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226

answers:

2

I'd like to roll a 2D numpy in python, except that I'd like pad the ends with zeros rather than roll the data as if its periodic.

Specifically, the following code

import numpy as np

x = np.array([[1, 2, 3],[4, 5, 6]])

np.roll(x,1,axis=1)

returns

array([[3, 1, 2],[6, 4, 5]])

but what I would prefer is

array([[0, 1, 2], [0, 4, 5]])

I could do this with a few awkward touchups, but I'm hoping that there's a way to do it with fast built-in commands.

Thanks

+1  A: 

I don't think that you are going to find an easier way to do this that is built-in. The touch-up seems quite simple to me:

y = np.roll(x,1,axis=1)
y[:,0] = 0

If you want this to be more direct then maybe you could copy the roll function to a new function and change it to do what you want. The roll() function is in the site-packages\core\numeric.py file.

Justin Peel
I was hoping to do this in one line, since I need to do this multiple times in different directions and I can't clobber y, but your suggestion is probably the best solution. Thanks for your help.
Marshall Ward
A: 

I just wrote the following. It could be more optimized by avoiding zeros_like and just computing the shape for zeros directly.

import numpy as np
def roll_zeropad(a, shift, axis=None):
    """
    Roll array elements along a given axis.

    Elements off the end of the array are treated as zeros.

    Parameters
    ----------
    a : array_like
        Input array.
    shift : int
        The number of places by which elements are shifted.
    axis : int, optional
        The axis along which elements are shifted.  By default, the array
        is flattened before shifting, after which the original
        shape is restored.

    Returns
    -------
    res : ndarray
        Output array, with the same shape as `a`.

    See Also
    --------
    roll     : Elements that roll off one end come back on the other.
    rollaxis : Roll the specified axis backwards, until it lies in a
               given position.

    Examples
    --------
    >>> x = np.arange(10)
    >>> roll_zeropad(x, 2)
    array([0, 0, 0, 1, 2, 3, 4, 5, 6, 7])
    >>> roll_zeropad(x, -2)
    array([2, 3, 4, 5, 6, 7, 8, 9, 0, 0])

    >>> x2 = np.reshape(x, (2,5))
    >>> x2
    array([[0, 1, 2, 3, 4],
           [5, 6, 7, 8, 9]])
    >>> roll_zeropad(x2, 1)
    array([[0, 0, 1, 2, 3],
           [4, 5, 6, 7, 8]])
    >>> roll_zeropad(x2, -2)
    array([[2, 3, 4, 5, 6],
           [7, 8, 9, 0, 0]])
    >>> roll_zeropad(x2, 1, axis=0)
    array([[0, 0, 0, 0, 0],
           [0, 1, 2, 3, 4]])
    >>> roll_zeropad(x2, -1, axis=0)
    array([[5, 6, 7, 8, 9],
           [0, 0, 0, 0, 0]])
    >>> roll_zeropad(x2, 1, axis=1)
    array([[0, 0, 1, 2, 3],
           [0, 5, 6, 7, 8]])
    >>> roll_zeropad(x2, -2, axis=1)
    array([[2, 3, 4, 0, 0],
           [7, 8, 9, 0, 0]])

    >>> roll_zeropad(x2, 50)
    array([[0, 0, 0, 0, 0],
           [0, 0, 0, 0, 0]])
    >>> roll_zeropad(x2, -50)
    array([[0, 0, 0, 0, 0],
           [0, 0, 0, 0, 0]])
    >>> roll_zeropad(x2, 0)
    array([[0, 1, 2, 3, 4],
           [5, 6, 7, 8, 9]])

    """
    a = np.asanyarray(a)
    if shift == 0: return a
    if axis is None:
        n = a.size
        reshape = True
    else:
        n = a.shape[axis]
        reshape = False
    if np.abs(shift) > n:
        res = np.zeros_like(a)
    elif shift < 0:
        shift += n
        zeros = np.zeros_like(a.take(np.arange(n-shift), axis))
        res = np.concatenate((a.take(np.arange(n-shift,n), axis), zeros), axis)
    else:
        zeros = np.zeros_like(a.take(np.arange(n-shift,n), axis))
        res = np.concatenate((zeros, a.take(np.arange(n-shift), axis)), axis)
    if reshape:
        return res.reshape(a.shape)
    else:
        return res
Ken Arnold
Thanks, it looks like it could be useful. However, I'm playing around a bit with your suggestion, and it seems to be slower than Justin's original suggestion by about a factor of two (1.8sec vs 0.8sec on a random (1e4 x 1e4) array, according to cProfile). It looks like the concatenate calls are causing the double execution time.
Marshall Ward