How can I filter an array based on the first letter of a contained property?

Question

I have the following JavaScript array of objects:

[{name: '4 Arn', isLetter: false},
{name: 'Abax', isLetter: false},
{name: 'Aramex', isLetter: false},
{name: 'Booking', isLetter: false},
{name: 'Dangerous', isLetter: false},
{name: 'Manali', isLetter: false}]

In the above array, I want to check the first letter of each item's name property. If it matches, I want to append a new object just before the object, as shown in the following examples:

[{letter: "#", isLetter: true},       // new object 
{name: '4 Arn', isLetter: false},
{letter: "A", isLetter: true},        // new Object
{name: 'Abax', isLetter: false},
{name: 'Aramex', isLetter: false},
{letter: "B", isLetter: true},        // new object
{name: 'Booking', isLetter: false},
{letter: "D", isLetter: true},        // new object
{name: 'Dangerous', isLetter: false},
{letter: "M", isLetter: true},        // new object
{name: 'Manali', isLetter: false}]

I tried the reduce() function, but I don't understand why it's giving me the wrong result:

var newArr = [];
list.reduce(function(prev, cur, index, originalArrray) {
    var previousCharcode = prev.name.toUpperCase().charCodeAt(0);
    currentCharCode = cur.name.toUpperCase().charCodeAt(0);

    newArr.push(prev);
    if(previousCharcode != currentCharCode) {
        newArr.splice(index, 0, {isLetter: true, letter: String.fromCharCode(currentCharCode)});
    }
    return cur;
});

Can I ask why you want to go for that structure? It is really not that good design. Better would be to put the nodes with the same starting letter as children of a letter-node, so that at the top level you only have letter nodes, and only at the second level the original nodes. — trincot, Aug 03 '16 at 12:02

score 4 · Accepted Answer · answered Aug 03 '16 at 12:41

There are at least two reasons why your code does not give the expected result:

The index you work with, points to the index in the original array. As your new array will have more elements, it makes no sense to use that index for a splice on the new array. It will be pointing to the wrong place eventually;
You only push the prev element, so the last element will never be pushed to the result array

I would suggest to use reduce with a accumulated value (first argument of the callback) that will build up the final array. To remember the last letter object that was introduced, I will pair this accumulated value with that letter. So I'll work with an array that contains two elements:

The final array being accumulated
The letter of the most recently added letter object

Then the final result will be taken from the first element of that array, ignoring the second value.

I suggest not to work with character codes, but just with the characters. It saves you from converting the code back to a character.

Here is the code:

var list = [
    {name: '4 Arn', isLetter: false},
    {name: 'Abax', isLetter: false},
    {name: 'Aramex', isLetter: false},
    {name: 'Booking', isLetter: false},
    {name: 'Dangerous', isLetter: false},
    {name: 'Manali', isLetter: false}
];

var newArr = list.reduce(function(collect, cur, index, originalArrray) {
    var currentChar = cur.name.toUpperCase().substr(0,1);
    if (currentChar < 'A') currentChar = '#';
    if (collect[1] != currentChar) {
        collect[0].push({isLetter: true, letter: currentChar});
        collect[1] = currentChar;
    }
    collect[0].push(cur);
    return collect;
}, [[], null])[0];

// output
console.log(newArr);

score 1 · Answer 2 · answered Aug 03 '16 at 12:05

Check this solution. Iterate the array and append it to a new array.

var names = [{
  name: '4 Arn',
  isLetter: false
}, {
  name: 'Abax',
  isLetter: false
}, {
  name: 'Aramex',
  isLetter: false
}, {
  name: 'Booking',
  isLetter: false
}, {
  name: 'Dangerous',
  isLetter: false
}, {
  name: 'Manali',
  isLetter: false
}];
var newNames = [];

for (var i in names) {
  var char = names[i].name.substring(0, 1);
  var isNumber = !isNaN(char);
  var entry = {
    letter: (isNumber ? '#' : char.toUpperCase()),
    isLetter: isNumber
  };
  newNames.push(entry);
  newNames.push(names[i]);
}

console.log(newNames);

I want to insert new object only once per each new first letter, not every time. — Wimal Weerawansa, Aug 03 '16 at 12:21

score 1 · Answer 3 · answered Aug 03 '16 at 12:25

Maybe with this approach you can resolve the problem

var list = [{name: '4 Arn', isLetter: false},
{name: 'Abax', isLetter: false},
{name: 'Aramex', isLetter: false},
{name: 'Booking', isLetter: false},
{name: 'Dangerous', isLetter: false},
{name: 'Manali', isLetter: false}];

var listResult = [];
list.map(function(item, index) {  
  if(index > 0) {
    var currentCharcode = item.name.toUpperCase().charCodeAt(0);
    var previousCharcode = list[index-1].name.toUpperCase().charCodeAt(0);
    if(previousCharcode != currentCharcode) {
      listResult.push({isLetter: true, letter: String.fromCharCode(currentCharcode)});
    }
    
  } else {
    listResult.push({isLetter: true, letter: String.fromCharCode(currentCharcode)});
  }
  listResult.push(item);
});


console.log(JSON.stringify(listResult));

Redu · Answer 4 · 2016-08-03T12:52:58.403

I guess you can also do in a functional way like this;

var arr = [{name: '4 Arn', isLetter: false},
           {name: 'Abax', isLetter: false},
           {name: 'Aramex', isLetter: false},
           {name: 'Booking', isLetter: false},
           {name: 'Dangerous', isLetter: false},
           {name: 'Manali', isLetter: false}],
  table = arr.reduce((p,c) => {var firstLetter = c.name[0];
                               isNaN(+firstLetter) ? p[firstLetter] ? p[firstLetter].push(c)
                                                                    : p[firstLetter] = [c]
                                                   : p["#"]         ? p["#"].push(c)
                                                                    : p["#"] = [c];
                               return p;
                              },{}),
 result = Object.keys(table).reduce((p,k) => p.concat({letter: k, isLetter: true},table[k]),[]);
 console.log(result);

Hints : +"A" returns NaN but +"4" returns 4 as number. So isNaN() is a very useful function to check the type.

Here is the version with conventional functions instead of arrows;

var arr = [{name: '4 Arn', isLetter: false},
           {name: 'Abax', isLetter: false},
           {name: 'Aramex', isLetter: false},
           {name: 'Booking', isLetter: false},
           {name: 'Dangerous', isLetter: false},
           {name: 'Manali', isLetter: false}],
  table = arr.reduce(function(p,c){
                       var firstLetter = c.name[0];
                       isNaN(+firstLetter) ? p[firstLetter] ? p[firstLetter].push(c)
                                                            : p[firstLetter] = [c]
                                           : p["#"]         ? p["#"].push(c)
                                                            : p["#"] = [c];
                       return p;
                     },{}),
 result = Object.keys(table).reduce(function(p,k){
                                      return p.concat({letter: k, isLetter: true},table[k]);
                                    },[]);
 console.log(result);

How can I filter an array based on the first letter of a contained property?

4 Answers4