在对象数组中查找最小值，最大值

Question

我有这个数组，里面有对象：

var pts = [
    { "X": 52.67528921580262,  "Y": 8.373513221740723 },
    { "X": 52.6759657545252,   "Y": 8.374114036560059 },
    { "X": 52.682574466310314, "Y": 8.37256908416748  },
    { "X": 52.68356308524067,  "Y": 8.373942375183105 },
    { "X": 52.68293869694087,  "Y": 8.375487327575684 },
    { "X": 52.67685044320001,  "Y": 8.376259803771973 },
    { "X": 52.6756535071859,   "Y": 8.379607200622559 },
    { "X": 52.676017795531436, "Y": 8.382096290588379 },
    { "X": 52.68101344348877,  "Y": 8.380722999572754 },
    { "X": 52.68351105322329,  "Y": 8.383641242980957 },
    { "X": 52.68,              "Y": 8.389             }
];

如何找到X的min，max和Y的min，max？

Answer 1

您可以调用Math.min和Math.max ，传入仅包含相关值的映射数组，如下所示：

function endProp( mathFunc, array, property ) {
    return Math[ mathFunc ].apply(array, array.map(function ( item ) {
        return item[ property ];
    }));
}

var maxY = endProp( "max", pts, "Y" ), // 8.389
    minY = endProp( "min", pts, "Y" ); // 8.37256908416748

Answer 2

排序数组可能是最简单的。

array.sort(function minXToMaxX(a,b){
   return a.x - b.x;
});

最低的x值在array[0] 。 x最高值是array[array.length - 1] ；

Answer 3

此代码可以正常工作。 有关演示，请参见摘录。

 var pts = [{ "X": 52.67528921580262, "Y": 8.373513221740723 }, { "X": 52.6759657545252, "Y": 8.374114036560059 }, { "X": 52.682574466310314, "Y": 8.37256908416748 }, { "X": 52.68356308524067, "Y": 8.373942375183105 }, { "X": 52.68293869694087, "Y": 8.375487327575684 }, { "X": 52.67685044320001, "Y": 8.376259803771973 }, { "X": 52.6756535071859, "Y": 8.379607200622559 }, { "X": 52.676017795531436, "Y": 8.382096290588379 }, { "X": 52.68101344348877, "Y": 8.380722999572754 }, { "X": 52.68351105322329, "Y": 8.383641242980957 }, { "X": 52.68, "Y": 8.389 }]; var xobj = []; var yobj = []; var len = pts.length; for (var i = 0; i < len; i++) { xobj.push(parseFloat(pts[i].X)); yobj.push(parseFloat(pts[i].Y)); } //height x value var value = xobj[0]; for (var n = 1; n < xobj.length; n++) { if (xobj[n] > value) { value = xobj[n]; } } //lowest x value var valuel = xobj[0]; for (var n = 1; n < xobj.length; n++) { if (xobj[n] < valuel) { valuel = xobj[n]; } } //height y value var valueY = yobj[0]; for (var n = 1; n < yobj.length; n++) { if (yobj[n] > valueY) { valueY = yobj[n]; } } //lowest x value var valuelY = yobj[0]; for (var n = 1; n < yobj.length; n++) { if (yobj[n] < valuelY) { valuelY = yobj[n]; } } document.getElementById("demo").innerHTML = valuel document.getElementById("demo1").innerHTML = value document.getElementById("demo2").innerHTML = valuelY document.getElementById("demo3").innerHTML = valueY 

 Minx: <p id="demo"></p> MaxX: <p id="demo1"></p> Miny: <p id="demo2"></p> MaxY: <p id="demo3"></p>