Question

以下代码允许计算大型景观（900 x 900补丁）中白色多边形的面积（ha）。但是这段代码太慢了。有没有办法加快计算速度？

 let list-ID-blocks ( [ID-polygon] of patches with [pcolor = white] )
 foreach list-ID-blocks [ 
 ask patches with [ID-polygon = ?] [ 
 set polygon-area ( count patches with [ID-polygon = ?] ) * 0.01 ] ]

提前感谢您的帮助。

Answer 1

问题是每个补丁都是独立计算区域的（count patches with [ ID-polygon = ? ]每次调用时都会遍历每个补丁）。相反，您可以计算每个多边形的面积一次，如下所示：

let list-ID-blocks remove-duplicates [ID-polygon] of patches with [pcolor = white]
foreach list-ID-blocks [
  let area 0.01 * count patches with [ ID-polygon = ? ]
  ask ask patches with [ ID-polygon = ? ] [ set polygon-area area ]
]

那应该快得多。如果它足够快，我就会这样做。但是，它仍然会为每个多边形查看每个补丁一次。如果多边形id只是介于0和某个数字之间的整数，那么您可以查看每个补丁一次，计算区域：

let max-id max [ ID-polygon ] of patches with [ pcolor = white ]
let areas n-values max-id [ 0 ]
ask patches with [ pcolor = white ] [
  let old-area item ID-polygon areas
  set areas replace-item ID-polygon areas (old-area + 1)
]
ask patches with [ pcolor = white ] [
  set polygon-area 0.01 * item polygon-ID areas
]

如何加快多边形面积的计算？

1 个答案: