還是不用要用'a'吧， 生成key值也麻煩， 用map, 索引做key

var newArr = [1,2,3].map(function(v, idx){
    var obj = {};
    obj[idx] = v;
    return obj;
});

長(zhǎng)度不確定，只能遍歷吧。但你長(zhǎng)度不確定，比如超過(guò)26，key已經(jīng)用完26個(gè)小寫(xiě)字母，接下來(lái)用什么？

如果能確定小于26，倒是可以直接charcode來(lái)++

let arr = [1,2,3,4,5];
let code = 'a'.charCodeAt();
let narr = {};
for(let v of arr){
    let k = String.fromCharCode(code++);
    narr[k] = v;
}
console.log(narr);

var arr = [1, 2, 3, 4, ……];
var arrString = ['a', 'b', 'c', 'd', ……];//與上面?zhèn)€數(shù)相同
var obj = {};
for(i in arr){
    obj[arrString[i]] = arr[i];
}

var rtn = {};
[ 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4, 1, 2, 3, 4 ].forEach( ( val, idx ) => {
    rtn[ Array( ~~( idx / 26 ) + 2 ).join( String.fromCharCode( 97 + idx % 26 ) ) ] = val;
} )

簡(jiǎn)單的處理下key重復(fù)的問(wèn)題