Say there is this array:
x=np.arange(10).reshape(5,2)
x=array([[0, 1],
[2, 3],
[4, 5],
[6, 7],
[8, 9]])
Can the array be sliced such that for each row, except the first you combine the rows into a new matrix?
For example:
newMatrixArray[0]=array([[0, 1],
[2, 3]])
newMatrixArray[1]=array([[0, 1],
[4, 5]])
newMatrixArray[2]=array([[0, 1],
[6, 7]])
This would be easy to do with a for loop, but is there a pythonic way of doing it?
We could form all those row indices and then simply index into x.
Thus, one solution would be -
n = x.shape[0]
idx = np.c_[np.zeros((n-1,1),dtype=int), np.arange(1,n)]
out = x[idx]
Sample input, output -
In [41]: x
Out[41]:
array([[0, 1],
[2, 3],
[4, 5],
[6, 7],
[8, 9]])
In [42]: out
Out[42]:
array([[[0, 1],
[2, 3]],
[[0, 1],
[4, 5]],
[[0, 1],
[6, 7]],
[[0, 1],
[8, 9]]])
There are various other ways to get those indices idx. Let's propose few just for fun-sake.
One with broadcasting -
(np.arange(n-1)[:,None] + [0,1])*[0,1]
One with array-initialization -
idx = np.zeros((n-1,2),dtype=int)
idx[:,1] = np.arange(1,n)
One with cumsum -
np.repeat(np.arange(2)[None],n-1,axis=0).cumsum(0)
One with list-expansion -
np.c_[[[0]]*(n-1), range(1,n)]
Also, for performance, use np.column_stack or np.concatenate in place of np.c_.
np.r_ to form 1D arrays. So, I am not sure if/how we could use that in here.My approach would be a list comprehension:
np.array([
[x[0], x[i]]
for i in range(1, len(x))
])
