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sirdaiz
Asked: 2020-04-12 01:54:27 +0800 CST 2020-04-12 01:54:27 +0800 CST

Get object from a list

  • 772

Suppose I have the following:

var ids = [1,5]
var persons = [{id:1, name:'stack'},{id:2, name:'stack2'},
              {id:3, name:'stack3'},{id:4, name:'stack4'},{id:5, name:'stack5'}]

So I need from the lists of "persons" all the ones that have id: 1 and 5. You could do a "for" of persons and check if the id exists inside the "ids" array, but of course if you had 100000 people I don't think it would be very efficient, is there any other better way?

var result = []
for (var co in persons) {
    if (id.indexOf(persons[co].id) > 0) {
        result.add(persons[co])
    }           
}
javascript

3 Answers

  1. Assuming you do the fetch once and then keep that list of items for a long time and you're going to have to fetch many times, it might be worth pre-processing the data and saving it to an object or a Map (not much of a difference in speed). of access) :

    const persons = [{id:1, name:'stack'},{id:2, name:'stack2'},
              {id:3, name:'stack3'},{id:4, name:'stack4'},{id:5, name:'stack5'}]

const personsMap= {}; // new Map();
persons.forEach(element => personsMap[element.id]=element);
// persons.forEach(element => personsMap.set(element.id,element));

function getPersonById(id) {
  return personsMap[id];
  //return personsMap.get(id);
}

    Another option would be to get that data already sorted, so you could do a binary search :

    1. You have N elements in an Array, sorted.
    2. You want to get the element with value X.
    3. You look for the element in the middle: M= N/2
    4. Compare with position M.
    5. If you've found the item, we're done.
    6. If the element at position M is smaller, then the one you're looking for will be in the larger half, from M to M *2, so we do M = M + M/2 and go to step 3.
    7. If the element in position M was greater, then M= M - M/2 and we go to step 3
    • 5
  2. Best Answer

    You could use the method filterto generate the list with the desired objects.

    According to the documentation, the method filter()of type Array, creates a new one Arraywith the elements that pass the filter condition.

    As for the efficiency, I couldn't tell you if it is more efficient than a loop or not for, it is probably exactly the same, but what is certain is that it looks more elegant.

    An example of how it would look:

    var ids = [1,5];
var persons = [{id:1, name:'stack'},{id:2, name:'stack2'},
              {id:3, name:'stack3'},{id:4, name:'stack4'},{id:5, name:'stack5'}];
var filteredPersons = persons.filter((person) => {
  var pass = false;
  ids.forEach((id) => {
    if(person.id == id) {
      pass = true;
    }
  });
  return pass;
});

console.log(filteredPersons);

    EDITION

    As @Kiko_L comments, if you use indexOf()it you save time mapand the code looks more concise. For example:

    var ids = [1,5];
var persons = [{id:1, name:'stack'},{id:2, name:'stack2'},
              {id:3, name:'stack3'},{id:4, name:'stack4'},{id:5, name:'stack5'}];
var filteredPersons = persons.filter((person) => {
  return ids.indexOf(person.id) >= 0 ? true : false;
});

console.log(filteredPersons);

    • 2

  3. You can useincludes()

    Translating from StackOverflow , you have different ways:

    For IE9 or higher:

    function include(arr,obj) {
    return (arr.indexOf(obj) != -1);
}

    Mozilla :

     if (!Array.prototype.indexOf)
  {

       Array.prototype.indexOf = function(searchElement /*, fromIndex */)

    {


    "use strict";

    if (this === void 0 || this === null)
      throw new TypeError();

    var t = Object(this);
    var len = t.length >>> 0;
    if (len === 0)
      return -1;

    var n = 0;
    if (arguments.length > 0)
    {
      n = Number(arguments[1]);
      if (n !== n)
        n = 0;
      else if (n !== 0 && n !== (1 / 0) && n !== -(1 / 0))
        n = (n > 0 || -1) * Math.floor(Math.abs(n));
    }

    if (n >= len)
      return -1;

    var k = n >= 0
          ? n
          : Math.max(len - Math.abs(n), 0);

    for (; k < len; k++)
    {
      if (k in t && t[k] === searchElement)
        return k;
    }
    return -1;
  };

}

    Daniel James :

    if (!Array.prototype.indexOf) {
  Array.prototype.indexOf = function (obj, fromIndex) {
    if (fromIndex == null) {
        fromIndex = 0;
    } else if (fromIndex < 0) {
        fromIndex = Math.max(0, this.length + fromIndex);
    }
    for (var i = fromIndex, j = this.length; i < j; i++) {
        if (this[i] === obj)
            return i;
    }
    return -1;
  };
}

    roosteronacid(broken link):

    Array.prototype.hasObject = (
  !Array.indexOf ? function (o)
  {
    var l = this.length + 1;
    while (l -= 1)
    {
        if (this[l - 1] === o)
        {
            return true;
        }
    }
    return false;
  } : function (o)
  {
    return (this.indexOf(o) !== -1);
  }
);

    I invite you to go through the web in English and also read how do i check if an array includes an object in javascript

    • 0