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Encontrar una cadena específica es impar o par de una matriz JavaScript

Estoy tratando de encontrar una cadena impar de una matriz dada.

Código:

 const friendArray = ["agdum", "bagdum", "chagdum", "lagdum", "jagdum", "magdum"]; function oddFriend(friendArray) { for (let i = 0; i < friendArray.length; i++) { if (friendArray[i].length % 2 != 0) { return friendArray[i]; } } } const myFriend = oddFriend(friendArray); console.log(myFriend);

about 4 years ago · Juan Pablo Isaza
3 answers
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0

Puedes aplicar esto:

 const friendArray = ["agdum","bagdum","chagdum","lagdum","jagdum","magdum",]; function oddFriend(friendArray) { if (friendArray.length % 2 != 0) { return friendArray; } } const myFriend = friendArray.filter(oddFriend); console.log(myFriend);
about 4 years ago · Juan Pablo Isaza Report

0

.flatMap() y una devolución de llamada ternaria hacen un filtro simple y legible. No está del todo claro si solo quería impar o impar, y aun así hay ambas versiones.

Cadenas de longitud impar

 const fArray = ["agdum", "bagdum", "chagdum", "lagdum", "jagdum", "magdum"]; let odd = fArray.flatMap(o => o.length % 2 === 1 ? [o] : []); console.log(odd);

Cuerdas de longitud par e impar

 const fArray = ["agdum", "bagdum", "chagdum", "lagdum", "jagdum", "magdum"]; let oe = [[], []]; fArray.forEach(o => o.length % 2 === 1 ? oe[0].push(o) : oe[1].push(o)); console.log(oe);

about 4 years ago · Juan Pablo Isaza Report

0

Puede usar el método Array.filter() que creará una nueva matriz con todos los amigos impares que pasen la prueba implementada por la función proporcionada.

 // Input array const friendArray = ["agdum", "bagdum", "chagdum", "lagdum", "jagdum", "magdum"]; // Filtered result console.log(friendArray.filter((friend) => friend.length % 2 !== 0));

about 4 years ago · Juan Pablo Isaza Report
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