Filter Object by array

Question

I have an object data coming from service, i want to filter out some items based on key-value, also it's not only one item but multiple items to remove;

var dataFromServce = [{
    kod: 1000, teur: 'do0',    
    kod: 1001, teur: 'do1',   
    kod: 1002, teur: 'do2',    
    kod: 1003, teur: 'do3'

}];
var removeItems = [1001, 1002, 1003];

if (removeItems != "") {

    // run thru dataFromServce loop

    for (var i = 0; i < dataFromServce.length; i++) {
        // if dataFromServce item match removeItems then remove it from dataFromServce
        // TODO: to match every item in removeItems
        if (dataFromServce[i].kod === removeItems[0]) {
            dataFromServce.splice(i, 1);

            i--;

        }

    }

this way it checks only one item (removeItems[0]) result should be

var dataFromServce = [{
    kod: 1000, teur: 'do0'     
}];

baao · Accepted Answer · 2020-04-27 09:49:33Z

2

You're looking for filter and includes

const dataFromService = [{
  kod: 1000,
  teur: 'do0'
}, {
  kod: 1001,
  teur: 'do1'
}, {
  kod: 1002,
  teur: 'do2'
}, {
  kod: 1003,
  teur: 'do3'
}];
const removeItems = [1001, 1002, 1003];

const res = dataFromService.filter(({
  kod
}) => !removeItems.includes(kod));
console.log(res);

The code in your question is obviously wrong since your dataFromService is invalid, I've fixed that in the answer.

answered Apr 27, 2020 at 9:49

baao

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3 Comments

Vasily Liaskovsky Over a year ago

This snippet has O(m*n) complexity, where m and n are sizes of input and filter. This can be improved to O(m*log n) by using hash instead of array for filter or using binary search for key (filter array should be sorted beforehand). Furthermore, if both input and filter come already sorted, improvement is possible to O(m+n)

yanivz Over a year ago

thank you for your answer @VasilyLiaskovsky , but there is one problem...its not working on internet explorer.

yanivz Over a year ago

needs refactor to dataFromService.filter(({kod}) => !removeItems.includes(kod));

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