reduce 方法是对数组中的每个元素执行一个由您提供的reducer函数(升序执行)，将其结果汇总为单个返回值。
reducer 函数接收4个参数:
arr.reduce(callback(accumulator, currentValue[, index[, array]])[, initialValue])
javascript
let names = ['Alice', 'Bob', 'Tiff', 'Bruce', 'Alice'];
let nameNum = names.reduce((res, cur)=>{
if (cur in res){
res[cur]++
} else{
res[cur] = 1
}
return res
},{})
// { Alice: 1 }
// { Alice: 1, Bob: 1 }
// { Alice: 1, Bob: 1, Tiff: 1 }
// { Alice: 1, Bob: 1, Tiff: 1, Bruce: 1 }
// { Alice: 2, Bob: 1, Tiff: 1, Bruce: 1 }
// {Alice: 2, Bob: 1, Tiff: 1, Bruce: 1}
console.log(nameNum);javascript
var result = [
{
subject: 'math',
score: 10
},
{
subject: 'chinese',
score: 20
},
{
subject: 'english',
score: 30
}
];
var sum = result.reduce(function(prev, cur) {
return cur.score + prev;
}, 0);
console.log(sum) // 60javascript
// 已知
let data = [
{id: 1, name: 'jack'},
{id: 2, name: 'jack'},
{id: 3, name: 'andy'},
];
/**
* 期望结果,name一样的话，合并重复的id
* [
{id: 1|2, name: 'jack'},
{id: 3, name: 'andy'},
]
*/
// 首先初始 res 为空数组
// cur 为当前元素，检查累加后的res是否存在重复的name，有的话，把id拼接起来
const result = data.reduce((res, cur) => {
const need = res.find(item => item.name === cur.name);
if (need && need.id) {
// 避免重复id，不拼接已有的
if (!(need.id + '').includes(cur.id)) {
need.id = need.id + '|' + cur.id;
}
} else {
res.push(cur)
}
return res;
}, []);
console.log(result)已知,源数据，三列分别表示 中心编号，受试者代码，受试结果
javascript
var data = [
['CHN001', 'CHN001014', true ],
['CHN002', 'CHN002001', true ],
['CHN002', 'CHN002001', false ],
['CHN002', 'CHN002002', true ],
['CHN002', 'CHN002002', false ],
['CHN002', 'CHN002003', true ],
['CHN002', 'CHN002004', true ],
['CHN002', 'CHN002005', true ],
['CHN002', 'CHN002007', false ],
['CHN002', 'CHN002008', false ],
['CHN003', 'CHN003001', true ],
['CHN003', 'CHN003001', false ],
['CHN005', 'CHN005001', true ],
['CHN005', 'CHN005001', false ],
['CHN005', 'CHN005002', true ],
['CHN005', 'CHN005003', true ],
['CHN005', 'CHN005004', true ],
['CHN007', 'CHN007001', true ],
['CHN007', 'CHN007001', false ],
['CHN007', 'CHN007001', false ],
['CHN007', 'CHN007002', true ],
['CHN007', 'CHN007003', true ],
['CHN007', 'CHN007003', false ],
['CHN007', 'CHN007004', true ],
['CHN007', 'CHN007004', true ],
['CHN007', 'CHN007004', false ],
['CHN007', 'CHN007007', true ],
['CHN007', 'CHN007008', true ],
['CHN007', 'CHN007009', false ]
]要求： 按照，中心编号和受试者代码分组， 比如 CHN005 下面包含了4个受试者，为CHN005001，CHN005002，CHN005003，CHN005004
javascript
['CHN005', 'CHN005001', true ],
['CHN005', 'CHN005001', false ],
['CHN005', 'CHN005002', true ],
['CHN005', 'CHN005003', true ],
['CHN005', 'CHN005004', true ],CHN005001 下面又有两条记录，只有同时都是true，才按累加1处理
javascript
['CHN005', 'CHN005001', true ],
['CHN005', 'CHN005001', false ],所以 CHN005001 为0 CHN005002，CHN005003，CHN005004都为1，最终
CHN005结果 为3
期望输出结果：
javascript
"CHN001" "1"
"CHN002" "3"
"CHN005" "3"
"CHN007" "4"求解：
javascript
function computeScore(list) {
const results = list.reduce((res, [no,code,result]) => {
res[no] = res[no]||{};
res[no][code] = (typeof res[no][code] == 'undefined' ? true : res[no][code])&&result;
return res;
}, {});
return Object.keys(results).reduce((res,key) => {
res[key] = Object.values(results[key]).reduce((acc,bol) => acc+bol,0)
return res;
}, {})
}第一步消重
第二步累加个数
此内容由惯性聚合(RSS阅读器)自动聚合整理，仅供阅读参考。 原文来自 — 版权归原作者所有。