Obtain the position of a value within an array that has been grouped together

Question

Obtain the position of a value within an array that has been grouped together

In my JavaScript code, I am dealing with two arrays. One array contains Ids:

[1,1,1,2,2,2,3,3,...]

The other array holds values:

[0,1,0,0,0,1,1,0,...]

What I need to do is determine the zero-based index of the group of Ids where the corresponding value is 1.

For example, for these arrays, the result could be:

[1,2,0,...]

This means that it is the index of the 1 in the Values array when grouped by the unique values in the Id array.

Keep in mind that there should only be a single 1 per group of Ids, even if the groups are not in sequential order like:

[1,1,2,1,2,2,3,3,...]. In this case, I still want the correct index for the grouped Ids.

I initially tried using a while loop but encountered duplicate values. Then I attempted to filter my array without complete success. Is there a way to achieve this task in JavaScript?

Examples:

Array 1 (IDs): [1,1,1,2,2,2,3,3,3,3]  
Array 2 (Values): [0,1,0,0,1,0,0,0,0,1]  
Result Array: [1,1,3]

Array 3 (IDs): [1,2,1,3,1,1,2,2,3]  
Array 4 (Values): [0,0,1,0,0,0,1,0,1]   
Result Array: [1,1,1]

javascript arrays indexing

Answer 1

Answer №1

If you assign an index counter to the same group, you can retrieve the value when a value of one is encountered.

Example 3:

[1, 2, 1, 3, 1, 1, 2, 2, 3]  ids
[0, 0, 1, 0, 0, 0, 1, 0, 1]  values
 0  0  1  0  2  3  1  2  1   indices by group
       ^           ^     ^   result

function getIndices(ids, values) {
    var map = new Map;
    return ids.reduce((r, v, i) => {
        if (values[i] === 1) r.push(map.get(v) || 0);
        map.set(v, (map.get(v) || 0) + 1);
        return r;
    }, []);
}

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

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

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

Instead of using a Map, you have the option to utilize an object as a hash table for indexing the values.

function getIndices(ids, values) {
    var hash = Object.create(null);
    return ids.reduce((r, v, i) => {
        hash[v] = hash[v] || 0;
        if (values[i] === 1) r.push(hash[v]);
        hash[v]++;
        return r;
    }, []);
}

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

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

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

Answer 2

If you assign an index counter to the same group, you can retrieve the value when a value of one is encountered.

Example 3:

[1, 2, 1, 3, 1, 1, 2, 2, 3]  ids
[0, 0, 1, 0, 0, 0, 1, 0, 1]  values
 0  0  1  0  2  3  1  2  1   indices by group
       ^           ^     ^   result

function getIndices(ids, values) {
    var map = new Map;
    return ids.reduce((r, v, i) => {
        if (values[i] === 1) r.push(map.get(v) || 0);
        map.set(v, (map.get(v) || 0) + 1);
        return r;
    }, []);
}

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

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

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

Instead of using a Map, you have the option to utilize an object as a hash table for indexing the values.

function getIndices(ids, values) {
    var hash = Object.create(null);
    return ids.reduce((r, v, i) => {
        hash[v] = hash[v] || 0;
        if (values[i] === 1) r.push(hash[v]);
        hash[v]++;
        return r;
    }, []);
}

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

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

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

Obtain the position of a value within an array that has been grouped together

Answer №1

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