Tengo algunos datos que deben transformarse en una matriz, uno por uno.
Estos son los datos de ejemplo (puede haber muchos más escuadrones):
var my_arr = []; var squad1 = [ { date: "2022-04-04", number_of_checkable_cells: 2 }, { date: "2022-04-05", number_of_checkable_cells: 2 } ]; var squad2 = [ { date: "2022-04-04", number_of_checkable_cells: 2 }, { date: "2022-04-05", number_of_checkable_cells: 1 }, { date: "2022-04-06", number_of_checkable_cells: 2 } ]; Qué sucede: my_arr recibe escuadrón1 para agregarlo, luego escuadrón2 se une a la fiesta.
Entonces, después de agregar escuadrón1, my_arr debería ser:
my_arr = [ { date: "2022-04-04", number_of_checkable_cells: 2 }, { date: "2022-04-05", number_of_checkable_cells: 2 } ];Luego se reciben los datos del escuadrón2 y quiero el siguiente resultado:
my_arr = [ { date: '2022-04-04', number_of_checkable_cells: 4 }, { date: '2022-04-05', number_of_checkable_cells: 3 }, { date: '2022-04-06', number_of_checkable_cells: 2 } ]El código que tengo:
function reduce_it(arr) { return arr.reduce((initiator, curr) => { var obj = initiator.find((i) => i.date == curr.date); if (obj) obj.number_of_checkable_cells += curr.number_of_checkable_cells; else initiator.push(curr); return initiator; }, []); } my_arr = [...my_arr, ...squad1]; my_arr = reduce_it(my_arr); my_arr = [...my_arr, ...squad2]; my_arr = reduce_it(my_arr);¿Hay una mejor manera de llegar al mismo resultado? Si hay, por favor guíame. :)
Mantendría la matriz de destino y la mutaría.
function add(target, source) { return source.reduce((t, o) => { const item = t.find(({ date }) => date == o.date); if (item) item.number_of_checkable_cells += o.number_of_checkable_cells; else t.push({ ...o }); return t; }, target); } const squad1 = [{ date: "2022-04-04", number_of_checkable_cells: 2 }, { date: "2022-04-05", number_of_checkable_cells: 2 }], squad2 = [{ date: "2022-04-04", number_of_checkable_cells: 2 }, { date: "2022-04-05", number_of_checkable_cells: 1 }, { date: "2022-04-06", number_of_checkable_cells: 2 }], my_arr = []; add(my_arr, squad1); console.log(my_arr); add(my_arr, squad2); console.log(my_arr);Lo que sugieres funciona, pero esto es lo que te sugiero que hagas. Básicamente, cada vez que llega un nuevo escuadrón, puede recorrerlo y actualizar su conjunto de escuadrones de la siguiente manera:
var my_arr = []; var squad1 = [ { date: "2022-04-04", number_of_checkable_cells: 2 }, { date: "2022-04-05", number_of_checkable_cells: 2 } ]; my_arr.push(...squad1); // Here you received the squad 2 var squad2 = [ { date: "2022-04-04", number_of_checkable_cells: 2 }, { date: "2022-04-05", number_of_checkable_cells: 1 }, { date: "2022-04-06", number_of_checkable_cells: 2 } ]; for (const squad of squad2) { const existingSquad = my_arr.find( (prevSquad) => prevSquad.date === squad.date ); if (existingSquad) { existingSquad.number_of_checkable_cells += squad.number_of_checkable_cells; } else { my_arr.push(squad); } } console.log(my_arr); /* (3) [{…}, {…}, {…}] 0: {date: '2022-04-04', number_of_checkable_cells: 4} 1: {date: '2022-04-05', number_of_checkable_cells: 3} 2: {date: '2022-04-06', number_of_checkable_cells: 2} length: 3 */Entonces, este ejemplo es el original, si desea envolverlo en un método, mire a continuación:
function updateSquads(squads, newSquads) { for (const squad of newSquads) { const existingSquad = squads.find( (prevSquad) => prevSquad.date === squad.date ); if (existingSquad) { existingSquad.number_of_checkable_cells += squad.number_of_checkable_cells; } else { squads.push(squad); } } } var my_arr = []; var squad1 = [ { date: "2022-04-04", number_of_checkable_cells: 2 }, { date: "2022-04-05", number_of_checkable_cells: 2 } ]; my_arr.push(...squad1); // Here you receive squad 2 var squad2 = [ { date: "2022-04-04", number_of_checkable_cells: 2 }, { date: "2022-04-05", number_of_checkable_cells: 1 }, { date: "2022-04-06", number_of_checkable_cells: 2 } ]; updateSquads(my_arr, squad2) console.log(my_arr); /* (3) [{…}, {…}, {…}] 0: {date: '2022-04-04', number_of_checkable_cells: 4} 1: {date: '2022-04-05', number_of_checkable_cells: 3} 2: {date: '2022-04-06', number_of_checkable_cells: 2} length: 3 */