I have a 3d array from which I am trying to a list of stabs. Put another way, given the array:
t = np.array([[[1,2],[3,4]],[[5,6],[7,8]],[[9,10],[11,12]]])
array([[[ 1, 2],
[ 3, 4]],
[[ 5, 6],
[ 7, 8]],
[[ 9, 10],
[11, 12]]])
I am trying to retrieve:
array([[ 1, 5, 9],
[ 2, 6, 10],
[ 3, 7, 11],
[ 4, 8, 12]])
np.ndarray.reshape seems to reorganise elements in a sequential order that precludes stabs.
numpy.lib.stride_tricks.as_strided might work, but I have yet to find the correct combination of values.
Transpose then reshape:
>>> t.transpose(1, 2, 0).reshape(4, 3)
array([[ 1, 5, 9],
[ 2, 6, 10],
[ 3, 7, 11],
[ 4, 8, 12]])
Edit: alternatively, you can reshape then transpose:
>>> t.reshape(3, 4).T
array([[ 1, 5, 9],
[ 2, 6, 10],
[ 3, 7, 11],
[ 4, 8, 12]])
Related
I have been going over this issue with numpy for a while and cant figure out if there is a intuitive way of converting the array while maintaining the position of the sub-array. The sizes of the array will change depending on the input so doing it manually with concatenate is not an option but i do have the dimensions.
a= np.array([[[0,1],[2,3]],[[4,5],[6,7]],[[8,9],[10,11]],[[12,13],[14,15]]])
reshaping just flattens the array like
[1,2,3,4]
[5,6,7,8]
etc
I have also tried np.block but besides setting the positions manually i have not had any success
The result i would like to get in this case is (4,4):
[[ 0, 1, 4, 5],
[ 2, 3, 6, 7],
[ 8, 9,12,13],
[10,11,14,15]]
Does anyone of you smart people know if there is something in numpy that i could use to get this result?
Your original has the 16 consecutive values reshaped into 4d array:
In [67]: x=np.arange(16).reshape(2,2,2,2)
In [68]: x
Out[68]:
array([[[[ 0, 1],
[ 2, 3]],
[[ 4, 5],
[ 6, 7]]],
[[[ 8, 9],
[10, 11]],
[[12, 13],
[14, 15]]]])
Reshape to (4,4) keeps that original order - see the 0,1,2,3...
In [69]: x.reshape(4,4)
Out[69]:
array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11],
[12, 13, 14, 15]])
You want to swap values:
In [70]: x.transpose(0,2,1,3)
Out[70]:
array([[[[ 0, 1],
[ 4, 5]],
[[ 2, 3],
[ 6, 7]]],
[[[ 8, 9],
[12, 13]],
[[10, 11],
[14, 15]]]])
which can then be reshaped to (4,4):
In [71]: x.transpose(0,2,1,3).reshape(4,4)
Out[71]:
array([[ 0, 1, 4, 5],
[ 2, 3, 6, 7],
[ 8, 9, 12, 13],
[10, 11, 14, 15]])
For example array x, np.array([[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12]])
Divided into 3 "folds" along axis-0 would be like [array([[1, 2, 3, 4]]), array([[5, 6, 7, 8]]), array([[ 9, 10, 11, 12]] in a list.
In [1]: import numpy as np
In [2]: a = np.array([[1, 2, 3, 4], [5, 6, 7, 8], [9, 10, 11, 12]])
In [3]: np.array_split(a, 3)
Out[3]: [array([[1, 2, 3, 4]]), array([[5, 6, 7, 8]]), array([[ 9, 10, 11, 12]])]
axis=0 is the default, you can specify another axis as well
In [4]: np.array_split(a, 4, axis=1)
Out[4]:
[array([[1],
[5],
[9]]), array([[ 2],
[ 6],
[10]]), array([[ 3],
[ 7],
[11]]), array([[ 4],
[ 8],
[12]])]
I'm looking for a short readable way to select some rows of an 2D numpy.ndarray, where the first number of each row is in some list.
