Python 具有周期性边界条件的numpy数组的边界框(环绕)
我想做一些类似于问题或其他问题的事情,但使用周期性边界条件(包装)。我来举个简单的例子 假设我有以下numpy数组:Python 具有周期性边界条件的numpy数组的边界框(环绕),python,arrays,numpy,bounding-box,Python,Arrays,Numpy,Bounding Box,我想做一些类似于问题或其他问题的事情,但使用周期性边界条件(包装)。我来举个简单的例子 假设我有以下numpy数组: 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 1 1 1 1 1 0
0 0 0 0 0 1 0 0 0 0 0
0 0 0 0 0 1 0 0 0 0 0
0 0 1 1 1 1 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
然后,通过使用两个链接问题中提出的方法之一,我能够提取非零值的边界框:
0 0 0 1 1 1 1 1
0 0 0 1 0 0 0 0
0 0 0 1 0 0 0 0
1 1 1 1 0 0 0 0
但是,如果非零元素“穿过”边界并返回到另一侧,如下图所示:
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
1 1 0 0 0 0 0 0 1 1 1
0 0 0 0 0 0 0 0 1 0 0
0 0 0 0 0 0 0 0 1 0 0
0 0 0 0 0 1 1 1 1 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
结果是:
1 1 0 0 0 0 0 0 1 1 1
0 0 0 0 0 0 0 0 1 0 0
0 0 0 0 0 0 0 0 1 0 0
0 0 0 0 0 1 1 1 1 0 0
这不是我想要的。我希望结果与前一个案例相同。我正试图找出一个聪明的方法来做这件事,但我被卡住了。有人有想法吗?我们可以这样修改答案:
import numpy as np
def wrapped_bbox(a):
dims = [*range(1,a.ndim)]
bb = np.empty((a.ndim,2),int)
i = 0
while True:
n = a.shape[i]
r = np.arange(1,2*n+1)
ai = np.any(a,axis=tuple(dims))
r1_a = np.where(ai,r.reshape(2,n),0).ravel()
aux = np.maximum.accumulate(r1_a)
aux = r-aux
idx = aux.argmax()
mx = aux[idx]
if mx > n:
bb[i] = 0,n
else:
bb[i] = idx+1, idx+1 - mx
if bb[i,0] >= n:
bb[i,0] -= n
elif bb[i,1] == 0:
bb[i,1] = n
if i == len(dims):
return bb
dims[i] -= 1
i += 1
# example
x = """
......
.x...-
..x...
.....x
"""
x = np.array(x.strip().split())[:,None].view("U1")
x = (x == 'x').view('u1')
print(x)
for r in range(x.shape[1]):
print(wrapped_bbox(np.roll(x,r,axis=1)))
运行:
[[0 0 0 0 0 0] # x
[0 1 0 0 0 0]
[0 0 1 0 0 0]
[0 0 0 0 0 1]]
[[1 4] # bbox vert
[5 3]] # bbox horz, note wraparound (left > right)
[[1 4]
[0 4]] # roll by 1
[[1 4]
[1 5]] # roll by 2
[[1 4]
[2 6]] # etc.
[[1 4]
[3 1]]
[[1 4]
[4 2]]