Applying a function to each object in a JavaScript array

JavascriptArrays

Javascript Problem Overview


[ {name:'hi',data:'1,2,3,4,5'} , {name:'hello',data:'5,4,3,2,1'} ]

What I need is to apply a split on data of each object in array so that the result would be:

[ {name:'hi',data:[1,2,3,4,5]} , {name:'hello',data:[5,4,3,2,1]} ]

I know I can loop through the array using for each and produce a new array, but is there a better, faster method?

var arr = [{
    name: 'hi',
    data: '1,2,3,4,5'
}, {
    name: 'hello',
    data: '5,4,3,2,1'
}];

var new_arr = [];
for (i in arr) {
    var temp = {};
    temp.name = arr[i].name;
    temp.data = arr[i].data.split(',');
    new_arr.push(temp);
}

Javascript Solutions


Solution 1 - Javascript

You could use Array.prototype.map:

var new_array = old_array.map(function(e) { 
  e.data = e.data.split(','); 
  return e;
});

As the comment said, this way changes the old_array. You could also return a new object in the callback function without changing the original array.

Solution 2 - Javascript

var data = [{
    name: 'hi',
    data: '1,2,3,4,5'
}, {
    name: 'hello',
    data: '5,4,3,2,1'
}];

You can use the Array.prototype.map on data to construct a new Array, like this

var result = data.map(function (currentObject) {
    return {
        name: currentObject.name,
        data: currentObject.data.split(",").map(Number)
    };
});

Here, we split the currentObject.data based on , and then we call Number function on all the split strings, so that you will get the result object's data as numbers, as you wanted in the question.

Output

[{    name: 'hi',    data: [1, 2, 3, 4, 5]
}, {
    name: 'hello',
    data: [5, 4, 3, 2, 1]
}]

let data = [{
    name: 'hi',
    data: '1,2,3,4,5'
  }, {
    name: 'hello',
    data: '5,4,3,2,1'
  }],
  result = data.map(function(currentObject) {
    return {
      name: currentObject.name,
      data: currentObject.data.split(",").map(Number)
    };
  });
console.log(result);

Attributions

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Content TypeOriginal AuthorOriginal Content on Stackoverflow
QuestionZalabozaView Question on Stackoverflow
Solution 1 - JavascriptxdazzView Answer on Stackoverflow
Solution 2 - JavascriptthefourtheyeView Answer on Stackoverflow