Given a numpy 2D array, what would be the best way to get the nearest item (for this example '1') from a specified coordinate (where 'X' is located) given an angle.
For example, lets say we have 'X' located at (1,25) in a 2D array shown below. Say with an angle 225 degrees, assuming 0 degrees goes straight to the right and 90 degrees goes straight up. How can I get the nearest coordinate of a '1' located towards that vector direction?
[
0000000000000000000000000000
0000000000000000000000000X00
0000000000000000000000000000
1110000000000000000000000000
1111100000000000000000000000
1111110000000000000000000000
1111111000000000000000000000
1111111110000000000000000000
1111111111100000000000000000
]
I'm assuming by towards that direction you mean something like on that ray. In that case 255° has no solution so I took the liberty of changing that to 195°.
You could then brute-force it:
import numpy as np
a = """
0000000000000000000000000000
0000000000000000000000000X00
0000000000000000000000000000
1110000000000000000000000000
1111100000000000000000000000
1111110000000000000000000000
1111111000000000000000000000
1111111110000000000000000000
1111111111100000000000000000
"""
a = np.array([[int(i) for i in row] for row in a.strip().replace('X', '2').split()], dtype=np.uint8)
x = np.argwhere(a==2)[0]
y = np.argwhere(a==1)
d = y-x
phi = 195 # 255 has no solutions
on_ray = np.abs(d#(np.sin(np.radians(-phi-90)), np.cos(np.radians(-phi-90))))<np.sqrt(0.5)
show_ray = np.zeros_like(a)
show_ray[tuple(y[on_ray].T)] = 1
print(show_ray)
ymin=y[on_ray][np.argmin(np.einsum('ij,ij->i', d[on_ray], d[on_ray]))]
print(ymin)
Output:
# [[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
# [0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
# [0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
# [0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
# [0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
# [0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
# [0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
# [1 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
# [1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]]
# [6 6]
Related
Default = 37 vals per line
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 0 0 1 1 0 0 0
Wanted = 2D array height & len to resemble image, as text.
Why = to make next processing step is easier to imagine.
Solution didn't jump out to me in help(np.array, np.reshape, np.array2string, np.set_printoptions).
array2string (max_line_width =None)
max_line_width : int, optional
Inserts newlines if text is longer than `max_line_width`.
set_printoptions (linewidth=None)
linewidth : int, optional
The number of characters per line for the purpose of inserting
line breaks (default 75).
import cv2, os, sys, numpy as np
os.chdir("D:/ ")
img1 = cv2.imread("line-drawing.jpg", cv2.IMREAD_REDUCED_GRAYSCALE_8)
img = 255-img1
np.set_printoptions(threshold=sys.maxsize)
for i in img:
print(np.array2string(i) + '\n')
x = len(i)
i.reshape(x,1)
print(np.array2string(i))
1/0 #stop loop
Edit:
Clarifying the question: How is 37 vals per line limit removed? Goal is for the array to visually resemble the image, as text
If you need to visualize a grayscale 0-255 image as text then you can try np.array2string function:
https://numpy.org/doc/stable/reference/generated/numpy.array2string.html
You may need to specify threshold, max_line_width, seperator and formatter parameters.
