您可以使用numpy.where
和numpy.diff
在第一列上找到索引值不同的索引:
>>> arr = np.array([(1.0, 3.0, 1, 427338.4297000002, 4848489.4332),
(1.0, 3.0, 2, 427344.7937000003, 4848482.0692),
(1.0, 3.0, 3, 427346.4297000002, 4848472.7469),
(1.0, 1.0, 7084, 427345.2709999997, 4848796.592),
(1.0, 1.0, 7085, 427352.9277999997, 4848790.9351),
(1.0, 1.0, 7086, 427359.16060000006, 4848787.4332)])
>>> np.split(arr, np.where(np.diff(arr[:,1]))[0]+1)
[array([[ 1.00000000e+00, 3.00000000e+00, 1.00000000e+00,
4.27338430e+05, 4.84848943e+06],
[ 1.00000000e+00, 3.00000000e+00, 2.00000000e+00,
4.27344794e+05, 4.84848207e+06],
[ 1.00000000e+00, 3.00000000e+00, 3.00000000e+00,
4.27346430e+05, 4.84847275e+06]]),
array([[ 1.00000000e+00, 1.00000000e+00, 7.08400000e+03,
4.27345271e+05, 4.84879659e+06],
[ 1.00000000e+00, 1.00000000e+00, 7.08500000e+03,
4.27352928e+05, 4.84879094e+06],
[ 1.00000000e+00, 1.00000000e+00, 7.08600000e+03,
4.27359161e+05, 4.84878743e+06]])]
首先,我们将在第二个第二列中获取项目:
>>> arr[:,1]
array([ 3., 3., 3., 1., 1., 1.])
现在要找出项目实际更改的位置,我们可以使用numpy.diff
:
>>> np.diff(arr[:,1])
array([ 0., 0., -2., 0., 0.])
任何非零的东西都意味着它旁边的项目是不同的,我们可以numpy.where
用来查找非零项目的索引,然后向其加1,因为该项目的实际索引比返回的索引大一:
>>> np.where(np.diff(arr[:,1]))[0]+1
array([3])