Multiply all even indexed integers by two - ruby

Question

Wanting to take a fixnum of integers and multiply all even(indexed) integers by two. I figured the best way to do this is first turn fixnum into an array. So lets say the following number of 16 digits: a = 4408041234567901
I know I could:
a.to_s.split('')
Which will return 'a' to an array of 'stringed' numbers. But then I cant follow up with:
a.map!.with_index {|i,n| i.even? n*2}
Guess I'm kinda stuck on how to create a method to do this. So my question may even be how to turn that group of numbers into an array of fixnums/integers instead of strings.

Answer 1

I would prefer to remove the conditional altogether from the loop, by creating an Enumerator that contains the coefficients you want to multiply by (2 for even indexes and 1 for odd.)
coef = [2, 1].cycle
This essentially creates an Enumerator that alternately returns 2 and 1 when next is called on it. You can then use this to simplify your map to:
a.to_s.each_char.map { |v| v.to_i * coef.next }

Answer 2

To change it to an Array, you could do
a = 4408041234567901
arr = a.to_s.chars.map(&:to_i)
# => [4, 4, 0, 8, 0, 4, 1, 2, 3, 4, 5, 6, 7, 9, 0, 1]
You can also multiply alternate numbers by 2
arr = a.to_s.chars.map.with_index {|n,i| i.even? ? n.to_i * 2 : n.to_i }
# => [8, 4, 0, 8, 0, 4, 2, 2, 6, 4, 10, 6, 14, 9, 0, 1]
Improving a little bit, you can use a Hash to find the number to be multiplied.
h = {true => 2, false => 1}
a.to_s.each_char.map.with_index {|n,i| n.to_i * h[i.even?]}
EDIT
I can explain each step, But it will be better if you can try to figure it out on your own. Open irb, type a.to_s and check the output. Then type a.to_s.chars and inspect the output and so on..

Answer 3

a = 4408041234567901
even_odd = [:even, :odd].cycle
#=> #<Enumerator: [:even, :odd]:cycle>
If the indexing starts with the highest-order (leftmost) digit:
a.to_s.each_char.map { |d|
(even_odd.next == :even) ? 2*d.to_i : d.to_i }
#=> [8, 4, 0, 8, 0, 4, 2, 2, 6, 4, 10, 6, 14, 9, 0, 1]
If the indexing starts with the ones digit:
s = a.to_s
even_odd.next if s.size.even?
s.each_char.map { |d| ( even_odd.next == :even) ? 2*d.to_i : d.to_i }
#=> [4, 8, 0, 16, 0, 8, 1, 4, 3, 8, 5, 12, 7, 18, 0, 2]
Here are the steps for the example when zero-based indexing starts with the highest-order digit.
Array#cycle converts the array [:even, :odd] to an enumerator:
even_odd = [:even, :odd].cycle
even_odd.next #=> :even
even_odd.next #=> :odd
even_odd.next #=> :even
even_odd.next #=> :odd
...
b = a.to_s
#=> "4408041234567901"
enum0 = b.each_char
#=> #<Enumerator: "4408041234567901":each_char>
The enumerator enum0 passes the digits of b to map. I could have instead written:
b = a.to_s.chars
# => ["4", "4", "0", "8", "0", "4", "1", "2",
# "3", "4", "5", "6", "7", "9", "0", "1"]
but that creates an intermediate array. The enumerator does not and therefore is more efficient. Continuing...
enum1 = enum0.map
#=> #<Enumerator: #<Enumerator: "4408041234567901":each_char>:map>
You can think of this as a "compound enumerator". We can see its contents by converting it to an array:
enum1.to_a
#=> ["4", "4", "0", "8", "0", "4", "1", "2",
# "3", "4", "5", "6", "7", "9", "0", "1"]
The method each will pass each element of the enumerator into the block. Proof:
enum1.each { |d| (enum.next == :even) ? 2*d.to_i : d.to_i }
# => [8, 4, 0, 8, 0, 4, 2, 2, 6, 4, 10, 6, 14, 9, 0, 1]
We can manually step through the elements of enum1 by using Enumerator#next. We will assign the value to the block variable d and perform the calculation in the block to map the digit d:
d = enum1.next
#=> "4"
(enum.next == :even) ? 2*d.to_i : d.to_i
#=> (:even == :even) ? 2*"4".to_i : "4".to_i
#=> (true) ? 8 : 4
#=> 8 ("4" is mapped to 8)
d = enum1.next
#=> "4"
(enum.next == :even) ? 2*d.to_i : d.to_i
#=> (:odd == :even) ? 2*"4".to_i : "4".to_i
#=> (false) ? 8 : 4
#=> 4 ("4" is mapped to 4)
d = enum1.next
#=> "0"
#=> (:even == :even) ? 2*"0".to_i : "0".to_i
#=> (true) ? 0 : 0
#=> 8 ("0" is mapped to 0)
and so on.