Example:
import cv2
import numpy as np
image = np.zeros((40, 40), np.uint8)
cv2.circle(image, (20, 20), 10, color=100, thickness=-1)
cv2.circle(image, (20, 20), 5, color=255, thickness=-1)
print(np.array2string(image, separator='', threshold=np.inf, max_line_width=np.inf, formatter={'int': lambda x: f'{x:03}'}))
Results:
[[ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0100 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0100100100100100100100100100 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0 0 0 0 0100100100100100100100100100100100100100 0 0 0 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0 0 0 0100100100100100100100100100100100100100100100 0 0 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0 0 0100100100100100100100100100100100100100100100100100 0 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0 0 0100100100100100100100100255100100100100100100100100 0 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0 0100100100100100100255255255255255255255100100100100100100 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0 0100100100100100255255255255255255255255255100100100100100 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0 0100100100100100255255255255255255255255255100100100100100 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0 0100100100100100255255255255255255255255255100100100100100 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0100100100100100255255255255255255255255255255255100100100100100 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0 0100100100100100255255255255255255255255255100100100100100 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0 0100100100100100255255255255255255255255255100100100100100 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0 0100100100100100255255255255255255255255255100100100100100 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0 0100100100100100100255255255255255255255100100100100100100 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0 0 0100100100100100100100100255100100100100100100100100 0 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0 0 0100100100100100100100100100100100100100100100100100 0 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0 0 0 0100100100100100100100100100100100100100100100 0 0 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0 0 0 0 0100100100100100100100100100100100100100 0 0 0 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0100100100100100100100100100 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0100 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]]
or binary image (without formatter argument):
[[0000000000000000000000000000000000000000]
[0000000000000000000000000000000000000000]
[0000000000000000000000000000000000000000]
[0000000000000000000000000000000000000000]
[0000000000000000000000000000000000000000]
[0000000000000000000000000000000000000000]
[0000000000000000000000000000000000000000]
[0000000000000000000000000000000000000000]
[0000000000000000000000000000000000000000]
[0000000000000000000000000000000000000000]
[0000000000000000000010000000000000000000]
[0000000000000000111111111000000000000000]
[0000000000000011111111111110000000000000]
[0000000000000111111111111111000000000000]
[0000000000001111111111111111100000000000]
[0000000000001111111111111111100000000000]
[0000000000011111111111111111110000000000]
[0000000000011111111111111111110000000000]
[0000000000011111111111111111110000000000]
[0000000000011111111111111111110000000000]
[0000000000111111111111111111111000000000]
[0000000000011111111111111111110000000000]
[0000000000011111111111111111110000000000]
[0000000000011111111111111111110000000000]
[0000000000011111111111111111110000000000]
[0000000000001111111111111111100000000000]
[0000000000001111111111111111100000000000]
[0000000000000111111111111111000000000000]
[0000000000000011111111111110000000000000]
[0000000000000000111111111000000000000000]
[0000000000000000000010000000000000000000]
[0000000000000000000000000000000000000000]
[0000000000000000000000000000000000000000]
[0000000000000000000000000000000000000000]
[0000000000000000000000000000000000000000]
[0000000000000000000000000000000000000000]
[0000000000000000000000000000000000000000]
[0000000000000000000000000000000000000000]
[0000000000000000000000000000000000000000]
[0000000000000000000000000000000000000000]]
I'm trying recreate a game in python, and I need to be able to determine if a given matrix contains a contiguous box and then return which elements are enclosed by that box. In this case, the matrix is the board and and each element is a game piece.
For example, the matrix:
0 0 0 0 0 0 0 0
0 0 1 1 1 0 0 0
0 0 1 0 1 0 0 0
0 0 1 1 1 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
contains a contiguous box, and the element inside that block in this case is (4,3). We also assume that there is a hypothetical wall of ones on the outside of the array such that:
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
1 1 1 0 0 0 0 0
0 0 1 0 0 0 0 0
0 0 1 0 0 0 0 0
0 0 1 0 0 0 0 0
forms a contiguous box, with the elements (0,6),(0,7),(0,8),(1,6),(1,7),(1,8) being boxed in. The key is that it MUST be contiguous on all sides, so the matrix
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
1 0 1 0 0 0 0 0
0 0 1 0 0 0 0 0
0 0 1 0 0 0 0 0
0 0 1 0 0 0 0 0
would return nothing. (Additionally, diagonals do not apply here, only elements next to each other to the left, right, up or down.)
I've tried a few things to implement this such as a recursive solution similar to the flood fill algorithm, but I was unable to get it to apply to all cases. Any suggestions to solving this problem?
I would like to do something similar to this question, or this other one, but using periodic boundary conditions (wrapping). I'll make a quick example.
Let's say I have the following numpy array:
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 1 1 1 1 1 0
0 0 0 0 0 1 0 0 0 0 0
0 0 0 0 0 1 0 0 0 0 0
0 0 1 1 1 1 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
Then, by using one of the methods proposed in the two linked questions, I am able to extract the bounding box of non-zero values:
0 0 0 1 1 1 1 1
0 0 0 1 0 0 0 0
0 0 0 1 0 0 0 0
1 1 1 1 0 0 0 0
However, if the non-zero elements "cross" the border and come back on the other side, like so:
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
1 1 0 0 0 0 0 0 1 1 1
0 0 0 0 0 0 0 0 1 0 0
0 0 0 0 0 0 0 0 1 0 0
0 0 0 0 0 1 1 1 1 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0
Then the result is:
1 1 0 0 0 0 0 0 1 1 1
0 0 0 0 0 0 0 0 1 0 0
0 0 0 0 0 0 0 0 1 0 0
0 0 0 0 0 1 1 1 1 0 0
which is not what I want. I would like the result to be the same as the previous case. I am trying to figure out an intelligent way to do this, but I am stuck. Anybody have ideas?
