Question

更新

感谢用户@Thomas，我能够修复我的脚本。您可以在下面的评论部分找到他的评论。

为了更容易，这里是最重要的引用：

[...]为什么你的向量包含字符串而不是数字？

我意外地继续在我的向量中使用字符串，从而破坏了正确的结果。使用数字时，请确保不要使用任何字符串。

此外，请确保您的结果排序正确：

排序依据：

min（|| house.min - client ||，|| house.max - client ||）

或

||（house.max + house.min）/ 2 - client ||

问题：

我有一个有3个房子的阵列，每个房子包含两个X，Y，Z向量。其中一个是最小点，而另一个是最大值。这两个向量代表房子所在的区域/区域。

此外，我有一个客户端，由一个向量表示。

现在我希望得到客户所在的最近的房子。优选地基于房屋的最小和最大矢量。

值：

var houses = [
    {
        "max": { x: "-7839.9688", y: "-7469.0312", z: "215.537" },
        "name": "House 1",
        "min": { x: "-7112.0312", y: "-7932.9692", z: "72.0312" }
    }, {
        "max": { x: "-6380.0312", y: "-9586.1963", z: "372.1475" },
        "name": "House 3",
        "min": { x: "-6764.2441", y: "-9189.0312", z: "96.3953" },
    }, {
        "max": { x: "-7747.1074", y: "-8659.9023", z: "422.2812" },
        "name": "House 2",
        "min": { x: "-6827.3853", y: "-9668.9688", z: "72.0312" },
    }
]

请注意：上面的数组不代表下图。该图仅用于说明目的。在这种特殊情况下，客户1离房子2很近。但是。在一个不同的用例中，他/她可能离房子3很近。

插图/图：

方法1：

＆＃13;

var houses = [
  {
    "max": { x: "-7839.9688", y: "-7469.0312", z: "215.537" },
    "name": "House 1",
    "min": { x: "-7112.0312", y: "-7932.9692", z: "72.0312" }
  }, 
  {
    "max": { x: "-6380.0312", y: "-9586.1963", z: "372.1475" },
    "name": "House 3",
    "min": { x: "-6764.2441", y: "-9189.0312", z: "96.3953" },
  },
  {
    "max": { x: "-7747.1074", y: "-8659.9023", z: "422.2812" },
    "name": "House 2",
    "min": { x: "-6827.3853", y: "-9668.9688", z: "72.0312" },
  }
]
    
var client = {}

client.x = '-7514.48'
client.y = '-9443.54'
client.z = '-183.969'

var distances = []

for (var i = 0; i < houses.length; i++) {
  
  var house = houses[i]
  
  var distance_between_house_min_max = Math.sqrt(((house.min.x - house.max.x)*(house.min.x - house.max.x)) + ((house.min.y - house.max.y)*(house.min.y - house.max.y)) + ((house.min.z - house.max.z)*(house.min.z - house.max.z)));
  
  // Here I would like to take house.max and house.min together in order to get the clients distance to a the house
  
  var distance_of_client_to_house = Math.sqrt(((client.x - house.max.x)*(client.x - house.max.x)) + ((client.y - house.max.y)*(client.y - house.max.y)) + ((client.z - house.max.z)*(client.z - house.max.z)));
  
  console.log("Distance to " + house.name + " is: " + distance_of_client_to_house)
  
  distances.push({ house: house, distance: distance_of_client_to_house })

}

// Order distances by lowest distance

distances = distances.sort(function (a, b) {
  return a.distance > b.distance
});

console.log("Distances:")
console.log(distances)

＆＃13;

方法2： 由于Stackoverflow不支持节点JS模块，因此我首先没有包含这种方法。这就是为什么我用字符串代替方法1而不是真正的＆＃39;矢量对象。

基本上，在这种方法中，我们使用vec3节点JS模块来提供向量操作。

var vector = require("vec3").Vec3

var houses = [
  {
    "max": new vector(-7839.9688, -7469.0312, 215.537),
    "name": "House 1",
    "min": new vector(-7112.0312, -7932.9692, 72.0312)
  }, 
  {
    "max": new vector(-6380.0312, -9586.1963, 372.1475),
    "name": "House 3",
    "min": new vector(-6764.2441, -9189.0312, 96.3953),
  },
  {
    "max": new vector(-7747.1074, -8659.9023, 422.2812),
    "name": "House 2",
    "min": new vector(-6827.3853, -9668.9688, 72.0312),
  }
]

var client = new vector(-7514.48, -9443.54, -183.969)

var distances = []

for (var i = 0; i < houses.length; i++) {

  var house = houses[i]

  var distance_between_house_min_max = house.min.distanceTo(house.max);

  // Here I would like to take house.max and house.min together in order to get the clients distance to a the house

  var distance_of_client_to_house = client.distanceTo(house.max);

  console.log("Distance to " + house.name + " is: " + distance_of_client_to_house)

  distances.push({ house: house, distance: distance_of_client_to_house })

}

// Order distances by lowest distance

distances = distances.sort(function (a, b) {
  return a.distance > b.distance
});

console.log("Distances:")
console.log(distances)

演示： https://tonicdev.com/stevemuster/57c3a52fee25751400967ddd

问题：我如何才能正确猜出客户所在的房屋？我愿意接受你的所有建议/提示。谢谢。

矢量：基于单个矢量获得最接近的矢量

0 个答案: