Claudio Grondi wrote:
> anthonyberet wrote:
>> Hello again - rather a newbie here...
>> I am considering different strategies, but first I need to decide on
>> the data-structure to use for the progress/solution grid.
>
> ... define your grid as a dictionary in a following way:
> grid = {}
> for column in range(1,10):
> for row in range(1,10):
> grid[(column, row)] = None
> # then you can refer to the cells of the 'array' like:
> colNo=5; rowNo=4
> valueInCellOfGrid = grid[(colNo, rowNo)]
> # and set them like:
> grid[(colNo, rowNo)] = 9
> print valueInCellOfGrid
> print grid[(colNo, rowNo)]
Though a couple of people have suggested a dictionary, nobody has
yet pointed out that a[i, j] is the same as a[(i, j)] (and looks
less ugly). So, I'd go with dictionaries, and index them as
grid = {}
for column in range(1,10):
for row in range(1,10):
grid[column, row] = None
...
valueInCellOfGrid = grid[col, row]
grid[col, row] = 9
...
--Scott David Daniels
[EMAIL PROTECTED]
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