你如何确定给定点是否在边界框内？
我的点数是 48.847172 , 2.386597.
边界框：
"48.7998602295",
"48.8198640442",
"2.46138595581",
"2.48138619423"
像往常一样做：
if( bb.ix <= p.x && p.x <= bb.ax && bb.iy <= p.y && p.y <= bb.ay ) {
// Point is in bounding box
}
bb(ix,iy)(ax,ay)p(x,y)此解决方案还考虑了 UI 发送一个跨越经度 180/-180 的框的情况（低缩放级别的地图视图，您可以看到整个世界，允许无限循环水平滚动，因此例如盒子的 bottomLeft.lng=170 while topRight.lng=-170(=190) 并且包括 20 度的范围。
def inBoundingBox(bl/*bottom left*/: Coordinates, tr/*top right*/: Coordinates, p: Coordinates): Boolean = {
// in case longitude 180 is inside the box
val isLongInRange =
if (tr.long < bl.long) {
p.long >= bl.long || p.long <= tr.long
} else
p.long >= bl.long && p.long <= tr.long
p.lat >= bl.lat && p.lat <= tr.lat && isLongInRange
}
如果你正在使用leaflet，你可以创建一个新的
LatLngBoundscontains()var bounds = new L.LatLngBounds(
new L.LatLng(gc.bbox['_northEast'].lat, gc.bbox['_northEast'].lng),
new L.LatLng(gc.bbox['_southWest'].lat, gc.bbox['_southWest'].lng));
return bounds.contains(new L.LatLng(pos.latitude, pos.longitude))
这应该更快。
function doesPointCollide(p,box) {
return !(p.x < box.left || p.x > box.right || p.y > box.bottom || p.y < box.top)
}
如果点在任何维度之外，我们知道它不在边界框中，否则它在边界框中，所以我们可以更快地忽略否定案例。
这个答案与上面kumetix的答案相同；然而，对于我的生活，我不明白那里发生了什么，因为它被浓缩了很多。这是带有详细解释的相同答案。另请注意，答案是在 Kotlin 中，而不是原始帖子要求的 JavaScript，但它具有足够的可读性，因此转换为您的语言应该不会太糟糕。
首先定义 Coordinate2D 和 BoundingBox 类：
data class Coordinate2D (val latitude: Double, val longitude: Double)
data class BoundingBox (val north: Double, val east: Double, val south: Double, val west: Double)
这里是判断一个点是否在任意边界框内的函数
fun isPointInBoundingBox(point: Coordinate2D, boundingBox: BoundingBox): Boolean {
//given the bounding box is an imaginary rectangle in a coordinate system
//bounding box has 4 sides - northLine, eastLine, southLine and westLine
//initially assume the point is not in our bounding box of interest as such:
var isPointEastOfWestLine = false
var isPointWestOfEastLine = false
var isPointSouthOfNorthLine = false
var isPointNorthOfSouthLine = false
if (boundingBox.east < boundingBox.west) {
//if west coordinate has a bigger value than the east,
//the bounding box must be crossing the dateline
//in other words, longitude 180 is inside the box
//let's see what's happening with the point
if (point.longitude >= boundingBox.west) {
//imagine a bounding box where westernmost longitude is +170 and easternmost longitude is -170
//if the point in question has a latitude of +171 as in the case expressed in the if
//statement, then we can conclude that point lies east of the west line
isPointEastOfWestLine = true
//we can also infer that the point must lie west of east line because the point's longitude is positive
//therefore, the point's position must be to the west of the easternmost longitude of the bounding box
isPointWestOfEastLine = true
}
if (point.longitude <= boundingBox.east) {
//imagine a bounding box where westernmost longitude is +170 and easternmost longitude is -170
//if the point in question has a latitude of -171 as in the case expressed in the if
//statement, then we can conclude that point lies west of the east line
isPointWestOfEastLine = true
//we can also infer that the point must lie east of the west line because the point's longitude is negative
//therefore, the point's position must be to the east of the westernmost longitude of the bounding box
isPointEastOfWestLine = true
}
} else {
//in the else case, bounding box does not cross the dateline, so comparisons are more straightforward
//longitudes range from -180 to +180; therefore, western side of a bounding box will always
//have lower value than eastern side
if (point.longitude >= boundingBox.west) {
//in this case, point's longitude has a higher value than the west side of the bounding box
//we can conclude that point lies to the east of the west line of the bounding box
isPointEastOfWestLine = true
}
if (point.longitude <= boundingBox.east) {
//in this case, point's longitude has a lower value than the east side of the bounding box
//we can conclude that point lies to the east of the west line of the bounding box
isPointWestOfEastLine = true
}
}
//comparing latitudes are little bit easier. latitude values range from -90 to +90 where
//-90 is the southern pole and +90 is the northern pole. The more north you go, higher the values.