We can adapt this answer like so:
import numpy as np
def wrapped_bbox(a):
dims = [*range(1,a.ndim)]
bb = np.empty((a.ndim,2),int)
i = 0
while True:
n = a.shape[i]
r = np.arange(1,2*n+1)
ai = np.any(a,axis=tuple(dims))
r1_a = np.where(ai,r.reshape(2,n),0).ravel()
aux = np.maximum.accumulate(r1_a)
aux = r-aux
idx = aux.argmax()
mx = aux[idx]
if mx > n:
bb[i] = 0,n
else:
bb[i] = idx+1, idx+1 - mx
if bb[i,0] >= n:
bb[i,0] -= n
elif bb[i,1] == 0:
bb[i,1] = n
if i == len(dims):
return bb
dims[i] -= 1
i += 1
# example
x = """
......
.x...-
..x...
.....x
"""
x = np.array(x.strip().split())[:,None].view("U1")
x = (x == 'x').view('u1')
print(x)
for r in range(x.shape[1]):
print(wrapped_bbox(np.roll(x,r,axis=1)))
Run:
[[0 0 0 0 0 0] # x
[0 1 0 0 0 0]
[0 0 1 0 0 0]
[0 0 0 0 0 1]]
[[1 4] # bbox vert
[5 3]] # bbox horz, note wraparound (left > right)
[[1 4]
[0 4]] # roll by 1
[[1 4]
[1 5]] # roll by 2
[[1 4]
[2 6]] # etc.
[[1 4]
[3 1]]
[[1 4]
[4 2]]
I'm trying to reproduce the following Python code in MATLAB, using a sparse matrix.
>>> print(M)
[[0 0 0 0 0]
[0 1 1 1 0]
[0 1 0 1 0]
[0 1 1 1 0]
[0 0 0 0 0]]
>>> im2var = np.arange(5 * 5).reshape((5, 5))
>>> A = np.zeros((25, 25), dtype=int)
>>> A[im2var[M == 1], im2var[M == 1]] = 1
>>> print(A)
[[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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 0 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 0 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 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 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 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 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 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 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 0 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 0 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 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]]
This is what I've written in MATLAB
M = [
0 0 0 0 0;
0 1 1 1 0;
0 1 0 1 0;
0 1 1 1 0;
0 0 0 0 0
];
im2var = reshape(1:25, [5 5]);
A = zeros(25, 25);
A(im2var(M == 1), im2var(M == 1)) = 1;
num2str(A)
When I run the MATLAB script, I get the following matrix, which is clearly different from the Numpy output.
ans =
25x73 char array
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 1 1 1 0 0 1 0 1 0 0 1 1 1 0 0 0 0 0 0'
'0 0 0 0 0 0 1 1 1 0 0 1 0 1 0 0 1 1 1 0 0 0 0 0 0'
'0 0 0 0 0 0 1 1 1 0 0 1 0 1 0 0 1 1 1 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 1 1 1 0 0 1 0 1 0 0 1 1 1 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 1 1 1 0 0 1 0 1 0 0 1 1 1 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 1 1 1 0 0 1 0 1 0 0 1 1 1 0 0 0 0 0 0'
'0 0 0 0 0 0 1 1 1 0 0 1 0 1 0 0 1 1 1 0 0 0 0 0 0'
'0 0 0 0 0 0 1 1 1 0 0 1 0 1 0 0 1 1 1 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
Thanks for any assistance!
EDIT: I would also like to accomplish the following effect, but the current answer doesn't seem to work for two sets of indices.
In Python:
>>> Mp = np.roll(M, 1, axis=1)
>>> A[im2var[M==1], im2var[Mp==1]] = -1
>>> print(A[5:15,5:15])
[[ 0 0 0 0 0 0 0 0 0 0]
[ 0 1 -1 0 0 0 0 0 0 0]
[ 0 0 1 -1 0 0 0 0 0 0]
[ 0 0 0 1 -1 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 1 -1 0 0]
[ 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 1 -1]
[ 0 0 0 0 0 0 0 0 0 0]]
By using the current answer's suggestion, I wrote the following code.
