Js - removing an object array element if value is 0

Question

I have this array object:

0:
  id: "123123"
  cost: 100
  quantity: 2
1:
  id: "112233"
  cost: 100
  quantity: 5
2:
  id: "112233"
  cost: 100
  quantity: 0
3:
  id: "126233"
  cost: 100
  quantity: 0

What I want is to scan the whole object array and delete the ones with 0 quantity value. How do I do this with javascript?

you could use a loop. many libs have map - reduce tricks too. Are you using any library? — Hogan
– Hogan, Commented Jun 21, 2020 at 4:53
stackoverflow.com/questions/15287865/… isn’t exactly the same question, but its top two answers apply. — Ry-
– Ry- ♦, Commented Jun 21, 2020 at 4:54
stackoverflow.com/questions/16491758/… isn’t exactly the same question, but its answers apply too. — Ry-
– Ry- ♦, Commented Jun 21, 2020 at 4:55

2 revs · Accepted Answer · 2020-06-21 05:35:21Z

3

Using the function from my answer at remove objects from array by object property,

filterInPlace(array, item => item.quantity !== 0);

const filterInPlace = (array, predicate) => {
    let end = 0;

    for (let i = 0; i < array.length; i++) {
        const obj = array[i];

        if (predicate(obj)) {
            array[end++] = obj;
        }
    }

    array.length = end;
};

const arr = [
    {
        id: "123123",
        cost: 100,
        quantity: 2,
    },
    {
        id: "112233",
        cost: 100,
        quantity: 5,
    },
    {
        id: "112233",
        cost: 100,
        quantity: 0,
    },
    {
        id: "126233",
        cost: 100,
        quantity: 0,
    },
];

filterInPlace(arr, item => item.quantity !== 0);

console.log(arr);

This does modify the array in place. If you don’t need that, the Array#filter equivalent is probably better.
This is linear-time, constant space, whereas solutions based splice have a worse worst case(s). (@Hogan’s answer is also linear-time; this is the same idea as it, but with the loops merged.)

edited Jun 21, 2020 at 5:35

community wiki

2 revs
Ry-

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1 Comment

Hogan Over a year ago

I don't know... I think it is one loop because the exit is based on the adding the two indexes together. Two loops makes it sound like O(N^2) or O(2N) when it is still O(N) --- thanks for the hat tip.

Amir Meimari · Accepted Answer · 2020-06-21 05:08:14Z

1

array.filter() creates new array with the filtered objects.

take a look at this:

const values = [
  {id: '123123', cost: 100, quantity: 2},
  {id: '112233', cost: 100, quantity: 5},
  {id: '112233', cost: 100, quantity: 0},
  {id: '126233', cost: 100, quantity: 0},
]

const filtered = values.filter((obj) => obj.quantity !== 0)

console.log(filtered)
// expected output: [ { id: '123123', cost: 100, quantity: 2 }, { id: '112233', cost: 100, quantity: 5 } ]

answered Jun 21, 2020 at 5:08

Amir Meimari

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4 Comments

Hogan Over a year ago

He specifically says he wants to scan an array and delete not create a new array.

Ry- Over a year ago

@Hogan: A lot of people don’t realize there’s a difference, so the OP might not necessarily care.

Hogan Over a year ago

@Ry- In my experience there is an expectation that answers do what the question asks -- not something else. Considering that filter is part of the standard API I think it more likely that a programmer would know what it did and be asking the question because that did not solve the problem. Why assume the questioner made a mistake?

Modus Tollens Over a year ago

@Hogan Imho it is a good thing to mention another way in answers because we try not to answer OP alone, but many users after them. Some of them might not know the API capability. Another reason: this might be an x/y problem.

Shubham Shukla · Accepted Answer · 2020-06-21 05:12:58Z

0

let data = [
 {id: "123123",
  cost: 100,
  quantity: 2},
 { id: "112233",
  cost: 100,
  quantity: 5}, 
 { id: "112233",
  cost: 100,
  quantity: 0},
 {id: "126233",
  cost: 100,
  quantity: 0} 
];

let i =0;
 while (i < data.length) {
    if (data[i].quantity === 0) {
      data.splice(i, 1);
    } else {
      ++i;
    }
  }

console.table(data);

answered Jun 21, 2020 at 5:12

Shubham Shukla

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Comments

mipsc · Accepted Answer · 2020-06-21 05:23:13Z

0

Just iterate over the elements of the array and check for the value of quantity and delete with splice:

let arr = [
    {
        id: "123123",
        cost: 100,
        quantity: 2,
    },
    {
        id: "112233",
        cost: 100,
        quantity: 5,
    },
    {
        id: "112233",
        cost: 100,
        quantity: 0
    },
    {
        id: "126233",
        cost: 100,
        quantity: 0
    }
];
let len = arr.length;
for( let i = 0; i < len; i++) {
    if(arr[i]['quantity'] === 0){
        arr.splice(i--,1);
        len--;
    }
}
console.log(arr);

output:

[
  { id: '123123', cost: 100, quantity: 2 },
  { id: '112233', cost: 100, quantity: 5 }
]

edited Jun 21, 2020 at 5:23

answered Jun 21, 2020 at 5:14

mipsc

1138 bronze badges

Comments

Nithish · Accepted Answer · 2020-06-21 05:27:51Z

0

I usually don't recommend mutating the input itself. But since you have this requirement, Just created this utility in order to fulfil your requirement.

let data = [
  { id: '123123', cost: 100, quantity: 2 },
  { id: '112233', cost: 100, quantity: 5 },
  { id: '112232', cost: 100, quantity: 0 },
  { id: '112234', cost: 200, quantity: 0 },
  { id: '112235', cost: 400, quantity: 1 }
]

const indexesToBeRemoved = [];
data.forEach((d, index) => {
  if(d.quantity === 0) indexesToBeRemoved.push(index)
})

const length = indexesToBeRemoved.length;
for(let i=length-1; i>=0; i--) {
  data.splice(indexesToBeRemoved[i],1)
}

console.log(data)

One another way of fulfilling your requirement.

let data = [
  { id: '123123', cost: 100, quantity: 2 },
  { id: '112233', cost: 100, quantity: 5 },
  { id: '112232', cost: 100, quantity: 0 },
  { id: '112234', cost: 100, quantity: 0 },
  { id: '112235', cost: 100, quantity: 1 }
]

let numberOfItemsToBeRemoved = 0;

data.forEach((d, index) => {
  if(d.quantity === 0) {
    data.unshift(...data.splice(index, 1))
    numberOfItemsToBeRemoved++;
  }
})

console.log("deleted items: ", data.splice(0, numberOfItemsToBeRemoved))
console.log(data)

Hope this helps.

edited Jun 21, 2020 at 5:27

answered Jun 21, 2020 at 5:13

Nithish

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4 Comments

Hogan Over a year ago

I don't think this works... the array changes every time you do the splice. Did you test this?

Nithish Over a year ago

@Hogan, Yes I have tested this and it works. Please do let me know if there's any issue with the existing one.

Hogan Over a year ago

It won't work -- just try a more complicated case. For example have an item to be deleted as the 2nd item and the last item. Then you will throw an exception. or try and delete something past the end of the array.

Nithish Over a year ago

I tried the scenario For example have an item to be deleted as the 2nd item and the last item and is working as expected. Please have a look here

Hogan · Accepted Answer · 2020-06-21 05:08:30Z

-1

If you really have an array you can't delete -- most libraries that do this functionally will create a new object. If you want to "delete" items what you really have to do is move other elements down to over write the ones you no longer want.

So I would use the following steps.

Check to see if the filter item appears at all -- if it does not there is nothing to do.

If it does appear then then let n be the index of the first item to filter. next is the next item out after it.

loop (index from n to index+next < size of array)
 while next is an item to filter increase next by 1
 copy from next to index
 --> increase next by 1 and back to start of loop

edited Jun 21, 2020 at 5:08

answered Jun 21, 2020 at 5:03

Hogan

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Comments

Sven.hig · Accepted Answer · 2020-06-21 05:35:41Z

-1

first get element indexes and then delete them

arr=[{
  'id': "123123",
  cost: 100,
  quantity: 2
},
{
  id: "112233",
  cost: 100,
  quantity: 5
},
{
  id: "112233",
  cost: 100,
  quantity: 0
},
{
  id: "126233",
  cost: 100,
  quantity: 0
}
]
index=[]
arr.forEach(x => {
 if (x.quantity==0) index.push(arr.indexOf(x))
});

    i=0
index.forEach(x=>{
  arr.splice(x+i,1)
  i--
})
console.log(arr)

edited Jun 21, 2020 at 5:35

answered Jun 21, 2020 at 5:13

Sven.hig

4,5292 gold badges10 silver badges18 bronze badges

9 Comments

Ry- Over a year ago

This is working on the example by coincidence. arr.indexOf(x) is the wrong operation and produces -1, which splices off the last element. If you replace that with x, it still won’t work, because splicing an element out changes the indexes of future elements.

Hogan Over a year ago

you get the element indexes and then start deleting -- after the first delete they are invalid. this solution is very broken.

Sven.hig Over a year ago

I have updated the solution do you think this code will work?

Hogan Over a year ago

No. You can't have the check in another loop -- there will always be ways it will fail because the location change with ever splice.

Ry- Over a year ago

… yes. I do think this code will work. :D But you can also do the splices directly instead of building an array first.

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