Create a new array from existing array and join each value into one unique value and check if a uniq-CodePudding

I am trying to create a form for uploading products with variants.

I am joining each value from originalArray into a unique value in a newArray. Example of joined values into one unique value: red/small/modern

However, I want check if there is already a unique (joined) value in the newArray. If true, keep exiting value and price in the newArray.

So if, value "red/small/modern" already exist with price 888, an array item of price of null should not be returned into the newArray.

let originalArray = [
  [
    { value: 'red', id: 99, price: null },
    { value: 'blue', id: 100, price: null },
  ],
  [
    { value: 'small', id: 101, price: null },
    { value: 'medium', id: 102, price: null },
  ],
  [
    { value: 'modern', id: 103, price: null },
    { value: 'classic', id: 104, price: null },
  ],
];
//
// existing array item
let newArray = [
  { value: 'red/small/modern', id: 1, price: 888 },
  { value: 'blue/medium/modern', id: 2, price: 100 },
];
//
console.log('example 1:', newArray); // [{…}, {…}]
//

newArray = originalArray
  .map((elem) => {
    return elem.map(({ value }) => value);
  })
  .reduce((acc, cur) => {
    return acc.flatMap((seq) => {
      return cur.map((part) => `${seq}/${part}`);
    });
  })
  .map((elem) => {
    return { value: elem, price: null };
  });
//
//
// price for value "red/small/modern" and "blue/medium/modern" should not be null, as they are already set in the newArray
console.log('example 2:', newArray);

Hope question make sense.

CodePudding user response：

You can filter the elements which are not already in the newArray, and push the new ones into it:

let originalArray=[[{value:"red",id:99,price:null},{value:"blue",id:100,price:null}],[{value:"small",id:101,price:null},{value:"medium",id:102,price:null}],[{value:"modern",id:103,price:null},{value:"classic",id:104,price:null}]];
let newArray=[{value:"red/small/modern",id:1,price:888},{value:"blue/medium/modern",id:2,price:100}];

const allCombinations = originalArray
  .map((elem) => {
    return elem.map(({ value }) => value);
  })
  .reduce((acc, cur) => {
    return acc.flatMap((seq) => {
      return cur.map((part) => `${seq}/${part}`);
    });
  })
  .map((elem) => ({ value: elem, price: null }));

newArray.push(
  ...allCombinations.filter(({ value }) => {
    return newArray.every(existing => existing.value !== value);
  })
);

console.log('example 2:', newArray);

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CodePudding user response：

just add a test ?:
if (-1 === newArray.findIndex(row=>row.value===value))

const originalArray = 
  [ [ { value: 'red',     id:  99, price: null } 
    , { value: 'blue',    id: 100, price: null } 
    ] 
  , [ { value: 'small',   id: 101, price: null } 
    , { value: 'medium',  id: 102, price: null } 
    ] 
  , [ { value: 'modern',  id: 103, price: null } 
    , { value: 'classic', id: 104, price: null } 
  ] ] 
, newArray = //  existing array item
  [ { value: 'red/small/modern',   id: 1, price: 888 } 
  , { value: 'blue/medium/modern', id: 2, price: 100 } 
  ]

originalArray
  .map((elem) => elem.map(({ value }) => value) ) 
  .reduce((acc, cur) => acc.flatMap(seq => cur.map(part => `${seq}/${part}`)))
  .forEach (value =>
    {
    if (-1 === newArray.findIndex(row=>row.value===value))
      newArray.push({value, price: null})
    })

console.log( newArray )

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otherwise, if you prefer to create an other newArray

const originalArray = 
  [ [ { value: 'red',     id:  99, price: null } 
    , { value: 'blue',    id: 100, price: null } 
    ] 
  , [ { value: 'small',   id: 101, price: null } 
    , { value: 'medium',  id: 102, price: null } 
    ] 
  , [ { value: 'modern',  id: 103, price: null } 
    , { value: 'classic', id: 104, price: null } 
  ] ] 
, newArray = //  existing array item
  [ { value: 'red/small/modern',   id: 1, price: 888 } 
  , { value: 'blue/medium/modern', id: 2, price: 100 } 
  ]
, otherNewArray = originalArray
  .map(elem => elem.map(({ value }) => value) ) 
  .reduce((acc, cur) => acc.flatMap(seq => cur.map(part => `${seq}/${part}`)))
  .map(value =>
    {
    let price = newArray.find(row=>row.value===value)?.price || null
    return({value, price})
    })

console.log( otherNewArray )

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