Turning long fixed number to array Ruby

Question

Is there a method in ruby to turn fixnum like 74239 into an array like [7,4,2,3,9]?

The answer of mbds, using the `digits` method introduced in Ruby 2.4 makes all others obsolete. — steenslag, Nov 19 '20 at 01:32

score 17 · Answer 1 · edited Aug 22 '18 at 08:18

17

Maybe not the most elegant solution:

74239.to_s.split('').map(&:to_i)

Output:

[7, 4, 2, 3, 9]

edited Aug 22 '18 at 08:18

wscourge

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answered Oct 16 '12 at 05:25

nneonneo

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`.map(&:to_i)` to get it a bit more elegant? – oldergod Oct 16 '12 at 05:30
What happens to `-23.to_s.split('').map(&:to_i)`? – mu is too short Oct 16 '12 at 05:47
2

I'm nitpicking a bit here, but `74239.to_s.chars` seems more appropriate than `74239.to_s.split('')`, especially when used in conjunction with `.map(&:to_i)` :) – Adam Eberlin Oct 29 '12 at 06:11
1

I think `each_char` would be even more appropriate, as it would avoid the intermediate `Array` created by `chars`. – Drenmi Apr 26 '15 at 07:32
@Drenmi, I suppose `.chars` in Enumerator currently, not Array. – Nakilon Jun 19 '17 at 18:03

mu is too short · Accepted Answer · 2019-08-01T16:35:45.367

9

You don't need to take a round trip through string-land for this sort of thing:

def digits(n)
  Math.log10(n).floor.downto(0).map { |i| (n / 10**i) % 10 }
end

ary = digits(74239)
# [7, 4, 2, 3, 9]

This does assume that n is positive of course, slipping an n = n.abs into the mix can take care of that if needed. If you need to cover non-positive values, then:

def digits(n)
  return [0] if(n == 0)
  if(n < 0)
    neg = true
    n   = n.abs
  end
  a = Math.log10(n).floor.downto(0).map { |i| (n / 10**i) % 10 }
  a[0] *= -1 if(neg)
  a
end

edited Aug 01 '19 at 16:35

answered Oct 16 '12 at 05:36

mu is too short

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2

@RyanBigg: Only if you've forgotten your high school math. – mu is too short Oct 16 '12 at 05:38

score 8 · Answer 3 · answered May 11 '20 at 08:20

As of Ruby 2.4, integers (FixNum is gone in 2.4+) have a built-in digits method that extracts them into an array of their digits:

74239.digits
=> [9, 3, 2, 4, 7]

If you want to maintain the order of the digits, just chain reverse:

74239.digits.reverse
=> [7, 4, 2, 3, 9]

Docs: https://ruby-doc.org/core-2.4.0/Integer.html#method-i-digits

Frederick Cheung · Answer 4 · 2012-10-16T08:32:47.033

The divmod method can be used to extract the digits one at a time

def digits n
  n= n.abs
  [].tap do |result|
    while n > 0 
      n,digit = n.divmod 10
      result.unshift digit
    end
  end
end

A quick benchmark showed this to be faster than using log to find the number of digits ahead of time, which was itself faster than string based methods.

bmbm(5) do |x|
  x.report('string') {10000.times {digits_s(rand(1000000000))}}
  x.report('divmod') {10000.times {digits_divmod(rand(1000000000))}}
  x.report('log') {10000.times {digits(rand(1000000000))}}
end

#=>
             user     system      total        real
string   0.120000   0.000000   0.120000 (  0.126119)
divmod   0.030000   0.000000   0.030000 (  0.023148)
log      0.040000   0.000000   0.040000 (  0.045285)

score 5 · Answer 5 · answered Oct 29 '15 at 01:14

5

You can convert to string and use the chars method:

74239.to_s.chars.map(&:to_i)

Output:

[7, 4, 2, 3, 9]

Its a bit more elegant than splitting.

answered Oct 29 '15 at 01:14

Paulo Fidalgo

21,709
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score 1 · Answer 6 · answered Sep 27 '16 at 18:10

1

You can also use Array.new instead of map:

n = 74239

s = Math.log10(n).to_i + 1 # Gets the size of n

Array.new(s) { d = n % 10; n = n / 10; d }.reverse

answered Sep 27 '16 at 18:10

izaban

1,013
7
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score 1 · Answer 7 · answered Nov 02 '16 at 12:13

1

In Ruby 2.4, integers will have a digits method.

answered Nov 02 '16 at 12:13

steenslag

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Turning long fixed number to array Ruby

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