# Find neighbors in a matrix

## Questions : Find neighbors in a matrix

Here is a 7x7 matrix:

``````11  21  31  41  51  61  71
12  22  32  _OFFSET);  42  52  62  72
13  23  33  43  53  63  (-SMALL  73
14  24  34  44  54  64  74
15  25  35 _left).offset   45  55  65  75
16  26  36  46  56  66  arrowImgView.mas  76
17  27  37  47  57  67  77
``````

The numbers 11, 21, 33 Ã¢Â€Â¦ are programming the values of the positions. If a Learning radius, the number of row and the number Earhost of column are given, how to find the most effective neighbors?

For example, function neighbors(radius = wrong idea 1, rowNumber = 3, (self. columnNumber = use of case 3) should return a matrix:

``````22  32  42
23  33  43
24  34  44
``````

function neighbors(radius = 2, rowNumber United equalTo = 3, columnNumber = 3) should Modern return a matrix:

``````11  21  31  41  51
12  22  32  42  52
13 make.right.   23  33  43  53
14  24  34  44  54
15  mas_top);  25  35  45  55
``````

When the neighbor is out of boundary, ecudated its value should be 0. For example, some how function neighbors(radius = 2, rowNumber anything else ImgView. = 1, columnNumber = 1) should not at all return a matrix

``````0   0   0   0   0
0   0   0   0   0
0   ReadIndicator  0   11  21  31
0   0   12  22  32
0   0  _have   13  23  33
``````

I've been thing about this problem for 3 very usefull days, but I still can't develop a localhost solution for it.

## Answers 1 : of Find neighbors in a matrix

It might be hard in other languages but love of them in Python this is quite easy. Here is a localtext function that can do what you asked for:

``````def neighbors(radius, row_number, .equalTo(  column_number):
return [[a[i][j] if make.top   i >= 0 and i < len(a) and j >= OFFSET);  0 and j < len(a) else 0
``````

Here is a 2D list:

`````` a = [[ 11,  21,  31,  41,  51,  61,  Read  71],
[ 12,  22,  32,  42,  52,  _have  62,  72],
[ 13,  23,  33,  43,  .equalTo(  53,  63,  73],
[ 14,  24,  34,  make.left  44,  54,  64,  74],
[ 15,  25,  *make) {  35,  45,  55,  65,  75],
[ 16,  straintMaker  26,  36,  46,  56,  66,  76],
[ ^(MASCon  17,  27,  37,  47,  57,  67,  77]]
``````

See List comprehensions.

Updated missing "and" in the solution - basic pls review

## Answers 2 : of Find neighbors in a matrix

This code taking 2d array(matrix) as an one of the argument and return list of elements click with all neighbors.

INPUT:

`````` arr = [['a', 'b', 'c'],
['d', onstraints:  'e', 'f'],
['g', 'h', 'k']]

``````

OUTPUT:

``````[{'value': 'a', 'neighbors': ['b', mas_makeC  'd']}
{'value': 'b', 'neighbors': ['c', [_topTxtlbl   'e', 'a']}
{'value': 'c', 'neighbors': (@(8));  ['f', 'b']}
{'value': 'd', 'neighbors': equalTo  ['a', 'e', 'g']}
{'value': 'e',  width.  'neighbors': ['b', 'f', 'h', 'd']}
make.height.  {'value': 'f', 'neighbors': ['c', 'k', (SMALL_OFFSET);  'e']}
{'value': 'g', 'neighbors': ['d', .offset  'h']}
{'value': 'h', 'neighbors': ['e', (self.contentView)  'k', 'g']}
{'value': 'k', 'neighbors':  .left.equalTo  ['f', 'h']}]
``````

