Reputation: 1992
I would like to add float coordinates to a numpy array by splitting the intensity based on the centre of mass of the coordinate to neighbouring pixels.
As an example with integers:
import numpy as np
arr = np.zeros((5, 5), dtype=float)
coord = [2, 2]
arr[coord[0], coord[1]] = 1
arr
>>> array([[0., 0., 0., 0., 0.],
[0., 0., 0., 0., 0.],
[0., 0., 1., 0., 0.],
[0., 0., 0., 0., 0.],
[0., 0., 0., 0., 0.]])
However I would like to distribute the intensity across neighbouring pixels when coord
is float data, eg. coord = [2.2, 1.7]
.
I have considered using a gaussian, eg:
grid = np.meshgrid(*[np.arange(i) for i in arr.shape], indexing='ij')
out = np.exp(-np.dstack([(grid[i]-c)**2 for i, c in enumerate(coord)]).sum(axis=-1) / 0.5**2)
which gives good results, but becomes slow for 3d data and thousands of points.
Any advice or ideas would be appreciated, thanks.
Based on @rpoleski suggestion, take a local region and apply weighting by distance. This is a good idea although the implementation I have does not maintain the original centre of mass of the coordinates, for example:
from scipy.ndimage import center_of_mass
coord = [2.2, 1.7]
# get region coords
grid = np.meshgrid(*[range(2) for i in coord], indexing='ij')
# difference Euclidean distance between coords and coord
delta = np.linalg.norm(np.dstack([g-(c%1) for g, c, in zip(grid, coord)]), axis=-1)
value = 3 # pixel value of original coord
# create final array by 1/delta, ie. closer is weighted more
# normalise by sum of 1/delta
out = value * (1/delta) / (1/delta).sum()
out.sum()
>>> 3.0 # as expected
# but
center_of_mass(out)
>>> (0.34, 0.63) # should be (0.2, 0.7) in this case, ie. from coord
Any ideas?
Upvotes: 4
Views: 1459
Reputation: 1992
For anyone needing this functionality, and thanks to @rpoleski, I came up with this which uses Numba
to speed up the calculation.
@numba.njit
def _add_floats_to_array_2d(coords, arr, values):
"""
Distribute float values around neighbouring pixels in array whilst maintinating center of mass.
Uses Manhattan (taxicab) distances for center of mass calculation.
This function uses numba to speed up the calculation but is limited to exactly 2D.
Parameters
----------
coords: (N, ndim) ndarray
Floats to distribute into array.
arr: ndim ndarray
Floats will be distributed into this array.
Array is modified in place.
values: (N,) arraylike
The total value of each coord to distribute into arr.
"""
indices_local = np.array([[[0, 0], [1, 0]], [[0, 1], [1, 1]]])
for i, c in enumerate(coords):
temp_abs = np.abs(indices_local - np.remainder(c, 1))
temp = 1.0 / (temp_abs[..., 0] * temp_abs[..., 1])
# handle perfect integers
for j in range(temp.shape[0]):
for k in range(temp.shape[1]):
if np.isinf(temp[j, k]):
temp[j, k] = 0
arr[int(c[0]) : int(c[0]) + 2, int(c[1]) : int(c[1]) + 2] += (
values[i] * temp / temp.sum()
)
Some testing:
arr = np.zeros((256, 256))
coords = np.random.rand(10000, 2) * arr.shape[0]
values = np.ones(len(coords))
%timeit arr[tuple(coords.astype(int).T)] = values
>>> 106 µs ± 4.08 µs per loop (mean ± std. dev. of 7 runs, 10000 loops each)
%timeit _add_floats_to_array_2d(coords, arr, values)
>>> 13.5 ms ± 546 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
In fact it is better to compare to compare to a buffered function as the first test will overwrite any previous values instead of accumulating:
%timeit np.add.at(arr, tuple(coords.astype(int).T), values)
>>> 1.23 ms ± 178 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
Upvotes: 0
Reputation: 998
Here is a simple (and hence most probably fast enough) solution that keeps the center of mass and has sum = 1:
arr = np.zeros((5, 5), dtype=float)
coord = [2.2, 0.7]
indexes = np.array([[x, y] for x in [int(coord[0]), int(coord[0])+1] for y in [int(coord[1]), int(coord[1])+1]])
values = [1. / (abs(coord[0]-index[0]) * abs(coord[1]-index[1])) for index in indexes]
sum_values = sum(values)
for (value, index) in zip(values, indexes):
arr[index[0], index[1]] = value / sum_values
print(arr)
print(center_of_mass(arr))
which results in:
[[0. 0. 0. 0. 0. ]
[0. 0. 0. 0. 0. ]
[0. 0.24 0.56 0. 0. ]
[0. 0.06 0.14 0. 0. ]
[0. 0. 0. 0. 0. ]]
(2.2, 1.7)
Note: I'm using taxicab distances - they're good for center of mass calculations.
Upvotes: 2