0
213
VistasPrint all even values in all object key end with odd value
I want to print all even values in all object key end with odd value but the coding I made just now is only specified for arr1, arr3, and arr5. Can anyone suggest me how to fix 'let oddArr' method (maybe in loop) so that when I changed arr1 into arr7, the result would be the same.
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 (e.g: 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
·
Juan Pablo Isaza
3 Respuestas
Responde la pregunta0
You might want to use
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 array
You can also make use of regex if the "arrX"-naming is not consistent:
+e[0].match(/\d+/) % 2 !== 0
See a working snippet below:
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());
·
Juan Pablo Isaza
Denunciar
0
You can try like below using for in loop and it works with any last character as odd.
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());
·
Juan Pablo Isaza
Denunciar
0
This also works.
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);
·
Juan Pablo Isaza
Denunciar
Responde la pregunta