@Kniffte: Warum hast du eine Liste von Arrays? Wenn du ein dreidimensionales Array hättest, könntest du das Array einfach mit der passenden Vertauschung transponieren:
Code: Alles auswählen
In [25]: list_of_points
Out[25]:
[array([[1, 2, 3],
[4, 5, 6],
[7, 8, 9]]), array([[1, 2, 3],
[4, 5, 6],
[7, 8, 9]]), array([[1, 2, 3],
[4, 5, 6],
[7, 8, 9]])]
In [26]: ps = np.array(list_of_points)
In [27]: ps
Out[27]:
array([[[1, 2, 3],
[4, 5, 6],
[7, 8, 9]],
[[1, 2, 3],
[4, 5, 6],
[7, 8, 9]],
[[1, 2, 3],
[4, 5, 6],
[7, 8, 9]]])
In [28]: ps.transpose(1, 0, 2)
Out[28]:
array([[[1, 2, 3],
[1, 2, 3],
[1, 2, 3]],
[[4, 5, 6],
[4, 5, 6],
[4, 5, 6]],
[[7, 8, 9],
[7, 8, 9],
[7, 8, 9]]])
In [32]: ps.transpose?
Docstring:
a.transpose(*axes)
Returns a view of the array with axes transposed.
For a 1-D array, this has no effect. (To change between column and
row vectors, first cast the 1-D array into a matrix object.)
For a 2-D array, this is the usual matrix transpose.
For an n-D array, if axes are given, their order indicates how the
axes are permuted (see Examples). If axes are not provided and
``a.shape = (i[0], i[1], ... i[n-2], i[n-1])``, then
``a.transpose().shape = (i[n-1], i[n-2], ... i[1], i[0])``.
Parameters
----------
axes : None, tuple of ints, or `n` ints
* None or no argument: reverses the order of the axes.
* tuple of ints: `i` in the `j`-th place in the tuple means `a`'s
`i`-th axis becomes `a.transpose()`'s `j`-th axis.
* `n` ints: same as an n-tuple of the same ints (this form is
intended simply as a "convenience" alternative to the tuple form)
Returns
-------
out : ndarray
View of `a`, with axes suitably permuted.
See Also
--------
ndarray.T : Array property returning the array transposed.
Examples
--------
>>> a = np.array([[1, 2], [3, 4]])
>>> a
array([[1, 2],
[3, 4]])
>>> a.transpose()
array([[1, 3],
[2, 4]])
>>> a.transpose((1, 0))
array([[1, 3],
[2, 4]])
>>> a.transpose(1, 0)
array([[1, 3],
[2, 4]])
Type: builtin_function_or_method
In [33]: