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我正在使用 Ramer Douglas Peucker 算法的实现来减少地图路线的点数。例如，如果我有超过 500 个点，我想以一个容差运行算法，该容差将点数减少到小于 500，同时尽可能接近它。到目前为止，我尝试过的效率非常低的方法如下：

simp = coordSimplify(data.tableData, 0)
while (simp.length > 400) {
    i += 0.0001;
    simp = coordSimplify(data.tableData, i);
}

但我意识到这会大大减慢整个过程。

我怎样才能使整个过程更有效率？我在考虑某种二进制斩波算法，但是我不确定每次如何计算上限和下限。

TIA

4

1 回答 1

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建议尝试以下几行，这实际上是一种寻找epsilon价值的二分搜索（使用术语https://en.wikipedia.org/wiki/Ramer%E2%80%93Douglas%E2%80%93Peucker_algorithm）在 和 的目标点长度范围simpTargetLo内simpTargetHi。

（请注意，我没有对此进行测试，因为我无权访问coordSimplify()，因此可能存在一些语法错误，但逻辑应该是合理的。）

// Set the acceptable result.
let simpTargetLo = 490;
let simpTargetHi = 510;

// Set the initial epsilon range which needs to be outside the estimated solution.
let epsilonLo = 0;
let epsilonHi = 1;
let epsilonMid;

// Calculate the initial low and high simp values.
let simpLo = coordSimplify(data.tableData, epsilonLo);
let simpHi = coordSimplify(data.tableData, epsilonHi);
let simpMid;

// Ensure that the initial simp values fall outside the target range.
if ( !( simpLo.length <= simpTargetLo && simpTargetHi <= simpHi.length ) ) {
  throw new Error( `Initial epsilon need expanding.\n  epsilonLo ${epsilonLo} returns ${simpLo.length}\n  epsilonHi ${epsilonHi} returns ${simpHi.length}` );
}

// Finally, let's ensure we don't get into an infinite loop in the event that
// their is no solution or the solution oscillates outside the target range.
let iterations = 0;
let maxIterations = 100;

do {
  
  // Calculate the simp at the midpoint of the low and high epsilon.
  epsilonMid = ( epsilonLo + epsilonHi ) / 2;
  simpMid = coordSimplify(data.tableData, epsilonMid );
  
  // Narrow the epsilon low and high range if the simp result is still outside
  // both the target low and high.
  if ( simpMid.length < simpTargetLo ) {
    epsilonLo = epsilonMid;
  } else if ( simpTargetHi < simpMid.length ) {
    epsilonHi = epsilonMid;
  } else {
    // Otherwise, we have a solution!
    break;
  }
  
  iterations++;
  
while( iterations < maxIterations );

if ( iterations < maxIterations ) {
  console.log( `epsilon ${epsilonMid} returns ${simpMid.length}` );
} else {
  console.log( `Unable to find solution.` );
}

请注意，这种缩小到解决方案的方法取决于正确选择初始epsilonLo和epsilonHi，此外还假设它coordSimplify()本质上是相当连续的......

于 2021-02-14T17:36:02.187 回答