如何获取具有相同属性且在数组中具有最大值的对象 我有像
这样的数据data = [{title: "test1", version: 1},
{title: "test2", version: 3},
{title: "test1", version: 2},
{title: "test2", version: 2},
{title: "test2", version: 1}];
我想要结果
result = [{title: "test1", version: 2},
{title: "test2", version: 3}];
有没有比我在这里做的更好的方法?
var titles = [...new Set(data.map(o=> o.title))];
var recentVersions = [];
for(var i = 0; i < titles.length; i++){
var recentVersion = null;
for(var j = 0; j < data.length; j++){
if(titles[i] === data[j].title){
if(!recentVersion){
recentVersion = data[j];
}else if (recentVersion.version < data[j].version){
recentVersion = data[j];
}
}
}
recentVersions.push(recentVersion);
}
答案 0 :(得分:1)
你应该查找像map,filter这样的数组方法,在这种情况下是reduce:
let result = data.reduce((r, x)=> {
if(!r[x.title] || r[x.title].version < x.version)
r[x.title] = x;
return r;
}, {});
(这当然是一个对象映射,根据您的用例,您需要通过调用Object.values(result)
来获取值,或者可以迭代对象本身
答案 1 :(得分:0)
您可以将Math.max
方法与filter
和map
方法结合使用。
let data = [{title: "test1", version: 1}, {title: "test2", version: 3}, {title: "test1", version: 2}, {title: "test2", version: 2}, {title: "test2", version: 1}];
let uniqueTitles = [...new Set(data.map(a => a.title))];
data = uniqueTitles.map(function(key){
return { title:key, version : Math.max(...data.filter(a=>a.title == key).map(a=>a.version)) };
});
console.log(data);
答案 2 :(得分:0)
您可以使用您的独特标题数组中的地图来获得您期望的结果:
const data = [
{title: "test1", version: 1},
{title: "test2", version: 3},
{title: "test1", version: 2},
{title: "test2", version: 2},
{title: "test2", version: 1}
];
const titles = [...new Set(data.map(d => d.title))];
const result = titles.map(title => ({
title,
version: Math.max(...data.filter(d => d.title === title).map(d => d.version))
}));
console.log(result);
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答案 3 :(得分:0)
您可以使用array#reduce
并使用查找对象获取给定标题的最高版本号。
var data = [{title: "test1", version: 1}, {title: "test2", version: 3}, {title: "test1", version: 2}, {title: "test2", version: 2}, {title: "test2", version: 1}],
result = Object.values(data.reduce((r,o) => {
r[o.title] = r[o.title] || {version: 0};
if(r[o.title].version < o.version)
r[o.title] = o;
return r;
},{}));
console.log(result);
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