Sort array of objects based on order of another array of objects

Question

I have two arrays of objects and I need to match the order of one of the arrays with the other. Here's an example:

const outOfOrderArray = [{field: 'foo'}, {field: 'bar'}, {field: 'bazz'}, {field: 'bizz'}];

const arrayInProperOrder = [{field: 'bizz'}, {field: 'bazz'}, {field: 'foo'}, {field: 'bar'}];

What can I do to make the outOfOrderArray match the order of objects in arrayInProperOrder?

Will these arrays have objects with unique values (bazz, bar, foo etc.) or these arrays can contain the duplicates too e.g. const outOfOrderArray = [{field: 'foo'}, {field: 'foo'}, {field: 'bar'}, {field: 'bazz'}, {field: 'bizz'}] where {field: 'foo'} is appearing 2 times, — hygull
– hygull, Commented May 11, 2018 at 14:17

31piy · Accepted Answer · 2018-05-11 14:11:53Z

4

Here is how you can do it in one less loop than the solution of @Faly. There is no need to sort the array. You may just loop over arrayInProperOrder and find matching element from outOfOrderArray so that the latter becomes in order.

const outOfOrderArray = [{field: 'foo'}, {field: 'bar'}, {field: 'bazz'}, {field: 'bizz'}];
const arrayInProperOrder = [{field: 'bizz'}, {field: 'bazz'}, {field: 'foo'}, {field: 'bar'}];

const newArray = [];

arrayInProperOrder.forEach(item => {
  const original = outOfOrderArray.find(i => i.field === item.field);

  if (original) {
    newArray.push(original);
  }
});

console.log(newArray);

If you need it in place, just assign newArray to outOfOrderArray.

answered May 11, 2018 at 14:11

31piy

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Eddie · Accepted Answer · 2018-05-11 14:43:59Z

1

You can create a temp object to store the order and use it as a basis on sort()

Use Infinity as the default value if some field is not found on temp object.

const outOfOrderArray = [{field: 'foo'}, {field: 'bar'}, {field: 'bazz'}, {field: 'bizz'}];

const arrayInProperOrder = [{field: 'bizz'}, {field: 'bazz'}, {field: 'foo'}, {field: 'bar'}];

//Make a temp object - This will store the order of the field. This nessasary so that no need to search every reiteration on sort()
const tempObj = arrayInProperOrder.reduce((c, v, i) => Object.assign(c, {[v.field]: i + 1}), {});

//Sort the array.
outOfOrderArray.sort((a, b) => (tempObj[a.field] || Infinity) - (tempObj[b.field] || Infinity));

console.log(outOfOrderArray);

Doc: sort(), Infinity

edited May 11, 2018 at 14:43

answered May 11, 2018 at 14:17

Eddie

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Comments

Faly · Accepted Answer · 2018-05-11 14:35:41Z

0

You can use array.prototype.sort and array.prototype.findIndex:

const outOfOrderArray = [{field: 'foo'}, {field: 'bar'}, {field: 'bazz'}, {field: 'bizz'}];
const arrayInProperOrder = [{field: 'bizz'}, {field: 'bazz'}, {field: 'foo'}, {field: 'bar'}];

outOfOrderArray.sort(
    (a, b) => arrayInProperOrder.findIndex(e => e.field === a.field) -  arrayInProperOrder.findIndex(e => e.field === b.field)
);

console.log(outOfOrderArray);

If you care about performance, you can build an object allowing you to get the index by the field value, from arrayInProperOrder array:

const outOfOrderArray = [{field: 'foo'}, {field: 'bar'}, {field: 'bazz'}, {field: 'bizz'}];
const arrayInProperOrder = [{field: 'bizz'}, {field: 'bazz'}, {field: 'foo'}, {field: 'bar'}];
const indexes = arrayInProperOrder.reduce((m, o, i) => (m[o.field] = i, m), {});

outOfOrderArray.sort(
    (a, b) => indexes[a.field] -  indexes[b.field]
);

console.log(outOfOrderArray);

edited May 11, 2018 at 14:35

answered May 11, 2018 at 14:07

Faly

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2 Comments

user2437417 Over a year ago

Performing a two linear searches at every step of a sort operation doesn't seem like a great idea. The .sort() method really isn't needed in the first place.

Faly Over a year ago

Performance does not always pose problem, depending on the quantity of data to be processed

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