Array destructuring with a ternary operator












1















I am trying to concat (with uniques values) two arrays and if the second array sometimes is a string.



Maybe it have a bug, but these are my three tryings:



let a = 'abcdefg'

// First try

[...new Set([..., ...(typeof(a) == 'string'? [a]: a))]

// Second try

[...new Set([..., [(typeof(a) == 'string'? ...[a]: ...a)]]

// Third try

[...new Set([..., (typeof(a) == 'string'? ...[a]: ...a)]





javascript arrays ecmascript-6 destructuring







share|improve this question


















  • 1





    please add the wanted result as well - and some use cases with their result.

    – Nina Scholz
    Nov 26 '18 at 12:01








  • 1





    ... makes no sense at all.

    – Bergi
    Nov 26 '18 at 12:22
















1















I am trying to concat (with uniques values) two arrays and if the second array sometimes is a string.



Maybe it have a bug, but these are my three tryings:



let a = 'abcdefg'

// First try

[...new Set([..., ...(typeof(a) == 'string'? [a]: a))]

// Second try

[...new Set([..., [(typeof(a) == 'string'? ...[a]: ...a)]]

// Third try

[...new Set([..., (typeof(a) == 'string'? ...[a]: ...a)]





javascript arrays ecmascript-6 destructuring







share|improve this question


















  • 1





    please add the wanted result as well - and some use cases with their result.

    – Nina Scholz
    Nov 26 '18 at 12:01








  • 1





    ... makes no sense at all.

    – Bergi
    Nov 26 '18 at 12:22














1












1








1








I am trying to concat (with uniques values) two arrays and if the second array sometimes is a string.



Maybe it have a bug, but these are my three tryings:



let a = 'abcdefg'

// First try

[...new Set([..., ...(typeof(a) == 'string'? [a]: a))]

// Second try

[...new Set([..., [(typeof(a) == 'string'? ...[a]: ...a)]]

// Third try

[...new Set([..., (typeof(a) == 'string'? ...[a]: ...a)]





javascript arrays ecmascript-6 destructuring







share|improve this question














I am trying to concat (with uniques values) two arrays and if the second array sometimes is a string.



Maybe it have a bug, but these are my three tryings:



let a = 'abcdefg'

// First try

[...new Set([..., ...(typeof(a) == 'string'? [a]: a))]

// Second try

[...new Set([..., [(typeof(a) == 'string'? ...[a]: ...a)]]

// Third try

[...new Set([..., (typeof(a) == 'string'? ...[a]: ...a)]





javascript arrays ecmascript-6 destructuring




javascript arrays ecmascript-6 destructuring






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked Nov 26 '18 at 12:00









tres.14159tres.14159

1701618




1701618








  • 1





    please add the wanted result as well - and some use cases with their result.

    – Nina Scholz
    Nov 26 '18 at 12:01








  • 1





    ... makes no sense at all.

    – Bergi
    Nov 26 '18 at 12:22














  • 1





    please add the wanted result as well - and some use cases with their result.

    – Nina Scholz
    Nov 26 '18 at 12:01








  • 1





    ... makes no sense at all.

    – Bergi
    Nov 26 '18 at 12:22








1




1





please add the wanted result as well - and some use cases with their result.

– Nina Scholz
Nov 26 '18 at 12:01







please add the wanted result as well - and some use cases with their result.

– Nina Scholz
Nov 26 '18 at 12:01






1




1





... makes no sense at all.

– Bergi
Nov 26 '18 at 12:22





... makes no sense at all.

– Bergi
Nov 26 '18 at 12:22












3 Answers
3






active

oldest

votes


















3














Instead of 



[...new Set([..., ...(typeof a === 'string' ? [a] : a))]


take, watch the round, square, round and squere closing brackets at the end.



[...new Set([..., ...(typeof a === 'string' ? [a] : a)])]

//                                                      ^







let a = 'abcdefg'



console.log([...new Set([..., ...(typeof a === 'string' ? [a] : a)])]);








share|improve this answer


























  • Thanks, my code had a bug. Sorry.