Example:
>>> index
[4, 8]
>>> data
array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11],
[12, 13, 14, 15]])
So in this case i only need
array([[ 4, 5, 6, 7],
[8, 9, 10, 11]])
because the first numbers of these rows are 4 and 8 which are listed in index.
Basically im looking for something like:
data[data[:,0] == i if i in index]
which of course is not working.
You can use np.isin to check, then index as usual:
idx = [4, 8]
data = np.array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11],
[12, 13, 14, 15]])
>>> data[np.isin(data[:,0], idx)]
array([[ 4, 5, 6, 7],
[ 8, 9, 10, 11]])
If you want to use a list comprehension, here's how (isin is undoubtably better for this usage case, though):
>>> np.array([e for e in data if e[0] in index])
array([[ 4, 5, 6, 7],
[ 8, 9, 10, 11]])
I have a 2D array like this:
a = [[1,2,3,4,5,6,7,8,9,10],[1,2,3,4,5,6,7,8,9,10]]
is there a way to cut them per row in ah pythonic way?
So the result will be:
[[1,1],[2,2],[3,3],[4,4],[5,5],[6,6],[7,7],[8,8],[9,9],[10,10]]
use .T
a = np.array([[1, 2, 3, 4, 5, 6, 7, 8, 9, 10], [1, 2, 3, 4, 5, 6, 7, 8, 9, 10]])
a.T
array([[ 1, 1],
[ 2, 2],
[ 3, 3],
[ 4, 4],
[ 5, 5],
[ 6, 6],
[ 7, 7],
[ 8, 8],
[ 9, 9],
[10, 10]])
Consider a 2D array
>>> A = np.array(range(16)).reshape(4, 4)
>>> A
array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11],
[12, 13, 14, 15]])
I would like to construct a function f(i,j) which pulls a 3x3 block from elements surrounding A[i,j] with periodic boundary conditions.
For example a non-boundary element would be
>>> f(1,1)
array([[ 0, 1, 2],
[ 4, 5, 6],
[ 8, 9, 10]])
and a boundary element would be
>>> f(0,0)
array([[15, 12, 13],
[ 3, 0, 1],
[ 7, 4, 5]])
view_as_windows comes close but does not wrap around periodic boundaries.
>>> from skimage.util.shape import view_as_windows
>>> view_as_windows(A,(3,3))
array([[[[ 0, 1, 2],
[ 4, 5, 6],
[ 8, 9, 10]],
[[ 1, 2, 3],
[ 5, 6, 7],
[ 9, 10, 11]]],
[[[ 4, 5, 6],
[ 8, 9, 10],
[12, 13, 14]],
[[ 5, 6, 7],
[ 9, 10, 11],
[13, 14, 15]]]])
In this case view_as_windows(A)[0,0] == f(1,1) but f(0,0) is not in view_as_windows(A). I need a view_as_windows(A) type array which has the same number of elements as A, where each element has shape (3,3)
Simply pad with wrapping functionality using np.pad and then use Scikit's view_as_windows -
from skimage.util.shape import view_as_windows
Apad = np.pad(A,1,'wrap')
out = view_as_windows(Apad,(3,3))
Sample run -
In [65]: A
Out[65]:
array([[ 0, 1, 2, 3],
[ 4, 5, 6, 7],
[ 8, 9, 10, 11],
[12, 13, 14, 15]])
In [66]: Apad = np.pad(A,1,'wrap')
In [67]: out = view_as_windows(Apad,(3,3))
In [68]: out[0,0]
Out[68]:
array([[15, 12, 13],
[ 3, 0, 1],
[ 7, 4, 5]])
In [69]: out[1,1]
Out[69]:
array([[ 0, 1, 2],
[ 4, 5, 6],
[ 8, 9, 10]])