How can I filter an array using the for structure?
The code is this:
const numeros = [1,2,3,4,5,55,190,355,747,1000,125];
I need to filter the numbers under 10.
I used it this way, but I was asked to create one using the for structure and I can not get out of 0.
const numeros = [1,2,3,4,5,55,190,355,747,1000,125];
const filterOne = x => x < 10;
const filterTwo = numeros.filter(filterOne);
console.log(filterTwo);Just do one condition, it can also be done like this:
numeros.forEach((n) => {
if (n < 10) {
filterTwo.push(n);
}
});
const numeros = [1,2,3,4,5,55,190,355,747,1000,125];
const filterOne = x => x < 10;
const filterTwo = [];
for (let i = 0; i < numeros.length; i++) {
// Verifica que o valor de "numeros" no índice "i" é menor que 10
if (numeros[i] < 10) {
// adiciona no array filterTwo
filterTwo.push(numeros[i]);
}
}
console.log(filterTwo); You can use another loop form, forEach , where the (e) parameter represents the value of each item in the array:
const numeros = [1,2,3,4,5,55,190,355,747,1000,125];
const result = [];
numeros.forEach((e)=>{
e < 10 && result.push(e);
});
console.log(result);Explanation of the # used in the above example:
Returns the second operand based on the value of the first operand. If the first one is false , the second operand is ignored.
e < 10 && result.push(e);
\____/ ↑ \____________/
1º op. | 2º op.
|
O 1º op. tem que ser true
e < 10 || result.push(e);
\____/ ↑ \____________/
1º op. | 2º op.
|
O 1º op. tem que ser false
does not meet the 1st op., similar to
? ):
e < 10 ? faz uma coisa : faz outra coisa;