compare 2 object arrays and return the elements in first array as true if it exists in second array, or as false if it does not

Question

I have 2 object arrays

const options = [
  { value: 'opt1', label: 'opt1' },
  { value: 'opt2', label: 'opt2' },
  { value: 'opt3', label: 'opt3' },
  { value: 'opt4', label: 'opt4' }
]

const selected = [
  { value: 'opt1', key: '1' },
  { value: 'opt2', key: '2' }
]

I need to compare these two arrays and get result as

result =
  { 'opt1', true },
  { 'opt2', true },
  { 'opt3', false },
  { 'opt4', false }
]

since opt1 and opt2 exists in second array. I know there are lots of methods, but what would be the shortest method?

Arfeo · Accepted Answer · 2019-03-01 00:43:43Z

1

The shortest one that I personally could imagine.

const result = options.map(o => ({ [o.value]: !!selected.find(s => s.value === o.value) }));

edited Mar 1, 2019 at 0:43

answered Mar 1, 2019 at 0:28

Arfeo

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Ele · Accepted Answer · 2019-03-01 00:30:43Z

1

You can use the function map to get the mapped values first of the selected and then a map the array options using a logic to ask for the existence of a value within the selected array. Finally, use computed-property-names to build the desired output.

const options = [  { value: 'opt1', label: 'opt1' },  { value: 'opt2', label: 'opt2' },  { value: 'opt3', label: 'opt3' },  { value: 'opt4', label: 'opt4' }],
      selected = [  { value: 'opt1', key: '1' },  { value: 'opt2',key: '2' }],
      mapped = selected.map(({value}) => value),
      result = options.map(({value}) => ({[value]: mapped.includes(value)}));

console.log(result);

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

answered Mar 1, 2019 at 0:30

Ele

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Comments

Joshua Paul Huang · Accepted Answer · 2019-03-01 00:32:43Z

1

return options.map(v => {
  let newObj = {};
  newObj[v.value] = selected.find(option => { return 
  option.value == v.value }) !== undefined;
  return newObj;
})

answered Mar 1, 2019 at 0:32

Joshua Paul Huang

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