Underscore Remove the Value without replace original variable

Question

I have an array of objects and I'm trying to find the value in the object and remove the value has been found in the exist object. For example,

Current JSON Object:

exist=[{"x":6811,"y":15551,"a":["aa","ab","ac"]},{"x":6811,"y":15552,"a":["bb"]},{"x":6812,"y":15551,"a":["aa","cc"]}]

I want to find the "a" Key with value aa

The last result is

exist= [{"x":6811,"y":15552,"a":["bb"]}]
found= [{"x":6811,"y":15551,"a":["aa","ab","ac"]},{"x":6812,"y":15551,"a":["aa","cc"]}]

Answer 1

You can do it without underscore:

 let exist= [ { x: 6811, y: 15551, a: ['aa', 'ab', 'ac'] }, { x: 6811, y: 15552, a: ['aa', 'bb'] }, { x: 6812, y: 15551, a: ['cc'] }, ]; const found = exist.filter(({ a }) => a.includes('aa')); exist = exist.filter(({ a }) => !a.includes('aa')); console.log('found:', found); console.log('exist:', exist); 

Answer 2

You can use .reject with .contains and _.result combination

 exist=[ {"x":6811,"y":15551,"a":["aa","ab","ac"]},{"x":6811,"y":15552,"a":["bb"]},{"x":6812,"y":15551,"a":["aa","cc"]}, {"x":6812,"y":15551} ] exist = _.reject(exist, function(item){ return _.contains(_.result(item, "a"), "aa"); }) console.log(exist); 

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