如何按日期对数组中的项目进行分组?
问题描述:
鉴于以下对象的数组:如何按日期对数组中的项目进行分组?
[
{
"notes": "Game was played",
"time": "2017-10-04T20:24:30+00:00",
"sport": "hockey",
"owner": "steve",
"players": "10",
"game_id": 1,
},
{
"notes": "Game was played",
"time": "2017-10-04T12:35:30+00:00",
"sport": "lacrosse",
"owner": "steve",
"players": "6",
"game_id": 2,
},
{
"notes": "Game was played",
"time": "2017-10-14T20:32:30+00:00",
"sport": "hockey",
"owner": "steve",
"players": "4",
"game_id": 3,
},
{
"notes": "Game was played",
"time": "2017-10-04T10:12:30+00:00",
"sport": "hockey",
"owner": "henry",
"players": "10",
"game_id": 4,
},
{
"notes": "Game was played",
"time": "2017-10-14T20:34:30+00:00",
"sport": "soccer",
"owner": "john",
"players": "12",
"game_id": 5,
}
]
我想创建一个新的阵列,这将创建一个对象,每个日期(如钥匙)和任何游戏在游戏上市当日播放(另一键)在同一个对象中。
新的数组应该是这样的:
[
{
date:'date string',
games:[array of games played on the date]
},
{
date: 'other date',
games:[games under the other date]
}
]...
这里是我做了什么:
let t = this.state.data; (data above)
let list = [];
for (let i = 0; i < t.length; i++) {
let dates = t[i].time.slice(0,10);
if (!list[dates]) {
list[dates] = [];
}
list[dates].push(t[i]);
}
我的版本返回与日期为值的数组,然后里面的那些的游戏列出,但我想列出它们在不同的键相同的对象。
任何帮助或指导将不胜感激。
答
以下是使用reduce
的解决方案。将time
拆分为日期字符串,然后为每个日期设置一个键。如果密钥存在其推入阵:
更新加入阵列格式版本
const data = [
{notes: 'Game was played', time: '2017-10-04T20:24:30+00:00', sport: 'hockey', owner: 'steve', players: '10', game_id: 1},
{ notes: 'Game was played', time: '2017-10-04T12:35:30+00:00', sport: 'lacrosse', owner: 'steve', players: '6', game_id: 2 },
{ notes: 'Game was played', time: '2017-10-14T20:32:30+00:00', sport: 'hockey', owner: 'steve', players: '4', game_id: 3 },
{ notes: 'Game was played', time: '2017-10-04T10:12:30+00:00', sport: 'hockey', owner: 'henry', players: '10', game_id: 4 },
{ notes: 'Game was played', time: '2017-10-14T20:34:30+00:00', sport: 'soccer', owner: 'john', players: '12', game_id: 5 }
];
// this gives an object with dates as keys
const groups = data.reduce((groups, game) => {
const date = game.time.split('T')[0];
if (!groups[date]) {
groups[date] = [];
}
groups[date].push(game);
return groups;
}, {});
// Edit: to add it in the array format instead
const groupArrays = Object.keys(groups).map((date) => {
return {
date,
games: groups[date]
};
});
console.log(groupArrays);
答
这是你在找什么???
var data = [
{
notes: 'Game was played',
time: '2017-10-04T20:24:30+00:00',
sport: 'hockey',
owner: 'steve',
players: '10',
game_id: 1
},
{
notes: 'Game was played',
time: '2017-10-04T12:35:30+00:00',
sport: 'lacrosse',
owner: 'steve',
players: '6',
game_id: 2
},
{
notes: 'Game was played',
time: '2017-10-14T20:32:30+00:00',
sport: 'hockey',
owner: 'steve',
players: '4',
game_id: 3
},
{
notes: 'Game was played',
time: '2017-10-04T10:12:30+00:00',
sport: 'hockey',
owner: 'henry',
players: '10',
game_id: 4
},
{
notes: 'Game was played',
time: '2017-10-14T20:34:30+00:00',
sport: 'soccer',
owner: 'john',
players: '12',
game_id: 5
}
];
function extract() {
var groups = {};
data.forEach(function(val) {
var date = val.time.split('T')[0];
if (date in groups) {
groups[date].push(val.sport);
} else {
groups[date] = new Array(val.sport);
}
});
console.log(groups);
return groups;
}
extract();
答
let a =[{
"notes": "Game was played",
"time": "2017-10-04T20:24:30+00:00",
"sport": "hockey",
"owner": "steve",
"players": "10",
"game_id": 1,
},
{
"notes": "Game was played",
"time": "2017-10-04T12:35:30+00:00",
"sport": "lacrosse",
"owner": "steve",
"players": "6",
"game_id": 2,
},
{
"notes": "Game was played",
"time": "2017-10-14T20:32:30+00:00",
"sport": "hockey",
"owner": "steve",
"players": "4",
"game_id": 3,
},
{
"notes": "Game was played",
"time": "2017-10-04T10:12:30+00:00",
"sport": "hockey",
"owner": "henry",
"players": "10",
"game_id": 4,
},
{
"notes": "Game was played",
"time": "2017-10-14T20:34:30+00:00",
"sport": "soccer",
"owner": "john",
"players": "12",
"game_id": 5,
}]
let finalObj = {}
a.forEach((games) => {
const date = games.time.split('T')[0]
if (finalObj[date]) {
finalObj[date].push(games);
} else {
finalObj[date] = [games];
}
})
console.log(finalObj)
你尝试过这么远吗?这里的解决方案非常简单。 –
嗨斯宾塞,我不确定如何在评论中发布代码,但我创建了一个空数组,并运行了for循环来遍历第一个信息。我剪切原始日期以获取(yyyy-mm-dd)格式,然后创建另一个for循环来检查当前空数组中是否存在该日期,如果没有,我将为此创建一个新的密钥日期,然后将任何游戏传入游戏密钥。起初,我得到一个无限循环,我的记忆崩溃了,然后我一直覆盖我以前的对象。 – Blake
您不想在评论中发布代码,而是将其作为编辑添加到问题中。 –