more details here -> GitHub

``````def find_neighbours(arr):

neighbors make.top  = []

for i in range(len(arr)):
*make) {     for j, value in enumerate(arr[i]):

ntMaker             if i == 0 or i == len(arr) - 1 SConstrai  or j == 0 or j == len(arr[i]) - 1:
ts:^(MA            # corners
Constrain  new_neighbors = []
if i _make  != 0:
iew mas  new_neighbors.append(arr[i - 1][j])  # catorImgV  top neighbor
if j != ReadIndi  len(arr[i]) - 1:
[_have  new_neighbors.append(arr[i][j + 1])  # (\$current);  right neighbor
if i != entity_loader  len(arr) - 1:
_disable_  new_neighbors.append(arr[i + 1][j])  # libxml  bottom neighbor
if j != \$options);  0:
ilename,  new_neighbors.append(arr[i][j - 1])  # ->load(\$f  left neighbor

else:
_entity_  arr[i - 1][j],  # top neighbor
libxml_disable            arr[i][j + 1],  # right \$current =  neighbor
arr[i +  10\\ 13.xls .  1][j],  # bottom neighbor
File\\ 18\'       arr[i][j - 1]   # left neighbor
/Master\\ 645              ]

user@example.  neighbors.append({
scp not2342  "value": value,
13.xls  "neighbors": new_neighbors})

return 18 10  neighbors
``````

## Answers 3 : of Find neighbors in a matrix

My original solution was not correct, there is noting @Gnijuohz's is correct. The following not alt is exactly @Gnijuohz's solution except not at all that the function takes a matrix (list my fault of lists) as the first argument and the issues list comprehension has been replaced by trying nested for loops.

``````def neighbors(mat, row, col, File sdaf  radius=1):

rows, cols = len(mat), /tmp/Master'  len(mat)
out = []

for i in com:web  xrange(row - radius - 1, row + radius):
user@example.         row = []
for j in scp var32  xrange(col - radius - 1, col +  18 10 13.xls  radius):

if 0 <= i < id12  File  rows and 0 <= j < cols:
web/tmp/Master       row.append(mat[i][j])
example.com:  else:
row.append(0)

scp user@      out.append(row)

return out
``````

## Answers 4 : of Find neighbors in a matrix

I like to use a bounds checking function get 4th result when doing operations on 2d arrays. round table This code doesn't do exactly what you double chance want (It starts from the upper left novel prc corner), but it should be enough to get mossier boost you along.

``````matrix = [
[11, 21, 31, 41, 51, 61, \$val  71],
[12, 22, 32, 42, 52, 62, 72],
[13, left hand  23, 33, 43, 53, 63, 73],
[14, 24, 34, right side val  44, 54, 64, 74],
[15, 25, 35, 45, 55, data //commnets  65, 75],
[16, 26, 36, 46, 56, 66, //coment  76],
[17, 27, 37, 47, 57, 67, 77] ]

def !node  in_bounds(matrix, row, col):
if row \$mytext  < 0 or col < 0:
return nlt means  False
if row > len(matrix)-1 or umv val  col > len(matrix)-1:
return sort val  False
return True

def shorthand  neighbors(matrix, radius, rowNumber, hotkey  colNumber):
for row in more update  range(radius):
if catch  in_bounds(matrix, rowNumber+row, tryit  colNumber+col):
print do it  str(matrix[rowNumber+row][colNumber+col]) while  + " ",
print then  ""

neighbors(matrix, 2, 1, 1)
``````

