我接受了挑战并且已经做了，但是我认为这不是一个好方法。 我认为做同一件事有很短的路要走。
我正在寻找MAP，REDUCE和FILTERS，但找不到一种好方法。
目标是：
说明： 有时，当向客户收费时，会创建重复的交易。我们需要找到那些交易，以便可以对其进行处理。交易的所有内容均应相同，但交易ID和发生的时间除外，因为可能会延迟一分钟。
findDuplicateTransactions（transactions）
查找具有相同sourceAccount，targetAccount，类别，金额且每次连续事务之间的时间差小于1分钟的所有事务。
输入 您可以假设所有参数将始终存在且有效。但是，不能保证传入的事务按任何特定顺序进行。
交易清单（交易[]） 输出 所有重复事务组的列表，按时间升序排列（Transaction [] []）。应按组中第一个事务的升序对组进行排序。 例 输入：
[
{
id: 3,
sourceAccount: 'A',
targetAccount: 'B',
amount: 100,
category: 'eating_out',
time: '2018-03-02T10:34:30.000Z'
},
{
id: 1,
sourceAccount: 'A',
targetAccount: 'B',
amount: 100,
category: 'eating_out',
time: '2018-03-02T10:33:00.000Z'
},
{
id: 6,
sourceAccount: 'A',
targetAccount: 'C',
amount: 250,
category: 'other',
time: '2018-03-02T10:33:05.000Z'
},
{
id: 4,
sourceAccount: 'A',
targetAccount: 'B',
amount: 100,
category: 'eating_out',
time: '2018-03-02T10:36:00.000Z'
},
{
id: 2,
sourceAccount: 'A',
targetAccount: 'B',
amount: 100,
category: 'eating_out',
time: '2018-03-02T10:33:50.000Z'
},
{
id: 5,
sourceAccount: 'A',
targetAccount: 'C',
amount: 250,
category: 'other',
time: '2018-03-02T10:33:00.000Z'
}
];
预期输出：
[
[
{
id: 1,
sourceAccount: "A",
targetAccount: "B",
amount: 100,
category: "eating_out",
time: "2018-03-02T10:33:00.000Z"
},
{
id: 2,
sourceAccount: "A",
targetAccount: "B",
amount: 100,
category: "eating_out",
time: "2018-03-02T10:33:50.000Z"
},
{
id: 3,
sourceAccount: "A",
targetAccount: "B",
amount: 100,
category: "eating_out",
time: "2018-03-02T10:34:30.000Z"
}
],
[
{
id: 5,
sourceAccount: "A",
targetAccount: "C",
amount: 250,
category: "other",
time: "2018-03-02T10:33:00.000Z"
},
{
id: 6,
sourceAccount: "A",
targetAccount: "C",
amount: 250,
category: "other",
time: "2018-03-02T10:33:05.000Z"
}
]
];
这是我的代码，但我不喜欢它。有一些不错的方法吗？
function findDuplicateTransactions (transactions = []) {
var result = [];
console.info("total itens :" + transactions.length);
//sort
transactions = transactions.sort((a,b)=> a.time.localeCompare(b.time))
//remove itens not duplicated
result = removeItens(transactions);
//group
result = groupBy(result, function(item){
return [item.sourceAccount, item.targetAccount, item.amount, item.category];
});
console.info(result);
//remove UniqueElements
result = removeUniqueElements(result);
return result;
}
function removeUniqueElements(array){
var filtered = array.filter(function(value, index, arr){
return value.length >= 2;
});
return filtered;
}
function removeItens(array){
var itensToBeRemoved = [];
for (var index = 0; index < array.length; index++) {
const element1 = array[index];
var cont = 0;
console.info("============== looking for: " + element1.id);
for (var index2 = 0; index2 < array.length; index2++) {
const element2 = array[index2];
if(element1.id != element2.id){
var date1 = new Date(element1.time);
var date2 = new Date(element2.time);
var timeDiff = Math.abs(date2.getTime() - date1.getTime());
console.info("comparing :" + element1.id + "<->" + element2.id + " diff: " + timeDiff);
if( timeDiff < 60000) {
//keep it - is similar
console.info("find one duplicated: " + element2.id);
break;
}else{
cont++;
}
}
}
//console.info("cont: " + cont)
if(cont == array.length-1){
//array.splice(index, 1);
console.info("possible duplicated: " + element1.id);
itensToBeRemoved.push(element1.id);
}
}
var filtered = [];
for(var i=0; i<itensToBeRemoved.length; i++){
console.info("remove item: " + itensToBeRemoved[i]);
array = arrayRemove(array, itensToBeRemoved[i]);
}
return array;
}
function arrayRemove(arr, value) {
return arr.filter(function(ele){
console.info("watching: " + ele.id);
console.info("index: " + value);
return ele.id != value;
});
}
function groupBy( array , f ){
