Question

我有一个对象数组，如下所示：

[
{"name":"Andrea","from":"USA","Food":"Candy"},
{"name":"Matt","from":"Taiwan","Food":"Chicken"},
{"name":"Roddy","from":"USA","Food":"Rice"},
{"name":"Andy","from":"Great Britain","Food":"Steak"},
];

有没有办法从上面的数组中获取所有国家/地区的列表，并摆脱重复的数据？

因此，从上面的列表中，我要获取的列表是：

["USA", "Taiwan", "Great Britain"]

谢谢！

Answer 1

只需循环遍历人员并在新数组中插入唯一的国家/地区。这是一个例子。

var countries = [];

var people = [
{"name":"Andrea","from":"USA","Food":"Candy"},
{"name":"Matt","from":"Taiwan","Food":"Chicken"},
{"name":"Roddy","from":"USA","Food":"Rice"},
{"name":"Andy","from":"Great Britain","Food":"Steak"},
];

for (var i = 0, l=people.length; i < l; i++) {
    if(people[i] && people[i].from) {//ensure country exists
        if (countries.indexOf(people[i].from) == -1) {//ensure unique
             countries.push(people[i].from);
        }
    }
}

Answer 2

reduce

的另一种变体

var arr = [
    {"name":"Andrea","from":"USA","Food":"Candy"},
    {"name":"Matt","from":"Taiwan","Food":"Chicken"},
    {"name":"Roddy","from":"USA","Food":"Rice"},
    {"name":"Andy","from":"Great Britain","Food":"Steak"},
];

var countries = arr.reduce(function(acc, cur){
                    if(!acc.map[cur.from]){
                        acc.map[cur.from]=true;
                        acc.result.push(cur.from);
                    }
                    return acc;
                }, {result:[], map:{}}).result;

＆＃13;

var arr = [
    {"name":"Andrea","from":"USA","Food":"Candy"},
    {"name":"Matt","from":"Taiwan","Food":"Chicken"},
    {"name":"Roddy","from":"USA","Food":"Rice"},
    {"name":"Andy","from":"Great Britain","Food":"Steak"},
];

var countries = arr.reduce(function(acc, cur){
                    if(!acc.map[cur.from]){
                        acc.map[cur.from]=true;
                        acc.result.push(cur.from);
                    }
                    return acc;
                }, {result:[], map:{}}).result;

document.getElementById('countries').innerHTML = countries.join();

＆＃13;

<span id="countries"></span>

＆＃13;

Answer 3

如果您已经使用了优秀的Lodash库，以下内容将在一行中整齐地为您完成：

var uniqueCountries = _(dataArray).pluck('from').unique().value();

UnderscoreJS使用链接具有类似的功能。

对于D3.js，以下将执行此操作：

var uniqueCountries = d3.set(dataArray.map(function (x) { return x.from; })).values();

如果不对服务器执行unique-ifying并单独返回该数据，则无法绕过所有记录至少一次来执行此操作。但是，对于1000条左右的记录，这仍然会非常快。对于普通的JS，请参阅其他答案。

Answer 4

我遍历数组并将国家/地区放入数组中，如果它还不在该数组中。

Javascript对象数组和唯一值

4 个答案: