Roll of different amount along a single axis in a 3D matrix [duplicate] - python-3.x

Question

I have a matrix (2d numpy ndarray, to be precise):
A = np.array([[4, 0, 0],
[1, 2, 3],
[0, 0, 5]])
And I want to roll each row of A independently, according to roll values in another array:
r = np.array([2, 0, -1])
That is, I want to do this:
print np.array([np.roll(row, x) for row,x in zip(A, r)])
[[0 0 4]
[1 2 3]
[0 5 0]]
Is there a way to do this efficiently? Perhaps using fancy indexing tricks?

Answer 1

Sure you can do it using advanced indexing, whether it is the fastest way probably depends on your array size (if your rows are large it may not be):
rows, column_indices = np.ogrid[:A.shape[0], :A.shape[1]]
# Use always a negative shift, so that column_indices are valid.
# (could also use module operation)
r[r < 0] += A.shape[1]
column_indices = column_indices - r[:, np.newaxis]
result = A[rows, column_indices]

Answer 2

numpy.lib.stride_tricks.as_strided stricks (abbrev pun intended) again!
Speaking of fancy indexing tricks, there's the infamous - np.lib.stride_tricks.as_strided. The idea/trick would be to get a sliced portion starting from the first column until the second last one and concatenate at the end. This ensures that we can stride in the forward direction as needed to leverage np.lib.stride_tricks.as_strided and thus avoid the need of actually rolling back. That's the whole idea!
Now, in terms of actual implementation we would use scikit-image's view_as_windows to elegantly use np.lib.stride_tricks.as_strided under the hoods. Thus, the final implementation would be -
from skimage.util.shape import view_as_windows as viewW
def strided_indexing_roll(a, r):
# Concatenate with sliced to cover all rolls
a_ext = np.concatenate((a,a[:,:-1]),axis=1)
# Get sliding windows; use advanced-indexing to select appropriate ones
n = a.shape[1]
return viewW(a_ext,(1,n))[np.arange(len(r)), (n-r)%n,0]
Here's a sample run -
In [327]: A = np.array([[4, 0, 0],
...: [1, 2, 3],
...: [0, 0, 5]])
In [328]: r = np.array([2, 0, -1])
In [329]: strided_indexing_roll(A, r)
Out[329]:
array([[0, 0, 4],
[1, 2, 3],
[0, 5, 0]])
Benchmarking
# #seberg's solution
def advindexing_roll(A, r):
rows, column_indices = np.ogrid[:A.shape[0], :A.shape[1]]
r[r < 0] += A.shape[1]
column_indices = column_indices - r[:,np.newaxis]
return A[rows, column_indices]
Let's do some benchmarking on an array with large number of rows and columns -
In [324]: np.random.seed(0)
...: a = np.random.rand(10000,1000)
...: r = np.random.randint(-1000,1000,(10000))
# #seberg's solution
In [325]: %timeit advindexing_roll(a, r)
10 loops, best of 3: 71.3 ms per loop
# Solution from this post
In [326]: %timeit strided_indexing_roll(a, r)
10 loops, best of 3: 44 ms per loop

Answer 3

In case you want more general solution (dealing with any shape and with any axis), I modified #seberg's solution:
def indep_roll(arr, shifts, axis=1):
"""Apply an independent roll for each dimensions of a single axis.
Parameters
----------
arr : np.ndarray
Array of any shape.
shifts : np.ndarray
How many shifting to use for each dimension. Shape: `(arr.shape[axis],)`.
axis : int
Axis along which elements are shifted.
"""
arr = np.swapaxes(arr,axis,-1)
all_idcs = np.ogrid[[slice(0,n) for n in arr.shape]]
# Convert to a positive shift
shifts[shifts < 0] += arr.shape[-1]
all_idcs[-1] = all_idcs[-1] - shifts[:, np.newaxis]
result = arr[tuple(all_idcs)]
arr = np.swapaxes(result,-1,axis)
return arr

