处理顺序布尔值的数组
处理顺序布尔值的数组
提问于 2018-12-21 00:10:40
很难为这本书设置标题和描述，但当你阅读代码和评论时，希望你会得到它。如果你们有更好的想法，请编辑。
对于这个特定的示例，我如何在javascript中获得每个true和false之间每天的时间戳和差？
作为数组的一般规则:它总是循环true - false。
我的实际数据有点复杂，但我似乎就是不能正确思考，即使对于这个简单的例子。
可以是reduce，可以是for循环，任何东西。谢谢!
const data = [
{ day: 'today', timestamp: 11, value: true },
{ day: 'today', timestamp: 13, value: false }, //here should be 13-11
{ day: 'today', timestamp: 14, value: true },
{ day: 'today', timestamp: 17, value: false }, //here should be 17-14
//the sum for today should be 5 (13-11 + 17-14)
{ day: 'tomorrow', timestamp: 9, value: true },
{ day: 'tomorrow', timestamp: 11, value: false }, //here should be 11-9
{ day: 'tomorrow', timestamp: 11, value: true },
{ day: 'tomorrow', timestamp: 16, value: false } //here should be 16-11
//the sum for today should be 7 (11-9 + 16-11)
]回答 3
Stack Overflow用户
回答已采纳
发布于 2018-12-21 00:20:01
一个简单的缩减就可以完成这项工作:只需添加较大的'false‘值，并从和中删除'true’值，就可以得到它;)
const data = [
{ day: 'today', timestamp: 11, value: true },
{ day: 'today', timestamp: 13, value: false }, //here should be 13-11
{ day: 'today', timestamp: 14, value: true },
{ day: 'today', timestamp: 17, value: false }, //here should be 17-14
//the sum for today should be 5 (13-11 + 17-14)
{ day: 'tomorrow', timestamp: 9, value: true },
{ day: 'tomorrow', timestamp: 11, value: false }, //here should be 11-9
{ day: 'tomorrow', timestamp: 11, value: true },
{ day: 'tomorrow', timestamp: 16, value: false } //here should be 16-11
//the sum for today should be 7 (11-9 + 16-11)
];
const result = data.reduce((acc, elt) => {
if(!acc[elt.day]) acc[elt.day] = 0;
if(!elt.value) {
acc[elt.day] += elt.timestamp;
} else {
acc[elt.day] -= elt.timestamp;
}
return acc;
},{});
console.log(result);
不要忘记在累加器中将每天初始化为0！希望这能有所帮助。
Stack Overflow用户
发布于 2018-12-21 00:26:08
您可以使用reduce轻松完成此操作
const data = [
{ day: 'today', timestamp: 11, value: true },
{ day: 'today', timestamp: 13, value: false },
{ day: 'today', timestamp: 14, value: true },
{ day: 'today', timestamp: 17, value: false },
{ day: 'tomorrow', timestamp: 9, value: true },
{ day: 'tomorrow', timestamp: 11, value: false },
{ day: 'tomorrow', timestamp: 11, value: true },
{ day: 'tomorrow', timestamp: 16, value: false }
]
const op = data.reduce((o,c)=>{
if(o[c['day']]){
o[c['day']]['timestamp'] += c.value ? -c.timestamp : c.timestamp;
} else {
o[c['day']] = {
'timestamp' : c.value ? -c.timestamp : c.timestamp
}
}
return o;
},{})
console.log(op)
Stack Overflow用户
发布于 2018-12-21 00:28:10
您可以检查结果集中最后插入的对象，并更新时间戳。
const
data = [{ day: 'today', timestamp: 11, value: true }, { day: 'today', timestamp: 13, value: false }, { day: 'today', timestamp: 14, value: true }, { day: 'today', timestamp: 17, value: false }, { day: 'tomorrow', timestamp: 9, value: true }, { day: 'tomorrow', timestamp: 11, value: false }, { day: 'tomorrow', timestamp: 11, value: true }, { day: 'tomorrow', timestamp: 16, value: false }],
result = data.reduce((r, { day, timestamp, value }) => {
var last = r[r.length - 1];
if (!last || last.day !== day) {
r.push(last = { day, timestamp: 0 });
}
last.timestamp += value ? -timestamp : +timestamp;
return r;
}, []);
console.log(result);
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原文链接：
https://stackoverflow.com/questions/53872263
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