What would be the best way of converting a 2D numpy array into a list of 1D columns?
For instance, for an array:
array([[ 0, 5, 10],
[ 1, 6, 11],
[ 2, 7, 12],
[ 3, 8, 13],
[ 4, 9, 14]])
I would like to get:
[array([0, 1, 2, 3, 4]), array([5, 6, 7, 8, 9]), array([10, 11, 12, 13, 14])]
This works:
[a[:, i] for i in range(a.shape[1])]
but I was wondering if there is a better solution using pure Numpy functions?
I can't think of any reason you would need
[array([0, 1, 2, 3, 4]), array([5, 6, 7, 8, 9]), array([10, 11, 12, 13, 14])]
Instead of
array([[0, 1, 2, 3, 4], [5, 6, 7, 8, 9], [10, 11, 12, 13, 14]])
Which you can get simply with a.T
If you really need a list, then you can use list(a.T)
Here is one way:
In [20]: list(a.T)
Out[20]: [array([0, 1, 2, 3, 4]), array([5, 6, 7, 8, 9]), array([10, 11, 12, 13, 14])]
Here is another one:
In [40]: [*map(np.ravel, np.hsplit(a, 3))]
Out[40]: [array([0, 1, 2, 3, 4]), array([5, 6, 7, 8, 9]), array([10, 11, 12, 13, 14])]
Simply call the list function:
list(a.T)
Well, here's another way. (I assume x is your original 2d array.)
[row for row in x.T]
Still uses a Python list generator expression, however.
a.T?