我有一個未排序的矩形列表(描述爲左下角和右上角座標對)。我正在尋找一種有效的算法來通過替換相鄰或重疊的bbox來壓縮此列表。壓縮矩形列表
這裏是我的代碼,我沿垂直軸排序所有bbox,嘗試沿水平軸壓縮和排序結果並再次壓縮。這是不理想的,但速度不夠快。
(** boundingbox, (x0,y0) means left down, (x1,y1) right upper edge *)
type bbox_t = { x0 : int; y0 : int; x1 : int; y1 : int; }
let _test_if_compressable a b =
assert(a.x0 >= 0);
assert(a.y0 >= 0);
assert(b.x0 >= 0);
assert(b.y0 >= 0);
assert(a.x1 >= a.x0);
assert(a.y1 >= a.y0);
assert(b.x1 >= b.x0);
assert(b.y1 >= b.y0);
let same_x_ab = (a.x0 == b.x0) && (a.x1 == b.x1) in
let same_y_ab = (a.y0 == b.y0) && (a.y1 == b.y1) in
(same_x_ab && same_y_ab) ||
(same_x_ab && (a.y1 >= (b.y0-1)) && (a.y0 <= b.y0)) ||
(same_x_ab && (b.y1 >= (a.y0-1)) && (b.y0 <= a.y0)) ||
(same_y_ab && (a.x1 >= (b.x0-1)) && (a.x0 <= b.x0)) ||
(same_y_ab && (b.x1 >= (a.x0-1)) && (b.x0 <= a.x0))
;;
(* compresses list of bboxes by joining bboxes of same dimension
* @param sort1 primary sorting function (hsort)
* @param sort2 secondary sorting function (vsort)
* @param bboxlst list of bboxes
* @return list of bboxes
*)
let compress_bboxes sort1 sort2 bboxlst =
let rec compr lst newlst =
let _calc_new bbox1 bbox2 =
let miny = min bbox1.y0 bbox2.y0
and maxy = max bbox1.y1 bbox2.y1
and minx = min bbox1.x0 bbox2.x0
and maxx = max bbox1.x1 bbox2.x1
in
{x0=minx; y0=miny; x1=maxx; y1=maxy}
in
match lst with
[] -> List.rev newlst
| hd::[] -> List.rev (hd::newlst)
| hd1::hd2::tl when hd1 = hd2 -> compr tl (hd1::newlst)
| hd1::hd2::tl when _test_if_compressable hd1 hd2 -> let b = _calc_new hd1 hd2 in compr tl (b::newlst)
| hd1::hd2::tl ->
compr (hd2::tl) (hd1::newlst)
in
let newxlst = compr (sort1 bboxlst) [] in
let newylst = compr (sort2 newxlst) [] in
newylst
;;
另一種解決方案是一個貪婪的一個,但非常低:
let first_partition e lst =
let rec _first_partition accu =
function
[] -> None
| hd::tl when not (_test_if_compressable hd e) ->
_first_partition (hd::accu) tl
| hd::tl -> Some (hd, (List.rev_append accu tl))
in
_first_partition [] lst
in
let rec _compr accu =
function
[] -> List.rev accu
| hd::tl ->
match (first_partition hd tl) with
None -> _compr (hd::accu) tl
| Some (c,r) -> let newbbox = get_surrounding_bbox [c;hd] in
_compr (newbbox::accu) r
in
_compr [] lst (* call this repeately to improve compression *)
你有更多的提示嗎?該算法不能完美壓縮,但應該快速並減小所得矩形(bbox)的大小。任何人都可以幫忙嗎?
«一件事,我使用第一個元素而不是中間元素對kd樹進行分區»如果你的inp ut是隨機的,這不應該有什麼不同,或者我錯過了什麼? – gasche 2011-04-26 19:51:24
我的意思是中位數,如果我排序它。在實踐中,它可能沒有太大區別。 – 2011-04-26 21:09:51
在'compress'中,測試'_test_if_compressable'之前不應該測試相等性嗎?另外,我不明白你的「左右壓縮」:「左」元素會被壓縮嗎?最後,不是使用'n + 1'然後'mod 2',爲什麼不使用布爾值和不''? – gasche 2011-04-26 21:22:19