这里有一个使用切片和 argmax
来获取边界的例子 -
def smallestbox(a):
r = a.any(1)
if r.any():
m,n = a.shape
c = a.any(0)
out = a[r.argmax():m-r[::-1].argmax(), c.argmax():n-c[::-1].argmax()]
else:
out = np.empty((0,0),dtype=bool)
return out
范例运行 -
In [142]: a
Out[142]:
array([[False, False, False, False, False, False],
[False, True, False, True, False, False],
[False, True, True, False, False, False],
[False, False, False, False, False, False]])
In [143]: smallestbox(a)
Out[143]:
array([[ True, False, True],
[ True, True, False]])
In [144]: a[:] = 0
In [145]: smallestbox(a)
Out[145]: array([], shape=(0, 0), dtype=bool)
In [146]: a[2,2] = 1
In [147]: smallestbox(a)
Out[147]: array([[ True]])
基准测试
其他方法 -
def argwhere_app(a):
coords = np.argwhere(a)
x_min, y_min = coords.min(axis=0)
x_max, y_max = coords.max(axis=0)
return a[x_min:x_max+1, y_min:y_max+1]
稀疏度不同程度的时间(大约10%,50%和90%) -
In [370]: np.random.seed(0)
...: a = np.random.rand(5000,5000)>0.1
In [371]: %timeit argwhere_app(a)
...: %timeit smallestbox(a)
1 loop, best of 3: 310 ms per loop
100 loops, best of 3: 3.19 ms per loop
In [372]: np.random.seed(0)
...: a = np.random.rand(5000,5000)>0.5
In [373]: %timeit argwhere_app(a)
...: %timeit smallestbox(a)
1 loop, best of 3: 324 ms per loop
100 loops, best of 3: 3.21 ms per loop
In [374]: np.random.seed(0)
...: a = np.random.rand(5000,5000)>0.9
In [375]: %timeit argwhere_app(a)
...: %timeit smallestbox(a)
10 loops, best of 3: 106 ms per loop
100 loops, best of 3: 3.19 ms per loop