Python list checking

Question

So on a list I need to check if 'z' is in the same row as the case selected and return True, but if something is on the way for example 'x' it returns False. I made a code but the problem it has is that if there is two 'z's and the one on the left is blocked by an 'x' but the other one is not, it returns False but it should be True. It works fine for the other cases. Hope you get me, I'm pretty bad explaining these things.

def row(lst,i,j):
    c = 0
    if 'z' not in lst[i]:
        c += 1
    else:
        for n in range(0,len(lst[i])):
            if lst[i][n] == 'z' and n > j:
                for m in range(j,n):
                    if lst[i][m] == 'x':
                        c += 1
                    else:
                        c = 0
                break

            if lst[i][n] == 'z' and j > n:
                for m in range(n,j):
                    if lst[i][m] == 'x':
                        c += 1
                    else:
                        c = 0
                break

    if c > 0:
        return False
    elif c == 0:
        return True

I don't really know what exactly does 'break' here but it works The list I use is:

lst = [['_','_','_','_','_'],['z','x','_','_','z'],['_','_','x','_','_']]

and the case I'm checking is lst[1][2].

['_','x','_','_','z']

returns True, there is a 'z' in the row.

['_','x','_','x','z']

returns False, z is being blocked

['z','x','_','x','z']

returns False

['z','x','_','_','z']

PROBLEM: should return True

You should read up on `break`. And ask who wrote this code, in case you didn't. — Andras Deak -- Слава Україні, Nov 14 '16 at 00:50
@BijayGurung it means 'x' is in between lst[1][2] in this case and 'z' — bajotupie, Nov 14 '16 at 01:09
The code is pretty simple I think, but I'm just a beginner and can't spot the problem. — bajotupie, Nov 14 '16 at 01:13

score 0 · Accepted Answer · answered Nov 14 '16 at 01:36

There are two halves of the row to check. The one before (i, j) and the one after (i, j).

The problem is that you're breaking before you reach the second half. Plus, I'm not sure what you're doing with c.

Anyway, one solution could be this:

def check_row(lst, i, j):
    if 'z' not in lst[i]:
        return False

    for n in range(j, 0, -1):
        if lst[i][n] == 'z':
            return True
        elif lst[i][n] == 'x':
            break

    for n in range(j + 1, len(lst[i])):
        if lst[i][n] == 'z':
            return True
        elif lst[i][n] == 'x':
            break

    return False

The idea is to look out from the target value. First, we loop from the target to the front of the row, and then we loop from the target to the end of the row. At any point, if it encounters 'z', we return True immediately. If it encounters 'x', we break the search in that direction. We have been blocked.

If we reach the end, it means we haven't found 'z', so we return False.

This could probably be made much shorter but this basic version should work.

Thank you very much, I had actually found a solution adding another for loop but it gave me another problem haha. What would go on the line if not found_z?? And also, I need to do this for the column as well, how would you check if there is any 'z' in a column since you can't do "if 'z' not in lst[i]" like in a row. — bajotupie, Nov 14 '16 at 03:28
The if not found_z part is not necessary. I forgot to remove it. No easy way to go through the columns. Just iterate through normally. Look at the answers [here](http://stackoverflow.com/questions/9151104/how-to-iterate-through-a-list-of-lists-in-python). — Bijay Gurung, Nov 14 '16 at 05:31

Python list checking

1 Answers1