如何在一个任务中收集两个 arrays 的交集和补码？

Question

Arrays -

const source = [1, 1, 1, 3, 4];
const target = [1, 2, 4, 5, 6];

Empty arrays -

const matches1 = [];
const matches2 = [];
const unmatches1 = [];

Loop -

for (const line of source) {
  const line2Match = target.find((line2) => {
    const isMatch = line === line2;
    return isMatch;
  });

  if (line2Match) {
    matches1.push(
      line
    );
    matches2.push(line2Match);
  } else {
    unmatches1.push(line);
  }
}

This is the output right now -

[ 1, 1, 1, 4 ] at matches1

[ 3 ] at unmatches1

[ 1, 1, 1, 4 ] at matches2

The desired output -

[ 1, 4 ] at matches1

[ 1, 1, 3 ] at unmatches1

[ 1, 4 ] at matches2

What I would like to add is when source has a match with target , the value will be deleted from the target array, how can I achive that?

Answer 1

The OP...

"What I would like to add is when source has a match with target, the value will be deleted from the target array, how can I achive that?"

One had to actively mutate the target array by slicing the common item from it which in the very end makes target become the relative complement of source in target ... commonly/vulgo... the target difference .

One of cause could apply a shallow copy of target as part of the initial value of a reduce task in order to not mutate the original target reference itself...

 function collectIntersectionAndComplements(collector, sourceItem) { const { targetDiff } = collector; const targetIndex = targetDiff.findIndex(targetItem => targetItem === sourceItem); if (targetIndex === -1) { // collect the relative complement of target // in source... vulgo... the source difference. (collector.sourceDiff??= []).push(sourceItem) } else { // collect the intersection of both source and target... (collector.intersection??= []).push( //... by rejecting the common item from the original // target, thus actively mutating the latter, which leads // to ending up additionally with the relative complement // of source in target... vulgo... the target difference. targetDiff.splice(targetIndex, 1)[0] ); } return collector; } const source = [1, 1, 1, 3, 4]; const target = [1, 2, 4, 5, 6]; const { intersection, sourceDiff, targetDiff, } = source.reduce(collectIntersectionAndComplements, { targetDiff: [...target] }); console.log({ source, target, intersection, sourceDiff, targetDiff });

 .as-console-wrapper { min-height: 100%;important: top; 0; }

@heyheyhey2 ... Btw, if one was talking about the Simple Theory of Sets then duplicate (identical) values within each array/list/set were not allowed. Sets always have to feature just unique values.

Answer 2

Just for matches1: (similar for matches2)

const source = [1, 1, 1, 3, 4];
const target = [1, 2, 4, 5, 6];

let matches1 = source.reduce((res, curr)  => {
    if (target.includes(curr) && !res.includes(curr)) {
        res.push(curr);
    }
    return res;
}, []
);

console.log(matches1)
//[1,4]

For unmatches1:

let unmatches1 = source.reduce((res, curr)  => {
    if ((target.includes(curr) && res[0].includes(curr)) || !target.includes(curr)) {    
        res[1].push(curr);
    }
    if (target.includes(curr)) {
        res[0].push(curr)
    }
    return res;
}, [[],[]]
)[1]

console.log(unmatches1)
//[1,1,3]

Answer 3

Note that for a single pass both arrays must have been sorted into ascending order.

function go(){
    const a = [0,1, 1, 1, 3, 4, 6,8,10];
    const b = [-3,-4,1, 2, 4, 5, 6,25,26];

    let nomatcha=[];
    let match=[];
    let nomatchb=[];
    let i=0,j=0;
    while(true){
        if(i>=a.length){
            for (;j<b.length;j++) nomatchb.push(b[j]);
            break;
        }
        else if (j>=b.length){
            for (;i<a.length;i++) nomatcha.push(a[j]);
            break;
        }
        else if(a[i]==b[j]){
            match.push(a[i++]);
            j++;
        }
        else if (a[i]<b[j]){
            let val=a[i++];
            if(nomatcha.length==0)nomatcha.push(val);
            else if(val != nomatcha[nomatcha.length-1])
                nomatcha.push(val);
        }
        else if (b[j]<a[i]){
            let val=b[j++];
            if(nomatchb.length==0)nomatchb.push(val);
            else if(val != nomatchb[nomatchb.length-1])
                nomatchb.push(val);
        }


    }

    console.log("match: "+match);
    console.log("nomatcha: "+nomatcha);
    console.log("nomatchb: "+nomatchb);

}

}

Output:

match: 1,4,6 nomatcha: 0,1,3,8,10 nomatchb: -3,-4,2,5,25,26