How to compare two vectors in java script [duplicate]

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How to compare two vectors in java script [duplicate]

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#1 by (2 votes)

0

I have a job to do on a topic about math, so I have to do about sets so I thought I would do an intersection that would show only the numbers that are repeated in both sets, but I'm not getting it, I was just able to collect the data and show the sets typed in.

var elemento = [];
var conjuntos = [];

numcon = prompt("Digite o Numero de Conjuntos:", "")
if (numcon == 0) {
  alert("Numero de Conjunto tem que ser maior que zero")
} else {

  for (var i = 1; i <= numcon; i++) {
    numelemt = prompt("digite o numero de elmentos do " + "  " + i + "º" + "  conjuntos:", "")
    conjuntos[i - 1] = [];
    if (numelemt == 0) {
      alert("Este conjunto sera nulo")
    } else {
      for (var cont1 = 1; cont1 <= numelemt; cont1++) {
        elemt = window.prompt("Digite o" + " " + cont1 + "º" + " " + "elemento do conjunto nº" + " " + i);

        conjuntos[i - 1].push(elemt)
      }
    }
  }
}

console.log(conjuntos)

for (var cont3 = 0; cont3 < conjuntos.length; cont3++) {
  document.write("<br>" + "conjunto " + (cont3 + 1) + " é {")
  for (var cont4 = 0; cont4 < conjuntos[cont3].length; cont4++) {
    if (cont4 == conjuntos[cont3].length - 1) {
      document.write(conjuntos[cont3][cont4]);
    } else {
      document.write(conjuntos[cont3][cont4] + ",");
    }
    //conjunto 1 é: {552655}
  }
  document.write("}");
}

javascript

asked by anonymous 03.07.2018 / 16:18

1 answer

C # - High memory consumption when calling method Error when user provides file name

score 2 · Answer 1

You can do this in a very simple way! The Array in JavaScript has a method called filter() and a method called indexOf() .

`filter()`

The filter() returns an array with elements that apply to a condition that you provide through a function.

`indexOf()`

The indexOf() returns the index of an element in the array. If it is not found, -1 is returned.

Solution

The solution to your problem would be:

conjuntos[i-1].filter(valor => -1 !== conjuntos[i].indexOf(valor));

See the snippet below:

var elemento = [];
var conjuntos = [];

numcon = prompt("Digite o Numero de Conjuntos:", "")
if (numcon == 0) {
  alert("Numero de Conjunto tem que ser maior que zero")
} else {

  for (var i = 1; i <= numcon; i++) {
    numelemt = prompt("digite o numero de elmentos do " + "  " + i + "º" + "  conjuntos:", "")
    conjuntos[i - 1] = [];
    if (numelemt == 0) {
      alert("Este conjunto sera nulo")
    } else {
      for (var cont1 = 1; cont1 <= numelemt; cont1++) {
        elemt = window.prompt("Digite o" + " " + cont1 + "º" + " " + "elemento do conjunto nº" + " " + i);
        conjuntos[i - 1].push(elemt)
      }
    }
  }
}

console.log(conjuntos)
for( let i = 1; i < conjuntos.length; i++) {
    const intersecao = conjuntos[i-1].filter(valor => -1 !== conjuntos[i].indexOf(valor));
    console.log('A interseção entre os conjuntos ${i} e ${i+1} é:', intersecao);

}

Response Reference: link