在numpy中的3维矩阵中插入未对齐元素 [英] Insertion of non aligned elements in 3-dimensional matrices in numpy
问题描述
我正在使用 numpy 1.9 和 python 2.7.5 处理3维矩阵. 这是一个示例:
I'm working with 3-dimensional matrices using numpy 1.9 and python 2.7.5. Here is an example:
>>> A
array([[[ 1., 1., 1.],
[ 1., 1., 1.],
[ 1., 1., 1.],
[ 1., 1., 1.],
[ 1., 1., 1.]],
[[ 1., 1., 1.],
[ 1., 1., 1.],
[ 1., 1., 1.],
[ 1., 1., 1.],
[ 1., 1., 1.]]])
>>> B
array([[[-1., -1., -1.],
[99., 100., 101.],
[-1., -1., -1.],
[-1., -1., -1.],
[-1., -1., -1.]],
[[-1., -1., -1.],
[-1., -1., -1.],
[102., 103., 104.],
[-1., -1., -1.],
[-1., -1., -1.]]])
>>> C
array([1, 2])
根据C，我想将B中的所有元素插入A中.
示例:c[0] = 1 => After A[0, 1, :] has to be inserted B[0, 1, :]
I'd like to insert in A all elements from B, according to C.
Example: c[0] = 1 => After A[0, 1, :] has to be inserted B[0, 1, :]
这是预期结果的一个例子
Here is an example of the expected result
>>> D
array([[[1., 1., 1.],
[1., 1., 1.],
[99., 100., 101.],
[1., 1., 1.],
[1., 1., 1.],
[1., 1., 1.]],
[[1., 1., 1.],
[1., 1., 1.],
[1., 1., 1.],
[102., 103., 104.]
[1., 1., 1.],
[1., 1., 1.]]])
我发现了这个stackoverflow问题除了解决方案仅适用于二维矩阵并且我正在使用3维之外，它与我的非常相似.
I found this stackoverflow question that is really similar to mine except that the solution are only for 2-dimensional matrix and I'm working with 3-dimensional.
这是我的解决方法，但结果不正确:
Here is my solution but I get incorrect results:
C2 = np.repeat(C, 3)
r1 = np.repeat(np.arange(A.shape[0]), 3)
r2 = np.tile(np.arange(3), A.shape[0])
index_map = np.ravel_multi_index((r1, C2, r2), A.shape) + 1
np.insert(A.ravel(), index_map, B.ravel()[index_map]).reshape(A.shape[0], A.shape[1] + 1, A.shape[2])
这是使用for循环的正确但缓慢的解决方案:
Here is a correct, but slow, solution using a for loop:
A_2 = np.zeros((A.shape[0], A.shape[1] + 1, A.shape[2]))
for j in xrange(np.size(C, 0)):
i = C[j]
A_2[j, :, :] = np.concatenate((A[j, 0:i + 1, :], [B[j, i, :]], A[j, i + 1:, :]))
有什么主意吗?
谢谢！
推荐答案
您的代码存在的问题是，当您需要插入多个 元素顺序排列，则需要将它们插入相同位置. 比较:
The problem with your code is that when you need to insert several elements sequentially, you need to insert them at the same position. Compare:
In [139]: x = np.ones(5); x
Out[139]: array([ 1., 1., 1., 1., 1.])
In [140]: np.insert(x, [1,2,3], 100)
Out[140]: array([ 1., 100., 1., 100., 1., 100., 1., 1.])
In [141]: np.insert(x, [1,1,1], 100)
Out[141]: array([ 1., 100., 100., 100., 1., 1., 1., 1.])
编辑:原始答案包括完整的整理/整形
返回，但是在3d模式下，您需要多加注意才能正确执行操作.有一个
考虑到np.insert和
np.take接受轴"参数，并允许多值插入.
这仍然需要进行一些重塑，但是并不能求助于
选择另外，请注意np.insert的mi+1参数要插入
之后，而不是所选行之前:
The original answer included full unravelling/reshaping
back, but in 3d you need a lot of care to do it right. There's an
easier solution that takes into account the fact that np.insert and
np.take accept "axis" parameter and allow multi-value insertion.
That still requires some reshaping, but it doesn't resort to
np.choose. Also, note the mi+1 argument to np.insert to insert
after, not before the chosen rows:
In [50]: mi = np.ravel_multi_index([np.arange(A.shape[0]), C], A.shape[:2]); mi
Out[50]: array([1, 7])
In [51]: bvals = np.take(B.reshape(-1, B.shape[-1]), mi, axis=0); bvals
Out[51]:
array([[ 99., 100., 101.],
[ 102., 103., 104.]])
In [52]: result = (np.insert(A.reshape(-1, A.shape[2]), mi + 1, bvals, axis=0)
.reshape(A.shape[0], -1, A.shape[2])); result
Out[52]:
array([[[ 1., 1., 1.],
[ 1., 1., 1.],
[ 99., 100., 101.],
[ 1., 1., 1.],
[ 1., 1., 1.],
[ 1., 1., 1.]],
[[ 1., 1., 1.],
[ 1., 1., 1.],
[ 1., 1., 1.],
[ 102., 103., 104.],
[ 1., 1., 1.],
[ 1., 1., 1.]]])
这是原始答案:
In [18]: ixs = np.repeat(np.array([np.arange(A.shape[0]),
C+1,
np.zeros(A.shape[0], dtype=np.int_)]),
A.shape[2], axis=1); ixs
....:
Out[18]:
array([[0, 0, 0, 1, 1, 1],
[2, 2, 2, 3, 3, 3],
[0, 0, 0, 0, 0, 0]])
In [19]: mi = np.ravel_multi_index(ixs, A.shape); mi
Out[19]: array([ 6, 6, 6, 24, 24, 24])
In [20]: result = (np.insert(A.ravel(), mi, bvals)
.reshape(A.shape[0], A.shape[1] +1, A.shape[2])); result
....:
Out[20]:
array([[[ 1., 1., 1.],
[ 1., 1., 1.],
[ 99., 100., 101.],
[ 1., 1., 1.],
[ 1., 1., 1.],
[ 1., 1., 1.]],
[[ 1., 1., 1.],
[ 1., 1., 1.],
[ 1., 1., 1.],
[ 102., 103., 104.],
[ 1., 1., 1.],
[ 1., 1., 1.]]])
In [21]: result = (np.insert(A.ravel(), mi, bvals)
.reshape(A.shape[0], A.shape[1] +1, A.shape[2])); result
....:
Out[21]:
array([[[ 1., 1., 1.],
[ 1., 1., 1.],
[ 99., 100., 101.],
[ 1., 1., 1.],
[ 1., 1., 1.],
[ 1., 1., 1.]],
[[ 1., 1., 1.],
[ 1., 1., 1.],
[ 1., 1., 1.],
[ 102., 103., 104.],
[ 1., 1., 1.],
[ 1., 1., 1.]]])
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