find an object with the latest timestamp string that is earlier than a reference value in an array-CodePudding

I have the following code to go through an array of objects and find the object with a timestamp that is still smaller of a reference value but the biggest of those

    for (const elem of objs) {
        if (elem._time >= failureTime) {//we are one too far
            break;          
        } else {
            temp = elem; //this will end up as the last element with timestamp before event
           }
        }

This assumes that objs is ordered by timestamp, which I can't predict so I need to reorder it every time. I tried to create a .reduce function to solve that but horribly failed. Is there a way to do it?

CodePudding user response：

If you're asking how to find the largest time less than failureTime when objs isn't sorted, you just don't stop when you see an out-of-range value. Instead, you ignore those, and keep track of the highest in-range value you've seen:

let temp = null;
for (const elem of objs) {
    if (elem._time < failureTime && (temp === null || elem._time > temp._time)) {
        temp = elem;
    }
}

Live Example:

const objs = [
    { _time: new Date("2022-01-02T00:00:00.000Z") },
    { _time: new Date("2022-01-04T00:00:00.000Z") },
    { _time: new Date("2022-01-03T00:00:00.000Z") },
    { _time: new Date("2022-01-06T00:00:00.000Z") },
    { _time: new Date("2022-01-01T00:00:00.000Z") },
];
const failureTime = new Date("2022-01-05T00:00:00.000Z");
let temp = null;
for (const elem of objs) {
    if (elem._time < failureTime && (temp === null || elem._time > temp._time)) {
        temp = elem;
    }
}
console.log(temp._time);

CodePudding user response：

You can use the reduce method;

//failureTime
arr.reduce((acc,key)=>{
if(acc._time <= failureTime){
 if(acc._time < key._time) return key;
 else return acc;
} 
},arr[0]);

Hope it helps.