    – tres.14159
    Nov 26 '18 at 13:46



















2














Instead of using spread, you can use Array.concat(), because it treats combine arrays and values in the same way:






const a = 'abcdefg'

console.log([...new Set(.concat(, a))])

console.log([...new Set(.concat(, [a]))])








share|improve this answer































    0














    If I understand correctly, if the a parameter is a string, and not a collection, searching unique values and the need for a Set is moot. Then you could short circuit as typeof a === 'string' ? [a] : [...new Set(a)]






    let a = 'abcdefg'
    

    
const createArr = a => typeof a === 'string' ? [a] : [...new Set(a)];
    

    
console.log(createArr(a));
    
console.log(createArr([a,a,'aa']));








    share|improve this answer
























    • For the record, I like the solution by @OriDrori better ;)

      – Me.Name
      Nov 26 '18 at 13:21












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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    3














    Instead of 



    [...new Set([..., ...(typeof a === 'string' ? [a] : a))]


    take, watch the round, square, round and squere closing brackets at the end.



    [...new Set([..., ...(typeof a === 'string' ? [a] : a)])]
    
//                                                      ^
    






    let a = 'abcdefg'
    

    
console.log([...new Set([..., ...(typeof a === 'string' ? [a] : a)])]);








    share|improve this answer


























    • Thanks, my code had a bug. Sorry.

      – tres.14159
      Nov 26 '18 at 13:46
















    3














    Instead of 



    [...new Set([..., ...(typeof a === 'string' ? [a] : a))]


    take, watch the round, square, round and squere closing brackets at the end.



    [...new Set([..., ...(typeof a === 'string' ? [a] : a)])]
    
//                                                      ^
    






    let a = 'abcdefg'
    

    
console.log([...new Set([..., ...(typeof a === 'string' ? [a] : a)])]);








    share|improve this answer


























    • Thanks, my code had a bug. Sorry.

      – tres.14159
      Nov 26 '18 at 13:46














    3












    3








    3







    Instead of 



    [...new Set([..., ...(typeof a === 'string' ? [a] : a))]


    take, watch the round, square, round and squere closing brackets at the end.



    [...new Set([..., ...(typeof a === 'string' ? [a] : a)])]
    
//                                                      ^
    






    let a = 'abcdefg'
    

    
console.log([...new Set([..., ...(typeof a === 'string' ? [a] : a)])]);








    share|improve this answer















    Instead of 



    [...new Set([..., ...(typeof a === 'string' ? [a] : a))]


    take, watch the round, square, round and squere closing brackets at the end.



    [...new Set([..., ...(typeof a === 'string' ? [a] : a)])]
    
//                                                      ^
    






    let a = 'abcdefg'
    

    
console.log([...new Set([..., ...(typeof a === 'string' ? [a] : a)])]);








    let a = 'abcdefg'
    

    
console.log([...new Set([..., ...(typeof a === 'string' ? [a] : a)])]);





    let a = 'abcdefg'
    

    
console.log([...new Set([..., ...(typeof a === 'string' ? [a] : a)])]);






    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited Nov 26 '18 at 12:24









    Bergi

    380k63581914




    380k63581914










    answered Nov 26 '18 at 12:08









    Nina ScholzNina Scholz

    195k15107178




    195k15107178













    • Thanks, my code had a bug. Sorry.

      – tres.14159
      Nov 26 '18 at 13:46



















    • Thanks, my code had a bug. Sorry.

      – tres.14159
      Nov 26 '18 at 13:46

















    Thanks, my code had a bug. Sorry.

    – tres.14159
    Nov 26 '18 at 13:46





    Thanks, my code had a bug. Sorry.