var lists = {};
array.forEach( function( o ){
var list = JSON.stringify( f(o) );
lists[list] = lists[list] || [];
lists[list].push( o );
});
return Object.keys(lists).map( function( list ){
return lists[list];
})
}
答案 0 :(得分：3)
您可以预先对数组进行排序，并通过查找具有相同类别的组来缩小数组。
var data = [{ id: 3, sourceAccount: 'A', targetAccount: 'B', amount: 100, category: 'eating_out', time: '2018-03-02T10:34:30.000Z' }, { id: 1, sourceAccount: 'A', targetAccount: 'B', amount: 100, category: 'eating_out', time: '2018-03-02T10:33:00.000Z' }, { id: 6, sourceAccount: 'A', targetAccount: 'C', amount: 250, category: 'other', time: '2018-03-02T10:33:05.000Z' }, { id: 4, sourceAccount: 'A', targetAccount: 'B', amount: 100, category: 'eating_out', time: '2018-03-02T10:36:00.000Z' }, { id: 2, sourceAccount: 'A', targetAccount: 'B', amount: 100, category: 'eating_out', time: '2018-03-02T10:33:50.000Z' }, { id: 5, sourceAccount: 'A', targetAccount: 'C', amount: 250, category: 'other', time: '2018-03-02T10:33:00.000Z' }],
result = data
.sort(({ time: a }, { time: b }) => a.localeCompare(b))
.reduce((r, o) => {
var temp = r.find(([{ category }]) => category === o.category);
if (!temp) r.push(temp = []);
temp.push(o);
return r;
}, []);
console.log(result);.as-console-wrapper { max-height: 100% !important; top: 0; }
带有哈希表。
const delta = (t1, t2) => Math.abs(new Date(t1) - new Date(t2));
var data = [{ id: 3, sourceAccount: 'A', targetAccount: 'B', amount: 100, category: 'eating_out', time: '2018-03-02T10:34:30.000Z' }, { id: 1, sourceAccount: 'A', targetAccount: 'B', amount: 100, category: 'eating_out', time: '2018-03-02T10:33:00.000Z' }, { id: 6, sourceAccount: 'A', targetAccount: 'C', amount: 250, category: 'other', time: '2018-03-02T10:33:05.000Z' }, { id: 4, sourceAccount: 'A', targetAccount: 'B', amount: 100, category: 'eating_out', time: '2018-03-02T10:36:00.000Z' }, { id: 2, sourceAccount: 'A', targetAccount: 'B', amount: 100, category: 'eating_out', time: '2018-03-02T10:33:50.000Z' }, { id: 5, sourceAccount: 'A', targetAccount: 'C', amount: 250, category: 'other', time: '2018-03-02T10:33:00.000Z' }],
keys = ['sourceAccount', 'targetAccount', 'amount', 'category'],
result = Object.values(data
.sort(({ time: a }, { time: b }) => a.localeCompare(b))
.filter((o, i, a) => {
while (a[--i] && delta(o.time, a[i].time) < 60000) {
if (keys.every(k => o[k] === a[i][k])) return;
}
return true;
})
.reduce((r, o) => ((r[o.category] = r[o.category] || []).push(o), r), {})
);
console.log(result);.as-console-wrapper { max-height: 100% !important; top: 0; }
答案 1 :(得分：0)
您可以先根据时间排序，然后减少
let data = [{id: 3,sourceAccount: 'A',targetAccount: 'B',amount: 100,category: 'eating_out',time: '2018-03-02T10:34:30.000Z'},{id: 1,sourceAccount: 'A',targetAccount: 'B',amount: 100,category: 'eating_out',time: '2018-03-02T10:33:00.000Z'},
{id: 6,sourceAccount: 'A',targetAccount: 'C',amount: 250,category: 'other',time: '2018-03-02T10:33:05.000Z'},
{id: 4,sourceAccount: 'A',targetAccount: 'B',amount: 100,category: 'eating_out',time: '2018-03-02T10:36:00.000Z'},
{id: 2,sourceAccount: 'A',targetAccount: 'B',amount: 100,category: 'eating_out',time: '2018-03-02T10:33:50.000Z'},{id: 5,sourceAccount: 'A',targetAccount: 'C',amount: 250,category: 'other',time: '2018-03-02T10:33:00.000Z'}];
let sortedData = data.sort((a,b)=> a.time.localeCompare(b.time))
let keyTrack = sortedData[0].sourceAccount + sortedData[0].targetAccount
let temp = []
let final = sortedData.reduce((op,inp) => {
let key = inp.sourceAccount + inp.targetAccount
if(keyTrack !== key){
keyTrack = key
op.push(temp)
temp = []
}
temp.push(inp)
return op
},[])
if(temp.length){
final.push(temp)
}
console.log(final)