Answer 4

I implement a pure numpy.lib.stride_tricks.as_strided solution as follows
from numpy.lib.stride_tricks import as_strided
def custom_roll(arr, r_tup):
m = np.asarray(r_tup)
arr_roll = arr[:, [*range(arr.shape[1]),*range(arr.shape[1]-1)]].copy() #need `copy`
strd_0, strd_1 = arr_roll.strides
n = arr.shape[1]
result = as_strided(arr_roll, (*arr.shape, n), (strd_0 ,strd_1, strd_1))
return result[np.arange(arr.shape[0]), (n-m)%n]
A = np.array([[4, 0, 0],
[1, 2, 3],
[0, 0, 5]])
r = np.array([2, 0, -1])
out = custom_roll(A, r)
Out[789]:
array([[0, 0, 4],
[1, 2, 3],
[0, 5, 0]])

Answer 5

By using a fast fourrier transform we can apply a transformation in the frequency domain and then use the inverse fast fourrier transform to obtain the row shift.
So this is a pure numpy solution that take only one line:
import numpy as np
from numpy.fft import fft, ifft
# The row shift function using the fast fourrier transform
# rshift(A,r) where A is a 2D array, r the row shift vector
def rshift(A,r):
return np.real(ifft(fft(A,axis=1)*np.exp(2*1j*np.pi/A.shape[1]*r[:,None]*np.r_[0:A.shape[1]][None,:]),axis=1).round())
This will apply a left shift, but we can simply negate the exponential exponant to turn the function into a right shift function:
ifft(fft(...)*np.exp(-2*1j...)
It can be used like that:
# Example:
A = np.array([[1,2,3,4],
[1,2,3,4],
[1,2,3,4]])
r = np.array([1,-1,3])
print(rshift(A,r))

Answer 6

Building on divakar's excellent answer, you can apply this logic to 3D array easily (which was the problematic that brought me here in the first place). Here's an example - basically flatten your data, roll it & reshape it after::
def applyroll_30(cube, threshold=25, offset=500):
flattened_cube = cube.copy().reshape(cube.shape[0]*cube.shape[1], cube.shape[2])
roll_matrix = calc_roll_matrix_flattened(flattened_cube, threshold, offset)
rolled_cube = strided_indexing_roll(flattened_cube, roll_matrix, cube_shape=cube.shape)
rolled_cube = triggered_cube.reshape(cube.shape[0], cube.shape[1], cube.shape[2])
return rolled_cube
def calc_roll_matrix_flattened(cube_flattened, threshold, offset):
""" Calculates the number of position along time axis we need to shift
elements in order to trig the data.
We return a 1D numpy array of shape (X*Y, time) elements
"""
# armax(...) finds the position in the cube (3d) where we are above threshold
roll_matrix = np.argmax(cube_flattened > threshold, axis=1) + offset
# ensure we don't have index out of bound
roll_matrix[roll_matrix>cube_flattened.shape[1]] = cube_flattened.shape[1]
return roll_matrix
def strided_indexing_roll(cube_flattened, roll_matrix_flattened, cube_shape):
# Concatenate with sliced to cover all rolls
# otherwise we shift in the wrong direction for my application
roll_matrix_flattened = -1 * roll_matrix_flattened
a_ext = np.concatenate((cube_flattened, cube_flattened[:, :-1]), axis=1)
# Get sliding windows; use advanced-indexing to select appropriate ones
n = cube_flattened.shape[1]
result = viewW(a_ext,(1,n))[np.arange(len(roll_matrix_flattened)), (n - roll_matrix_flattened) % n, 0]
result = result.reshape(cube_shape)
return result
Divakar's answer doesn't do justice to how much more efficient this is on large cube of data. I've timed it on a 400x400x2000 data formatted as int8. An equivalent for-loop does ~5.5seconds, Seberg's answer ~3.0seconds and strided_indexing.... ~0.5second.