CODEWITHSUNDEEP
javascriptarrays
ubuntuser
ubuntuser

Reputation: 45

search for matching values in two arrays with repeated values in javascript

I have two arrays:

a = [2,2,3,0,6]
b = [6,3,2,2,0]

I am trying use for loop to match values and get the index of a in a new array c. How can we do this? Notice that there are multiple values which match and so I think the previous match must be skipped.

Upvotes: 0

Views: 73

Answers (5)

Nina Scholz
Nina Scholz

Reputation: 386550

This is a proposal which respects the last index and looks further.

How it works:

It uses Array#map for iterating array b with a callback. map gets an own this space with an really empty object Object.create(null).

The callback has on parameter bb which is one element of `b.

Next is to find the element is in array a with a Array#indexOf and a fromIndex, based on the former searches. The former index is stored in the this object, as long as the result is not -1, because this would reset the fromIndex to zero.

If there is no this[bb] or a falsy value of this[bb] take zero as fromIndex.

Later, a found index is incremented and stored in this[bb].

At least, the index is returned.

var a = [2, 2, 3, 0, 6],
    b = [6, 3, 2, 2, 0],
    c = b.map(function (bb) {
        var index = a.indexOf(bb, this[bb] || 0);
        if (index !== -1) {
            this[bb] = index + 1;
        }				
        return index;
    }, Object.create(null));

console.log(c);

Another solution could be first generate an object with all indices of a and use it in the iteration of b for returning the indices.

The example is a bit extended, to show what happen if there is no more than two indices (2) and one without being in a (7).

The content of aObj with all indices of a:

{
    "0": [3],
    "2": [0, 1],
    "3": [2],
    "6": [4]
}


var a = [2, 2, 3, 0, 6],
    b = [6, 3, 2, 2, 0, 7, 2],
    aObj = Object.create(null),
    c;

a.forEach(function (aa, i) {
    aObj[aa] = aObj[aa] || [];
    aObj[aa].push(i);
});
c = b.map(function (bb) {
    return aObj[bb] && aObj[bb].length ? aObj[bb].shift() : -1;
});

console.log(c);

Upvotes: 1

JYoThI
JYoThI

Reputation: 12085

try something like this 

var a = [2,2,3,0,6];
var b = [6,3,2,2,0];

var arrayLength_a = a.length;
var arrayLength_b = b.length;
var new_array=[];

 for (var i = 0; i < arrayLength_a; i++)
 {
    for (var j = 0; j < arrayLength_b; j++) 
    {
       if (a[i] == b[j]) 
       {
          if(new_array.indexOf(a[i]) === -1)
          {  
             new_array.push(a[i]);
          }
       }     
    }
 }

Upvotes: 0

Mahedi Sabuj
Mahedi Sabuj

Reputation: 2944

As far I Understand, You can try this:

var a = [2,2,3,0,6];
var b = [6,3,2,2,0];

var c = new Array();

for(i = 0; i < b.length; i++)
{
    for(j = 0; j < a.length; j++)
    {  
        if(b[i] === a[j] && c.indexOf(j) < 0)
        {
             c.push(j);
             break;
        }
    }
}

console.log(c); // [4, 2, 0, 1, 3]

FIDDLE DEMO HERE

Upvotes: 0

Jack jdeoel
Jack jdeoel

Reputation: 4584

loop .map function and check same value by indexOf

indexOf will return a number,representing the position where the specified search value occurs for the first time, or -1 if it never occurs

var arr = [];    
a.map(function(v){      
    if(b.indexOf(v) > -1){
        arr.push(v);
    }
});
console.log(arr);

Upvotes: 0

Artem Markov
Artem Markov

Reputation: 413

If I understand correct. 

let c = a.map(i => b.indexOf(i))

or

var c = a.map(function(i) { return b.indexOf(i); });

Upvotes: 0

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