javascript中(多维)数组中的反向项

javascript中(多维)数组中的反向项,javascript,multidimensional-array,openstreetmap,reverse,leaflet,Javascript,Multidimensional Array,Openstreetmap,Reverse,Leaflet,我正在使用传单JS在OpenStretchMap上显示一些多边形 我有一个外部数据源,它给我多边形的坐标。不幸的是,这个数组的坐标顺序错误 例如: 我得到: [10.5254913,52.2734311],[10.5258872,52.2734632]] 我需要: [52.2734311,10.5254913],[52.2734632,10.5258872]] 所以我自己编写了一个小函数,它遍历数组并反转条目: var polCoords = []; $.each(value.polygon[0

我正在使用传单JS在OpenStretchMap上显示一些多边形

我有一个外部数据源,它给我多边形的坐标。不幸的是,这个数组的坐标顺序错误

例如: 我得到:

[10.5254913,52.2734311],[10.5258872,52.2734632]]

我需要:

[52.2734311,10.5254913],[52.2734632,10.5258872]]

所以我自己编写了一个小函数,它遍历数组并反转条目:

var polCoords = [];
$.each(value.polygon[0], function(key,value){
polCoords[key] = [value[1],value[0]]; 
});
这个很好用。但是现在我发现一些多边形的数组是多维的!我有一个这样的数组:

[[[10.5261828,52.2726556],[10.5263222,52.2726767],[10.5263578,52.2726821],[10.5263637,52.2726677],[10.5263738,52.2726428],[10.5264042,52.2725678],[10.526186,52.2725346],[10.5261395,52.272649],[10.5261828,52.2726556]],[[10.5261828,52.2726556],[10.5261713,52.2726821],[10.5261621,52.2727047],[10.5259248,52.2726687],[10.5257879,52.2726479],[10.5257435,52.2727573],[10.5258014,52.2727661],[10.5257967,52.2727777],[10.5260173,52.2728113],[10.5261107,52.2728254],[10.5260641,52.2729403],[10.5259711,52.2729262],[10.5259526,52.2729234],[10.5258101,52.2732746],[10.5258697,52.2732837],[10.5260636,52.2733132],[10.5261371,52.2733243],[10.5262746,52.2729854],[10.5262888,52.2729876],[10.526312,52.2729304],[10.5262636,52.2729231],[10.5262239,52.272917],[10.5263222,52.2726767],[10.5261828,52.2726556]],[[10.5260636,52.2733132],[10.5260595,52.2733365],[10.5260575,52.2733486],[10.5258607,52.2733326],[10.5258631,52.2733195],[10.5258697,52.2732837],[10.5260636,52.2733132]]]
它似乎由多个多边形组成


如何反转此多维数组的每个条目?

假设
数组
是您的数据,请尝试以下操作:

var reversed = array.map(function reverse(item) {
    return Array.isArray(item) && Array.isArray(item[0]) 
               ? item.map(reverse) 
               : item.reverse();
});
var reversed = $.map(array, function reverse(item) {
    return $.isArray(item) && $.isArray(item[0])
               ? $.map(item, reverse)
               : item.reverse();
});
这是香草JavaScript。我认为在jquery中应该是这样的:

var reversed = array.map(function reverse(item) {
    return Array.isArray(item) && Array.isArray(item[0]) 
               ? item.map(reverse) 
               : item.reverse();
});
var reversed = $.map(array, function reverse(item) {
    return $.isArray(item) && $.isArray(item[0])
               ? $.map(item, reverse)
               : item.reverse();
});
我没有试过,但他们两个都可以。让我知道。


<script type="text/javascript">
        //given array
        var arr = [
                    [
                        [10.5261828, 52.2726556],
                        [10.5263222, 52.2726767],
                        [10.5263578, 52.2726821],
                        [10.5263637, 52.2726677],
                        [10.5263738, 52.2726428],
                        [10.5264042, 52.2725678],
                        [10.526186, 52.2725346],
                        [10.5261395, 52.272649],
                        [10.5261828, 52.2726556]
                    ],
                    [
                        [10.5261828, 52.2726556],
                        [10.5261713, 52.2726821],
                        [10.5261621, 52.2727047],
                        [10.5259248, 52.2726687],
                        [10.5257879, 52.2726479],
                        [10.5257435, 52.2727573],
                        [10.5258014, 52.2727661],
                        [10.5257967, 52.2727777],
                        [10.5260173, 52.2728113],
                        [10.5261107, 52.2728254],
                        [10.5260641, 52.2729403],
                        [10.5259711, 52.2729262],
                        [10.5259526, 52.2729234],
                        [10.5258101, 52.2732746],
                        [10.5258697, 52.2732837],
                        [10.5260636, 52.2733132],
                        [10.5261371, 52.2733243],
                        [10.5262746, 52.2729854],
                        [10.5262888, 52.2729876],
                        [10.526312, 52.2729304],
                        [10.5262636, 52.2729231],
                        [10.5262239, 52.272917],
                        [10.5263222, 52.2726767],
                        [10.5261828, 52.2726556]
                    ],
                    [
                        [10.5260636, 52.2733132],
                        [10.5260595, 52.2733365],
                        [10.5260575, 52.2733486],
                        [10.5258607, 52.2733326],
                        [10.5258631, 52.2733195],
                        [10.5258697, 52.2732837],
                        [10.5260636, 52.2733132]
                    ]
                ];        

