我有一系列对象：
var arr = [
{
timemark: "2017-03-01",
db_total: c1,
db_used: d1,
hosts: e1,
items: f1
},{
timemark: "2017-03-02",
db_total: c2,
db_used: d2,
hosts: e2,
items: f2
},{
timemark: "2017-03-03",
db_total: c3,
db_used: d3,
hosts: e3,
items: f3
},..]
我正在思考如何将它转换为另一个阵列，但结构不同：
var result = [
{
topic: "db_total",
data: [
{
x: "2017-03-01",
y: c1
},{
x: "2017-03-02",
y: c2
},{
x: "2017-03-03",
y: c3
},...]
},{
topic: "db_used",
data: [
{
x: "2017-03-01",
y: d1
},{
x: "2017-03-02",
y: d2
},{
x: "2017-03-03",
y: d3
},...]
},{
topic: "hosts",
data: [
{
x: "2017-03-01",
y: e1
},{
x: "2017-03-02",
y: e2
},{
x: "2017-03-03",
y: e3
},...]
},{
topic: "items",
data: [
{
x: "2017-03-01",
y: f1
},{
x: "2017-03-02",
y: f2
},{
x: "2017-03-03",
y: f3
},...]
},...];
我知道我必须做这样的事情：
//convert
var result = [];
for (var i=0; i<arr.length; i++) {
result[i]=[arr[i].timemark];
}
创建数组数组：
[
[2017-03-01],
[2017-03-02],
[2017-03-03]
]
几个小时之后，这是一个开始。但我找不到如何开始在数组内创建对象而不是数组的方法？试着走宝贝步骤：）
但是我在理解代码片段时遇到了问题，并且可能使用错误的语法无法使其工作。
在这种情况下，有人可以解释如何正确使用循环吗？
答案 0 :(得分：2)
你可以做这样的逻辑;映射每个分组，并编译最终结果对象;
var arr = [
{
timemark: "2017-03-01",
db_total: "c1",
db_used: "d1",
hosts: "e1",
items: "f1"
},{
timemark: "2017-03-02",
db_total: "c2",
db_used: "d2",
hosts: "e2",
items: "f2"
},{
timemark: "2017-03-03",
db_total: "c3",
db_used: "d3",
hosts: "e3",
items: "f3"
}];
var result = [];
Object.keys(arr[0])
.filter(field => field != "timemark")
.forEach(field => result.push(finalObj(field, arr.map(e => xy(e.timemark, e[field])))));
console.log(result);
function xy(x, y) {
return { x : x, y : y };
}
function finalObj(name, arr) {
return { topic : name, data : arr };
}
由于您建议您有更多字段，因此在最终对象中有更多topic s，如果是这种情况，我已修改，以便您添加的字段越多，它就会自动显示在最终产生的对象。 （timemark字段除外）
答案 1 :(得分：1)
您可以使用Array.prototype的{{3}}函数将数组中的每个元素映射到另一个元素，从而节省数组的长度。
例如：
var result = arr.map(e => ({x: e.timemark, y:db_total}));
var input = [{
timemark: "2017-03-01",
db_total: 1,
db_used: 1,
hosts: 1,
items: 1
},{
timemark: "2017-03-02",
db_total: 1,
db_used: 1,
hosts: 1,
items: 1
},{
timemark: "2017-03-03",
db_total: 1,
db_used: 1,
hosts: 1,
items: 1
}];
var output = [{
topic:'db_total',
data: input.map(e=> ({x:e.timemark, y:e.db_total}))
},{
topic:'db_used',
data: input.map(e=> ({x:e.timemark, y:e.db_used}))
},{
topic:'hosts',
data: input.map(e=> ({x:e.timemark, y:e.hosts}))
},{
topic:'items',
data: input.map(e=> ({x:e.timemark, y:e.items}))
}]
console.log(output)
答案 2 :(得分：1)
var arr = [{
timemark: "2017-03-01",
db_total: 'c1',
db_used: 'd1',
hosts: 'e1',
items: 'f1'
},{
timemark: "2017-03-02",
db_total: 'c2',
db_used: 'd2',
hosts: 'e2',
items: 'f2'
},{
timemark: "2017-03-03",
db_total: 'c3',
db_used: 'd3',
hosts: 'e3',
items: 'f3'
}];
console.log(_getConvertedArray(arr))
function _getConvertedArray(array){
var res = [];
array.forEach(function(obj){
Object.keys(obj).forEach(function(key){
res.push({
topic: key,
data: _getTopicData(arr, key)
});
});
});
return res;
}
