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My task is to write a function that swaps the value of an element with the value at the same location within the second object.

{placeOne:10,placeTwo:20},{ten:"firstPlace",twenty:"secondPlace"}   

{placeOne:"firstPlace",placeTwo:"secondPlace"},{ten:10,twenty:20} // should equal this

I wanted to try an approach that pushed the objects value into an array, then loop through the object and set each position to the position within the array.

But I had trouble looping through the object and the array at the same time so I could not solve it that way.

Heres what I have so far.

function swapObj(obj1,obj2){
  let obj1Arr = [];
  let obj2Arr = [];

  for(var i in obj1) {
    obj1Arr.push(obj1[i]);
  }

  for(var k in obj2) {
    obj2Arr.push(obj2[k])
  }

swapObj({placeOne:10,placeTwo:20,placeThree:30,}, 
        {ten:"firstPlace",twenty:"secondPlace",thirty:"thirdPlace"}
)
user7400006
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  • what is the problem? you don't start with an array so what do you mean by swap? your code just places those 2 objects into 2 arrays inside swapObj – George Apr 09 '18 at 21:27
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    the order of keys in an object is arbitrary, so you might create `{ten:10,twenty:20}` but may also create `{ten:20,twenty:10}` (see: https://stackoverflow.com/questions/5525795/does-javascript-guarantee-object-property-order) – c2huc2hu Apr 09 '18 at 21:28
  • @user3080953 looks like the question is not asking about swapping keys inside an object, but the objects themselves inside an array, although it is ambiguous. – George Apr 09 '18 at 21:31

1 Answers1

1

If I understood your problem correctly, this should do it (each step explained with a comment):

const swapValues = (a, b) => {
    // obtain arrays of entries ([key, value] pairs) of input objects
    // assuming the entries come in insertion order,
    // which is true in practice for all major JS engines
    const entriesA = Object.entries(a)
    const entriesB = Object.entries(b)

    // output is a pair of objects:
    // first with keys from a, but values from b
    //      at corresponding entry indices
    // second with keys from b, but values from a
    //      at corresponding entry indices
    // assuming both objects have the same number of entries
    //      (might want to check that)
    return entriesA.reduce(
        // for each entry from a with keyA, valueA and index
        (acc, [keyA, valueA], index) => {
            // get corresponding entry from b
            const entryB = entriesB[index]
            // with keyB and valueB
            const [keyB, valueB] = entryB
            // put valueB at keyA in the first output object
            acc[0][keyA] = valueB
            // put valueA at keyB in the second output object
            acc[1][keyB] = valueA

            return acc
        },
        // initially the output objects are empty:
        [{}, {}]
    )
}

console.log(swapValues(
    {placeOne: 10, placeTwo: 20},
    {ten: "a", twenty: "b"}
)) // -> [ { placeOne: 'a', placeTwo: 'b' }, { ten: 10, twenty: 20 } ]

You may want to adapt this to your version of JS. Note that no mutation of the input objects is taking place -- you get two brand new objects (may share some structure with your input objects if they have nested objects as values).

  • what do [ ] the brackets do around keyA and valueA in the reduce? Is that to put keyA and valueA into an array? – user7400006 Apr 09 '18 at 23:56
  • They destructure[0] the second argument of the reduce's callback. You could write: `(acc, entry, index) => { const keyA = entry[0]; const valueA = entry[1]; /* ... */ }` and you'd get the same thing. [0] https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Destructuring_assignment – Darius JJ Chuck Apr 10 '18 at 17:57