我需要将minutes和seconds_left加起来并按id_project分组。
0: {id_project: 2, minutes: 12, seconds_left: NaN}
1: {id_project: 3, minutes: 15, seconds_left: 11}
2: {id_project: 4, minutes: 0, seconds_left: 11}
3: {id_project: 5, minutes: 0, seconds_left: 2}
4: {id_project: 0, minutes: 0, seconds_left: NaN}
5: {id_project: 1, minutes: 0, seconds_left: NaN}
6: {id_project: 4, minutes: 200, seconds_left: 6}
7: {id_project: 6, minutes: 43, seconds_left: NaN}
8: {id_project: 5, minutes: 100, seconds_left: NaN}
9: {id_project: 2, minutes: 123, seconds_left: NaN}
10: {id_project: 3, minutes: 454, seconds_left: NaN}
11: {id_project: 2, minutes: 89, seconds_left: NaN}
12: {id_project: 5, minutes: 23, seconds_left: NaN}
13: {id_project: 4, minutes: 0, seconds_left: NaN}
14: {id_project: 4, minutes: 11, seconds_left: NaN}
15: {id_project: 3, minutes: 66, seconds_left: NaN}
16: {id_project: 1, minutes: 676, seconds_left: NaN}
我想到了地图＆amp;减少，但我不能这样做。
我的问题有解决办法吗？
感谢您的回答：）
答案 0 :(得分：1)
首先关闭所有沟渠NaN或在执行前将它们转换为0，否则算术将全部具有NaN值。
现在这就是我所做的：我创建了一个函数来将您的数组映射到具有唯一项目ID的新对象，其中每个都具有分钟和秒的总和：
var arr = [
{ id_project: 2, minutes: 12, seconds_left: 1 },
{ id_project: 3, minutes: 15, seconds_left: 11 },
{ id_project: 4, minutes: 0, seconds_left: 11 },
{ id_project: 5, minutes: 0, seconds_left: 2 },
{ id_project: 0, minutes: 0, seconds_left: 2 },
{ id_project: 1, minutes: 0, seconds_left: 3 },
{ id_project: 4, minutes: 200, seconds_left: 6 },
{ id_project: 6, minutes: 43, seconds_left: 3 },
{ id_project: 5, minutes: 100, seconds_left: 3 },
{ id_project: 2, minutes: 123, seconds_left: 10 },
{ id_project: 3, minutes: 454, seconds_left: 2 },
{ id_project: 2, minutes: 89, seconds_left: 2 },
{ id_project: 5, minutes: 23, seconds_left: 3 },
{ id_project: 4, minutes: 0, seconds_left: 4 },
{ id_project: 3, minutes: 66, seconds_left: 5 },
{ id_project: 1, minutes: 676, seconds_left: 5 }
];
var _flatten = function (arr) {
var struct = {}; // We'll use a new flattened structure.
var len = arr.length;
for (var i = 0; i < len; i++) {
var pID = arr[i].id_project;
if (undefined !== struct[pID] && null !== struct[pID]) {
// Already assigned. Sum minutes and seconds to already existing values.
struct[pID].sum =
struct[pID].sum +
(arr[i].minutes * 60) +
arr[i].seconds_left;
}
else {
// Current project ID has not been assigne to new structure. Assign.
