Question

我想对数组的元素值求和，它们具有相同的年份。

var newData = [];
var data = new Array(['2013', 0], ['2013', 0],['2016', 15], ['2017', 2], ['2015', 1], ['2013', 0], ['2016', 12], ['2014', 0], ['2017', 3], ['2015', 1], ['2013', 21], ['2013', 2]);

var sum = data.reduce(function (a, b,i) {
    if (b[0][i]==b[0][i+1]) {
         return newData.push(b[0],a + b[1]);
    		 
    }    
 }, 0);
 console.log(newData);
 
 //return to:
 
[
["2013",23],
["2014",0],
["2015",2],
["2016",27],
["2017",5]
]

有可能吗？谢谢你的帮助

Answer 1

您可以创建自己的简单且更加理解的自定义逻辑，例如

将JSON对象作为响应。

var newData = [];
var data = new Array(['2013', 0], ['2013', 0],['2016', 15], ['2017', 2], ['2015', 1], ['2013', 0], ['2016', 12], ['2014', 0], ['2017', 3], ['2015', 1], ['2013', 21], ['2013', 2]);

var res = {};
data.forEach((item)=>{
  if(Object.keys(res).includes(item[0])){
    res[item[0]] += item[1];
  } else {
    res[item[0]] = item[1];
  }   
});
 console.log(res);

.as-console-wrapper { max-height: 100% !important; top: 0; }

将JSON数组作为响应。

var newData = [];
var data = new Array(['2013', 0], ['2013', 0],['2016', 15], ['2017', 2], ['2015', 1], ['2013', 0], ['2016', 12], ['2014', 0], ['2017', 3], ['2015', 1], ['2013', 21], ['2013', 2]);

var res = [];
data.forEach((item)=>{
  var found = false;
  for(var i=0; i<res.length; i++){
    if(res[i][0] === item[0]){
       res[i][1] += item[1];
       found = true;
       break;
    }
  }
  if(!found){
    res.push(item);
  }
});
console.log(res);

.as-console-wrapper { max-height: 100% !important; top: 0; }

Answer 2

您可以将ES6 Map与reduce方法一起使用，以按年份汇总值并展开语法...，以便从地图中获取数组。

＆＃13;

var data = new Array(['2013', 0], ['2013', 0], ['2016', 15], ['2017', 2], ['2015', 1], ['2013', 0], ['2016', 12], ['2014', 0], ['2017', 3], ['2015', 1], ['2013', 21], ['2013', 2]);

const result = data.reduce((r, [y, v]) => {
  r.set(y, (r.get(y) || 0) + v);
  return r;
}, new Map)

console.log([...result]);

＆＃13;

Answer 3

如果我理解你的需要，那就试试吧

var sums=data.reduce(function(a,b){
    a[b[0]]=(a[b[0]] || 0)+b[1];
},{});

Answer 4

执行此工作的一种非常快速的方法是使用JS的稀疏数组并选择非空插槽;

＆＃13;

var data    = [['2013', 0], ['2013', 0],['2016', 15], ['2017', 2], ['2015', 1], ['2013', 0], ['2016', 12], ['2014', 0], ['2017', 3], ['2015', 1], ['2013', 21], ['2013', 2]],
    interim = data.reduce((r,e) => (r[+e[0]] && (r[+e[0]][1] += e[1]) || (r[+e[0]] = e), r), []),
    result  = Object.keys(interim)
                    .map(k => interim[k]);

console.log(result);

＆＃13;

.as-console-wrapper {
height     : 100%;
max-height : 100% !important
}

＆＃13;

因此Object.keys(interim)部分只是以极快的方式过滤掉未定义键的插槽。所以没什么可担心的。看看这个;

＆＃13;

var a = [],
    b = [];
a[1000000000] = "west";
a[500000000]  = "test";
a[0]          = "best";
console.time("test");
b = Object.keys(a)
          .map(k => a[k]);
console.timeEnd("test");
console.log(b);

＆＃13;

具有条件的数组的总和

4 个答案: