Discover the absent day in an array of dates using JavaScript

Question

Discover the absent day in an array of dates using JavaScript

Upon receiving an array of day dates from an API, the following structure is observed:

0:{date: "2016-11-17T00:00:00",…}
1:{date: "2016-11-18T00:00:00",…}
2:{date: "2016-11-19T00:00:00",…}
3:{date: "2016-11-21T00:00:00",…}
4:{date: "2016-11-22T00:00:00",…}
5:{date: "2016-11-23T00:00:00",…}

In this particular instance, a date seems to be absent from the array:

{date: "2016-11-20T00:00:00",…}

What would be the most effective method in order to identify a missing day within an array of dates using Javascript or Angular?

This information may prove valuable when configuring a datepicker by setting disabled days accordingly.

javascript angularjs arrays

Answer 1

Answer №1

Take a look at this solution:

To start, ensure the array is sorted using Array.prototype.sort
Next, utilize Array.prototype.reduce along with a hash table to identify any missing dates

See the demonstration provided in the snippet below:

var array=[
  {date:"2016-11-17T00:00:00"},
  {date:"2016-11-19T00:00:00"},
  {date:"2016-11-18T00:00:00"},
  {date:"2016-11-21T00:00:00"},
  {date:"2016-11-22T00:00:00"},
  {date:"2016-11-23T00:00:00"},
  {date:"2016-11-27T00:00:00"}
];

var result = array.sort(function(a,b){
   return Date.parse(a.date) - Date.parse(b.date);
}).reduce(function(hash){
  return function(p,c){
    var missingDaysNo = (Date.parse(c.date) - hash.prev) / (1000 * 3600 * 24);
    if(hash.prev && missingDaysNo > 1) {
      for(var i=1;i<missingDaysNo;i++)
        p.push(new Date(hash.prev+i*(1000 * 3600 * 24)));
    }
    hash.prev = Date.parse(c.date);
    return p;
  };
}(Object.create(null)),[]);

console.log(result);

.as-console-wrapper{top:0;max-height:100%!important;}

Answer 2

Take a look at this solution:

To start, ensure the array is sorted using Array.prototype.sort
Next, utilize Array.prototype.reduce along with a hash table to identify any missing dates

See the demonstration provided in the snippet below:

var array=[
  {date:"2016-11-17T00:00:00"},
  {date:"2016-11-19T00:00:00"},
  {date:"2016-11-18T00:00:00"},
  {date:"2016-11-21T00:00:00"},
  {date:"2016-11-22T00:00:00"},
  {date:"2016-11-23T00:00:00"},
  {date:"2016-11-27T00:00:00"}
];

var result = array.sort(function(a,b){
   return Date.parse(a.date) - Date.parse(b.date);
}).reduce(function(hash){
  return function(p,c){
    var missingDaysNo = (Date.parse(c.date) - hash.prev) / (1000 * 3600 * 24);
    if(hash.prev && missingDaysNo > 1) {
      for(var i=1;i<missingDaysNo;i++)
        p.push(new Date(hash.prev+i*(1000 * 3600 * 24)));
    }
    hash.prev = Date.parse(c.date);
    return p;
  };
}(Object.create(null)),[]);

console.log(result);

.as-console-wrapper{top:0;max-height:100%!important;}

Answer 3

Answer №2

Initialize a new array called missingDates[]

Go through each element in the provided array by using a for loop

for (i = 0; i < array.length; i++){
    var date1 = convert the element at index i in the array to a date
    var date2 = convert the element at index i + 1 in the array to a date (keeping in mind that index i + 1 should not exceed array.length)

    //calculate the number of days between the two dates, if it is greater than 1, then there is a missing date
    var missingDate = create the missing date by adding one day to your date1 variable

    //add the missingDate to the missingDates[] array
    missingDates.push(missingDate)
}

Answer 4

Initialize a new array called missingDates[]

Go through each element in the provided array by using a for loop

for (i = 0; i < array.length; i++){
    var date1 = convert the element at index i in the array to a date
    var date2 = convert the element at index i + 1 in the array to a date (keeping in mind that index i + 1 should not exceed array.length)

    //calculate the number of days between the two dates, if it is greater than 1, then there is a missing date
    var missingDate = create the missing date by adding one day to your date1 variable

    //add the missingDate to the missingDates[] array
    missingDates.push(missingDate)
}

