deferred.map: return nothing in some iteration

[Bitliner]

I execute a map on an array.

I want to return only some (transformed) values of this array, only those that satisfies a condition.

But if i return only when condition is satisfied the result will contain undefined values.

Is there a way to return only some values?

deferred.map(item, function(item){
        if (item=='string'){
            return ngram
        }
}).then(function(res){
            // res is [undefined,'string',undefined] - I would like it as ['string']
})

[Mariusz Nowak]

@Bitliner you should just fallback to regular `filter` for arrays.

You can do it either via invoke

deferred.map(item, function () { ...}).invoke('filter', Boolean).then(function (res) {

  // res is ['string'];
});

or (as it's sync operation) you can just filter your result directly when you have it:

deferred.map(item, function () { ...}).then(function (res) {

  res = res.filter(Boolean);

  // res is ['string'];
});

Mind that `Boolean` will filter all falsy values (so also empty strings and 0 etc.) if you want to filter just undefined and null, use `function (val) { return val != null; }` instead.

Hope that helps.

[Giovanni Gaglione]

Thanks. I already used filtering, I  was imagining a map function that automatically rejects all values that is not returned.

 

2013/5/12 Mariusz Nowak <mar...@medikoo.com>

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Giovanni Gaglione                           

Telefono: +39 388 - 75.06.756
Email: giovanni...@gmail.com

Skype: bitliner 

[Mariusz Nowak]

Map is map, there's no filtering involved, and composability is much better when map and filter are separate functions.