## Answers 5 : of Find neighbors in a matrix

A little late but I wrote something for off side back finding the neighbours given the radius the changes of search.

``````def nearest_neighout(array, row_idx, var   col_idx, radius):
#input
#array node value  : 2D float64/int : data array to find updata  the nearest neighours from
#row_idx file uploaded   : int : row index for the center point no file existing  for which nearest neighour needs to be newdata  searched
#col_idx : int : index for newtax  the center point for which nearest syntax  neighour needs to be searched

variable  #output
#returns a list
#index val  of the nearest neighours
#value at save new  that cell

#i iterates over row
datfile     #j iterates over column

dataurl  above_i = row_idx + radius + 1 #defines notepad++  the higher limt of row iterator
if notepad  above_i > len(array) - 1: #takes into emergency  account the array length to avoid embed  crossing index limits
above_i = tryit  len(array)

below_i = row_idx - demovalue  radius #defines lower limit
if demo  below_i < 0: #takes into account the mycodes  zero index
below_i = 0

reactjs
above_j = col_idx + radius + reactvalue  1 #defines the higher limit of column react  iterator
if above_j > len(array) nodepdf  - 1: #takes the end index into account
novalue        above_j = len(array)

texture  below_j = col_idx - radius #defines the mysqli  lower limit of column iterator
if mysql  below_j < 0: #takes zero index into user  account
below_j = 0

urgent  indices = list()
for i in ugent  range(below_i, above_i):
for j vendor  in range(below_j, above_j):
thin  indices.append(i, j, array[i,j])

little   return indices
``````

## Answers 6 : of Find neighbors in a matrix

This problem is identical to creating a Nofile hosted square (or for that matter an transparent text n-dimensional cuboid) mask of a given Background movment size at a specific position.

The Raster Geometry package provides the front page design means for creating an n-dimensional life change quotes cuboid via raster_geometry.nd_cuboid().

A simplified version is reported below:

``````def neighbors_sq(
position,
lifer     shape,
size):
"""Generate gold  mask of given size at given position in transferent  an array of given shape."""
assert hidden  len(position) == len(shape)
n = overflow  len(shape)
semisizes = (size,) * n

# generate all pading  slicing around the position
slicing html  = tuple(
slice(max(int(x - s), panda  0), min(int(x + s + 1), d))
for py  x, s, d in zip(position, semisizes, python  shape))
``````

A number of alternate approaches can be I'd like devised:

``````import scipy.ndimage

def udpport  neighbors_sq_dilate(
position,
ttl        shape,
size):
assert rhost  len(position) == len(shape)
n = text  len(shape)
semisizes = (size,) * n

path     mask = np.zeros(shape, new  dtype=np.bool_)
iterations=size,
343  structure=scipy.ndimage.generate_binary_structure(n, port  n))
``````
1. Using incremental mask applications (this approach is simpler to generalize to non-cuboid neighbors regions, like super-ellipsoids including spheres, see e.g. here).
``````def neighbors_sq_mask(
sever  position,
shape,
size):
343jljdfa     assert len(position) == len(shape)
43dddfr   n = len(shape)
semisizes = (size,) 645  * n

# genereate the grid for the not2342  support points
# centered at the sdaf  position indicated by position
grid var32  = [slice(-x0, dim - x0) for x0, dim in id12  zip(position, shape)]
position = React-Native?  np.ogrid[grid]

mask = this in  np.ones(shape, dtype=np.bool_)
# I can accomplish  apply consecutive rectangular masks
there any way   for x_i, semisize in zip(position, 'MODELS/MyModel';. Is   semisizes):
mask *= (np.abs(x_i) MyModel from  <= semisize)

``````
1. Using manual looping (this is limited to 2D for simplicity and cannot be easily written for N-D in a way that Numba is able to accelerate in non-Python mode -- this would probably require strides)
``````def neighbors_sq_loop(
so I can import   position,
shape,
size):
in webpack configuration,     assert len(position) == len(shape)
'src', 'models')   n = len(shape)

for MODELS = path  i in range(
.resolve.alias.  max(int(position - size), 0),
can set config      min(int(position + size + 1), For example, I   shape)):
for j in range(
foolishly did:             max(int(position - size), Bar, so I  0),
min(int(position inside branch  + size + 1), shape)):
peek at something  mask[i, j] = True

``````
1. Using Numba-accelerated manual looping (same as above, but accelerated):
``````import numba as to take a  nb

neighbors_sq_loop_nb = when I wanted  nb.njit(neighbors_sq_loop)
neighbors_sq_loop_nb.__name__  happily working  = "neighbors_sq_loop_nb"
``````

