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Find closest date in array with JavaScript

I have an array with days in it. Each day is an object, for example:

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

I have also added a timestamp attribute to each day object, by using:

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 days in the array are arbitrary, so there is no real logic to them.

Now I want to find the two closest days to any given date. So if the array with days contains

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

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

I have tried using an answer from another question, that 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, this returns unidentified.

What would be the most efficient (least processor/memory intensive way) to achieve this?

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

I'm now using the following to generate an array of dates:

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()});
}

The strange part is, using the code below, this only works for an array of 10 dates or shorter. Whenever I use an array of 11 or more dates, the results become unexpected.

For instance: using an array of 15 dates, starting on August 6, 2012, to August 21, 2012. If I then call findClosest(full_day_array, new Date("30/07/2012"); you would expect it to return {nextIndex: 0, prevIndex: -1}. However, it returns {nextIndex: 7, prevIndex: -1}. Why?

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};
}
like image 213
Rein Avatar asked Aug 03 '12 11:08

Rein


2 Answers

You can easily use the sort function with a custom comparator function:

// assuming you have an array of Date objects - everything else is crap:
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; // sort a before b when the distance is smaller
});

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

To get only results before or after the diffdate, you can filter the array for that:

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

If you have your custom array with the {the_date_object: new Date(...)} objects, you will need to adapt the sort algorithm with

    var distancea = Math.abs(diffdate - a.the_date_object);
    var distanceb = Math.abs(diffdate - b.the_date_object);
like image 195
Bergi Avatar answered Sep 19 '22 11:09

Bergi


If you use an array of Date objects instead of your self-defined structure it can be achieved very easily in O(N):

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 best date will be days[bestDate] */

If the array is sorted it can be achieved in O(log N) with binary search.

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){
   // If currDiff is negative, then testDate is more in the past than days[i].
   // This means, that from testDate's point of view, days[i] is in the future
   // and thus by a candidate for the next date.
       bestNextDate = i;
       bestNextDiff = currDiff;
   }
   if(currDiff > 0 && currDiff < bestPrevDiff){
   // If currDiff is positive, then testDate is more in the future than days[i].
   // This means, that from testDate's point of view, days[i] is in the past
   // and thus by a candidate for the previous date.
       bestPrevDate = i;
       bestPrevDiff = currDiff;
   }   

}
/* days[bestPrevDate] is the best previous date, 
   days[bestNextDate] is the best next date */
like image 34
Zeta Avatar answered Sep 20 '22 11:09

Zeta