问题描述
[{
name: "aaa",date: "2020",time: [20,10,30],meal: ["before","after","before"],amount: [1,2,3]
},{
name: "bbb",date: "2021",time: [200,100,300],amount: [10,20,30]
}]
[{
name: "aaa",time: 20,meal: "before",amount: 1
},{
name: "aaa",time: 10,meal: "after",amount: 2
},time: 30,amount: 3
},time: 200,amount: 10
},time: 100,amount: 20
},time: 300,amount: 30
}]
然后我要按时间和用餐分组。 我尝试了很多事情,但无法得到结果。
解决方法
如果time
,meal
和amount
的长度始终相同,则可以使用flatMap创建具有相同属性的新对象并将其平整为新数组。
const data=[{name:"aaa",date:"2020",time:[20,10,30],meal:["before","after","before"],amount:[1,2,3]},{name:"bbb",date:"2021",time:[200,100,300],amount:[10,20,30]}];
const combos = data.flatMap(({time,meal,amount,...rest}) => {
return time.map((num,index) => ({
...rest,time: num,meal: meal[index],amount: amount[index],}));
});
console.log(combos);
.as-console-wrapper{min-height:100%;}
,
如果说您的数组位于变量x中,则可以简单地执行操作,而无需使用下划线:
var y = [];
x.forEach(z => {
z.meal.forEach((val,index) => {
y.push({
name: z.name,date: z.date,time: z.time[index],meal: val,amount: z.amount[index]
});
});
});
y
将包含您的新数组。
如果您随后想按某个字段分组或执行其他聚合,并且知道SQL,那么使用alasql是一种有趣的方法。包括/要求它,然后您就可以这样做:
alasql('SELECT name,date,time,SUM(amount) FROM ? GROUP BY time,meal',[y]);