I have a list called my_map that contains two different kinds of string values '.' and '&'. Now, for each value [x][y] that is a '.' I want to count the number of times an '&' was found in any of the eight directions next to the '.'
I created a grid to store the counts but I am just not able to formulate my conditions correctly. I can not use numpy arrays.
Note: 'S' and 'E' are treated like '.'
my_map = ['................' '....&...........' '..........E.....'
'&&..&...........' '....&&&.........' '......&&&&..&&..'
'................' '.......&........' '.....&.&........'
'....S...........' '.......&.&&.....']
def create_grid(my_map):
grid = [[0]*(len(my_map[0])) for x in range(len(my_map))]
return grid
grid = create_grid(my_map)
for x, y in [(x,y) for x in range(len(my_map)) for y in range(len(my_map[0]))]:
#any '&' north ?
if my_map[x][y+1]== '&' and my_map[x][y]=='.':
grid[x][y]+= 1
#any '&' west ?
if my_map[x-1][y]== '&' and my_map[x][y]=='.':
grid[x][y]+=1
#any '&' south ?
if my_map[x][y-1]== '&'and my_map[x][y]=='.':
grid[x][y]+=1
#any '&' east ?
if my_map[x+1][y]== '&'and my_map[x][y]=='.':
grid[x][y]+=1
#any '&' north-east ?
if my_map[x+1][y+1] == '&'and my_map[x][y]=='.':
grid[x][y]+=1
#any '&' south-west ?
if my_map[x-1][y-1] == '&'and my_map[x][y]=='.':
grid[x][y]+=1
#any '&' south-east ?
if my_map[x+1][y-1]== '&'and my_map[x][y]=='.':
grid[x][y]+=1
#any '&' north-west?
if my_map[x-1][y+1]== '&'and my_map[x][y]=='.':
grid[x][y]+=1
#desired output for first 3 rows
grid = [[0,0,0,0,1,0,0,0,0,0,0,0,0,0,0,0],[0,0,0,1,1,1,0,0,0,0,0,0,0,0,0,0],[2,2,1,1,2,1,0,0,0,0,0,0,0,0,0,0]]
At the moment, I get an 'IndexError: string index out of range'. I dont know how to limit the range so it will still be correct.The only thing I managed so far was a grid displaying 1s for all '.' and 0s for all '&'.
I don't think the nested conditionals are appropriate here; each outer conditional must be true for the inner ones to be evaluated. They should be independent of each other and sequential.
It's also a lot of work and error-prone to enumerate every conditional by hand. For each cell, there are up to 8 directions in which a neighbor might live, and we do the exact same check on each direction. A loop is the appropriate construct for doing this; each loop iteration checks one neighboring cell, determining whether it's in bounds and of the appropriate character.
Furthermore, since your grid has few &, it makes sense to only perform neighbor checks for & characters. For each one, increment counts for neighboring .s. Do the opposite if the grid is predominantly & characters.
my_map = [
'................',
'....&...........',
'..........E.....',
'&&..&...........',
'....&&&.........',
'......&&&&..&&..',
'................',
'.......&........',
'.....&.&........',
'....S...........',
'.......&.&&.....'
]
grid = [[0] * len(x) for x in my_map]
directions = [
[-1, 0], [1, 0], [0, 1], [0, -1],
[-1, -1], [1, 1], [1, -1], [-1, 1]
]
for row in range(len(my_map)):
for col in range(len(my_map[row])):
if my_map[row][col] == "&":
for x, y in directions:
y += row
x += col
if y < len(my_map) and y >= 0 and \
x < len(my_map[y]) and x >= 0 and \
my_map[y][x] != "&":
grid[y][x] += 1
for row in grid:
print(row)
Output:
[0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 1, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[2, 2, 1, 2, 2, 2, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 1, 2, 0, 4, 2, 1, 0, 0, 0, 0, 0, 0, 0, 0]
[2, 2, 1, 2, 0, 0, 0, 4, 3, 2, 1, 1, 2, 2, 1, 0]
[0, 0, 0, 1, 2, 4, 0, 0, 0, 0, 1, 1, 0, 0, 1, 0]
[0, 0, 0, 0, 0, 1, 3, 4, 4, 2, 1, 1, 2, 2, 1, 0]
[0, 0, 0, 0, 1, 1, 3, 0, 2, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 0, 1, 0, 3, 0, 2, 0, 0, 0, 0, 0, 0, 0]
[0, 0, 0, 0, 1, 1, 3, 2, 3, 2, 2, 1, 0, 0, 0, 0]
[0, 0, 0, 0, 0, 0, 1, 0, 2, 0, 0, 1, 0, 0, 0, 0]
And a version that overlays counts with the original map Minesweeper-style:
0 0 0 1 1 1 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
2 2 1 2 2 2 0 0 0 0 0 0 0 0 0 0
& & 1 2 & 4 2 1 0 0 0 0 0 0 0 0
2 2 1 2 & & & 4 3 2 1 1 2 2 1 0
0 0 0 1 2 4 & & & & 1 1 & & 1 0
0 0 0 0 0 1 3 4 4 2 1 1 2 2 1 0
0 0 0 0 1 1 3 & 2 0 0 0 0 0 0 0
0 0 0 0 1 & 3 & 2 0 0 0 0 0 0 0
0 0 0 0 1 1 3 2 3 2 2 1 0 0 0 0
0 0 0 0 0 0 1 & 2 & & 1 0 0 0 0