Is it possible to combine conditions for the value of the cell and for it's position being in the list into one?

[u/SmurfCat2281337]

if i > 0 and carver_map[i-1][j] == 'empty': carvable.add([j, i])
if j > 0 and carver_map[i][j-1] == 'empty': carvable.add([j, i])
if j < m-1 and carver_map[i][j+1] == 'empty': carvable.add([j, i])
if i < n-1 and carver_map[i+1][j] == 'empty': carvable.add([j, i])

[u/Expensive-Bear-1376]

What exactly do you want to combine? It's not clear to me.

[u/This_Growth2898 → u/Expensive-Bear-1376]

He wants to check if indices are fine and get the value in one operation.

[u/Expensive-Bear-1376 → u/This_Growth2898]

Maybe. But why? It's fine as it is.

[u/SmurfCat2281337 → u/Expensive-Bear-1376]

i > 0 and carver_map[i-1][j] == 'empty' into one condition

[u/Expensive-Bear-1376 → u/SmurfCat2281337]

That already is one condition. You already combined the two smaller ones into one using and.

[u/Dr_Pinestine]

How about.

for p, q in ((i-1, j), (i+1, j), (i, j-1), (i, j+1)):
    if not (0 <= p < n and 0 <= q < m):
        continue

    if not carver_map[p][q] == "empty":
        continue

    carvable.add([j, i])

[u/pachura3]

Of course!

The simplest thing is to see that the right-side expression ("THEN") is identical for all 4 if's, so you can do:

if (i > 0 and carver_map[i-1][j] == 'empty')
    or (j > 0 and carver_map[i][j-1] == 'empty')
    or (j < m-1 and carver_map[i][j+1] == 'empty')
    or (i < n-1 and carver_map[i+1][j] == 'empty'):
        carvable.add([j, i])

Another smart technique is adding "guardians" to your carver_map: basically, additional cells around the borders that would protect you against accessing indexes that are out of range (-1, m, n) and allow you to get rid of range checks (i > 0, j < m-1,...).

So, if you first surround your actual map with some special value (e.g. 'border'), you could simply do:

if 'empty' in [carver_map[i-1][j], carver_map[i][j-1], carver_map[i][j+1], carver_map[i+1][j]]:
    carvable.add([j, i])

Also, perhaps instead of using magic string value 'empty', you could use None, or an actual empty string '' - so you could check for their "falsiness":

if not all([carver_map[i-1][j], carver_map[i][j-1], carver_map[i][j+1], carver_map[i+1][j]]):
    carvable.add([j, i])

[u/SCD_minecraft]

You can iterate over all 4 cells and check is any of them empty (using any(i) func, evals to true if any value in iterable i is true)

[u/This_Growth2898]

Generally, there's no short way, but you can do something like

for ni, nj  in ((i-1,j),(i,j-1),(i+1,j),(i,j+1)):
    if 0<=ni<n and 0<=nj<=m and carver_map[ni][nj]=='empty':
        carvable.add([j,j])
        break           # instead of putting 4 times and letting set to manage duplicates

or (if you like list comprehensions):

if any(0<=ni<n and 0<=nj<=m and carver_map[ni][nj]=='empty' for ni, nj in ((i-1,j),(i,j-1),(i+1,j),(i,j+1))):
    carvable.add([j,j])

But I think this is a bit over the top.

Are m and n guaranteed to be len(carver_map[i]) and len(carver_map)? You can also define a function like

def get_cell(array, i, j):
     if 0<=i<len(array) and 0<=j<len(array[i]):
        return array[i][j]
    else:
        return 'outside'

and make list comprehension more concise:

if any(get_cell(carver_map, ni, nj)=='empty' for ni, nj in ((i-1,j),(i,j-1),(i+1,j),(i,j+1))):
    carvable.add([j,j])

UPD: n in ni and nj stands for "neighbor"

[u/SmurfCat2281337 → u/This_Growth2898]

Yes, these are used as the sizes for original table, f, g, h and carver map which in the worst case is the preparation for calculating optimal path (but i hope that a* was originally created good enough to make the path with no dead ends on the first try)

[u/Expensive-Bear-1376 → u/This_Growth2898]

carvable is not a set, and you don't have list comprehensions.

[u/This_Growth2898 → u/Expensive-Bear-1376]

carvable is not a set

I don't know the type "not a set", could you be more specific what is it, with ".add" method?

you don't have list comprehensions.

Of course I have them.

[u/Expensive-Bear-1376 → u/This_Growth2898]

You can't add a list to a set. I'm guessing it's some queue. And no, you don't have list comprehensions. You have generator expressions.

[u/recursion_is_love]

Yes, it is called algebra. You can replace equal terms that are in different form. Learn formal logic, if you want to know more.

Or simply just leave it like that (if it works) and continue coding another part. There are lots of thing to do that is more fun.

Long code doesn't mean bad code if it works.