How can I remove duplicate member from array of objects based on object property values?

Question

The array looks like:

var test = [
           {
            Time: new Date(1000), 
            psi:100.0
           }, 
           {
            Time: new Date(1000), 
            psi:200.0
           }, 
           {
            Time: new Date(2000), 
            psi:200.0
           }
          ]

The function looks like (the function was copied from some online resource, couldn't find the exact reference.)

function uniqTimetable(nums){
  console.log(nums); //log#1
  var length = nums.length;
  var count = 0;
  for (var i =0; i< length-1; i++){
    if (nums[count].Time.getTime() !== nums[i+1].Time.getTime()){
      count ++;
      nums[count] = nums[i+1];
    }
  }
  nums.length = count + 1;
  console.log(nums); // log #2
}

uniqTimetable(test);
console.log(test);// log #3

Are there any problems of

copy one object to the other array member by this linenums[count] = nums[i+1]
rescale the array length by this linenums.length = count + 1 ?

With electron/node.js, the output looks a bit weird.

In log#1, it shows the array length is 2 instead of 3. Something seems wrong with the function. Any suggestions welcome.

Thanks in advance.

Wired though. Mine is older (12.13.0), but it shows correct about the length of test. — Panwen Wang
– Panwen Wang, Commented Jul 10, 2020 at 18:24

ZER0 · Accepted Answer · 2020-07-10 22:37:26Z

2

If I understood the problem, you can solve it in one line function:

const test = [{
    Time: new Date(1000),
    psi: 100.0
  },
  {
    Time: new Date(1000),
    psi: 200.0
  },
  {
    Time: new Date(2000),
    psi: 200.0
  }
]

const unique = test.filter(({Time}, i, array) => 
  !array.find((item, j) => item.Time.getTime() === Time.getTime() && j > i));

console.log(unique);

(I assumed psi is not relevant)

It uses find and filter array's methods.

answered Jul 10, 2020 at 22:37

ZER0

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1 Comment

Akin Okegbile Over a year ago

Love it, but i work on a legacy team that would get confused by the chaining :D

Akin Okegbile · Accepted Answer · 2020-07-10 18:32:51Z

1

function removeDuplicates(myArr, prop) {
    return myArr.filter((obj, pos, arr) => {
        return arr.map(mapObj => mapObj[prop]).indexOf(obj[prop]) === pos;
    });
}
removeDuplicates(test,"Time")

Using the filter function of javascript, to return the List of only unique elements.

edited Jul 10, 2020 at 18:32

answered Jul 10, 2020 at 18:24

Akin Okegbile

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1 Comment

Akin Okegbile Over a year ago

Changed based on your comment @soumyasunny

Carsten Massmann · Accepted Answer · 2020-07-11 06:11:10Z

I like @Akin's answer! But unfortunately the .indexOf() function does not work for Date objects with the same value, since each object is still considered to be different and the .indexOf() function will always find the index of the Date object itself and never another "copy". To overcome this I convert the Date object back to milliseconds before collecting the values in an array (tst). I do this with myArr.map(mapObj => mapObj[prop].getTime()) before I go into the .filter function.

As the .getTime() method will only work for Date objects it does not make sense to keep the property prop as an argument. Instead I hardcoded the .Time property into the code.

Edit:

By coercing a numerical data type with a unary operator + for the .Time property I can leave out the .getTime() method which will be applied implicitly.

var test = [{Time: new Date(1000), psi:100.0}, 
            {Time: new Date(1000), psi:200.0}, 
            {Time: new Date(2000), psi:200.0}];

function Cars10m_remDups(myArr) {
  let tst=myArr.map(o=>+o.Time);
  return myArr.filter((o, i)=>tst.indexOf(+o.Time)===i);
}
// for comparison: --> will list all three objects!
function Akin_remDups(myArr, prop) {
  return myArr.filter((obj, pos, arr) => {
   return arr.map(mapObj => mapObj[prop]).indexOf(obj[prop]) === pos;
  });
}
// Zero's one-liner works too, here: my shortened version
const Zero_remDups = myArr => myArr.filter(({Time}, i, array) => 
  !array.find((item, j) => +item.Time-Time==0&&i>j));
// also: with "i>j" I pick the first unique (Zero chose the last)


console.log('Cars10m:',Cars10m_remDups(test));
console.log('Akin:',Akin_remDups(test,"Time"));
console.log('Zero:',Zero_remDups(test));

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

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