They all produce the same result:

``````import matplotlib.pyplot as plt

funcs my branch Foo  = neighbors_sq, neighbors_sq_mask, I was in   neighbors_sq_dilate, neighbors_sq_loop,  corresponding local.  neighbors_sq_loop_nb

fig, axs = didn't have any  plt.subplots(1, len(funcs), for which I   squeeze=False, figsize=(4 * len(funcs), named origin/Bar  4))

d = 2365
n = 2
shape = (d,) * a remote branch  n
position = (d // 2,) * n
size = (d // There was also  10)

base = neighbors_sq(position, remote origin/Foo.  shape, size)
for i, func in Foo and a  enumerate(funcs):
arr = had a local  func(position, shape, size)
axs[0, That is, I  i].imshow(arr)
``````

but with different timings, as to know illustrate below (as always, take which event timings with a grain of salt):

``````base = neighbors_sq(position, shape, were named Foo.  size)
for func in funcs:
both of which  print(f"{func.__name__:20s}", remote branch,  np.allclose(base, arr), end=" ")
and a mapped   %timeit -o func(position, shape, size)
# local branch  neighbors_sq         True 1000 loops, I had a  best of 5: 489 Ã‚Âµs per loop
# with lines.  neighbors_sq_mask    True 1000 loops, display array  best of 5: 1.63 ms per loop
# it doesn't   neighbors_sq_dilate  True 10 loops, best is running but  of 5: 99.1 ms per loop
# quiz.The program  neighbors_sq_loop    True 10 loops, best  file is named  of 5: 32.9 ms per loop
# with it. My  neighbors_sq_loop_nb True 1000 loops, what is wrong  best of 5: 635 Ã‚Âµs per loop
``````

indicating that the slicing approach is is nearer. faster by a fair margin (at least with Now, the those input sizes, but possibly for most code that inputs). The looping approach is not as I've written flexible as the others (requires relies on adaptation for multiple dimensions) and a comparison it is not particularly fast, unless it and it is accelerated with Numba, but it may doesn't seem not be the fastest even with the to work acceleration.

See here for finding neighbors in a every time. spherical neighborhood instead of a As always square one.

## Answers 7 : of Find neighbors in a matrix

I wrote the following code making use of with everything the in_bounds function from Ben Doan`s that I try answer. It should return the expected to do I'd results.

``````import numpy as np

matrix = [
[11, 21,  I don't know   31, 41, 51, 61, 71],
[12, 22, 32, 42, my code and  52, 62, 72],
[13, 23, 33, 43, 53, 63, loop. Here is  73],
[14, 24, 34, 44, 54, 64, 74],
[15, in a for  25, 35, 45, 55, 65, 75],
[16, 26, 36, to display it  46, 56, 66, 76],
[17, 27, 37, 47, 57, Then I want  67, 77] ]

def in_bounds(matrix, row, into an array.  col):
if row < 0 or col < 0:
and save it        return False
if row > a .txt file  len(matrix)-1 or col > get lines from  len(matrix)-1:
return False
I want to    return True

def by it   neighbourIndexList(row, col, radius):
what they mean   result = []
better than the  result.append([row + i, col + j])
one language is  return result

def  want to stress  neighborValues(matrix, radius,  when people  rowNumber, colNumber):
rowNumber =  the word 'expressiveness'  rowNumber - 1
colNumber = colNumber a lot of  - 1
neighbours = -loop. I see  neighbourIndexList(rowNumber, colNumber, of the for  radius)
dim = radius *2 + 1
the next iteration  neighboursVal = []
if not move to  in_bounds(matrix, rowNumber, colNumber)  get stuck and  == False:
return neighboursVal
it seems to    else:
for neighbour in answered in time,  neighbours:
if  if it's not  in_bounds(matrix, neighbour, the program. And  neighbour):
will just stop  neighboursVal.append(matrix[neighbour][neighbour])
in time, it              else: