JavaScript Sorting Array Based On Condition

Question

If I have an array of below items and I want to display all the IsImportant=true items sorted first and then IsImportant=false items sorted next.

var sampleArray = [{
  "Name": "A",
  "IsImportant": true
}, {
  "Name": "B",
  "IsImportant": true
}, {
  "Name": "C",
  "IsImportant": false
}, {
  "Name": "D",
  "IsImportant": false
}, {
  "Name": "E",
  "IsImportant": false
},, {
  "Name": "F",
  "IsImportant": true
},
, {
  "Name": "G",
  "IsImportant": true
}, {
  "Name": "H",
  "IsImportant": false
}];

I was trying to write a compareFunction like below to reach the result but didn't succeed. Would there be any better alternative?

function comparator(a, b) {
  //console.log(a,b); 
  if (a.IsImportant || b.IsImportant) {
    if (a.Name > b.Name) {
      return 1;
    }
    if (a.Name < b.Name) {
      return -1;
    }
  }
  return 0;
};

Fiddle

return b.IsImportant - a.IsImportant || a.Name.localeCompare(b.Name); — melpomene
– melpomene, Commented Nov 6, 2016 at 14:09

Pointy · Accepted Answer · 2016-11-06 14:13:49Z

2

What you're trying to do is sort on 2 keys: first, the "IsImportant" key, and second, the names. The comparator should check each key in order and only move on to subsequent keys when the higher-priority keys are equal.

Thus:

function comparator(a, b) {
  //console.log(a,b); 
  if (a.IsImportant && !b.IsImportant)
    return 1;

  if (!a.IsImportant && b.IsImportant)
    return -1;

  return a.Name.localeCompare(b.Name);
};

(The .localeCompare() function is a handy way to compare strings for sorting.)

You could take advantage of the fact that JavaScript will treat the Boolean constants true and false as 1 and 0 when they're used with (some) math operators, so you could shorten that a little:

function comparator(a, b) {
  var diff = b.IsImportant - a.IsImportant;
  if (diff) return diff;
  return a.Name.localeCompare(b.Name);
}

Either way, the comparator doesn't bother to compare the names when the "IsImportant" flag is different. When both of those flags are true or false for the two items being compared, then the names are checked. That way, you'll get all the important ones first in name order, and then all the other important ones.

edited Nov 6, 2016 at 14:13

answered Nov 6, 2016 at 14:08

Pointy

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2 Comments

Pointy Over a year ago

@melpomene uhh ... maybe ... no wait, yes of course. Thanks.

Pointy Over a year ago

@melpomene you can easily imagine what 90% of my debugging time is spent doing.

kevin ternet · Accepted Answer · 2016-11-06 14:42:03Z

0

And why not just this :

var sampleArray = [{
  "Name": "A",
  "IsImportant": true
}, {
  "Name": "B",
  "IsImportant": true
}, {
  "Name": "C",
  "IsImportant": false
}, {
  "Name": "D",
  "IsImportant": false
}, {
  "Name": "E",
  "IsImportant": false
},, {
  "Name": "F",
  "IsImportant": true
},
, {
  "Name": "G",
  "IsImportant": true
}, {
  "Name": "H",
  "IsImportant": false
}];

var results = sampleArray.filter(x => x.IsImportant).concat(sampleArray.filter(x => !x.IsImportant));
console.log(results);

answered Nov 6, 2016 at 14:42

kevin ternet

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