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VistasImprima todos los valores pares en todos los extremos de la clave de objeto con valor impar
Quiero imprimir todos los valores pares en todos los extremos de la clave de objeto con valor impar, pero la codificación que hice ahora solo se especifica para arr1, arr3 y arr5. ¿Alguien puede sugerirme cómo arreglar el método 'let oddArr' (quizás en bucle) para que cuando cambie arr1 a arr7, el resultado sea el mismo?
var num = { arr1 : [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], arr2 : [11, 12, 13, 14, 15, 16, 17, 18, 19, 20], arr3 : [21, 22, 23, 24, 25, 26, 27, 28, 29, 30], arr4 : [31, 32, 33, 34, 35, 36, 37, 38, 39, 40], arr5 : [41, 42, 43, 44, 45, 46, 47, 48, 49, 50], }; let oddArr = [...num.arr1, ...num.arr3, ...num.arr5] //need some correction here let evenNum = oddArr.filter(number => number % 2 == 0); console.log(evenNum.toString()); //help me fix 'let oddArr' (maybe in loop method) so that when I changed the object of the array (eg: arr1 -> arr7) it would come out with the same result //the result/output should be 2,4,6,8,10,22,24,26,28,30,42,44,46,48,50 based on var num
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Juan Pablo Isaza
3 Respuestas
Responde la pregunta0
Es posible que desee utilizar
let oddArr = Object.entries(num).filter( // filter key names e => +e[0].replace("arr", '') % 2 !== 0 // replace "arr" and check if X in arrX is odd ).map(e => e[1]).flat() // merge values and flattern arrayTambién puede utilizar expresiones regulares si el nombre "arrX" no es consistente:
+e[0].match(/\d+/) % 2 !== 0Vea un fragmento de trabajo a continuación:
var num = { arr1: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], arr2: [11, 12, 13, 14, 15, 16, 17, 18, 19, 20], arr3: [21, 22, 23, 24, 25, 26, 27, 28, 29, 30], arr4: [31, 32, 33, 34, 35, 36, 37, 38, 39, 40], arr5: [41, 42, 43, 44, 45, 46, 47, 48, 49, 50], }; let oddArr = Object.entries(num).filter( e => +e[0].replace("arr", '') % 2 !== 0 ).map(e => e[1]).flat() let evenNum = oddArr.filter(number => number % 2 == 0); console.log(evenNum.toString());
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Juan Pablo Isaza
Denunciar
0
Puede probar como a continuación usando for in loop y funciona con cualquier último carácter como impar.
var num = { arr1: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], arr2: [11, 12, 13, 14, 15, 16, 17, 18, 19, 20], arr3: [21, 22, 23, 24, 25, 26, 27, 28, 29, 30], arr4: [31, 32, 33, 34, 35, 36, 37, 38, 39, 40], arr7: [41, 42, 43, 44, 45, 46, 47, 48, 49, 50] }; let oddArr = []; for (let key in num) { if (key.charAt(key.length - 1) % 2 !== 0) { oddArr = [...oddArr, ...num[key]]; } } let evenNum = oddArr.filter((number) => number % 2 === 0); console.log(evenNum.toString()); 
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Juan Pablo Isaza
Denunciar
0
Esto también funciona.
var num = { arr1: [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], arr2: [11, 12, 13, 14, 15, 16, 17, 18, 19, 20], arr3: [21, 22, 23, 24, 25, 26, 27, 28, 29, 30], arr4: [31, 32, 33, 34, 35, 36, 37, 38, 39, 40], arr5: [41, 42, 43, 44, 45, 46, 47, 48, 49, 50], arr6: [51, 52, 53, 55, 55, 56, 57, 58, 59, 60], arr7: [61, 62, 63, 66, 65, 66, 67, 68, 69, 70], }; var evenNums = Object.keys(num).filter((item) => { itemNum = item.replace("arr", ""); return itemNum % 2 !== 0; }).map((o) => num[o]).flat().filter((x) => x % 2 == 0); console.log(evenNums);
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Juan Pablo Isaza
Denunciar
Responde la pregunta