Discovering the highest value within an array of objects

Question

Discovering the highest value within an array of objects

I have a collection of peaks in the following format:

peaks = 
0: {intervalId: 7, time: 1520290800000, value: 54.95125000000001}
1: {intervalId: 7, time: 1520377200000, value: 49.01083333333333}

and so on.

I am looking to determine the peak with the highest value. I attempted to achieve this using the following code:

this.loadPeak = peaks.map(a => Math.max(a.value));

However, instead of getting the maximum value along with the intervalId and time, I only received an array of values from all peaks without obtaining the maximum value itself.

**I appreciate all the solutions provided. Although I cannot accept them all, they were all helpful. Thank you!**

javascript typescript

Answer 1

Answer №1

Sorting an array can lead to unnecessary iterations, especially as the size of the array increases. This slows down the process significantly as elements are constantly moved around. By utilizing the reduce() method, we can efficiently handle this task by simply replacing the previous value with the current element if it is greater:

var peaks = [
  {intervalId: 7, time: 1520290800000, value: 54.95125000000001},
  {intervalId: 7, time: 1520377200000, value: 49.01083333333333}
];

const maxPeak = peaks.reduce((p, c) => p.value > c.value ? p : c);

console.log(maxPeak);

Answer 2

Sorting an array can lead to unnecessary iterations, especially as the size of the array increases. This slows down the process significantly as elements are constantly moved around. By utilizing the reduce() method, we can efficiently handle this task by simply replacing the previous value with the current element if it is greater:

var peaks = [
  {intervalId: 7, time: 1520290800000, value: 54.95125000000001},
  {intervalId: 7, time: 1520377200000, value: 49.01083333333333}
];

const maxPeak = peaks.reduce((p, c) => p.value > c.value ? p : c);

console.log(maxPeak);

Answer 3

Answer №2

If you want to focus solely on the value aspect of Math.max, you can do so.

var peaks = [{ intervalId: 7, time: 1520290800000, value: 54.95125000000001 }, { intervalId: 7, time: 1520377200000, value: 49.01083333333333 }]
    max = Math.max(...peaks.map(({ value }) => value)),
    object = peaks.find(({ value }) => value === max);

console.log(max);
console.log(object);

Answer 4

If you want to focus solely on the value aspect of Math.max, you can do so.

var peaks = [{ intervalId: 7, time: 1520290800000, value: 54.95125000000001 }, { intervalId: 7, time: 1520377200000, value: 49.01083333333333 }]
    max = Math.max(...peaks.map(({ value }) => value)),
    object = peaks.find(({ value }) => value === max);

console.log(max);
console.log(object);

Answer 5

Answer №3

The straightforward approach involves using a loop:

this.highestPeak = null;
for (const peak of peaks) {
    if (!this.highestPeak || peak.value > this.highestPeak.value) {
        this.highestPeak = peak;
    }
}

Example in Action:

const peaks = [
  {intervalId: 7, time: 1520290800000, value: 54.95125000000001},
  {intervalId: 7, time: 1520377200000, value: 49.01083333333333}
];

let highestPeak = null;
for (const peak of peaks) {
    if (!highestPeak || peak.value > highestPeak.value) {
        highestPeak = peak;
    }
}
console.log(highestPeak);

Just like with any array operation, you can also use the reduce method:

this.highestPeak = peaks.reduce((maxPeak, peak) => !maxPeak || maxPeak.value < peak.value ? peak : maxPeak, null);

const peaks = [
  {intervalId: 7, time: 1520290800000, value: 54.95125000000001},
  {intervalId: 7, time: 1520377200000, value: 49.01083333333333}
];

const highestPeak = peaks.reduce((maxPeak, peak) => !maxPeak || maxPeak.value < peak.value ? peak : maxPeak, null);
console.log(highestPeak);

I discovered that this was a duplicate question. I have identified the duplicate and tagged it accordingly. This is now a community wiki answer.

Answer 6

The straightforward approach involves using a loop:

this.highestPeak = null;
for (const peak of peaks) {
    if (!this.highestPeak || peak.value > this.highestPeak.value) {
        this.highestPeak = peak;
    }
}

Example in Action:

const peaks = [
  {intervalId: 7, time: 1520290800000, value: 54.95125000000001},
  {intervalId: 7, time: 1520377200000, value: 49.01083333333333}
];

let highestPeak = null;
for (const peak of peaks) {
    if (!highestPeak || peak.value > highestPeak.value) {
        highestPeak = peak;
    }
}
console.log(highestPeak);

Just like with any array operation, you can also use the reduce method:

this.highestPeak = peaks.reduce((maxPeak, peak) => !maxPeak || maxPeak.value < peak.value ? peak : maxPeak, null);

const peaks = [
  {intervalId: 7, time: 1520290800000, value: 54.95125000000001},
  {intervalId: 7, time: 1520377200000, value: 49.01083333333333}
];

const highestPeak = peaks.reduce((maxPeak, peak) => !maxPeak || maxPeak.value < peak.value ? peak : maxPeak, null);
console.log(highestPeak);

I discovered that this was a duplicate question. I have identified the duplicate and tagged it accordingly. This is now a community wiki answer.

Answer 7

Answer №4

To find the highest value in your list of peaks, you can arrange them in descending order and select the first element.

let peaks = [{
    intervalId: 7,
    time: 1520290800000,
    value: 54.95125000000001
  },
  {
    intervalId: 7,
    time: 1520377200000,
    value: 49.01083333333333
  }
]

let sortedPeaks = peaks.sort((a, b) => b.value - a.value)

let highestPeak = sortedPeaks[0];

console.log(highestPeak);

Answer 8

To find the highest value in your list of peaks, you can arrange them in descending order and select the first element.

let peaks = [{
    intervalId: 7,
    time: 1520290800000,
    value: 54.95125000000001
  },
  {
    intervalId: 7,
    time: 1520377200000,
    value: 49.01083333333333
  }
]

let sortedPeaks = peaks.sort((a, b) => b.value - a.value)

let highestPeak = sortedPeaks[0];

console.log(highestPeak);

Discovering the highest value within an array of objects

Answer №1

Answer №2

Answer №3

Answer №4

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