Determining the Nearest Date in an Array Using JavaScript

Question

Determining the Nearest Date in an Array Using JavaScript

In my array, I store information about each day as an object. For example:

{day_year: "2012", day_month: "08", day_number: "03", day_name: "mon"}

To enhance the day objects, I have included a timestamp attribute using the following function:

function convertDays() {
    var max_i = days.length;
    for(var i = 0; i < max_i; i++) {
        var tar_i = days[i];
        tar_i.timestamp = new Date(tar_i.day_year, tar_i.day_month, tar_i.day_number);
    }
}

The order of the days in the array is random without any specific logic.

Now, my goal is to identify the two closest dates to a given date. So, for instance, if my array contains:

August 2, 2012
August 4, 2012
August 23, 2012

And I search for August 11, 2012, I expect it to return August 4, 2012 and August 23, 2012.

I attempted to use a solution from another source which looks like this:

function findClosest(a, x) {
    var lo, hi;
    for(var i = a.length; i--;) {
        if (a[i] <= x && (lo === undefined || lo < a[i])) lo = a[i];
        if (a[i] >= x && (hi === undefined || hi > a[i])) hi = a[i];
    }
    return [lo, hi];
}

However, I encountered issues as it returned unidentified.

What would be the most efficient way to achieve this while minimizing processor and memory usage?

Edit: "However, how are those results "strange"? Could you provide an example of your code and data?"

I now generate an array of dates using the following approach:

var full_day_array = [];
for(var i = 0; i < 10; i++) {
    var d = new Date();
    d.setDate(d.getDate() + i);
    full_day_array.push({day_year: d.getFullYear().toString(), day_month: (d.getMonth() + 1).toString(), day_number: d.getDate().toString()});
}

Anomalies arise when working with arrays containing more than 10 dates. For example, with a range from August 6, 2012, to August 21, 2012 consisting of 15 dates. When applying

findClosest(full_day_array, new Date("30/07/2012"));

one might predict a result of {nextIndex: 0, prevIndex: -1}. Surprisingly, it returns {nextIndex: 7, prevIndex: -1}. Why does this discrepancy occur?

function findClosest(objects, testDate) {
    var nextDateIndexesByDiff = [],
        prevDateIndexesByDiff = [];

    for(var i = 0; i < objects.length; i++) {
        var thisDateStr = [objects[i].day_month, objects[i].day_number, objects[i].day_year].join('/'),
            thisDate    = new Date(thisDateStr),
            curDiff     = testDate - thisDate;

        curDiff < 0
            ? nextDateIndexesByDiff.push([i, curDiff])
            : prevDateIndexesByDiff.push([i, curDiff]);
    }

    nextDateIndexesByDiff.sort(function(a, b) { return a[1] < b[1]; });
    prevDateIndexesByDiff.sort(function(a, b) { return a[1] > b[1]; });

    var nextIndex;
    var prevIndex;

    if(nextDateIndexesByDiff.length < 1) {
        nextIndex = -1;
    } else { 
        nextIndex = nextDateIndexesByDiff[0][0];
    }
  
    if(prevDateIndexesByDiff.length < 1) {
        prevIndex = -1;
    } else {    
        prevIndex = prevDateIndexesByDiff[0][0];
    }
   
    return {nextIndex: nextIndex, prevIndex: prevIndex};
}

javascript arrays search

Answer 1

Answer №1

To utilize the sort function effectively, you can implement a customized comparator function:

// assuming you are working with an array of Date objects - everything else is irrelevant:
var arr = [new Date(2012, 7, 1), new Date(2012, 7, 4), new Date(2012, 7, 5), new Date(2013, 2, 20)];
var diffdate = new Date(2012, 7, 11);

arr.sort(function(a, b) {
    var distancea = Math.abs(diffdate - a);
    var distanceb = Math.abs(diffdate - b);
    return distancea - distanceb; // arrange a before b when the distance is shorter
});

// result:
[2012-08-05, 2012-08-04, 2012-08-01, 2013-03-20]

If you only want results prior to or after the diffdate, you can filter the array accordingly:

var beforedates = arr.filter(function(d) {
    return d - diffdate < 0;
}),
    afterdates = arr.filter(function(d) {
    return d - diffdate > 0;
});

In case your custom array contains objects like {the_date_object: new Date(...)}, you will need to modify the sort algorithm with

    var distancea = Math.abs(diffdate - a.the_date_object);
    var distanceb = Math.abs(diffdate - b.the_date_object);

Answer 2

To utilize the sort function effectively, you can implement a customized comparator function:

// assuming you are working with an array of Date objects - everything else is irrelevant:
var arr = [new Date(2012, 7, 1), new Date(2012, 7, 4), new Date(2012, 7, 5), new Date(2013, 2, 20)];
var diffdate = new Date(2012, 7, 11);

arr.sort(function(a, b) {
    var distancea = Math.abs(diffdate - a);
    var distanceb = Math.abs(diffdate - b);
    return distancea - distanceb; // arrange a before b when the distance is shorter
});

// result:
[2012-08-05, 2012-08-04, 2012-08-01, 2013-03-20]