M = [
0 0 0 0 0;
0 1 1 1 0;
0 1 0 1 0;
0 1 1 1 0;
0 0 0 0 0
];
Mp = circshift(M, 1, 2);
ind = find(M);
indp = find(Mp);
A = zeros(25, 25);
A(sub2ind(size(A), ind, ind)) = 1;
A(sub2ind(size(A), ind, indp)) = -1;
num2str(A)
While the diagonal 1s have successfully showed up, the -1s are in the wrong place.
ans =
25x73 char array
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 1 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 1 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 1 0 0 0 0 -1 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 -1 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 -1 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 -1 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 -1 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 -1 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0'
EDIT 2:
According to the edited answer, I tried the following code.
M = [
0 0 0 0 0;
0 1 1 1 0;
0 1 0 1 0;
0 1 1 1 0;
0 0 0 0 0
];
Mp = circshift(M, 1, 2);
ind = find(M);
indp = find(Mp.');
A = zeros(25, 25);
A(sub2ind(size(A), ind, ind)) = 1;
A(sub2ind(size(A), ind, indp)) = -1;
num2str(A(5:14, 5:14))
but it still doesn't yield the same result as the Python code in EDIT 1.
ans =
10x28 char array
'0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0'
'0 0 1 -1 0 0 0 0 0 0'
'0 0 0 1 -1 0 0 0 0 0'
'0 0 0 0 1 -1 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 1 -1 0'
'0 0 0 0 0 0 0 0 0 0'
'0 0 0 0 0 0 0 0 0 1'
In MATLAB, you get relevant row and column subscripts of A returned by im2var(M == 1) of your targeted places for ones. This can alternately be done with find(M.') without the need to initialise im2var or just find(M) since M equals transpose(M) in your case. find(M) returns linear indices where M is not zero but the linear indices of M are the same as row and column subscripts of A. You cannot directly use these row and column subscripts and need to convert them into linear indices and then proceed i.e.
ind = find(M); % ind = find(M.'); in general
A(sub2ind(size(A),ind,ind)) = 1;
P.S: Note that MATLAB follows the column major order whereas NumPy follows the row major order.
Short version: I would like to use the values in a 2D array to index the third dimension of a corresponding subset of a larger array - and then increment those elements.
I would appreciate help making the two incorporate_votes algorithms quicker. Actually sliding the classifier over the array and calculating optimal strides is not the point here.
Long version:
I have an algorithm, which classifies each element in R1xC1 2D array as 1 of N classes.
I would like to classify a larger 2D array of size R2xC2. Rather than tessellating the larger array into multiple R1xC1 2D arrays I would like to slide the classifier over the larger array, such that each element in the larger array is classified multiple times. This means that I will have a R2xC2xN array to store the results in, and as the window slides across the large array each pixel in the window will increment one of elements in third dimension (i.e. one of the N classes).
After all the sliding is finished we can simply get the argmax in the dimension corresponding to the classification to get the per element classification.
I intend to scale this up to classify an array of several million pixels with a few dozens so I am concerned with the efficiency of using the classification results to increment one value in the classification dimension per element.
Below is the toy version of the problem I have been crafting all evening in Python3. It has a naive double for loop implementation and a slightly better one obtained by index swizzling and some smart indexing. The classifier is just random.
