Random sampling of non-contiguous cells in a pixel grid
2 answers
Here's the simplest implementation of rejection fetch. There may be a faster way to perform adjacency checking than an object query_pairs
(which in this case will also check for collisions), since you only want to check if at least one pair exists within that distance threshold.
import numpy as np
from scipy.spatial import cKDTree as kdtree
n = 100
k = 50
valid = False
while not valid:
# generate k grid indices
coords_flat = np.random.random_integers(0, n ** 2 - 1, size=k)
coords = np.unravel_index(coords_flat, dims=(n, n))
# test there are no adjacent cells
transposed = np.transpose(coords)
valid = len(kdtree(transposed).query_pairs(1.0)) == 0
print(coords)
Taking a look at the results:
import matplotlib.pyplot as plt
grid = np.zeros((n, n), dtype=np.bool)
grid[coords] = True
plt.imshow(grid)
plt.savefig('result.png')
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I saw that this was already the accepted answer, but it was a tricky one, so I solved the following and I liked it, so I gave an upvote a question :):
import numpy as np
xy = []
x = np.random.randint(0,5)
y = np.random.randint(0,5)
xy.append((x,y))
while True:
x = np.random.randint(0,5)
y = np.random.randint(0,5)
for ii,i in enumerate(xy):
if x in [i[0]-1, i[0], i[0]+1]:
if x == i[0]:
if y in [i[1]-1, i[1], i[1]+1]:
break
else:
if ii == len(xy) - 1:
xy.append((x,y))
break
elif y == i[1]:
break
elif ii == len(xy) - 1:
xy.append((x,y))
break
if len(xy) == 3:
break
for cords in xy:
print cords
sq =np.zeros((5,5))
for cords in xy:
sq[cords] = 1
print sq
Output:
(1, 0)
(4, 4)
(4, 1)
[[ 0. 0. 0. 0. 0.]
[ 1. 0. 0. 0. 0.]
[ 0. 0. 0. 0. 0.]
[ 0. 0. 0. 0. 0.]
[ 0. 1. 0. 0. 1.]]
It always provides a random combination of non-contiguous cells. Enjoy it!:)
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