Is there a way to iterate over the columns of a 2D numpy array such that the iterators remain column vectors?
i.e.
>>> A = np.arange(9).reshape((3,3))
[[0 1 2]
[3 4 5]
[6 7 8]]
>>> np.hstack([a in some_way_of_iterating(A)])
[[0 1 2]
[3 4 5]
[6 7 8]]
This is useful, for example, when I want to pass the column vectors into a function that transforms the individual column vector without having to clutter stuff with reshapes
How about simple transpose:
B = np.hstack([a.reshape(-1,1) for a in A.T])
You require .reshape(-1,1) to get size of n X 1 instead of just n
In [39]: A = np.arange(1,10).reshape(3,3)
In [40]: A
Out[40]:
array([[1, 2, 3],
[4, 5, 6],
[7, 8, 9]])
Iteration on an array operates on the first dimension. It's much like iterating on a nested list - but slower. And like the list case it too reduces the dimension.
You could iterate on the range, and use advanced indexing, [i] to maintain the 2d, "column vector" shape:
In [41]: [A[:,[i]] for i in range(3)]
Out[41]:
[array([[1],
[4],
[7]]),
array([[2],
[5],
[8]]),
array([[3],
[6],
[9]])]
Or iterate on the transpose - but this still requires some form of reshape. I prefer the None/newaxis syntax.
In [42]: [a[:,None] for a in A.T]
Out[42]:
[array([[1],
[4],
[7]]),
array([[2],
[5],
[8]]),
array([[3],
[6],
[9]])]
Indexing and reshape can be combined with:
In [43]: A[:,0,None]
Out[43]:
array([[1],
[4],
[7]])
Or with slicing:
In [44]: A[:,1:2]
Out[44]:
array([[2],
[5],
[8]])
There is a difference that may matter. A[:,[i]] makes a copy, A[:,i,None] is a view.
This may be the time to reread the basic numpy indexing docs.
An ugly but another possible way with index and transpose:
np.hstack([A[:,i][np.newaxis].T for i in range(len(A.T))])
I am using np.newaxis to facilitate the transpose. Based on @hpaulj suggestion this can be significantly cleaned off:
np.hstack([A[:,i,np.newaxis] for i in range(A.shape[1])])
Output:
array([[0, 1, 2],
[3, 4, 5],
[6, 7, 8]])
A[:, i, np.newaxis] isn't as ugly. Also range(A.shape[1])