function _getTopicData(array, topic){
var res = [];
array.forEach(function(obj){
res.push({
x: obj.timemark,
y: obj[topic]
});
});
return res;
}
答案 3 :(得分：1)
您可以使用reduce来执行此操作
var arr = [{timemark: "2017-03-01",db_total: 'c1',db_used: 'd1',hosts: 'e1',items: 'f1'}, {timemark: "2017-03-02",db_total: 'c2',db_used: 'd2',hosts: 'e2',items: 'f2'}, {timemark: "2017-03-03",db_total: 'c3',db_used: 'd3',hosts: 'e3',items: 'f3'}];
let res = arr.reduce((a, b) => {
for (let key in b) {
if (b.hasOwnProperty(key) && key !== 'timemark' && ! a.find(v => v.topic === key)) {
a.push({
topic: key,
data: arr.map(o => ({
x: o.timemark,
y: o[key]
}))
});
}
}
return a;
}, []);
console.log(res);
只是为了玩地图的乐趣 - 下面会给出你想要的另一个结果集，但是依赖于你对数据的进一步使用，它可能会有所帮助（而且创建的代码也要短得多）：
var arr = [{timemark: "2017-03-01",db_total: 'c1',db_used: 'd1',hosts: 'e1',items: 'f1'}, {timemark: "2017-03-02",db_total: 'c2',db_used: 'd2',hosts: 'e2',items: 'f2'}, {timemark: "2017-03-03",db_total: 'c3',db_used: 'd3',hosts: 'e3',items: 'f3'}];
let r = arr.reduce((a,b) => {
for (let key in b) {
if (b.hasOwnProperty(key) && key !== 'timemark')
a.set(key, (a.get(key) || [] ).concat({x: b.timemark, y: b[key]}))
}
return a;
}, new Map());
console.log(r);
// USAGE
// get the data array for db_total:
console.log(r.get("db_total"));
// or for hosts:
console.log(r.get("hosts"));
答案 4 :(得分：0)
作为一个相当简单但非常具体的函数，您可以创建一个“空”结果数组并将其填入：
var arr = [
{
timemark: "2017-03-01",
db_total: 'c1',
db_used: 'd1',
hosts: 'e1',
items: 'f1'
},{
timemark: "2017-03-02",
db_total: 'c2',
db_used: 'd2',
hosts: 'e2',
items: 'f2'
},{
timemark: "2017-03-03",
db_total: 'c3',
db_used: 'd3',
hosts: 'e3',
items: 'f3'
}
];
var result = [{topic: "db_total",data: []},
{topic: "db_used",data: []},
{topic: "hosts",data: []},
{topic: "items",data: []}
];
arr.forEach(function (obj) {
result[0].data.push({x:obj.timemark,y: obj.db_total});
result[1].data.push({x:obj.timemark,y: obj.db_used});
result[2].data.push({x:obj.timemark,y: obj.hosts});
result[3].data.push({x:obj.timemark,y: obj.items});
});
console.log(result);
或者你可以选择一种更通用的方法，它只取决于 timemark 属性，并根据提供的其他属性构建结果：
var arr = [{
timemark: "2017-03-01",
db_total: 'c1',
db_used: 'd1',
hosts: 'e1',
items: 'f1'
}, {
timemark: "2017-03-02",
db_total: 'c2',
db_used: 'd2',
hosts: 'e2',
items: 'f2'
}, {
timemark: "2017-03-03",
db_total: 'c3',
db_used: 'd3',
hosts: 'e3',
items: 'f3'
}]
// Collect keys but exclude timemark
var keys = Object.keys(arr[0]).filter(function(v) {
return v != 'timemark';
});
// Build result: loop over every object in arr
var result = arr.reduce(function(result, obj) {
// Build object to insert into result array
Object.keys(obj).forEach(function(p) {
var i = keys.indexOf(p);
// Create object for key if it's not timemark and doesn't exist
// And push the data
if (i != -1) {
result[i] = result[i] || {topic:p, data:[]};
result[i].data.push({x:obj.timemark,y:obj[p]});
}
})
return result;
}, []);
console.log(result);