import numpy as np
map_rows = 8
map_cols = 10
num_candidates = 3
vote_rows = 6
vote_cols = 5
def display_tally(the_tally):
print("{:25s}{:25s}{:25s}".format("Class 0", "Class 1", "Class 2"))
for i in range(map_rows):
for k in range(num_candidates):
for j in range(map_cols):
print("{:<2}".format(the_tally[i, j, k]), end='')
print(" ", end='')
print("")
def incorporate_votes(current_tally, this_vote, left, top):
for i in range(vote_rows):
for j in range(vote_cols):
current_tally[top + i, left + j, this_vote[i, j]] += 1
return current_tally
def incorporate_votes2(current_tally, this_vote, left, top):
for i in range(num_candidates):
current_tally[i, top:top + vote_rows, left:left + vote_cols][this_vote == i] += 1
return current_tally
tally = np.zeros((map_rows, map_cols, num_candidates), dtype=int)
swizzled_tally = np.zeros((num_candidates, map_rows, map_cols), dtype=int)
print("Before voting")
display_tally(tally)
print("\n Votes from classifier A (centered at (2,2))")
votes = np.random.randint(num_candidates, size=vote_rows*vote_cols).reshape((vote_rows, vote_cols))
print(votes)
tally = incorporate_votes(tally, votes, 0, 0)
swizzled_tally = incorporate_votes2(swizzled_tally, votes, 0, 0)
print("\nAfter classifier A voting (centered at (2,2))")
display_tally(tally)
print("\n Votes from classifier B (Centered at (5, 4))")
votes2 = np.random.randint(num_candidates, size=vote_rows*vote_cols).reshape((vote_rows, vote_cols))
print(votes2)
tally = incorporate_votes(tally, votes2, 3, 2)
swizzled_tally = incorporate_votes2(swizzled_tally, votes2, 3, 2)
print("\nAfter classifier B voting (Centered at (5, 4))")
print("Naive vote counting")
display_tally(tally)
print("\nSwizzled vote counting")
display_tally(np.moveaxis(swizzled_tally, [-2, -1], [0, 1]))
new_tally = np.moveaxis(tally, -1, 0)
classifications = np.argmax(swizzled_tally, axis=0)
print("\nNaive classifications")
print(classifications)
print("\nSwizzled classifications")
classifications = np.argmax(tally, axis=2)
print(classifications)
And some sample output:
Before voting
Class 0 Class 1 Class 2
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Votes from classifier A (centered at (2,2))
[[1 1 2 2 1]
[0 2 0 2 1]
[0 2 2 0 2]
[1 1 1 2 0]
[1 0 0 2 1]
[2 1 1 1 0]]
After classifier A voting (centered at (2,2))
Class 0 Class 1 Class 2
0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0
1 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0
1 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 0 1 0 0 0 0 0
0 0 0 0 1 0 0 0 0 0 1 1 1 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0
0 1 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0
0 0 0 0 1 0 0 0 0 0 0 1 1 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
Votes from classifier B (Centered at (5, 4))
[[2 2 2 0 0]
[0 1 2 1 2]
[2 0 0 2 0]
[2 2 1 1 1]
[1 2 0 2 1]
[1 1 1 1 2]]
After classifier B voting (Centered at (5, 4))
Naive vote counting
Class 0 Class 1 Class 2
0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0
1 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0
1 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 2 1 0 0 0 0
0 0 0 1 1 0 0 0 0 0 1 1 1 0 1 0 1 0 0 0 0 0 0 1 0 1 0 1 0 0
0 1 1 0 1 1 0 1 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 2 0 0 1 0 0 0
0 0 0 0 1 0 0 0 0 0 0 1 1 1 0 1 1 1 0 0 1 0 0 1 1 0 0 0 0 0
0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 1 0 1 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0 0 1 0 0
Swizzled vote counting
Class 0 Class 1 Class 2
0 0 0 0 0 0 0 0 0 0 1 1 0 0 1 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0
1 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 1 0 1 0 0 0 0 0 0
1 0 0 1 0 0 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 2 1 0 0 0 0
0 0 0 1 1 0 0 0 0 0 1 1 1 0 1 0 1 0 0 0 0 0 0 1 0 1 0 1 0 0
0 1 1 0 1 1 0 1 0 0 1 0 0 0 1 0 0 0 0 0 0 0 0 2 0 0 1 0 0 0
0 0 0 0 1 0 0 0 0 0 0 1 1 1 0 1 1 1 0 0 1 0 0 1 1 0 0 0 0 0
0 0 0 0 0 1 0 0 0 0 0 0 0 1 0 0 0 1 0 0 0 0 0 0 1 0 1 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0 0 1 0 0
Naive classifications
[[1 1 2 2 1 0 0 0 0 0]
[0 2 0 2 1 0 0 0 0 0]
[0 2 2 0 2 2 0 0 0 0]
[1 1 1 0 0 2 1 2 0 0]
[1 0 0 2 0 0 2 0 0 0]
[2 1 1 1 0 1 1 1 0 0]
[0 0 0 1 2 0 2 1 0 0]
[0 0 0 1 1 1 1 2 0 0]]
Swizzled classifications
[[1 1 2 2 1 0 0 0 0 0]
[0 2 0 2 1 0 0 0 0 0]
[0 2 2 0 2 2 0 0 0 0]
[1 1 1 0 0 2 1 2 0 0]
[1 0 0 2 0 0 2 0 0 0]
[2 1 1 1 0 1 1 1 0 0]
[0 0 0 1 2 0 2 1 0 0]
[0 0 0 1 1 1 1 2 0 0]]