I have a nested object where I'd like to convert an array of a single value group, into a single value:
real_data = [
[
{x: 1, group: [1]},
{x: 9, group: [2]},
{x: 3, group: [3]}
],
[
{x: 2, group: [1]},
{x: 4, group: [2]},
{x: 10, group: [3]}
],
[
{x: 1, group: [1]},
{x: 5, group: [2]},
{x: 12, group: [3]}
],
[],
[
{x: 1, group: [1]},
{x: 5, group: [2]},
{x: 12, group: [3]}
]
]
I've tried using map here but have been struggling due to the nested nature of the array. I also tried and struggled with some nested for loops. The desired output is:
desired_output = [
[
{x: 1, group: 1},
{x: 9, group: 2},
{x: 3, group: 3}
],
[
{x: 2, group: 1},
{x: 4, group: 2},
{x: 10, group: 3}
],
[
{x: 1, group: 1},
{x: 5, group: 2},
{x: 12, group: 3}
],
[],
[
{x: 1, group: 1},
{x: 5, group: 2},
{x: 12, group: 3}
]
]
Any help appreciated!!
It is simple.
group in inner arrayreal_data = [
[
{x: 1, group: [1]},
{x: 9, group: [2]},
{x: 3, group: [3]}
],
[
{x: 2, group: [1]},
{x: 4, group: [2]},
{x: 10, group: [3]}
],
[
{x: 1, group: [1]},
{x: 5, group: [2]},
{x: 12, group: [3]}
],
[],
[
{x: 1, group: [1]},
{x: 5, group: [2]},
{x: 12, group: [3]}
]
];
var desiredOutput = real_data.map( elements => {
let newElements = elements.map(element => {
element.group = element.group[0];
return element;
});
return newElements;
})
You can achieve the desired result without modifying the original input -
const input =
[[{x:1,group:[1]},{x:9,group:[2]},{x:3,group:[3]}],[{x:2,group:[1]},{x:4,group:[2]},{x:10,group:[3]}],[{x:1,group:[1]},{x:5,group:[2]},{x:12,group:[3]}],[],[{x:1,group:[1]},{x:5,group:[2]},{x:12,group:[3]}]]
const output =
input.map(x => x.map(({ group: [group], ...y }) => ({ ...y, group })))
console.log(output)[
[
{x:1,group:1},
{x:9,group:2},
{x:3,group:3}
],
[
{x:2,group:1},
{x:4,group:2},
{x:10,group:3}
],
[
{x:1,group:1},
{x:5,group:2},
{x:12,group:3}
],
[],
[
{x:1,group:1},
{x:5,group:2},
{x:12,group:3}
]
]
based on the desired output you can use the following
real_data.map(o=>{
return o.map(p=>{
return {...p, group:parseInt(p.group.toString(),10)}
})
})