Grouping elements in array by multiple properties是与我的问题最接近的匹配,因为它确实通过数组中的多个键对对象进行分组.问题是这个解决方案没有总结属性值然后删除重复项,而是将所有重复项嵌入二维数组中.
@H_502_2@预期的行为
@H_502_2@我有一个对象数组,必须按形状和颜色分组.
var arr = [
{shape: 'square',color: 'red',used: 1,instances: 1},{shape: 'square',used: 2,{shape: 'circle',color: 'blue',used: 0,instances: 0},used: 4,instances: 4},instances: 5},instances: 1}
];
@H_502_2@仅当它们的形状和颜色相同时,此数组中的对象才被视为重复.如果是,我想分别总结其used和实例值,然后删除重复项.
@H_502_2@因此,在此示例中,结果数组可能只包含四种组合：方形红色,方形蓝色,圆形红色,圆形蓝色
@H_502_2@问题
@H_502_2@我在这里尝试了一种更简单的方法：
var arr = [
{shape: 'square',instances: 2}
];
result = [];
arr.forEach(function (a) {
if ( !this[a.color] && !this[a.shape] ) {
this[a.color] = { color: a.color,shape: a.shape,instances: 0 };
result.push(this[a.color]);
}
this[a.color].used += a.used;
this[a.color].instances += a.instances;
},Object.create(null));
console.log(result);
@H_502_2@但它输出
[{shape: "square",color: "red",used: 11,instances: 9},{shape: "circle",color: "blue",instances: 4}]
@H_502_2@而不是预期的结果：
[{shape: "square",used: 5,instances: 3},{shape: "square",instances: 0}]
@H_502_2@如何使我的功能按形状和颜色正确分组对象？即总结他们的价值并删除重复？
解决方法
将
Array#reduce与辅助对象一起使用以对类似对象进行分组.对于每个对象,检查帮助器中是否存在组合的形状和颜色.如果没有,请使用
Object#assign添加到帮助程序以创建对象的副本,然后推送到该数组.如果是,请将其值添加到used和instances.
var arr = [{"shape":"square","color":"red","used":1,"instances":1},{"shape":"square","used":2,{"shape":"circle","color":"blue","used":0,"instances":0},"used":4,"instances":4},"instances":5},"instances":1}];
var helper = {};
var result = arr.reduce(function(r,o) {
var key = o.shape + '-' + o.color;
if(!helper[key]) {
helper[key] = Object.assign({},o); // create a copy of o
r.push(helper[key]);
} else {
helper[key].used += o.used;
helper[key].instances += o.instances;
}
return r;
},[]);
console.log(result);
@H_502_2@如果您可以使用ES6,则使用Map收集值,然后在spreading Map#values将其转换回数组：
const arr = [{"shape":"square","instances":1}];
const result = [...arr.reduce((r,o) => {
const key = o.shape + '-' + o.color;
const item = r.get(key) || Object.assign({},o,{
used: 0,instances: 0
});
item.used += o.used;
item.instances += o.instances;
return r.set(key,item);
},new Map).values()];
console.log(result);