If you only want results prior to or after the diffdate, you can filter the array accordingly:

var beforedates = arr.filter(function(d) {
    return d - diffdate < 0;
}),
    afterdates = arr.filter(function(d) {
    return d - diffdate > 0;
});

In case your custom array contains objects like {the_date_object: new Date(...)}, you will need to modify the sort algorithm with

    var distancea = Math.abs(diffdate - a.the_date_object);
    var distanceb = Math.abs(diffdate - b.the_date_object);

Answer 3

Answer №2

Instead of using a self-defined structure, you can achieve the desired result easily in O(N) by utilizing an array of Date objects:

var testDate = new Date(...);
var bestDate = days.length;
var bestDiff = -(new Date(0,0,0)).valueOf();
var currDiff = 0;
var i;

for(i = 0; i < days.length; ++i){
   currDiff = Math.abs(days[i] - testDate);
   if(currDiff < bestDiff){
       bestDate = i;
       bestDiff = currDiff;
   }   
}

/* The closest match will be found at days[bestDate] */

If the array is sorted, achieving this in O(log N) through binary search becomes possible.

Edit: "it is crucial that I find both the closest match before and after the date"

var testDate = new Date(...);

var bestPrevDate = days.length;
var bestNextDate = days.length;

var max_date_value = Math.abs((new Date(0,0,0)).valueOf());

var bestPrevDiff = max_date_value;
var bestNextDiff = -max_date_value;

var currDiff = 0;
var i;

for(i = 0; i < days.length; ++i){
   currDiff = testDate - days[i].the_date_object;
   if(currDiff < 0 && currDiff > bestNextDiff){
       bestNextDate = i;
       bestNextDiff = currDiff;
   }
   if(currDiff > 0 && currDiff < bestPrevDiff){
       bestPrevDate = i;
       bestPrevDiff = currDiff;
   }   

}
/* days[bestPrevDate] represents the best previous date,
   days[bestNextDate] represents the best next date */

Answer 4

Instead of using a self-defined structure, you can achieve the desired result easily in O(N) by utilizing an array of Date objects:

var testDate = new Date(...);
var bestDate = days.length;
var bestDiff = -(new Date(0,0,0)).valueOf();
var currDiff = 0;
var i;

for(i = 0; i < days.length; ++i){
   currDiff = Math.abs(days[i] - testDate);
   if(currDiff < bestDiff){
       bestDate = i;
       bestDiff = currDiff;
   }   
}

/* The closest match will be found at days[bestDate] */

If the array is sorted, achieving this in O(log N) through binary search becomes possible.

Edit: "it is crucial that I find both the closest match before and after the date"

var testDate = new Date(...);

var bestPrevDate = days.length;
var bestNextDate = days.length;

var max_date_value = Math.abs((new Date(0,0,0)).valueOf());

var bestPrevDiff = max_date_value;
var bestNextDiff = -max_date_value;

var currDiff = 0;
var i;

for(i = 0; i < days.length; ++i){
   currDiff = testDate - days[i].the_date_object;
   if(currDiff < 0 && currDiff > bestNextDiff){
       bestNextDate = i;
       bestNextDiff = currDiff;
   }
   if(currDiff > 0 && currDiff < bestPrevDiff){
       bestPrevDate = i;
       bestPrevDiff = currDiff;
   }   

}
/* days[bestPrevDate] represents the best previous date,
   days[bestNextDate] represents the best next date */

Answer 5

Answer №3

Give this a shot

let datesArray = [
'May 7, 2001 12:45:00',
'June 10, 2001 12:45:00',
'July 11, 2001 12:45:00',
'August 5, 2001 12:45:00',
'September 2, 2001 12:45:00',
'October 6, 2001 12:45:00',
]

let timeDifference = datesArray.map(date => Math.abs(new Date() - new Date(date).getTime()));
let shortestTimeIndex = timeDifference.indexOf(Math.min(...timeDifference));
console.log(datesArray[shortestTimeIndex]);

Answer 6

Give this a shot

let datesArray = [
'May 7, 2001 12:45:00',
'June 10, 2001 12:45:00',
'July 11, 2001 12:45:00',
'August 5, 2001 12:45:00',
'September 2, 2001 12:45:00',
'October 6, 2001 12:45:00',
]

let timeDifference = datesArray.map(date => Math.abs(new Date() - new Date(date).getTime()));
let shortestTimeIndex = timeDifference.indexOf(Math.min(...timeDifference));
console.log(datesArray[shortestTimeIndex]);

Answer 7

Answer №4

Omega's response is top-notch, but my curiosity lies in exploring the approach for finding the nearest N objects in both directions. Here is my take on it:

var items = [
    { day_year: "2012",
      day_month: "08",
      day_number: "02"
    },
    { day_year: "2012",
      day_month: "08",
      day_number: "04"
    },
    { day_year: "2012",
      day_month: "08",
      day_number: "23"
    }
];

var testingDate = new Date('08/11/2012'),
    nextIndexesByDifference = [],
    prevIndexesByDifference = [];

for(var j = 0; j < items.length; j++) {
    var currentDateStr = [items[j].day_month, items[j].day_number, items[j].day_year].join('/'),
        currentDate    = new Date(currentDateStr),
        currentDiff     = testingDate - currentDate;

    currentDiff < 0
        ? nextIndexesByDifference.push([j, currentDiff])
        : prevIndexesByDifference.push([j, currentDiff]);
}

nextIndexesByDifference.sort(function(x, y) { return x[1] < y[1]; });
prevIndexesByDifference.sort(function(x, y) { return x[1] > y[1]; });

console.log(['closest future date', items[nextIndexesByDifference[0][0]]]);
console.log(['closest past date', items[prevIndexesByDifference[0][0]]]);

