Question

我有一个来自数据库的对象数组，我想要折叠以获得简洁。我已经导入了underscore.js以帮助完成此过程，并且尝试使用＆＃39; groupBy＆＃39;提供的功能是为了压缩以下列表：

[
    {ID:2570,name:"jim",latitude:59.4,longitude:-7.29},
    {ID:2573,name:"joe",latitude:54.4,longitude:-7.36},
    {ID:2573,name:"joe",latitude:54.3,longitude:-7.37},
    {ID:2574,name:"bob",latitude:58.4,longitude:-7.31},
    {ID:2574,name:"bob",latitude:58.6,longitude:-7.38},
    {ID:2574,name:"bob",latitude:58.8,longitude:-7.39},
    {ID:2575,name:"mary",latitude:54.1,longitude:-7.30},
]

格式：

[
    {ID:2570, name:"jim", locs : [[59.4,-7.29]]},
    {ID:2573, name:"joe", locs : [[54.4,-7.36], [54.3,-7.37]]} //etc...
]

Answer 1

遵循您的直觉使用_.groupBy，

＆＃13;

var data = [
    {ID:2570,name:"jim",latitude:59.4,longitude:-7.29},
    {ID:2573,name:"joe",latitude:54.4,longitude:-7.36},
    {ID:2573,name:"joe",latitude:54.3,longitude:-7.37},
    {ID:2574,name:"bob",latitude:58.4,longitude:-7.31},
    {ID:2574,name:"bob",latitude:58.6,longitude:-7.38},
    {ID:2574,name:"bob",latitude:58.8,longitude:-7.39},
    {ID:2575,name:"mary",latitude:54.1,longitude:-7.30},
];
    
var groups = _.groupBy(data, function(locationObject) { return locationObject.ID; });
    
var result = _.map(groups, function(group) {
    var condensed = {ID: group[0].ID, name: group[0].name};
    condensed.locs = _.map(group, function(row) { 
        return [row.latitude, row.longitude]; 
    }); 
    return condensed;
});

document.getElementById('log').innerHTML = JSON.stringify(result, null, '  ');

＆＃13;

<script src="http://underscorejs.org/underscore-min.js"></script>
<pre id="log"></pre>

＆＃13;

Answer 2

我认为您可以这样使用_.reduce方法：

_.reduce(data, function(prev, curr) {
    var obj = _.findWhere(prev, {ID: curr.ID});
    if (obj) {
        obj.locs.push([curr.latitude, curr.longitude]);
    }
    else {
        curr.locs = [[curr.latitude, curr.longitude]];
        prev.push(_.omit(curr, ['latitude', 'longitude']));
    }
    return prev;
}, []);

演示：http://jsfiddle.net/yd1Lc3ws/1/

Answer 3

var 
    input = [
        {ID:2570,name:"jim",latitude:59.4,longitude:-7.29},
        {ID:2573,name:"joe",latitude:54.4,longitude:-7.36},
        {ID:2573,name:"joe",latitude:54.3,longitude:-7.37},
        {ID:2574,name:"bob",latitude:58.4,longitude:-7.31},
        {ID:2574,name:"bob",latitude:58.6,longitude:-7.38},
        {ID:2574,name:"bob",latitude:58.8,longitude:-7.39},
        {ID:2575,name:"mary",latitude:54.1,longitude:-7.30},
    ]
    , output = {} //Handles filtering
    , outputArray = [] //Not mandatory, if you want output to be an array
    ;
for(var key,i=-1;++i<input.length;){
    key = input[i].ID; //if ID is unique and ID/name combination is always same
    //key = input[i].ID + input[i].name; //if not
    if(!output[key]){
        output[key] = {ID : input[i].ID, name : input[i].name, locs : []};
        outputArray.push(output[key]);
    }
    output[key].locs.push([input[i].latitude, input[i].longitude]);
}

console.dir(output);
console.dir(outputArray);

UnderscoreJS Group对象列表

3 个答案:

演示：http://jsfiddle.net/yd1Lc3ws/1/