Better poi index computation#109
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gberaudo
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Aug 28, 2023
| flatCoordinates[offset], | ||
| flatCoordinates[offset + 1] | ||
| ); | ||
| if (squaredDistance < minSquaredDistance) { |
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This condition is always true.
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| ((Math.sqrt(squaredDistance) - Math.sqrt(minSquaredDistance)) / | ||
| maxDelta) | |
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Is this an integer?
Also, instead of max we could do xxxx | 1 and that would also avoid te infinite loop.
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| let totalSquaredDistance = 0; | ||
| for (i = offset + stride; i < minIndex; ++i) { | ||
| totalSquaredDistance += squaredDx( | ||
| flatCoordinates[i - stride], | ||
| flatCoordinates[i - stride + 1], | ||
| flatCoordinates[i], | ||
| flatCoordinates[i + 1] | ||
| ) | ||
| } | ||
| totalSquaredDistance += squaredDx( | ||
| flatCoordinates[minIndex], | ||
| flatCoordinates[minIndex + 1], | ||
| closestPoint[0], | ||
| closestPoint[1] | ||
| ); | ||
|
|
||
| return { | ||
| totalSquaredDistance: totalSquaredDistance, | ||
| squaredDistance: minSquaredDistance, | ||
| closestPoint: closestPoint, | ||
| } |
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I think this part can be computed by the algorithm automatically:
- from the initial coordinates;
- add 1 dimension for the distanceFromTheStartToThisPoint;
- then the assignClosest function will automatically compute the distance from the start of the track to the projected point.
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| minSquaredDistance = squaredDistance; | ||
| minIndex = index; | ||
| for (i = 0; i < stride; ++i) { | ||
| closestPoint[i] = tmpPoint[i]; | ||
| } | ||
| closestPoint.length = stride; |
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I think we could avoid these copies by defining 2 points and swapping between them appropriately:
let closestPoint = [];
let assignedPoint = [];
...
assignedPoint = assignClosest(..., assignedClosest);
if (squaredDistance < minSquaredDistance) {
minSquaredDistance = squaredDistance;
minIndex = index;
[closestPoint, assignedPoint] = [assignedPoint, closestPoint]; // swap variables
...
}
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