javascript - 处理一组顺序 bool 值

Question

很难为这个设置标题和描述，但希望您在阅读代码和注释时能够理解。如果你们有更好的主意，请编辑。

对于这个特定示例，我如何在 javascript 中获取每天每个 true 和 false 之间的时间戳总和差异？

作为数组的一般规则:它是否总是循环 true - false。

我的真实数据有点复杂，但我似乎无法正确思考，即使对于这个简化的示例也是如此。

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)
]

Answer 1

一个简单的 reduce 就可以完成这项工作:只需添加更大的“假”值，然后从总和中删除“真”值即可；)

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);

不要忘记每天将累加器中的值初始化为 0!希望这会有所帮助。