How do I iterate through the digits of an integer?

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Turning long fixed number to array Ruby

Well, I have to iterate over the digits of a integer in Ruby. Right now I was just splitting it up into an array, and then iterating over that. However I was wondering if there was a faster way to do this?

What do you mean by faster? More performant or more concise? — Alex Peattie, Oct 26 '12 at 17:28
@link Btw, is it an integer as String or Fixnum ? (1233 or "1233") — Roberto Decurnex, Oct 26 '12 at 17:30
please show how are you doing it now so we have a reference. — tokland, Oct 26 '12 at 17:38
@robertodecurnex, it is a Fixnum, which is converted to an array of strings, which is converted to integers. — Rivasa, Oct 26 '12 at 18:10
That's a very interesting question. The "normal" way would be (as already said in one answer) `number.to_s.each_char(&:to_i)` or `number.to_s.chars.map(&:to_i)` . But if you want to have some speed... nice answers in this thread! — musicmatze, Oct 26 '12 at 21:45

tokland · Accepted Answer · 2016-11-05T23:28:44.817

47

The shortest solution probably is:

1234.to_s.chars.map(&:to_i)
#=> [1, 2, 3, 4]

A more orthodox mathematical approach:

class Integer
  def digits(base: 10)
    quotient, remainder = divmod(base)
    quotient == 0 ? [remainder] : [*quotient.digits(base: base), remainder]
  end
end

0.digits #=> [0]
1234.digits #=> [1, 2, 3, 4]
0x3f.digits(base: 16) #=> [3, 15]

edited Nov 05 '16 at 23:28

answered Oct 26 '12 at 17:37

tokland

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To just iterate through the digits, you can also use array slice methods, right? n = 12.to_s sum = (n[0..1].to_i + n[1..2].to_i) product = (n[0,1].to_i * n[1,1].to_i) – Linju Jun 08 '17 at 16:09

user229044 · Answer 2 · 2012-10-26T18:18:06.730

15

You can use the old trick of modulus/divide by 10, but this won't be measurably faster unless you have huge numbers, and it will give the digits to you backwards:

i = 12345

while i > 0 
  digit = i % 10
  i /= 10
  puts digit
end

Output:

edited Oct 26 '12 at 18:18

answered Oct 26 '12 at 17:29

user229044

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score 5 · Answer 3 · answered Oct 26 '12 at 18:23

5

split=->(x, y=[]) {x < 10 ? y.unshift(x) : split.(x/10, y.unshift(x%10))}

split.(1000) #=> [1,0,0,0]
split.(1234) #=> [1,2,3,4]

answered Oct 26 '12 at 18:23

Roberto Decurnex

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steenslag · Answer 4 · 2012-10-26T22:26:16.200

Ruby has divmod, which will calculate both x%10and x/10 in one go:

class Integer
  def split_digits
    return [0] if zero?
    res = []
    quotient = self.abs #take care of negative integers
    until quotient.zero? do
      quotient, modulus = quotient.divmod(10) #one go!
      res.unshift(modulus) #put the new value on the first place, shifting all other values
    end
    res # done
  end
end

p 135.split_digits #=>[1, 3, 5]

For things like Project Euler, where speed is of some importance, this is nice to have. Defining it on Integer causes it to be available on Bignum too.

Patrick Oscity · Answer 5 · 2022-07-19T22:15:28.690

5

I like to use enumerators for this purpose:

class Integer
  def digits
    to_s.each_char.lazy.map(&:to_i)
  end
end

This gives you access to all the good Enumerator stuff:

num = 1234567890

# use each to iterate over the digits
num.digits.each do |digit|
  p digit
end

# make them into an array
p num.digits.to_a     # => [1, 2, 3, 4, 5, 6, 7, 8, 9, 0]

# or take only some digits
p num.digits.take(5)  # => [1, 2, 3, 4, 5]

# ...

edited Jul 19 '22 at 22:15

answered Oct 27 '12 at 11:31

Patrick Oscity

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Maybe `to_s.chars.each { |c| x << c.to_i }` is more orthodox as the block performs a side-effect? – tokland Jul 18 '22 at 21:20
1

@tokland thanks good point, I edited my answer to use a lazy enumerator which gets rid of potential side effects. – Patrick Oscity Jul 19 '22 at 22:16

score 2 · Answer 6 · answered Oct 26 '12 at 17:28

2

Try mod by 10 (will give you the last digit), then divide by 10 (will give you the rest of digits), repeat this until you're down to the final digit. Of course, you'll have to reverse the order if you want to go through the digits from left to right.

answered Oct 26 '12 at 17:28

Mohamed Nuur

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How do I iterate through the digits of an integer?

6 Answers6