Answer 8

Omega's response is top-notch, but my curiosity lies in exploring the approach for finding the nearest N objects in both directions. Here is my take on it:

var items = [
    { day_year: "2012",
      day_month: "08",
      day_number: "02"
    },
    { day_year: "2012",
      day_month: "08",
      day_number: "04"
    },
    { day_year: "2012",
      day_month: "08",
      day_number: "23"
    }
];

var testingDate = new Date('08/11/2012'),
    nextIndexesByDifference = [],
    prevIndexesByDifference = [];

for(var j = 0; j < items.length; j++) {
    var currentDateStr = [items[j].day_month, items[j].day_number, items[j].day_year].join('/'),
        currentDate    = new Date(currentDateStr),
        currentDiff     = testingDate - currentDate;

    currentDiff < 0
        ? nextIndexesByDifference.push([j, currentDiff])
        : prevIndexesByDifference.push([j, currentDiff]);
}

nextIndexesByDifference.sort(function(x, y) { return x[1] < y[1]; });
prevIndexesByDifference.sort(function(x, y) { return x[1] > y[1]; });

console.log(['closest future date', items[nextIndexesByDifference[0][0]]]);
console.log(['closest past date', items[prevIndexesByDifference[0][0]]]);

Answer 9

Answer №5

Here's how we approach it:

In this specific function, we are targeting the closest date within an array to a given date parameter (dateToCompare).

By iterating through each item in the array and comparing their respective dates to dateToCompare, we aim to return the element with the date closest to the comparison date.

getClosestDateInArray (array, dateParam, dateToCompare) {
  let minDiff = null;
  let mostAccurateDate = array[0];
  array.map((item) => {
    const diff = Math.abs(moment(dateToCompare).diff(item[dateParam], 'minutes', true));
    if (!minDiff || diff < minDiff) {
      minDiff = diff;
      mostAccurateDate = item
    }
  });
  return mostAccurateDate;
}

Note: This solution relies on the momentJS library for accurate date comparisons.

Answer 10

Here's how we approach it:

In this specific function, we are targeting the closest date within an array to a given date parameter (dateToCompare).

By iterating through each item in the array and comparing their respective dates to dateToCompare, we aim to return the element with the date closest to the comparison date.

getClosestDateInArray (array, dateParam, dateToCompare) {
  let minDiff = null;
  let mostAccurateDate = array[0];
  array.map((item) => {
    const diff = Math.abs(moment(dateToCompare).diff(item[dateParam], 'minutes', true));
    if (!minDiff || diff < minDiff) {
      minDiff = diff;
      mostAccurateDate = item
    }
  });
  return mostAccurateDate;
}

Note: This solution relies on the momentJS library for accurate date comparisons.

Answer 11

Answer №6

This function efficiently finds the closest date in an array regardless of its length:

function efficientDateFinder(dates, testDate) {
    var before = [];
    var after = [];
    var max = dates.length;
    for(var i = 0; i < max; i++) {
        var tar = dates[i];
        var arrDate = new Date(tar.day_year, tar.day_month, tar.day_number);
        var diff = (arrDate - testDate) / (3600 * 24 * 1000); // Convert milliseconds to days.
        if(diff > 0) {
            before.push({diff: diff, index: i});
        } else {
            after.push({diff: diff, index: i});
        }
    }
    before.sort(function(a, b) {
        return a.diff - b.diff;
    });

    after.sort(function(a, b) {
        return b.diff - a.diff;
    });
    return {datesBefore: before, datesAfter: after};
}

Answer 12

This function efficiently finds the closest date in an array regardless of its length:

function efficientDateFinder(dates, testDate) {
    var before = [];
    var after = [];
    var max = dates.length;
    for(var i = 0; i < max; i++) {
        var tar = dates[i];
        var arrDate = new Date(tar.day_year, tar.day_month, tar.day_number);
        var diff = (arrDate - testDate) / (3600 * 24 * 1000); // Convert milliseconds to days.
        if(diff > 0) {
            before.push({diff: diff, index: i});
        } else {
            after.push({diff: diff, index: i});
        }
    }
    before.sort(function(a, b) {
        return a.diff - b.diff;
    });

    after.sort(function(a, b) {
        return b.diff - a.diff;
    });
    return {datesBefore: before, datesAfter: after};
}

Determining the Nearest Date in an Array Using JavaScript

Answer №1

Answer №2

Edit: "it is crucial that I find both the closest match before and after the date"

Answer №3

Answer №4

Answer №5

Answer №6

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