Building and filling a fixed-width array in JavaScript

Question

Building and filling a fixed-width array in JavaScript

I have a collection of data in the following format:

[[timestamp, jobid, time to completion],[..]]

This data is sourced from a SQL database and is grouped by timestamp and jobid. The array looks like this:

[  
[1, 30, 400],  
[1, 31, 200],  
[2, 29, 300],  
..  
]

My goal is to rearrange this data into a new array where each row corresponds to a timestamp and each column represents a specific jobid.

To achieve this, I initially attempted iterating through the existing array and populating a new one. However, the resulting array ended up being variable in width, making it difficult to identify which values corresponded to each job ID.

What I desire is an output that resembles the following structure:

timestamp, jobid29, jobid30, jobid31  
[  
[1, 0, 400, 200],  
[2, 300, 0, 0],  
..  
]

Unfortunately, outputting as a map is not an option for me.

How would you suggest achieving this? My current plan involves first identifying all distinct jobids within the input and then mapping each jobid to a specific position in the output array. Is there a better approach?

Appreciate your insights.

javascript arrays json

Answer 1

Answer №1

My approach involves utilizing two arrays and two maps with numerous possibilities for enhancement opportunities. Allow me to outline the general plan:

Begin by sorting your array based on the jobID. (let's refer to this as the "jobList")
Maintain a "map" that is organized by unique timestamps. ("timeStampMap"). Each unique timestamp should have an entry in this map.
Create another array consisting of unique timestamps, which should also be sorted. ("timeList")
Every item in the "timeStampMap" contains another map storing jobs associated with that particular timestamp value
Iterate through each value in the timestamp list.

During each iteration, go through every job in the jobList. If the job exists in the corresponding timestamp map, output the job's completion time.

If not found, output zero.

There are two potential optimizations that can be applied to the code provided below: 1. It may be possible to utilize the input array directly as the "jobList" without requiring duplication. 2. Combining the map of maps into a single large map might also be feasible.

function jobSorter(a,b) {
    return a.jobID - b.jobID;
}

function numberSort(a,b) {
    return a - b;
}


function buildSortedArray(yourArray) {

    var index, j, item;
    var jobList = []; // array of {jobID, timeStamp, timeComp};
    var timeStampMap = {};
    var timeStampList = [];
    var key, jobkey;
    var output = [];

    // loop through every item in the array
    for (index = 0; index < yourArray.length; index++) {
        item = yourArray[index];         

        // push each item into a "job list"
        job = {jobID: item[1], timeStamp: item[0], timeComp: item[2]};
        jobList.push(job);

        // now use both a map and a list to keep track of all the unique timestamps
        key = "$timestamp$" + job.timeStamp.toString();

        if (timeStampMap[key] === undefined) {
            timeStampMap[key] = {};
            timeStampMap[key].jobMap = {};

            timeStampList.push(job.timeStamp); // keep another timestamp array that we can sort on
        }

        // add this job to the timestamp
        jobkey = "$jobid$" + job.jobID.toString();
        timeStampMap[key].jobMap[jobkey] = job;
    }

    // let's do some sorting
    timeStampList.sort(numberSort); // timeStampList is now in ascending order
    jobList.Sort(jobSorter);        // jobList is now in ascending order


    for (index = 0; index < timeStampList.length; index++) {
        item = [];

        item.push(timeStampList[index]);
        for (j = 0; j < jobList.length; j++) {

            key = "$timestamp$" + timeStampList[index].toString();
            jobkey = "$jobid$"  + jobList[j].toString();

            if (timeStampMap[key].jobMap[jobkey]) {
                item.push(timeStampMap[key].jobMap[jobkey].timeComp);
            }
            else {
              item.push(0);
            }
        }

        output.push(item);
    }


    return output;
}

Answer 2

My approach involves utilizing two arrays and two maps with numerous possibilities for enhancement opportunities. Allow me to outline the general plan:

Begin by sorting your array based on the jobID. (let's refer to this as the "jobList")
Maintain a "map" that is organized by unique timestamps. ("timeStampMap"). Each unique timestamp should have an entry in this map.
Create another array consisting of unique timestamps, which should also be sorted. ("timeList")
Every item in the "timeStampMap" contains another map storing jobs associated with that particular timestamp value
Iterate through each value in the timestamp list.

During each iteration, go through every job in the jobList. If the job exists in the corresponding timestamp map, output the job's completion time.

If not found, output zero.

There are two potential optimizations that can be applied to the code provided below: 1. It may be possible to utilize the input array directly as the "jobList" without requiring duplication. 2. Combining the map of maps into a single large map might also be feasible.

function jobSorter(a,b) {
    return a.jobID - b.jobID;
}

function numberSort(a,b) {
    return a - b;
}


function buildSortedArray(yourArray) {

    var index, j, item;
    var jobList = []; // array of {jobID, timeStamp, timeComp};
    var timeStampMap = {};
    var timeStampList = [];
    var key, jobkey;
    var output = [];

    // loop through every item in the array
    for (index = 0; index < yourArray.length; index++) {
        item = yourArray[index];         

        // push each item into a "job list"
        job = {jobID: item[1], timeStamp: item[0], timeComp: item[2]};
        jobList.push(job);

        // now use both a map and a list to keep track of all the unique timestamps
        key = "$timestamp$" + job.timeStamp.toString();

        if (timeStampMap[key] === undefined) {
            timeStampMap[key] = {};
            timeStampMap[key].jobMap = {};

            timeStampList.push(job.timeStamp); // keep another timestamp array that we can sort on
        }

        // add this job to the timestamp
        jobkey = "$jobid$" + job.jobID.toString();
        timeStampMap[key].jobMap[jobkey] = job;
    }

    // let's do some sorting
    timeStampList.sort(numberSort); // timeStampList is now in ascending order
    jobList.Sort(jobSorter);        // jobList is now in ascending order


    for (index = 0; index < timeStampList.length; index++) {
        item = [];

        item.push(timeStampList[index]);
        for (j = 0; j < jobList.length; j++) {

            key = "$timestamp$" + timeStampList[index].toString();
            jobkey = "$jobid$"  + jobList[j].toString();

            if (timeStampMap[key].jobMap[jobkey]) {
                item.push(timeStampMap[key].jobMap[jobkey].timeComp);
            }
            else {
              item.push(0);
            }
        }

        output.push(item);
    }


    return output;
}

Building and filling a fixed-width array in JavaScript

Answer №1

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