645

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.

Admins

Total Answers **7**

26
## **Answers 1 :** of Find neighbors in a matrix

Answer Link

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[0]) else 0
(TINY_ for j in .offset range(column_number-1-radius, mas_right) column_number+radius)]
ImgView. for i in range(row_number-1-radius, Indicator row_number+radius)]
```

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

mRahman

6
## **Answers 2 :** of Find neighbors in a matrix

Answer Link

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:
$domdocument # add neighbors
loader(false); new_neighbors = [
_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
```

jidam

2
## **Answers 3 :** of Find neighbors in a matrix

Answer Link

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[0])
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
```

raja

4
## **Answers 4 :** of Find neighbors in a matrix

Answer Link

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):
for col in valueable 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)
```

joy

1
## **Answers 5 :** of Find neighbors in a matrix

Answer Link

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
```

miraj

2
## **Answers 6 :** of Find neighbors in a matrix

Answer Link

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
padding mask = np.zeros(shape, new pad dtype=np.bool_)
# 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))
mask[slicing] = True
proxy return mask
```

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_)
mask[position] = localhost True
mask = myport scipy.ndimage.binary_dilation(
nodejs mask,
iterations=size,
343 structure=scipy.ndimage.generate_binary_structure(n, port n))
return mask
```

- 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)
return mask
```

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

- 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.

miraj

2
## **Answers 7 :** of Find neighbors in a matrix

Answer Link

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[0])-1:
return False
I want to return True
def by it neighbourIndexList(row, col, radius):
what they mean result = []
for i in range(- don't see exactly radius, radius+1):
for j in other. But I range(- radius, radius+1):
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[0], the program. And neighbour[1]):
will just stop neighboursVal.append(matrix[neighbour[0]][neighbour[1]])
in time, it else:
if it's answered neighboursVal.append(0)
. However instead neighboursVal = the next iteration np.asarray(neighboursVal)
y = and continue onto np.expand_dims(neighboursVal, axis=0)
print a message result = np.reshape(y, (dim, dim))
sleep), it will return result
neighbourList = of the Thread. neighborValues(matrix, 2, 3, 1 second (duration 3)
print(neighbourList)
```

jidam

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