从对象数组中提取对象

Question

I need to split an object array into two variables. The first variable ( main ) should get the object with a title . The second one ( content ) should get all other objects.

Sample data

[
  { _id: '1', title: 'Main' },
  { _id: '2', foo: 'bar' },
  { _id: '2', foo: 'bar' }
]

I did it with find() / filter() commands, but is it really necessary to find twice?

const main = data.find(doc => doc.title)
const content = data.filter(doc => !doc.title)

Is it possible to extract the main object instead of finding it?

Answer 1

You could take a single loop approach and an object for same named arrays with a check if the property exist with in operator .

 var data = [{ _id: '1', title: 'Main' }, { _id: '2', foo: 'bar' }, { _id: '2', foo: 'bar' }], main = [], content = [], temp = { main, content }; data.forEach(doc => temp['title' in doc ? 'main' : 'content'].push(doc)); console.log(main); console.log(content); 

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

With just one object with title, you could assign the object directly and collec the other objects in an array.

 var data = [{ _id: '1', title: 'Main' }, { _id: '2', foo: 'bar' }, { _id: '2', foo: 'bar' }], main, content = []; data.forEach(doc => 'title' in doc ? main = doc : content.push(doc) ); console.log(main); console.log(content); 

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

Answer 2

const content = data.find(doc => !doc.title)

This doesn't work, because find returns one object only (the first found), not all the objects that don't match the condition. You'll have to use a filter for that

const content = data.filter(doc => !doc.title)

So yes, you'll have to iterate twice, one time to find the matching object, and a second time to build an array excluding the matching object.

Answer 3

var docs = [
  { _id: '1', title: 'Main' },
  { _id: '2', foo: 'bar' },
  { _id: '2', foo: 'bar' }
];

var doc = [],main = [];

for(_doc in docs){
    if(typeof docs[_doc].title != 'undefined') {
        doc.push(docs[_doc]);
        continue;
  }

  main.push(docs[_doc]);
}

Working fiddle: https://jsfiddle.net/andreitodorut/fcLkata6/

Answer 4

I would do this a bit differently. The notion of partitioning a list by a predicate is worth extracting. So I might do this:

 const docs = [{_id: '1', title: 'Main'},{_id: '2', foo: 'bar'},{_id: '2', foo: 'bar'}] const partition = pred => data => data.reduce( ([matches, mismatches], item) => pred(item) ? [matches.concat(item), mismatches] : [matches, mismatches.concat(item)], [[], []] ) const withTitle = partition(doc => 'title' in doc) console.log(withTitle(docs)) 

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

Of course it's quite possible that withTitle should be inlined, if it's not being reused. It's also possible that the currying I put in, my usual default, is not necessary for you, and it should read (pred, data) => data.reduce(...)