範例:findClosestLocatioin()

// 傳回'array'當中與指定得 latitude/longitude 最接近的位置
// 假設地表為完美的平面
var findClosestLocatioin = function (lat ,lng ,array) {
    var closest;
    var closest_dist = Number.MAX_VALUE;
    for (var i = 0;i < array.length;i += 1) {
        // 將2個點都轉換成為弧度量
        var lat_rad = radians(lat);
        var lng_rad = radians(lng);
        var lat2_rad = radians(array[i].latitude);
        var lng2_rad = radians(array[i].longitude);
        
        //使用球面餘弦定理公式
        var dist = Math.acos(Math.sin(lat_rad) * Math.sin(lat2_rad) +
                             Math.cos(lat_rad) * Math.cos(lat2_rad) *
                             Math.cos(lng2_rad - lng_rad));
        if (dist < closest_dist) {
            closest = array[i];
            closest_dist = dist;
        }
    }
    return closest;
}

迴圈中大多數程式碼在處理格不相關子問題：「計算2個點之間的球面座標值。」有理由將之抽離：

var spherical_distance = function (lat1,lng1,lat2,lng2) {
     var lat_rad = radians(lat1);
     var lng_rad = radians(lng1);
     var lat2_rad = radians(lat2);
     var lng2_rad = radians(lng2);
     
     //使用球面餘弦定理公式
        var dist = Math.acos(Math.sin(lat_rad) * Math.sin(lat2_rad) +
                             Math.cos(lat_rad) * Math.cos(lat2_rad) *
                             Math.cos(lng2_rad - lng_rad));
};

剩下程式碼就成為：

var findClosestLocatioin = function (lat ,lng ,array) {
    var closest;
    var closest_dist = Number.MAX_VALUE;
    for (var i = 0;i < array.length;i += 1) {
        var dist = spherical_distance(lat,lng,array[i].latitude,array[i].longitude);
        if (dist < closest_dist) {
        closest = array[i];
        closest_dist = dist;
        }
     }
    return closest;
};

Reference

• <<易讀程式之美學－提升程式碼可讀性的簡單法則 >>(The Art of Readable Code)