    – tres.14159
    Nov 26 '18 at 13:46













    2














    Instead of using spread, you can use Array.concat(), because it treats combine arrays and values in the same way:






    const a = 'abcdefg'
    
console.log([...new Set(.concat(, a))])
    
console.log([...new Set(.concat(, [a]))])








    share|improve this answer




























      2














      Instead of using spread, you can use Array.concat(), because it treats combine arrays and values in the same way:






      const a = 'abcdefg'
      
console.log([...new Set(.concat(, a))])
      
console.log([...new Set(.concat(, [a]))])








      share|improve this answer


























        2












        2








        2







        Instead of using spread, you can use Array.concat(), because it treats combine arrays and values in the same way:






        const a = 'abcdefg'
        
console.log([...new Set(.concat(, a))])
        
console.log([...new Set(.concat(, [a]))])








        share|improve this answer













        Instead of using spread, you can use Array.concat(), because it treats combine arrays and values in the same way:






        const a = 'abcdefg'
        
console.log([...new Set(.concat(, a))])
        
console.log([...new Set(.concat(, [a]))])








        const a = 'abcdefg'
        
console.log([...new Set(.concat(, a))])
        
console.log([...new Set(.concat(, [a]))])





        const a = 'abcdefg'
        
console.log([...new Set(.concat(, a))])
        
console.log([...new Set(.concat(, [a]))])






        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered Nov 26 '18 at 12:22









        Ori DroriOri Drori

        81.4k138997




        81.4k138997























            0














            If I understand correctly, if the a parameter is a string, and not a collection, searching unique values and the need for a Set is moot. Then you could short circuit as typeof a === 'string' ? [a] : [...new Set(a)]






            let a = 'abcdefg'
            

            
const createArr = a => typeof a === 'string' ? [a] : [...new Set(a)];
            

            
console.log(createArr(a));
            
console.log(createArr([a,a,'aa']));








            share|improve this answer
























            • For the record, I like the solution by @OriDrori better ;)

              – Me.Name
              Nov 26 '18 at 13:21
















            0














            If I understand correctly, if the a parameter is a string, and not a collection, searching unique values and the need for a Set is moot. Then you could short circuit as typeof a === 'string' ? [a] : [...new Set(a)]






            let a = 'abcdefg'
            

            
const createArr = a => typeof a === 'string' ? [a] : [...new Set(a)];
            

            
console.log(createArr(a));
            
console.log(createArr([a,a,'aa']));








            share|improve this answer
























            • For the record, I like the solution by @OriDrori better ;)

              – Me.Name
              Nov 26 '18 at 13:21














            0












            0








            0







            If I understand correctly, if the a parameter is a string, and not a collection, searching unique values and the need for a Set is moot. Then you could short circuit as typeof a === 'string' ? [a] : [...new Set(a)]






            let a = 'abcdefg'
            

            
const createArr = a => typeof a === 'string' ? [a] : [...new Set(a)];
            

            
console.log(createArr(a));
            
console.log(createArr([a,a,'aa']));








            share|improve this answer













            If I understand correctly, if the a parameter is a string, and not a collection, searching unique values and the need for a Set is moot. Then you could short circuit as typeof a === 'string' ? [a] : [...new Set(a)]






            let a = 'abcdefg'
            

            
const createArr = a => typeof a === 'string' ? [a] : [...new Set(a)];
            

            
console.log(createArr(a));
            
console.log(createArr([a,a,'aa']));








            let a = 'abcdefg'
            

            
const createArr = a => typeof a === 'string' ? [a] : [...new Set(a)];
            

            
console.log(createArr(a));
            
console.log(createArr([a,a,'aa']));





            let a = 'abcdefg'
            

            
const createArr = a => typeof a === 'string' ? [a] : [...new Set(a)];
            

            
console.log(createArr(a));
            
console.log(createArr([a,a,'aa']));






            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered Nov 26 '18 at 13:10









            Me.NameMe.Name

            10.2k22039




            10.2k22039













            • For the record, I like the solution by @OriDrori better ;)

              – Me.Name
              Nov 26 '18 at 13:21



















            • For the record, I like the solution by @OriDrori better ;)

              – Me.Name
              Nov 26 '18 at 13:21

















            For the record, I like the solution by @OriDrori better ;)

            – Me.Name
            Nov 26 '18 at 13:21





            For the record, I like the solution by @OriDrori better ;)

            – Me.Name
            Nov 26 '18 at 13:21


















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