Format Ruby string to 3 characters with an integer

Question

I'd like for this script

 (1..134).each do |x|
   puts "0#{x}" # ????
 end

to output:

001
002
...
011
...
134

Is this possible without doing a bunch of if statements using just native format? It doesn't need to handle greater than 3 digits.

FYI, Ruby can also iterate numerical strings: ('001'..'134').each { |s| puts s } — Stefan
– Stefan, Commented Apr 26, 2018 at 18:58
What @Stefan said although I prefer the readability and ephemeral nature of #upto e.g. '001'.upto('134') — engineersmnky
– engineersmnky, Commented Apr 26, 2018 at 19:01

Zoran · Accepted Answer · 2018-04-26 18:22:16Z

5

One way to achieve the zero padding you want is to use #rjust:

(1..134).each do |x|
  puts x.to_s.rjust(3, '0')
end

Hope this helps!

answered Apr 26, 2018 at 18:22

Zoran

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Comments

Igor Drozdov · Accepted Answer · 2018-04-26 18:27:23Z

3

Sure. It can be done using the following formatter:

'%03d' % 1   # 001
'%03d' % 10  # 010
'%03d' % 100 # 100

The loop is going to look like this:

(1..134).each { |x| puts '%03d' % x }

There's also Kernel#format method, which does exactly this, but self-explanatory:

(1..134).each { |x| puts format('%03d', x) }

answered Apr 26, 2018 at 18:21

Igor Drozdov

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This works but it's going to be completely mystifying to a novice who wants to know how it works.

Alternatively: 1.upto(134) { |i| printf("%03d\n", i) }

@JordanRunning where is @mudasobwa to pull out a (1..134).each(&method(:printf).curry(2).call("%03d\n"))

The documentation for the % operator is here: String#%. The format string is specified in Kernel#sprintf, of which Kernel#format is an alias.