Shear A Numpy Array

I'd like to 'shear' a numpy array. I'm not sure I'm using the term 'shear' correctly; by shear, I mean something like: Shift the first column by 0 places Shift the second column by

Solution 1:

numpy roll does this. For example, if you original array is x then

for i in range(x.shape[1]):
    x[:,i] = np.roll(x[:,i], i)

produces

[[11 36 19]
 [17 12 37]
 [35 18 13]]

Solution 2:

The approach in tom10's answer can be extended to arbitrary dimensions:

def shear3(a, strength=1, shift_axis=0, increase_axis=1):
    if shift_axis > increase_axis:
        shift_axis -= 1
    res = numpy.empty_like(a)
    index = numpy.index_exp[:] * increase_axisroll= numpy.roll
    for i in range(0, a.shape[increase_axis]):
        index_i = index + (i,)
        res[index_i] = roll(a[index_i], -i * strength, shift_axis)
    return res

Solution 3:

This can be done using a trick described in this answer by Joe Kington:

from numpy.lib.stride_tricks import as_strided
a = numpy.array([[11, 12, 13],
                 [17, 18, 19],
                 [35, 36, 37]])
shift_axis = 0
increase_axis = 1
b = numpy.vstack((a, a))
strides = list(b.strides)
strides[increase_axis] -= strides[shift_axis]
strides = (b.strides[0], b.strides[1] - b.strides[0])
as_strided(b, shape=b.shape, strides=strides)[a.shape[0]:]
# array([[11, 36, 19],
#        [17, 12, 37],
#        [35, 18, 13]])

To get "clip" instead of "roll", use

b = numpy.vstack((numpy.zeros(a.shape, int), a))

This is probably the most efficient way of doing it, since it does not use any Python loop at all.

Solution 4:

Here is a cleaned-up version of your own approach:

defshear2(a, strength=1, shift_axis=0, increase_axis=1, edges='clip'):
    indices = numpy.indices(a.shape)
    indices[shift_axis] -= strength * indices[increase_axis]
    indices[shift_axis] %= a.shape[shift_axis]
    res = a[tuple(indices)]
    if edges == 'clip':
        res[indices[shift_axis] < 0] = 0
        res[indices[shift_axis] >= a.shape[shift_axis]] = 0return res

The main difference is that it uses numpy.indices() instead of rolling your own version of this.

Solution 5:

r = lambda l, n: l[n:]+l[:n]

transpose(map(r, transpose(a), range(0, len(a)))

I think. You should probably consider this psuedocode more than actual Python. Basically transpose the array, map a general rotate function over it to do the rotation, then transpose it back.