struct[pID] = { sum: (arr[i].minutes * 60) + arr[i].seconds_left };
}
}
return struct;
}
输出（总和以秒为单位）：
_flatten(arr);
0 : {sum: 2}
1 : {sum: 40568}
2 : {sum: 13453}
3 : {sum: 32118}
4 : {sum: 12021}
5 : {sum: 7388}
6 : {sum: 2583}
答案 1 :(得分：1)
您可以将哈希表作为对具有相同id_project的对象的引用，并收集结果的所有值。
var array = [{ id_project: 2, minutes: 12, seconds_left: NaN }, { id_project: 3, minutes: 15, seconds_left: 11 }, { id_project: 4, minutes: 0, seconds_left: 11 }, { id_project: 5, minutes: 0, seconds_left: 2 }, { id_project: 0, minutes: 0, seconds_left: NaN }, { id_project: 1, minutes: 0, seconds_left: NaN }, { id_project: 4, minutes: 200, seconds_left: 6 }, { id_project: 6, minutes: 43, seconds_left: NaN }, { id_project: 5, minutes: 100, seconds_left: NaN }, { id_project: 2, minutes: 123, seconds_left: NaN }, { id_project: 3, minutes: 454, seconds_left: NaN }, { id_project: 2, minutes: 89, seconds_left: NaN }, { id_project: 5, minutes: 23, seconds_left: NaN }, { id_project: 4, minutes: 0, seconds_left: NaN }, { id_project: 4, minutes: 11, seconds_left: NaN }, { id_project: 3, minutes: 66, seconds_left: NaN }, { id_project: 1, minutes: 676, seconds_left: NaN }],
temp = {},
result = [];
array.forEach(function (o) {
if (!temp[o.id_project]) {
temp[o.id_project] = { id_project: o.id_project, minutes: 0, seconds_left: 0 };
result.push(temp[o.id_project]);
}
temp[o.id_project].minutes += o.minutes || 0;
temp[o.id_project].seconds_left += o.seconds_left || 0;
});
console.log(result);.as-console-wrapper { max-height: 100% !important; top: 0; }
答案 2 :(得分：0)
如果可以遍历此数组，请创建一个包含数组总和的新对象
像这样的东西
var myArray = [{id_project: 2, minutes: 12, seconds_left: 30},{id_project: 2, minutes: 12, seconds_left: 50},{id_project: 3, minutes: 12, seconds_left: 12}]
var result = {};
myArray.forEach(function (element) {
if (!result.hasOwnProperty(element.id_project))
result[element.id_project] = {minutes : 0, seconds_left : 0};
result[element.id_project].minutes += element.minutes;
result[element.id_project].seconds_left += element.seconds_left;
});
console.log(result);
答案 3 :(得分：0)
以下是你要求的，你可以使用map和reduce。我只使用map替换NaN为0并减少两次，一次获得总秒数，一次创建小时，分钟，秒对象：
var tmpTotal = [
{id_project: 2, minutes: 12, seconds_left: NaN}
,{id_project: 3, minutes: 15, seconds_left: 11}
,{id_project: 4, minutes: 0, seconds_left: 11}
,{id_project: 5, minutes: 0, seconds_left: 2}
,{id_project: 0, minutes: 0, seconds_left: NaN}
,{id_project: 1, minutes: 0, seconds_left: NaN}
,{id_project: 4, minutes: 200, seconds_left: 6}
,{id_project: 6, minutes: 43, seconds_left: NaN}
,{id_project: 5, minutes: 100, seconds_left: NaN}
,{id_project: 2, minutes: 123, seconds_left: NaN}
,{id_project: 3, minutes: 454, seconds_left: NaN}
,{id_project: 2, minutes: 89, seconds_left: NaN}
,{id_project: 5, minutes: 23, seconds_left: NaN}
,{id_project: 4, minutes: 0, seconds_left: NaN}
,{id_project: 4, minutes: 11, seconds_left: NaN}
,{id_project: 3, minutes: 66, seconds_left: NaN}
,{id_project: 1, minutes: 676, seconds_left: NaN}
]
.map(//set seconds_left to 0 when NaN
item=>
isNaN(item.seconds_left)
? Object.assign({},item,{seconds_left:0})
: item
)
.reduce(//group totals by id
(acc,item)=>{
acc[item.id_project]=acc[item.id_project] || {sum:0};
acc[item.id_project].sum =
acc[item.id_project].sum + item.minutes * 60;
acc[item.id_project].sum =
acc[item.id_project].sum + item.seconds_left;
return acc;
}
,{}
);
console.log(
JSON.stringify(
Object.keys(tmpTotal)//create hours,minutes,seconds from sum object
.reduce(
(acc,key)=>{
acc[key] = {
hours: Math.floor(tmpTotal[key].sum/3600),
minutes: Math.floor((tmpTotal[key].sum%3600)/60),
seconds: tmpTotal[key].sum%60
}
return acc;
}
,{}
)
,undefined
,2
)
);