Out[113]: (1, 5) In [114]: %timeit nearest_nonzero_idx(a,x,y) 10000 loops, best of 3: 23.1 µs per loop In [115]: %timeit nearest_nonzero_idx_v2(a,x,y) 100000 loops, best of 3: 4.85 µs per loop