Question 291863 in Javascript Question posted in
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Javascript – How remove duplicate values from the result of a generator function

I have three generators function the third one is delegate function

function* genNumbers() {
    yield* [1, 2, 2, 2, 3, 4, 5];
  }

  function* genLetters() {
    yield* ["A", "B", "B", "B", "C", "D"];
  }

  function* genAll() {
    yield* genNumbers();
    yield* genLetters();
  }

  let generator = genAll();

I want to print the all values by using the generator.next() But **Remove **the duplicate value, like that:

  console.log(generator.next()); // {value: 1, done: false}
  console.log(generator.next()); // {value: 2, done: false}
  console.log(generator.next()); // {value: 3, done: false}
  console.log(generator.next()); // {value: 4, done: false}
  console.log(generator.next()); // {value: 5, done: false}
  console.log(generator.next()); // {value: "A", done: false}
  console.log(generator.next()); // {value: "B", done: false}
  console.log(generator.next()); // {value: "C", done: false}
  console.log(generator.next()); // {value: "D", done: false}
Tags:

3

Answers


  1. 0 
    function* genNumbers() {
  yield* [1, 2, 2, 2, 3, 4, 5];
}

function* genLetters() {
  yield* ["A", "B", "B", "B", "C", "D"];
}

function* genAll() {
  const uniqueValues = new Set();
  
  for (const value of genNumbers()) {
    if (!uniqueValues.has(value)) {
      uniqueValues.add(value);
      yield value;
    }
  }

  for (const value of genLetters()) {
    if (!uniqueValues.has(value)) {
      uniqueValues.add(value);
      yield value;
    }
  }
}

let generator = genAll();

console.log(generator.next()); // {value: 1, done: false}
console.log(generator.next()); // {value: 2, done: false}
console.log(generator.next()); // {value: 3, done: false}
console.log(generator.next()); // {value: 4, done: false}
console.log(generator.next()); // {value: 5, done: false}
console.log(generator.next()); // {value: "A", done: false}
console.log(generator.next()); // {value: "B", done: false}
console.log(generator.next()); // {value: "C", done: false}
console.log(generator.next()); // {value: "D", done: false}

    To achieve printing unique values without duplicates from the generators, you can modify the genAll function to use a Set to keep track of values that have already been yielded. Here’s an updated version of your code:

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  2. 0

    Keeping things generator based, you could just create a generator that takes your returned generator that does the de-duping. Doing this you could then keep this as a util you can use in other places.

    Generators that take generators as source is a very memory efficient way of processing huge amounts of data, as it avoid intermediate memory storage.

    function* genNumbers() {
  yield* [1, 2, 2, 2, 3, 4, 5];
}

function* genLetters() {
  yield* ["A", "B", "B", "B", "C", "D"];
}

function* dedupe(gen) {
  const done = new Set();
  for (const v of gen) {
    if (done.has(v)) continue;
    yield v;
    done.add(v);
  }
}

function* genAll() {
  yield* genNumbers(); 
  yield* genLetters();
}

for (const g of dedupe(genAll())) console.log(g);
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  3. 0

    You need two common iterator idioms: chain that combines multiple iterators sequentially and uniq which returns only unique values from an iterator:

    function* genNumbers() {
    yield* [1, 'A', 2, 2, 3, 4, 5];
}

function* genLetters() {
    yield* ["A", "B", "B", "B", "C", "D"];
}

function* chain(...iters) {
    for (let it of iters)
        yield *it
}

function* uniq(it) {
    const done = new Set();

    for (let x of it) {
        if (!done.has(x)) {
            yield x;
            done.add(x);
        }
    }
}


unqiues = uniq(chain(genNumbers(), genLetters()))
console.log(...unqiues);

    This assumes you want to uniquify both iterators as a whole, if you want each to be handled separately, it should be the other way around:

     chain(uniq(first), uniq(second))
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