I have an object like below
var score = { 'A': 1, 'B': 2, 'C': 1 };
I want to create another object with key, value swapped and same valued keys grouped together, like below.
{ 1:['A', 'C'], 2: 'B' };
I dont want the single values, like 'B', inside an array. I want them alone. Please help.
Hi thanks for the answers. Finally to remove the array syntax if there is only one element i did the following.
$.each(obj, function(key, arr){
if(arr.length == 1){
obj[key] = arr[0]
}
});
Possible solution, using Array#forEach.
var score = { 'A': 1, 'B': 2, 'C': 1 }, hash = {};
Object.keys(score).forEach(v => (hash[score[v]] || (hash[score[v]] = [])).push(v));
console.log(JSON.stringify(hash));You can do this:
var score = { 'A': 1, 'B': 2, 'C': 1 };
var obj = {};
for(var i in score) {
if(!obj[score[i]]) {
obj[score[i]] = [i];
} else {
obj[score[i]].push(i);
}
}
console.log(obj);EDIT:
In the case when you don't want array for a single element, you can use this:
var score = { 'A': 1, 'B': 2, 'C': 1, 'D':1 };
var obj = {};
for (var i in score) {
if (!obj[score[i]]) {
obj[score[i]] = i;
} else {
var temp = obj[score[i]];
if (Object.prototype.toString.call(temp) === "[object Array]") {
obj[score[i]].push(i);
} else {
obj[score[i]] = [];
obj[score[i]].push(temp);
obj[score[i]].push(i);
}
}
}
console.log(obj);
var score = {
'A': 1,
'B': 2,
'C': 1
};
var group = Object.keys(score).reduce(function(a, b) {
var vKey = score[b];
a[vKey] = a[vKey] || [];
a[vKey].push(b);
return a;
}, {});
console.log(group);The following solution uses some more recent Javascript language features: object spread, array spread, computed property names, and the array concat method. Note that this method does not return an array for a single value as the question specifies.
const s = { 'A': 1, 'B': 2, 'C': 1 };
const o = Object.keys(s).reduce((a, k) =>
({...a, ...{[s[k]]: a[s[k]] ? [...a[s[k]]].concat(k) : k}}), {});
console.log(o);The same solution is shown below, just fleshed out a little for more readability. It also includes an additional value ('D': 1) in the score which, if you pull apart the answer, shows the needs for the array spread. The comments show how the variable values change in each iteration through the loop.
const score = { 'A': 1, 'B': 2, 'C': 1, 'D': 1 };
const invertedScore = Object.keys(score).reduce((acc, key) => {
// iter'n#: 0 1 2 3
// key: 'A' 'B' 'C' 'D'
const val = score[key]; // val: 1 2 1 1
const prev = acc[val]; // prev: undef undef 'A' ['A','C']
const chgd = prev ? [...prev].concat(key) : key;
// chgd: 'A' 'B' ['A','C'] ['A','C','D']
return ({...acc, ...{[val]: chgd}});
}, {});
console.log(invertedScore);