        var arrModified = new Array();
        var intDepth = 0;

        //array depth calculation
        function arrayDepth(arr) {
            for(item in arr) {
                if (Object.prototype.toString.call(arr) === '[object Array]') {
                    intDepth++;
                    arrayDepth(item);
                }
            }
        }
        arrayDepth(arr);

        //for depth 1
        if (intDepth == 1) {            
                arrModified = arr.reverse();
        }

        //for depth 2
        if (intDepth == 2) {
            for (item in arr) {
                arrModified[item] = arr[item].reverse();
            }
        }

        //for depth 3
        if (intDepth == 3) {
            for (i in arr) {
                var arrTemp = arr[i];
                var arrTemp1 = new Array();
                for (j in arrTemp) {
                    arrTemp1[j] = arrTemp[j].reverse();
                }
                arrModified[i] = arrTemp1;                
            }            
        }

        //so on for other depths
        //arrModified is the required array

    </script>
//给定数组 var arr=[ [ [10.5261828, 52.2726556], [10.5263222, 52.2726767], [10.5263578, 52.2726821], [10.5263637, 52.2726677], [10.5263738, 52.2726428], [10.5264042, 52.2725678], [10.526186, 52.2725346], [10.5261395, 52.272649], [10.5261828, 52.2726556] ], [ [10.5261828, 52.2726556], [10.5261713, 52.2726821], [10.5261621, 52.2727047], [10.5259248, 52.2726687], [10.5257879, 52.2726479], [10.5257435, 52.2727573], [10.5258014, 52.2727661], [10.5257967, 52.2727777], [10.5260173, 52.2728113], [10.5261107, 52.2728254], [10.5260641, 52.2729403], [10.5259711, 52.2729262], [10.5259526, 52.2729234], [10.5258101, 52.2732746], [10.5258697, 52.2732837], [10.5260636, 52.2733132], [10.5261371, 52.2733243], [10.5262746, 52.2729854], [10.5262888, 52.2729876], [10.526312, 52.2729304], [10.5262636, 52.2729231], [10.5262239, 52.272917], [10.5263222, 52.2726767], [10.5261828, 52.2726556] ], [ [10.5260636, 52.2733132], [10.5260595, 52.2733365], [10.5260575, 52.2733486], [10.5258607, 52.2733326], [10.5258631, 52.2733195], [10.5258697, 52.2732837], [10.5260636, 52.2733132] ] ]; var arrrmodified=新数组(); var-intDepth=0; //阵列深度计算 函数arrayDepth(arr){ 用于(arr中的项目){ if(Object.prototype.toString.call(arr)='[Object Array]'){ intDepth++; 阵列深度(项目); } } } arrayDepth(arr); //深度1 如果(intDepth==1){ arrModified=arr.reverse(); } //深度2 if(intDepth==2){ 用于(arr中的项目){ arrModified[item]=arr[item].reverse(); } } //深度3 if(intDepth==3){ 对于(我在arr中){ var arrTemp=arr[i]; var arrTemp1=新数组(); 用于(ARRTEM中的j){ arrTemp1[j]=arrTemp[j]。反向(); } arrModified[i]=arrTemp1; } } //其他深度也是如此 //这是所需的数组
注意:此方法将更改原始数组


香草javaScript也应该可以工作。不需要jQuery。我正在测试这个,但是我得到了“UncaughtTypeError:Cannotreadproperty'length'of undefined”,我为我的问题找到了一个解决方案,但它不是这个问题的答案。以防其他人遇到同样的问题:我现在在传单中使用geoJSON,它以我接收gps坐标的方式接受gps坐标。我没有用您编写的数据测试jQuery坐标,而是普通的坐标,它工作了——而且,那里没有访问
长度的权限,所以不可能是那个代码。不管怎样,很高兴你用另一种方式解决了。硬编码的深度级别?为什么不使用递归函数呢?