if (point.latitude >= boundingBox.south) {
//point's latitude is higher, therefore, point must lie to the north of the south line
isPointNorthOfSouthLine = true
}
if (point.latitude <= boundingBox.north) {
//point's latitude is higher, therefore, point must lie to the north of the south line
isPointSouthOfNorthLine = true
}
return isPointEastOfWestLine &&
isPointWestOfEastLine &&
isPointNorthOfSouthLine &&
isPointSouthOfNorthLine
}
CGRect 和 CGPoint 有非常好的实用方法（假设您不介意它们使用 CGFloat 存储坐标的事实 - 并且查看您的值，您不会:-)）。
你可以这样做：
// Create bounding box
CGRect area = CGRectMake(x, y, width, height);
// Define point
CGPoint point = CGPointMake(pX, pY);
/Check
BOOL isInside = CGRectContainsPoint(area, point);
对c plus plus使用这个函数来检查一个点是否存在于矩形内
struct Box{
Vec2 corner1;
Vec2 corner2;
Vec2 corner3;
Vec2 corner4;
};
bool boxContainsPoint(Vec2 point, Box box){
//Calculate distances from corner to corner
float abL = box.corner1.getDistance(box.corner2);////////////////////
float bcL = box.corner2.getDistance(box.corner3);////////////////////
float cdL = box.corner3.getDistance(box.corner4);////////////////////
float daL = box.corner4.getDistance(box.corner1);////////////////////
//Calculate corner touch distance//////////////////////////////////
float apL = box.corner1.getDistance(point);/////////////////////////
float bpL = box.corner2.getDistance(point);/////////////////////////
float cpL = box.corner3.getDistance(point);/////////////////////////
float dpL = box.corner4.getDistance(point);/////////////////////////
//Here we calculate the touch area
//Heron's Formula
///////////////////////////////////////////////////////////////////
float area1 = (abL + apL + bpL) / 2;///////////////////////////////
float area2 = (bcL + bpL + cpL) / 2;///////////////////////////////
float area3 = (cdL + cpL + dpL) / 2;///////////////////////////////
float area4 = (daL + dpL + apL) / 2;///////////////////////////////
float a1 = sqrtf(area1 * (area1 - abL)*(area1 - apL)*(area1 - bpL));
float a2 = sqrtf(area2 * (area2 - bcL)*(area2 - bpL)*(area2 - cpL));
float a3 = sqrtf(area3 * (area3 - cdL)*(area3 - cpL)*(area3 - dpL));
float a4 = sqrtf(area4 * (area4 - daL)*(area4 - dpL)*(area4 - apL));
//Calculate the rectangle area
float A = roundf(abL*bcL);
//Check to see if the point contains the polygon rect
if(A ==roundf(a1 + a2 + a3 + a4)) return true;
return false;
}
我认为这是最简单的方法。
function pointInBB(px,py,pz,bx,by,bz) {
return px<bx&&py<by&&pz<bz
}
基于 kumetix 解决方案，我编写了与 React Native Maps 完美配合的相同功能
Typescript 代码，您可以删除类型以使其也适用于 javascript
export function isPointInBoundaries(p: LatLng, bl: LatLng, tr: LatLng) {
const isLongInRange = () => {
if (tr.longitude < bl.longitude) return p.longitude >= bl.longitude || p.longitude <= tr.longitude;
else return p.longitude >= bl.longitude && p.longitude <= tr.longitude;
};
return p.latitude >= bl.latitude && p.latitude <= tr.latitude && isLongInRange();
}
可以像这样使用：
const p: LatLng = {longitude: 0, latitude: 0};
const mapBoundaries = mapRef.current.getMapBoundaries();
const condition = isPointInBoundaries(p, mapBoundaries.southWest, mapBoundaries.northEast)
检查点是否在可见边界内的最短和最简单的方法（或一般的任何边界，只需使用
LngLatBounds.contains()return map.getBounds().contains(new mapboxgl.LngLat(LNG, LAT))