iPhone硬计算和缓存 [英] iPhone hard computation and caching

问题描述

我有问题。我的数据库中有 500k 条记录。每个记录存储纬度，经度，动物种类，观察日期。我必须在mapkit视图上绘制网格（15x10），显示在这个网格单元中的物种的集中。每个单元格为32x32框。

I have problem. I have database with 500k records. Each record store latitude, longitude, specie of animal,date of observation. I must draw grid(15x10) above mapkit view, that show the concentration of specie in this grid cell. Each cell is 32x32 box.

如果我在运行时计算，这是非常慢。

If I calculate in run-time it is very slow. Have somebody idea how to cache it?In memory or in database.

数据结构：

观察：

• 纬度


• 经度


• 日期

• Latitude
• Longitude
• Date
• Specie
• some other unimportant data

屏幕示例：

alt text http://img6.imageshack.us/img6/7562/20091204201332.png

每个红色方框都显示该区域的物种数。

Each red box opocasity show count of species in this region.

现在使用的代码：
data - >从数据库中选择，在地图区域中查看

Code that i use now: data -> select from database, it is all observation in map region

for (int row = 0; row < rows; row++)
 { 
  for (int column = 0; column < columns; column++)
  {
   speciesPerBox=0;
   minG=boxes[row][column].longitude;
   if (column!=columns-1) {
    maxG=boxes[row][column+1].longitude;
   } else {
    maxG=buttomRight.longitude;
   }

   maxL=boxes[row][column].latitude;
   if (row!=rows-1) {
    minL=boxes[row+1][column].latitude;
   } else {
    minL=buttomRight.latitude;
   }

   for (int i=0; i<sightingCount; i++) {
    l=data[i].latitude;
    g=data[i].longitude;

    if (l>=minL&&l<maxL&&g>=minG&&g<maxG) {
     for (int j=0; j<speciesPerBox; j++) {
       if (speciesCountArray[j]==data[i].specie) {
        hasSpecie=YES;
       }
      }

      if (hasSpecie==NO) {
       speciesCountArray[speciesPerBox]=data[i].specie;
       speciesPerBox++;
      }

      hasSpecie=NO;


     }
    }
   }


   mapData[row][column].count = speciesPerBox;
  }
 }

推荐答案

你的数据是静态的，你可以预先计算每个网格的每个物种，并将其存储在数据库中，而不是所有的位置坐标。

Since you data is static, you can pre-compute each species for each grid and store it in the database instead of all the location coordinates.

由于你有15 x 10 = 150个单元格，你最终会在数据库中有150 * [num of species]条记录，这应该是一个小得多的数字。

Since you have 15 x 10 = 150 cells, you'll end up with 150 * [num of species] records in the database, which should be a much smaller number.

索引在适当的列。否则，您的查询将必须一次扫描每个记录。

Also, make sure you have indexes on the proper columns. Otherwise, your queries will have to scan every single record over and over again.

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