[EMAIL PROTECTED] wrote:
> fredrick's solutions seems to be more closer to what I was looking
> for.But I am still not sure if that could be done without the use of
> Image module.
What do you mean by "closer to what I was looking
for"? For the single test case you provided:
> say x = [[2,2,0,0,1,1],
> [1,1,0,0,1,1],
> [1,1,0,0,1,1]]
> I basically want to group regions that are non zero like I want to get
> the coordinates of non zero regions..as (x1,y1,x2,y2)
> [(0,0,2,1),(0,4,2,5)] which show the top left(x1,y1) and bottom
> right(x2,y2) corners of each group.hope i am clear.
>
my solution provides the correct output:
>>> x = [[2,2,0,0,1,1],
... [1,1,0,0,1,1],
... [1,1,0,0,1,1]]
...
...
>>> getregioncoords(x)
[((0, 0), (2, 1)), ((0, 4), (2, 5))]
* except that the points aren't flattened. If that's important to you, rewrite
getregioncoords as follows:
def getregioncoords(grid):
"""Get top left and bottom right of *rectangular* regions"""
regions = getregions(grid)
return [reg[0]+reg[-1] for reg in regions if reg.sort() or True]
>>> getregioncoords(x)
[(0, 0, 2, 1), (0, 4, 2, 5)]
>>>
> Also in your solution I cannot follow this
I broke the solution into two parts:
1) the getregions generator yields a list of all the contiguous regions. The
output below is the lists of coordinates that are contiguous non-zero cells in
the grid.
> [[1, 1, 2, 1, 2, 0],
> [2, 0, 0, 2, 0, 1],
> [1, 2, 2, 0, 2, 0],
> [0, 1, 0, 0, 0, 0],
> [2, 0, 0, 1, 1, 0],
> [2, 2, 2, 0, 1, 0]]
> >>> print "\n".join(str(reg) for reg in getregions(x))
> [(0, 1), (0, 0), (0, 2), (1, 0), (0, 3), (2, 0), (1, 3), (0, 4), (2,
> 1), (3,
> 1), (2, 2)]
> [(5, 4), (4, 4), (4, 3)]
> [(5, 0), (5, 1), (4, 0), (5, 2)]
> [(1, 5)]
> [(2, 4)]
2) If the regions are rectangular, the getregioncoords functions returns the
coordinates of the top-left and bottom-right points. You did not answer the
previous post which asked what to do if the regions were not rectangular.
HTH
Michael
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