I wanted to be able to transform the age into a single array, so I would already know how to filter, then apply the mapping to only show people over 18 years old, in addition, to present the abbreviated names of the people.
Example:
name: "Raul", age: 27,
name: "Jose", age: 14,
name: "Maria", age: 52,
name: "Jesus", age: 17,
name: "Neo", age: 2
[Ra, Ma] -> Those are above 18, and their names go abbreviated
Here what i tried to do:
const amantesGatos = {
name: "Raul", age: 27,
name: "Jose", age: 14,
name: "Maria", age: 52,
name: "Jesus", age: 17,
name: "Neo", age: 2
};
// this needs to be filtered
const idade = amantesGatos.age
//i tried to destructuring this array
const nomeAbrev = [amantesGatos.map(n=>n[0] n[1])]
//Tried to abbreviated names
//Filter above 18
const array = [-3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 444, 17];
function above18(num) {
for (let i = 18; num < i; i--) {
if (num < i) {
return false;
}
}
return num;
}
console.log(array.filter(above18));
CodePudding user response:
If I'm understanding correctly the desired outcome is [Ra, Ma]
?
If so you can .filter.map.
const peeps = [
{name: "Raul", age: 27,},
{name: "Jose", age: 14,},
{name: "Maria", age: 52,},
{name: "Jesus", age: 17,},
{name: "Neo", age: 2}
]
const ofAgePeeps = peeps.filter(({age}) => age > 18)
const shortNames = ofAgePeeps.map(({name}) => name.substring(0,2))
You can also chain these...
peeps.filter(({age}) => age > 18).map(({name}) => name.substring(0,2))
That said your amantesGatos
is an object with a bunch of duplicate keys and not an array. Which means it's really an object with only the last name and age. For example...
const obj = {
name: 'Tom', age: 2, name: 'Bob', age: 100
}
console.log(obj) // {name: 'Bob', age: 100}
CodePudding user response:
You can do as following
let users = [
{ name: "Raul", age: 27,},
{name: "Jose", age: 14,},
{name: "Maria", age: 52,},
{name: "Jesus", age: 17,},
{name: "Neo", age: 2},
]
let data = users.filter(user => user.age > 18).map(user => user.name.slice(0, 2))
data would be [ 'Ra', 'Ma' ]