Answer 5

Answer №3

To determine a missing date, you can create a method called getMissingDate. This method will return null if there is no missing date, or it will return the corresponding Date object if there is a gap of more than one day between two dates:

var arr1 = [{date: "2016-11-17T00:00:00"}, {date: "2016-11-18T00:00:00"}, {date: "2016-11-19T00:00:00"}, {date: "2016-11-21T00:00:00"}, {date: "2016-11-22T00:00:00"}, {date: "2016-11-23T00:00:00"}],
    arr2 = [{date: "2016-11-17T00:00:00"}, {date: "2016-11-18T00:00:00"}, {date: "2016-11-19T00:00:00"}, {date: "2016-11-20T00:00:00"}, {date: "2016-11-21T00:00:00"}, {date: "2016-11-22T00:00:00"}, {date: "2016-11-23T00:00:00"}],
    getMissingDate = function(arr) {
      var result = null;
      for (var i = 0, l = arr.length - 1; i < l; i++) {
        var current = new Date(arr[i].date),
            next = new Date(arr[i + 1].date);

        if (1 < Math.ceil(Math.abs(next.getTime() - current.getTime()) / (1000 * 3600 * 24))) {
          result = new Date(current.setDate(current.getDate() + 1));
          break;
        } 
      }

      return result;
    };

console.log('arr1:', getMissingDate(arr1));
console.log('arr2:', getMissingDate(arr2));

Answer 6

To determine a missing date, you can create a method called getMissingDate. This method will return null if there is no missing date, or it will return the corresponding Date object if there is a gap of more than one day between two dates:

var arr1 = [{date: "2016-11-17T00:00:00"}, {date: "2016-11-18T00:00:00"}, {date: "2016-11-19T00:00:00"}, {date: "2016-11-21T00:00:00"}, {date: "2016-11-22T00:00:00"}, {date: "2016-11-23T00:00:00"}],
    arr2 = [{date: "2016-11-17T00:00:00"}, {date: "2016-11-18T00:00:00"}, {date: "2016-11-19T00:00:00"}, {date: "2016-11-20T00:00:00"}, {date: "2016-11-21T00:00:00"}, {date: "2016-11-22T00:00:00"}, {date: "2016-11-23T00:00:00"}],
    getMissingDate = function(arr) {
      var result = null;
      for (var i = 0, l = arr.length - 1; i < l; i++) {
        var current = new Date(arr[i].date),
            next = new Date(arr[i + 1].date);

        if (1 < Math.ceil(Math.abs(next.getTime() - current.getTime()) / (1000 * 3600 * 24))) {
          result = new Date(current.setDate(current.getDate() + 1));
          break;
        } 
      }

      return result;
    };

console.log('arr1:', getMissingDate(arr1));
console.log('arr2:', getMissingDate(arr2));

Answer 7

Answer №4

var data=[
  {timestamp:"2016-01-01T00:00:00"},
  {timestamp:"2016-03-01T00:00:00"},
  {timestamp:"2016-04-01T00:00:00"},
  {timestamp:"2016-07-01T00:00:00"},
  {timestamp:"2016-09-01T00:00:00"},
  {timestamp:"2016-11-01T00:00:00"},
  {timestamp:"2016-12-01T00:00:00"}
];

var sortedData = data.sort(function(a,b){
   return Date.parse(a.timestamp) - Date.parse(b.timestamp);
}).reduce(function(hash){
  return function(previous, current){
    var missingDaysNo= (Date.parse(current.timestamp) - hash.prev) / (1000 * 3600 * 24);
    if(hash.prev && missingDaysNo > 1) {
      for(var i=1; i<missingDaysNo; i++)
        previous.push(new Date(hash.prev+i*(1000 * 3600 * 24)));
    }
    hash.prev = Date.parse(current.timestamp);
    return previous;
  };
}(Object.create(null)),[]);

console.log(sortedData);

.as-console-wrapper{top:0;max-height:100%!important;}

Answer 8

var data=[
  {timestamp:"2016-01-01T00:00:00"},
  {timestamp:"2016-03-01T00:00:00"},
  {timestamp:"2016-04-01T00:00:00"},
  {timestamp:"2016-07-01T00:00:00"},
  {timestamp:"2016-09-01T00:00:00"},
  {timestamp:"2016-11-01T00:00:00"},
  {timestamp:"2016-12-01T00:00:00"}
];

var sortedData = data.sort(function(a,b){
   return Date.parse(a.timestamp) - Date.parse(b.timestamp);
}).reduce(function(hash){
  return function(previous, current){
    var missingDaysNo= (Date.parse(current.timestamp) - hash.prev) / (1000 * 3600 * 24);
    if(hash.prev && missingDaysNo > 1) {
      for(var i=1; i<missingDaysNo; i++)
        previous.push(new Date(hash.prev+i*(1000 * 3600 * 24)));
    }
    hash.prev = Date.parse(current.timestamp);
    return previous;
  };
}(Object.create(null)),[]);

console.log(sortedData);

.as-console-wrapper{top:0;max-height:100%!important;}

Discover the absent day in an array of dates using JavaScript

Answer №1

Answer №2

Answer №3

Answer №4

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