How to use map function inside an array of tuples in scala?

Question

I tried executing the below code in Scala shell:

 var chars = ('a' to 'z').toArray.zipWithIndex
 chars: Array[(Char, Int)] = Array((a,0), (b,1), (c,2), (d,3), (e,4), (f,5), (g,6),
(h,7), (i,8), (j,9), (k,10), (l,11), (m,12), (n,13), (o,14), (p,15), (q,16), 
(r,17), (s,18), (t,19), (u,20), (v,21), (w,22), (x,23), (y,24), (z,25))

Now I want the index of every character in the array of tuples mentioned above to be updated by 1 i.e. 'a' at index 1 and z at index 26. How do I achieve this using the map function ?

Do you want to solve this specific problem (of mapping letters to their 1-based index), or is this just an example of the kind of thing you need to solve? You have answers going in both directions. — Ethan
– Ethan, Commented Oct 5, 2018 at 19:13

Lasf · Accepted Answer · 2018-10-05 15:21:32Z

1

Like this

('a' to 'z').toArray.zipWithIndex.map(t => (t._1, t._2 + 1))

answered Oct 5, 2018 at 15:21

Lasf

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2 Comments

jub0bs Over a year ago

... or destructure the tuple, to avoid resorting to those unsightly _1 and _2 methods: .map { case (c, i) => (c, i + 1) }

Lasf Over a year ago

@jubobs totally agree. I hate those methods.

stack0114106 · Accepted Answer · 2018-10-05 15:37:22Z

1

Use zip instead of zipWithIndex. sample below

scala> var chars = ('a' to 'z').toArray.zip(Stream from 1)
chars: Array[(Char, Int)] = Array((a,1), (b,2), (c,3), (d,4), (e,5), (f,6), (g,7), (h,8), (i,9), (j,10), (k,11), (l,12), (m,13), (n,14), (o,15), (p,16), (q,17), (r,18), (s,19), (t,20), (u,21), (v,22), (w,23), (x,24), (y,25), (z,26))

scala>
scala>  var chars = ('a' to 'z').toArray.zip(Stream from 100)
chars: Array[(Char, Int)] = Array((a,100), (b,101), (c,102), (d,103), (e,104), (f,105), (g,106), (h,107), (i,108), (j,109), (k,110), (l,111), (m,112), (n,113), (o,114), (p,115), (q,116), (r,117), (s,118), (t,119), (u,120), (v,121), (w,122), (x,123), (y,124), (z,125))

scala>

answered Oct 5, 2018 at 15:37

stack0114106

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Comments

Leo C · Accepted Answer · 2018-10-05 15:37:50Z

1

You could also do this:

('a' to 'z') zip (Stream from 1)

which would produce a Vector. If you want an Array, simply apply toArray as well.

answered Oct 5, 2018 at 15:37

Leo C

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Comments

RAGHHURAAMM · Accepted Answer · 2018-10-07 12:37:24Z

1

Use,

 var chars = ('a' to 'z').toArray.zipWithIndex
 chars: Array[(Char, Int)] = Array((a,0), (b,1), (c,2), (d,3), (e,4), (f,5), (g,6),
(h,7), (i,8), (j,9), (k,10), (l,11), (m,12), (n,13), (o,14), (p,15), (q,16), 
(r,17), (s,18), (t,19), (u,20), (v,21), (w,22), (x,23), (y,24), (z,25))


    chars.map(x=>{val (a,b)=x;(a,b+1)}) // Here x is each `tuple` in `chars` array.
// `var (a,b) = x` de-structures(extracts) the tuple into `a and b` where `b`
// starts from `0`. To make it start from `1`, we use `b+1` for `each tuple` in
// `map` function

Or

val alpha = 'a' to 'z'
val s1 = alpha.zip(1 to alpha.size)

edited Oct 7, 2018 at 12:37

answered Oct 5, 2018 at 15:39

RAGHHURAAMM

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2 Comments

Sarvagya Dubey Over a year ago

Thanks alot both of your answers work equally alike, the second one is a lot easier to understand. Could you please once elaborate the first one. I am learning so didn't quite get the first answer. Thanks again

RAGHHURAAMM Over a year ago

please check the explanation which may be clear to understand now.

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