Create new array with deleted values (React Native)

Question

I have an array like this one :

[
Object {
"hex": "#00b2b9",
"label": "text",
"value": "364",
},
Object {
"hex": "#50690e",
"label": "text",
"value": "354",
},
Object {
"hex": "#925fa3",
"label": "text",
"value": "355"
}]

I have another array with this :

Array [
"355",
"356"
]

I'm looking to create a the first array but without the objects containing value 355 and 356. I tried with .filter()... but I'm a newbie with JS & React Native :-)

I tried some stuffs but almost every time I recreate my Array only with the values (i'm losing the objects inside)...

What I want to do is : If I found 355 and 356 in my First Array, I delete the objects with them and I recreate my array with the only object remaining (value 364)

I was thinking about sth like that : myFirstArray.filter(item => item.value != mySecondArray.value) but it's not a success ...

Thanks for help

have you try lodash?

slashsharp
– slashsharp

2018-09-04 10:48:31 +00:00
Commented Sep 4, 2018 at 10:48 — slashsharp
– slashsharp, Commented Sep 4, 2018 at 10:48

Adam · Accepted Answer · 2018-09-04 10:56:42Z

2

The previous answers involve iterating over your id's array many times.

A more performant solution would be to store the id's in a set and then use that combined with a filter to produce your new array.

const arr = [
     {
    "hex": "#00b2b9",
    "label": "text",
    "value": "364",
    },
     ...,
     {
    "hex": "#925fa3",
    "label": "text",
    "value": "355"
    }];

const ids = ["354", "355"];

const idSet = new Set(ids);
const output = arr.filter(e => !idSet.has(e.value));

answered Sep 4, 2018 at 10:56

Adam

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Comments

Lazar Nikolic · Accepted Answer · 2018-09-04 10:51:32Z

1

var firstArray = [{
"hex": "#00b2b9",
"label": "text",
"value": "364",
},
{
"hex": "#50690e",
"label": "text",
"value": "354",
},
{
"hex": "#925fa3",
"label": "text",
"value": "355"
}]

var secondArray = [
"355",
"356"
]

var thirdArray = firstArray.filter(item => secondArray.includes(item.value))

console.log(thirdArray)

answered Sep 4, 2018 at 10:51

Lazar Nikolic

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1 Comment

Clément CREUSAT Over a year ago

thank you Lazar! It's working. You all came with the same idea. So cool

Luca Kiebel · Accepted Answer · 2018-09-04 10:50:05Z

1

You are nearly there, just use Array#includes() to determine whether or not the value of an item is in the second array:

myFirstArray.filter(item => !mySecondArray.includes(item.value))

let myFirstArray = [{
    "hex": "#00b2b9",
    "label": "text",
    "value": "364",
  },
  {
    "hex": "#50690e",
    "label": "text",
    "value": "354",
  },
  {
    "hex": "#925fa3",
    "label": "text",
    "value": "355"
  }
]

let mySecondArray = [
  "355",
  "356"
]

console.log(
  myFirstArray.filter(item => !mySecondArray.includes(item.value))
)

answered Sep 4, 2018 at 10:50

Luca Kiebel

10.1k7 gold badges34 silver badges47 bronze badges

1 Comment

Clément CREUSAT Over a year ago

Thank you Luca ! It' working. I was so close but it's better to understand the concept now !

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