使用 google s2 库 - 查找圆圈内某个级别的所有 s2 单元，给定 lat/lng 和以英里/公里为单位的半径

Question

我应该使用什么 S2Region 以及如何使用它使用谷歌的 s2 库在给定纬度、经度和以英里/公里为单位的半径的圆圈内获取所有单元格？

S2Region region = ?
S2RegionCoverer coverer = new S2RegionCoverer();
coverer.setMinLevel(17);
coverer.setMaxCells(17);
S2CellUnion covering = coverer.getCovering(region_cap);

谢谢

Answer 1

这是来自 npm 包 s2geometry-node 的 C++ 示例。代码是从取景器/viewfinder.cc 复制过来的

const double kEarthCircumferenceMeters = 1000 * 40075.017;

double EarthMetersToRadians(double meters) {
  return (2 * M_PI) * (meters / kEarthCircumferenceMeters);
}

string CellToString(const S2CellId& id) {
  return StringPrintf("%d:%s", id.level(), id.ToToken().c_str());
}

// Generates a list of cells at the target s2 cell levels which cover
// a cap of radius 'radius_meters' with center at lat & lng.
vector<string> SearchCells(double lat, double lng, double radius_meters,
                           int min_level, int max_level) {
  const double radius_radians = EarthMetersToRadians(radius_meters);
  const S2Cap region = S2Cap::FromAxisHeight(
      S2LatLng::FromDegrees(lat, lng).Normalized().ToPoint(),
      (radius_radians * radius_radians) / 2);
  S2RegionCoverer coverer;
  coverer.set_min_level(min_level);
  coverer.set_max_level(max_level);

  vector<S2CellId> covering;
  coverer.GetCovering(region, &covering);
  vector<string> v(covering.size());
  for (size_t i = 0; i < covering.size(); ++i) {
    v[i] = CellToString(covering[i]);
  }
  return v;
}

Answer 2

我遇到了同样的任务，并使用S2RegionCoverer.getCovering(S2Region region, ArrayList<S2CellId> covering) 方法解决了它。

S2RegionCoverer.getCovering(S2Region region) 文档中描述了您获得不同级别单元格的问题：

/**
   * Return a normalized cell union that covers the given region and satisfies
   * the restrictions *EXCEPT* for min_level() and level_mod(). These criteria
   * cannot be satisfied using a cell union because cell unions are
   * automatically normalized by replacing four child cells with their parent
   * whenever possible. (Note that the list of cell ids passed to the cell union
   * constructor does in fact satisfy all the given restrictions.)
   */